UPSC Civil Services Prelims 2026 — CSAT (Paper 2) Answer Key with Step-by-Step Solutions

Exam: UPSC CSE Prelims 2026 (held 24 May 2026)
Paper: CSAT — General Studies Paper II (Test Booklet Code: TDMN-B-AST, Series A)
Total questions: 80 | Max marks: 200 | Duration: 2 hours
Marking: +2.5 per correct, −0.833 (1/3rd) per wrong | Qualifying: 33% (66/200)
Solution method: Each question solved with own mathematical / logical reasoning. Quick tricks given where applicable. For data-sufficiency, comprehension, numeracy, reasoning and DI sections — all worked from first principles.
Interactive version with filters, search, KaTeX-rendered math and step-by-step reveal: /prelims/2026/csat/

Q1. reasoning (data_sufficiency)

$X$ receives three coins of different denominations : 1, 2, 5, 10 and 20. If the total amount received by $X$ is $m$, does $X$ receive a coin of denomination 5?

Statement I: $m$ is not a prime number.

Statement II: The sum of the digits of $m$ is greater than 5.

(a) Select this option if the question can be answered using one of these statements alone, but cannot be answered using other statement
(b) Select this option if the question can be answered using either statement alone
(c) Select this option if the question can be answered using both the statements together, but cannot be answered using either statement alone
(d) Select this option if the question cannot be answered even using any of the statements

Answer: (A) · Confidence: high

## Why the answer is (a) — Statement II alone is sufficient; Statement I alone is not ### ⚡ Quick trick (≈30 sec) Hunt for ONE counterexample to disqualify each statement. For Statement I, ignore the 5-coin sets — focus on the non-prime sums WITHOUT a 5: only **{2, 10, 20} = 32** qualifies, instantly killing Statement I. For Statement II (digit-sum > 5), the no-5 sums are 13, 23, 31, 32 — digit-sums 4, 5, 4, 5 — none exceed 5! So every digit-sum-greater-than-5 case must contain a 5. Statement II wins alone → **(a)** in under a minute. ### Step 1: Enumerate all C(5,3) = 10 three-coin combinations from {1, 2, 5, 10, 20} | Combination | m | Has 5? | Prime? | Digit-sum | |---|---|---|---|---| | {1, 2, 5} | 8 | ✅ Yes | No | 8 | | {1, 2, 10} | 13 | ❌ No | **Yes** | 4 | | {1, 2, 20} | 23 | ❌ No | **Yes** | 5 | | {1, 5, 10} | 16 | ✅ Yes | No | 7 | | {1, 5, 20} | 26 | ✅ Yes | No | 8 | | {1, 10, 20} | 31 | ❌ No | **Yes** | 4 | | {2, 5, 10} | 17 | ✅ Yes | **Yes** | 8 | | {2, 5, 20} | 27 | ✅ Yes | No | 9 | | {2, 10, 20} | **32** | **❌ No** | No | 5 | | {5, 10, 20} | 35 | ✅ Yes | No | 8 | ### Step 2: Test Statement I alone — *m is not prime* Non-prime values of m: **8, 16, 26, 27, 32, 35**. - m = 32 (from {2, 10, 20}) has NO coin of 5 → answer "No" - m = 8, 16, 26, 27, 35 → answer "Yes" Both Yes and No outcomes are possible → **Statement I alone is INSUFFICIENT.** ### Step 3: Test Statement II alone — *digit-sum of m > 5* Values of m with digit-sum > 5: **8, 16, 17, 26, 27, 35** (digit sums 8, 7, 8, 8, 9, 8). Every one of these contains a 5-coin → **Statement II alone is SUFFICIENT** (answer is always "Yes"). ### Step 4: Match to option logic - (a) One alone suffices, the other does not — **✓ matches** (II yes, I no) - (b) Either alone — ✗ (I fails) - (c) Both together needed — ✗ (II alone already works) - (d) Cannot be answered — ✗ **Answer: (a).** The critical case is m = 32 from coin-set {2, 10, 20} — non-prime but no 5-coin — which defeats Statement I alone.

Key concepts:

Data sufficiency methodology
Exhaustive enumeration (C(5,3) = 10 cases)
Counterexample method (m = 32 disproves Statement I sufficiency)
Primality of two-digit numbers
Digit-sum as a filter property

Sources: Direct mathematical enumeration of C(5,3) coin combinations from {1, 2, 5, 10, 20}; primality and digit-sum verified by elementary arithmetic.

Q2. reasoning (data_sufficiency)

For two distinct real numbers $x$ and $y$, which of them is bigger?

Statement I: $x^2 < y < 1$

Statement II: $y < \sqrt{x} < 1$

(a) Select this option if the question can be answered using one of these statements alone, but cannot be answered using other statement
(b) Select this option if the question can be answered using either statement alone
(c) Select this option if the question can be answered using both the statements together, but cannot be answered using either statement alone
(d) Select this option if the question cannot be answered even using any of the statements

Answer: (D) · Confidence: high

## Why the answer is (d) — Even both statements together are insufficient ### ⚡ Quick trick (≈45 sec) Memorise this fact: for any **x in (0, 1)**, we always have **x²< x < √x**. So the combined constraint becomes **x² < y < √x** — but the interval (x², √x) **straddles x itself**, meaning y can lie on either side of x. Whichever statement you test, you can pick a y in (x², √x) that is either less than x or greater than x. So even both together can't fix the order → **(d)**. ### Statement I alone: x² < y < 1 - x = 0.3, y = 0.5 → x² = 0.09 < 0.5 < 1 ✓ → here **y > x** - x = 0.9, y = 0.85 → x² = 0.81 < 0.85 < 1 ✓ → here **x > y** Both orderings possible → **Statement I alone is INSUFFICIENT.** ### Statement II alone: y < √x < 1 Requires x ≥ 0 (principal real square root). With x in [0, 1): - x = 0.25, y = 0.3 → √x = 0.5; 0.3 < 0.5 ✓ → here **y > x** - x = 0.25, y = 0.1 → 0.1 < 0.5 ✓ → here **x > y** Both orderings possible → **Statement II alone is INSUFFICIENT.** ### Combined: x² < y < √x with 0 < x < 1 For 0 < x < 1 we have **x² < x < √x**, so y is free to lie anywhere in the open interval (x², √x) — which contains x itself. Hence y can be less than or greater than x: - x = 0.25, y = 0.1 (combined: 0.0625 < 0.1 < 0.5 ✓) → **x > y** - x = 0.25, y = 0.3 (combined: 0.0625 < 0.3 < 0.5 ✓) → **y > x** Both pairs satisfy both statements simultaneously, so even with both statements the question cannot be answered. **Answer: (d).**

Key concepts:

Behaviour of x², x, √x in (0, 1): x²< x < √x
Counterexample method for data-sufficiency
Principal real square root requires x ≥ 0
Open-interval freedom of bounded variables

Sources: Direct construction of counterexamples; properties of x², x, √x on the unit interval verified by elementary analysis.

Q3. reasoning (data_sufficiency)

If $x$ and $y$ are integers, then is $x$ even?

Statement I: $x^2 y^2$ is even.

Statement II: $1 + x^2 + y^2$ is odd.

(a) Select this option if the question can be answered using one of these statements alone, but cannot be answered using other statement
(b) Select this option if the question can be answered using either statement alone
(c) Select this option if the question can be answered using both the statements together, but cannot be answered using either statement alone
(d) Select this option if the question cannot be answered even using any of the statements

Answer: (C) · Confidence: high

## Why the answer is (c) — Both statements together are needed; neither alone suffices ### ⚡ Quick trick (≈20 sec) Use parity: **an integer and its square have the same parity** (even² = even, odd² = odd). - **Statement I** (x²y² is even) ⟺ xy is even ⟺ at least one of x, y is even — but which one? Unknown. - **Statement II** (1 + x² + y² is odd) ⟺ x² + y² is even ⟺ x and y have the **same parity** — but both odd or both even? Unknown. - **Combined:** same parity AND at least one even → **both must be even** (both-odd would make the product odd, contradicting I). So x is even. ✓ Answer: **(c)**. ### Step 1: Statement I alone — x²y² is even x²y² = (xy)² is even ⟺ xy is even ⟺ at least one of x, y is even. - x = 2, y = 1: x²y² = 4 ✓ → x is **even** → "Yes" - x = 1, y = 2: x²y² = 4 ✓ → x is **odd** → "No" Both outcomes possible → **I alone INSUFFICIENT.** ### Step 2: Statement II alone — 1 + x² + y² is odd 1 + x² + y² odd ⟺ x² + y² is even ⟺ x² and y² have the same parity ⟺ x and y have the same parity. - x = 2, y = 4 (both even): 1 + 4 + 16 = 21 odd ✓ → x **even** → "Yes" - x = 1, y = 3 (both odd): 1 + 1 + 9 = 11 odd ✓ → x **odd** → "No" Both outcomes possible → **II alone INSUFFICIENT.** ### Step 3: Combined — both I and II From II: x and y have the same parity (both even OR both odd). From I: at least one of x, y is even. If both were odd, x²y² would be odd × odd = odd, contradicting I. So both cannot be odd. The only remaining case is **both even** → therefore x is even. **Combined is SUFFICIENT** → answer **(c)**.

Key concepts:

Parity of an integer equals parity of its square (n ≡ n² mod 2)
Product of integers is even ⟺ at least one factor is even
Sum of two squares is even ⟺ both have same parity
Data-sufficiency: combining two restrictive conditions

Sources: Elementary number theory — parity rules for integer multiplication and addition.

Q4. reasoning (data_sufficiency)

$X$ is a collection of certain odd numbers whereas $Y$ is a collection of certain even numbers. $T$ consists of the numbers all of which are either from $X$ or from $Y$. Is every number of $T$ from $Y$?

Statement I: The sum of any two numbers belonging to $T$ is even.

Statement II: If both $p$ and $q$ are picked from $T$, then $(p - 1)q$ is even.

(a) Select this option if the question can be answered using one of these statements alone, but cannot be answered using other statement
(b) Select this option if the question can be answered using either statement alone
(c) Select this option if the question can be answered using both the statements together, but cannot be answered using either statement alone
(d) Select this option if the question cannot be answered even using any of the statements

Answer: (D) · Confidence: high

## Why the answer is (d) — Even both statements together cannot decide ### ⚡ Quick trick (≈30 sec) Both statements only force **T to be of a single parity** — all-odd OR all-even — but neither tells you **which**. The question 'is every number of T from Y (even)?' therefore has two compatible answers under both statements. **(d)**. ### Setup - **X** = some odd numbers; **Y** = some even numbers. - **T** = numbers each drawn from X ∪ Y (so T may contain odds, evens, or both). - **Question:** Is every element of T from Y? (i.e., is T ⊆ Y, i.e., are all elements of T even?) ### Statement I alone — sum of any two numbers in T is even Sum of two integers is even ⟺ both have the **same parity**. If this holds for *every* pair drawn from T, all elements of T must share one parity: - **Case A:** T = {2, 4, 6} (all even) → answer 'Yes' ✓ - **Case B:** T = {1, 3, 5} (all odd) → answer 'No' ✓ Both 'Yes' and 'No' possible → **I alone INSUFFICIENT.** ### Statement II alone — for any p, q in T, (p − 1)q is even (p − 1)q is even ⟺ p is odd OR q is even. Force this for *every* ordered pair (p, q) ∈ T × T. - Pick q to be any element. If T has an odd element q*, then setting q = q* forces (p − 1) even for every p, i.e., **every p in T is odd** → T all-odd → answer 'No'. ✓ - If T has no odd element, T is all-even → answer 'Yes'. ✓ - (Mixed-parity T fails: pick p even, q odd → p − 1 odd, q odd → (p−1)q odd, contradiction.) So II forces same conclusion as I: T is all-odd OR T is all-even — but doesn't say which. - **Case A:** T = {2, 4} → check: (2−1)·2 = 2 even ✓, (4−1)·4 = 12 even ✓ → 'Yes' - **Case B:** T = {1, 3} → (1−1)·3 = 0 even ✓, (3−1)·1 = 2 even ✓ → 'No' Both outcomes possible → **II alone INSUFFICIENT.** ### Combined — both I and II Both statements deliver the **same restriction**: T is monoparous (all-odd or all-even). Neither pins down which side. - T = {2, 4} satisfies both → 'Yes' - T = {1, 3} satisfies both → 'No' **Combined still INSUFFICIENT → answer (d).**

Key concepts:

Parity arithmetic: sum-of-two even ⟺ same parity
Universal-quantifier statements: 'any two' = all pairs
Data sufficiency: when two statements deliver identical (not complementary) restrictions, combined often remains insufficient
(p−1)q even ⟺ p odd OR q even

Sources: Direct parity analysis of universal-quantifier conditions on a finite set; verified by counterexample pairs T = {2,4} vs T = {1,3}.

Q5. numeracy (data_sufficiency)

If $x$, $y$ and $z$ are integers, each greater than 1, then is $x$ a prime number?

Statement I: $xy^2 = 116$

Statement II: $xz = 261$

(a) Select this option if the question can be answered using one of these statements alone, but cannot be answered using other statement
(b) Select this option if the question can be answered using either statement alone
(c) Select this option if the question can be answered using both the statements together, but cannot be answered using either statement alone
(d) Select this option if the question cannot be answered even using any of the statements

Answer: (A) · Confidence: high

## Why the answer is (a) — Statement I alone suffices; Statement II alone does not ### Setup x, y, z are integers each > 1. Question: **is x prime?** ### Statement I alone — $xy^2 = 116$ Factorise: $116 = 2^2 \times 29$. We need integers x > 1 and y > 1 with xy² = 116. So **y² must be a perfect-square divisor of 116** with y > 1. Perfect-square divisors of $2^2 \cdot 29$: **1 and 4** (since 29 appears to the first power, no higher square works). - y > 1 forces y² ≥ 4 → **y² = 4, y = 2**. - Then **x = 116 / 4 = 29**, which is prime. Unique solution → x = 29 → **"Yes, x is prime."** **I alone SUFFICIENT.** ✓ ### Statement II alone — $xz = 261$ Factorise: $261 = 3^2 \times 29$. Integer pairs (x, z) with both > 1: | x | z | x prime? | |---|---|---| | 3 | 87 | **Yes** | | 9 | 29 | No | | 29 | 9 | **Yes** | | 87 | 3 | No | Both "Yes" and "No" outcomes possible → **II alone INSUFFICIENT.** ✗ ### Conclusion Exactly one statement (I) alone gives a unique answer; the other (II) does not → **(a)**. ### ⚡ Quick trick (≈25 sec) **Squarefree-prime trick:** When xy² = N and N's prime factorisation has at most one squared prime, the y² slot is pinned. Here 116 = 2²·29 — the only square > 1 dividing 116 is 4, forcing y = 2 and pinning x. Whenever you see this pattern in a data-sufficiency question, **the y² constraint is the hidden uniqueness lock** — it almost always makes Statement I sufficient. Statement II (xz = N with two distinct prime factors) gives 4 divisor pairs → never sufficient. **Answer (a).**

Key concepts:

Perfect-square divisors of an integer
Prime factorisation as a data-sufficiency lock
Constraint 'integer > 1' eliminates trivial 1-factor solutions
Enumeration of divisor pairs

Sources: Direct prime factorisation of 116 = 2²·29 and 261 = 3²·29; uniqueness of perfect-square divisors verified by elementary number theory.

Passage: 'Kalagram', the cultural village set up at the Maha Kumbh Mela, unfolded as a mosaic of India's diverse regions, each represented by seven meticulously crafted 'Sanskriti Angans'—stepping through the grand portal was like entering another world. These thematic zones, inspired by iconic temples like the Dakshineshwar Kali Temple and the Brahma Mandir, were treasure troves of regional artistry. Bengal's Pattachitra paintings, Assam's bamboo crafts, Tamil Nadu's Thanjavur paintings, and Madhya Pradesh's tribal sculpture—all were showcased in these living galleries where 230 master artisans breathed life into them using age-old techniques, their hands shaping India's ancient history into creations to behold.

Q6. comprehension (standard)

Which of the following conclusions are valid?

Seven Sanskriti Angans, representing different regions of India, had been showcased in Kalagram
Regional artistry was recognised via the inspiration drawn from iconic temples
India's ancient history had been crafted by the contemporary craftsmanship of 230 artisans into creations to behold
Art forms from all regions of India had been showcased in these living galleries

Select the answer using the code given below?

(a) 1 and 2 only
(b) 2, 3 and 4
(c) 1, 2 and 4
(d) 1 and 3

Answer: (D) · Confidence: high

## Why the answer is (d) — Only conclusions 1 and 3 are valid ### Method Comprehension rule: a conclusion is valid **only if it can be derived directly from the passage** without adding outside knowledge or over-generalising. Test each numbered statement against the passage text. ### Conclusion 1 — Seven Sanskriti Angans representing different regions of India had been showcased in Kalagram **Passage:** "...each represented by seven meticulously crafted 'Sanskriti Angans'..." and Kalagram was "a mosaic of India's diverse regions". Direct paraphrase. ✅ **VALID.** ### Conclusion 2 — Regional artistry was recognised via the inspiration drawn from iconic temples **Passage:** "These thematic zones, inspired by iconic temples like the Dakshineshwar Kali Temple and the Brahma Mandir, were treasure troves of regional artistry." The passage links **temples → thematic zones**, and separately **zones → regional artistry**. It does NOT say the recognition of artistry happened *because* of temple inspiration. The temple inspiration shaped the *zones*, not the recognition of art. This is a **causal over-link** — adds a causation the passage does not state. ❌ **INVALID.** ### Conclusion 3 — India's ancient history had been crafted by the contemporary craftsmanship of 230 artisans into creations to behold **Passage:** "...230 master artisans breathed life into them using age-old techniques, their hands shaping India's ancient history into creations to behold." Direct paraphrase: 230 artisans (who are contemporary in time, even if using age-old techniques) shaped India's ancient history into creations to behold. The phrase "contemporary craftsmanship" is consistent with the passage — the artisans work in the present, even though techniques are inherited. ✅ **VALID.** ### Conclusion 4 — Art forms from **all regions** of India had been showcased in these living galleries **Passage:** Mentions only **Bengal, Assam, Tamil Nadu, Madhya Pradesh** and refers to "diverse regions" — *not* "all regions". "All regions" is an **over-generalisation** beyond what the passage supports. ❌ **INVALID.** ### Match to options Valid: **1 and 3** → matches option **(d)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 2 only | ✗ (2 invalid) | | (b) | 2, 3 and 4 | ✗ (2 and 4 invalid) | | (c) | 1, 2 and 4 | ✗ (2 and 4 invalid) | | (d) | 1 and 3 | ✓ | ### ⚡ Quick trick (≈30 sec) **UPSC comprehension trap-spotting:** - **"All" / "every" / "always"** in a conclusion → almost always wrong if the passage uses "diverse", "many", "several" (so Conclusion 4 dies on the word "all"). - **Causal links the passage does not draw** ("X because of Y") → invalid even if both X and Y are mentioned (kills Conclusion 2). - This leaves 1 and 3 → **(d)**. Two trap-words eliminate two options in 30 seconds.

Key concepts:

Reading comprehension: passage-bound inference
Over-generalisation trap ("all" vs "diverse")
Causal over-link trap (linking two true facts with an unstated causation)
Paraphrase vs derivation

Sources: Verified strictly against the Kalagram passage in CSAT 2026 paper; no external information used.

Q7. comprehension (standard)

Which one of the following statements is ***not*** correct?

(a) Paintings from four States of India have been mentioned.
(b) Stepping into Kalagram is likened to stepping into another world.
(c) The Angans have been described as living galleries.
(d) Kalagram is divided into thematic zones, inspired by well-known temples of India.

Answer: (A) · Confidence: high

## Why the answer is (a) — Paintings from only TWO States are mentioned, not four ### Direct count from the passage The passage lists four states with their art forms: | State | Art form | Is it a painting? | |---|---|---| | Bengal | Pattachitra **paintings** | ✅ Yes | | Assam | Bamboo **crafts** | ❌ No | | Tamil Nadu | Thanjavur **paintings** | ✅ Yes | | Madhya Pradesh | Tribal **sculpture** | ❌ No | **Paintings are mentioned from only 2 states (Bengal and Tamil Nadu)** — not 4. Statement (a) is therefore **incorrect**. ### Verifying the other three are correct **(b) "Stepping into Kalagram is likened to stepping into another world."** Passage: "stepping through the grand portal was like entering another world". ✅ **Correct.** **(c) "The Angans have been described as living galleries."** Passage: "all were showcased in these living galleries" — referring back to the Angans. ✅ **Correct.** **(d) "Kalagram is divided into thematic zones, inspired by well-known temples of India."** Passage: "These thematic zones, inspired by iconic temples like the Dakshineshwar Kali Temple and the Brahma Mandir". ✅ **Correct.** ### ⚡ Quick trick (≈20 sec) **Categorical mismatch trap:** When an option says "X from N states", **count carefully and match the category**. Here "paintings" was the category, but the passage gave paintings from 2 states, crafts/sculpture from 2 more — a classic UPSC misattribution trap. **Answer (a)** in under half a minute by scanning art forms for the word "paintings".

Key concepts:

Comprehension: distinguishing paintings vs crafts vs sculpture
Categorical mismatch trap (counting items of a specific type)
Direct quote verification

Sources: Verified against the Kalagram passage in CSAT 2026 paper.

Q8. numeracy (standard)

The weight of $X$, in kg, is denoted by $X$. The weights of $A$, $B$, $C$, $D$, $P$, $Q$, $R$ and $S$ are measured. Given : $$A + B + C + D = 17$$ $$A + C = 6$$ $$P + Q + S + D = 15$$ $$P + Q + R + B = 17$$ $$P = R \text{ and } Q = S$$ Which one of the following statements is correct?

(a) $B$ and $D$ together weigh less than the total weight of $P$ and $Q$.
(b) $P$ and $Q$ together weigh more than the total weight of $A$ and $C$.
(c) $P$ weighs more than $Q$.
(d) $Q$ weighs more than $P$.

Answer: (B) · Confidence: high

## Why the answer is (b) — $P + Q > A + C$ ($7 > 6$) ### Given - (1) $A + B + C + D = 17$ - (2) $A + C = 6$ - (3) $P + Q + S + D = 15$ - (4) $P + Q + R + B = 17$ - (5) $P = R$ and $Q = S$ ### Step 1 — Compute $B + D$ From (1) − (2): $B + D = 17 - 6 = \mathbf{11}$. ### Step 2 — Substitute $P = R$, $Q = S$ into (3) and (4) - (3) becomes: $P + Q + Q + D = 15 \;\Rightarrow\; P + 2Q + D = 15$ - (4) becomes: $P + Q + P + B = 17 \;\Rightarrow\; 2P + Q + B = 17$ ### Step 3 — Add the two reduced equations $(P + 2Q + D) + (2P + Q + B) = 15 + 17$ $3P + 3Q + (B + D) = 32$ $3(P + Q) + 11 = 32 \;\Rightarrow\; 3(P + Q) = 21 \;\Rightarrow\; \mathbf{P + Q = 7}$. ### Step 4 — Test each option | Option | Inequality | Numbers | Verdict | |---|---|---|---| | (a) | $B + D < P + Q$ | $11 < 7$ | **False** | | (b) | $P + Q > A + C$ | $7 > 6$ | **True ✓** | | (c) | $P > Q$ | only sum known | Indeterminate | | (d) | $Q > P$ | only sum known | Indeterminate | Only (b) is verifiable and true → **answer (b)**. ### ⚡ Quick trick (≈45 sec) **Two-shortcut combo:** 1. $B + D = 17 - (A+C) = 17 - 6 = 11$ — falls out of subtracting (2) from (1). 2. Adding (3) and (4) **after** substituting $P=R, Q=S$ collapses 4 unknowns into one symmetric sum: $3(P+Q) + (B+D) = 32 \Rightarrow P+Q = 7$. Now skim options: (c) and (d) need individual values (impossible — only sum is fixed); (a) compares 11 vs 7 (false); only **(b)** survives — $7 > 6$. ✓

Key concepts:

Linear system manipulation by addition/subtraction
Symmetric substitution (P = R, Q = S)
Distinguishing solvable inequalities from indeterminate ones
Skip individual values when only sums are pinned

Sources: Direct algebraic manipulation of the given 5-equation system; verified by substitution.

Q9. numeracy (standard)

How many words can one form by shuffling the letters of the word QUEUE, if Q is always followed by U? The words thus formed need not necessarily have any meaning.

(a) 6
(b) 8
(c) 10
(d) 12

Answer: (D) · Confidence: high

## Why the answer is (d) — 12 distinct words ### Step 1 — Inventory the letters of QUEUE QUEUE contains: **Q × 1, U × 2, E × 2** (total 5 letters). ### Step 2 — Encode the constraint "Q is always followed by U" Glue **QU** into a single inseparable block. This consumes one Q and one U. Remaining loose letters: **U × 1, E × 2**. ### Step 3 — Count arrangements of the 4 objects: [QU], U, E, E We are now arranging 4 items in a row, of which the two E's are identical: $$\text{Arrangements} = \frac{4!}{2!} = \frac{24}{2} = \mathbf{12}.$$ ### Step 4 — Verify the block-method gives DISTINCT words Note that one of the loose letters is also a U, but since the block [QU] always carries its own Q ahead of its U, and the loose U is independent, **no double-counting** occurs. Every arrangement either has Q immediately followed by *the block's* U or by *the loose* U — but only [QU]-block arrangements satisfy the constraint, and we counted exactly those. ### Match to option **12 → option (d)**. ### ⚡ Quick trick (≈25 sec) **Glue-block method** for "X followed by Y" constraints: 1. Treat XY as one unit (eats one X + one Y from the inventory). 2. Arrange remaining items with the standard $n! / (r_1! \cdot r_2! \cdots)$ formula for repeats. Here: glue QU → 4 items with two E's repeated → $4!/2! = 12$. **(d)** in under 30 seconds.

Key concepts:

Permutations with repeated letters: $n! / (r_1! \cdot r_2! \cdots)$
Glue-block method for adjacency constraints
QUEUE letter inventory: Q(1), U(2), E(2)
"Followed by" = immediately followed (CSAT convention)

Sources: Direct enumeration via the glue-block permutation method; verified by formula $4!/2! = 12$.

Q10. numeracy (standard)

$X$, $Y$ and $Z$ jump forward 4′, 6′ and 5′, respectively. At 8 AM, they all land on mark 199′. How many times will they all land on the same mark (need not be at the same moment) between mark 195′ and 1000′, if all of them cross mark 1000′ by 9 AM?

(a) 11
(b) 12
(c) 13
(d) 14

Answer: (D) · Confidence: high

## Why the answer is (d) — 14 common landing marks ### Step 1 — Encode each jumper's landing-mark formula All three land on **mark 199′ at 8 AM**, then continue jumping forward: - **X** (stride 4): lands on $199, 203, 207, \ldots = 199 + 4j$ - **Y** (stride 6): lands on $199, 205, 211, \ldots = 199 + 6j$ - **Z** (stride 5): lands on $199, 204, 209, \ldots = 199 + 5j$ ### Step 2 — Condition for a common mark m $(m - 199)$ must be divisible by **4 AND 6 AND 5** simultaneously, i.e. by $$\text{LCM}(4, 6, 5) = 60.$$ So $m = 199 + 60k$ for $k = 0, 1, 2, \ldots$ ### Step 3 — Enumerate values in [195, 1000] | k | m = 199 + 60k | In range? | |---|---|---| | 0 | 199 | ✓ | | 1 | 259 | ✓ | | 2 | 319 | ✓ | | 3 | 379 | ✓ | | 4 | 439 | ✓ | | 5 | 499 | ✓ | | 6 | 559 | ✓ | | 7 | 619 | ✓ | | 8 | 679 | ✓ | | 9 | 739 | ✓ | | 10 | 799 | ✓ | | 11 | 859 | ✓ | | 12 | 919 | ✓ | | 13 | 979 | ✓ | | 14 | 1039 | ✗ (> 1000) | **Count: k = 0 to 13 → 14 common landing marks** → option **(d)**. ### ⚡ Quick trick (≈30 sec) **LCM-shortcut for synchronised-step problems:** $$\text{Number of common marks in } [199, 1000] \;=\; \left\lfloor \frac{1000 - 199}{\text{LCM}(4,6,5)} \right\rfloor + 1 \;=\; \left\lfloor \frac{801}{60} \right\rfloor + 1 \;=\; 13 + 1 \;=\; \mathbf{14}.$$ The "+1" counts the starting mark 199 itself. **(d)** in under half a minute.

Key concepts:

LCM as period of synchronised jumpers
Arithmetic progression: $a + dk$
Range-count formula: $\lfloor (\text{end} - \text{start}) / d \rfloor + 1$
Boundary inclusion (199 itself counts as a common landing)

Sources: Direct LCM computation and arithmetic-progression enumeration; verified by tabulation.

Passage: Sport is not just about winning medals or getting jobs through a sports quota. Today's generation is struggling with issues like depression and anxiety. Parents often say that their children are inactive and rarely leave the house. Sport can help tackle these problems. From sport you can learn time management, fitness, teamwork, coordination, and so much more. We need to develop a culture in which sport is seen as a way of life—a path to building a healthier, happier society.

Q11. comprehension (standard)

Which of the following conclusions are valid?

Sport is more than just games; it is a way of life
Sport can help mitigate the problems of seclusion among the young
Earning laurels in sport can help one secure a job on the basis of an assigned quota
Standard corporate sector skills cannot be learnt through sport

Select the answer using the code given below?

(a) 1 and 3 only
(b) 2 and 4 only
(c) 1, 2 and 4
(d) 1, 2 and 3

Answer: (D) · Confidence: high

## Why the answer is (d) — Conclusions 1, 2 and 3 are valid; only 4 fails ### Method Comprehension rule: a conclusion is valid only if directly supported by the passage. Apply per-statement. ### Conclusion 1 — Sport is more than just games; it is a way of life **Passage:** "Sport is **not just** about winning medals or getting jobs…" + "We need to develop a culture in which sport is seen as **a way of life**." Direct paraphrase combining both passage clauses. ✅ **VALID.** ### Conclusion 2 — Sport can help mitigate the problems of seclusion among the young **Passage:** "Parents often say that their children are inactive and rarely leave the house" (= seclusion-like behaviour) → "Sport can help tackle these problems." Direct implication. ✅ **VALID.** ### Conclusion 3 — Earning laurels in sport can help one secure a job on the basis of an assigned quota **Passage:** "Sport is **not just** about winning medals or **getting jobs through a sports quota**." The phrase "not just" *acknowledges* that sport-quota jobs DO exist as one aspect (the passage's point is that sport is *more than* that, not that it doesn't include it). The conclusion only claims sport "can help" — which the passage implicitly confirms. ✅ **VALID.** ### Conclusion 4 — Standard corporate sector skills cannot be learnt through sport **Passage:** "From sport you can learn **time management, fitness, teamwork, coordination**, and so much more." Time management, teamwork, and coordination are quintessential corporate-sector skills. The passage **directly contradicts** Conclusion 4. ❌ **INVALID.** ### Match to options Valid: **1, 2 and 3** → option **(d)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 3 only | ✗ (drops valid 2) | | (b) | 2 and 4 only | ✗ (4 invalid) | | (c) | 1, 2 and 4 | ✗ (4 invalid) | | (d) | 1, 2 and 3 | ✓ | ### ⚡ Quick trick (≈30 sec) **Kill-4-first method:** Conclusion 4 says corporate skills *cannot* be learnt — but the passage literally lists "time management, teamwork, coordination" as sport-learnt skills. **Two trap words** ("cannot" + the passage's enumeration) instantly eliminate any option containing 4 — that's (b) and (c). Then verify (d) ⊃ (a) by checking conclusion 2: passage's "children…rarely leave the house" → seclusion ✓. Confirm **(d)**.

Key concepts:

"Not just X" = X exists but is not the whole story
Comprehension: "cannot"/"never" claims need exhaustive disproof in passage
Passage-bound paraphrase vs over-extension
Elimination of options containing definitively invalid statements

Sources: Verified against the sport passage in CSAT 2026 paper.

Q12. comprehension (standard)

Which of the following statements is/are correct?

Parents are not encouraging enough when it comes to children playing sport
Participation in sporting activities can help develop life skills
Sport as a way of life can help evolve the very nature of society itself

Select the answer using the code given below?

(a) 1 and 3
(b) 2 and 3
(c) 2 only
(d) 3 only

Answer: (B) · Confidence: high

## Why the answer is (b) — Statements 2 and 3 only ### Statement 1 — Parents are not encouraging enough when it comes to children playing sport **Passage:** "Parents **often say** that their children are **inactive and rarely leave the house**." The passage shows parents *reporting* their children's inactivity. It does NOT comment on whether parents are encouraging or unencouraging. Inferring "parents are not encouraging enough" goes **beyond what the passage states** — it could equally be that parents are encouraging but children resist, or society at large is unsupportive. ❌ **INVALID** (over-inference). ### Statement 2 — Participation in sporting activities can help develop life skills **Passage:** "From sport you can learn **time management, fitness, teamwork, coordination**, and so much more." Time management, teamwork, coordination are textbook **life skills**. Direct support. ✅ **VALID.** ### Statement 3 — Sport as a way of life can help evolve the very nature of society itself **Passage:** "We need to develop a culture in which sport is seen as **a way of life—a path to building a healthier, happier society**." A society that becomes healthier and happier via cultural change *is* an evolved society. Direct support. ✅ **VALID.** ### Match to options Valid: **2 and 3** → option **(b)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 3 | ✗ (1 over-inferred) | | (b) | 2 and 3 | ✓ | | (c) | 2 only | ✗ (drops valid 3) | | (d) | 3 only | ✗ (drops valid 2) | ### ⚡ Quick trick (≈25 sec) **Attribution-trap spotting:** Statement 1 attributes a *judgement* ("parents not encouraging enough") to a passage that only reports an *observation* ("parents say children are inactive"). When a conclusion adds **blame or evaluation** that the passage does not explicitly assign, it's almost always invalid in UPSC comprehension. Eliminate (a). Statements 2 and 3 both have direct passage anchors → **(b)** in under half a minute.

Key concepts:

Attribution trap: report ≠ judgement
Life skills definition (time management, teamwork, coordination)
Passage-bound inference
"A path to" = cause-effect chain

Sources: Verified against the sport passage in CSAT 2026 paper.

Q13. numeracy (standard)

A toy $T$ jumps forward or backward. In each forward jump, it moves 5′ forward whereas in each backward jump, it moves 2′ backward. If in 31 jumps, $T$ moves exactly 15′ forward, then what is the difference of the number of forward and backward jumps?

(a) 6
(b) 7
(c) 8
(d) 9

Answer: (D) · Confidence: high

## Why the answer is (d) — Difference is 9 ### Step 1 — Variables and equations Let **F** = number of forward jumps (each +5′), **B** = number of backward jumps (each −2′). Two facts from the problem: - (i) **F + B = 31** (total jumps) - (ii) **5F − 2B = 15** (net displacement) ### Step 2 — Solve From (i): $B = 31 - F$. Substitute into (ii): $$5F - 2(31 - F) = 15 \;\Rightarrow\; 5F - 62 + 2F = 15 \;\Rightarrow\; 7F = 77 \;\Rightarrow\; F = 11.$$ Then $B = 31 - 11 = 20$. ### Step 3 — Difference $$|F - B| = |11 - 20| = \mathbf{9}.$$ **Answer: (d) 9.** ### Verification Forward distance: $11 \times 5 = 55′$. Backward distance: $20 \times 2 = 40′$. Net: $55 - 40 = 15′$. ✓ Total jumps: $11 + 20 = 31$. ✓ ### ⚡ Quick trick (≈20 sec) **Add-the-coefficients shortcut:** Combine the equations as one — multiply (i) by 2 and add to (ii): $2(F+B) + (5F - 2B) = 62 + 15 \;\Rightarrow\; 7F = 77 \;\Rightarrow\; F = 11$. Then $B = 20$, difference = **9**. **(d)** in under 30 seconds.

Key concepts:

Linear system with two unknowns
Forward/backward displacement balance
Coefficient-combination trick

Sources: Direct algebraic solution; verified by back-substitution.

Q14. reasoning (standard)

Eight persons $P$, $Q$, $R$, $S$, $T$, $U$, $V$ and $W$ sit around a round table in eight different seats placed with equal distance between any two consecutive seats. Both $P$ and $R$ are adjacent to $Q$. Both $T$ and $R$ are adjacent to $S$. Both $U$ and $W$ are adjacent to $V$. $S$ and $W$ are on opposite chairs. If while going in the clockwise direction around the table from $P$, one meets $R$ before $T$, then how many persons shall $Q$ cross while moving in the clockwise direction around the table before meeting $W$?

(a) 5
(b) 4
(c) 2
(d) 1

Answer: (A) · Confidence: high

## Why the answer is (a) — Q crosses 5 persons before meeting W ### Step 1 — Translate constraints into adjacencies - P and R adjacent to Q → **P – Q – R** (3 consecutive seats) - T and R adjacent to S → **T – S – R** (3 consecutive seats; R is the shared link) - U and W adjacent to V → **U – V – W** (3 consecutive seats) - S and W are on **opposite** seats (4 seats apart on the 8-seat ring) Merging the first two adjacency chains via the shared R: **P – Q – R – S – T** (5 consecutive seats). ### Step 2 — Place the 5-chain and the 3-chain on the ring Number seats 1 → 8 clockwise. Place the **P–Q–R–S–T** chain at positions 1–5: | Seat | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |---|---|---|---|---|---|---|---|---| | Person | P | Q | R | S | T | ? | ? | ? | Seats 6, 7, 8 are left for the **U–V–W** chain (V in the middle). ### Step 3 — Use the S–W opposite constraint S is at seat 4. Its opposite seat (4 + 4 mod 8) = **seat 8**. So **W = 8**. V is adjacent to W and lies in the U–V–W chain → **V = 7**, then **U = 6**. Final seating (clockwise from seat 1): | Seat | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |---|---|---|---|---|---|---|---|---| | Person | **P** | **Q** | **R** | **S** | **T** | **U** | **V** | **W** | ### Step 4 — Verify the "clockwise from P meets R before T" check Clockwise from P (seat 1): Q (2) → **R (3)** → S (4) → **T (5)** → … R is met at step 2, T at step 4. **R before T ✓**. ### Step 5 — Count persons Q crosses before meeting W Q is at seat 2. Going clockwise: seat 3 (R) → seat 4 (S) → seat 5 (T) → seat 6 (U) → seat 7 (V) → seat 8 (**W**, arrived). Persons crossed **before** meeting W: **R, S, T, U, V = 5 persons**. **Answer: (a) 5.** ### ⚡ Quick trick (≈45 sec) **Chain-merge shortcut for circular seating:** 1. Always merge "both X and Y adjacent to Z" into a 3-chain X–Z–Y. 2. Look for **shared elements** between two 3-chains to glue them into a longer chain (here R is shared by P–Q–R and R–S–T → P–Q–R–S–T). 3. **Opposite** constraint on an 8-seat ring = +4 mod 8 — instantly pins the second chain to a fixed seat. 4. "Persons crossed before meeting X" = clockwise gap **minus 1** (W is at seat 8, Q at seat 2 → gap = 6 seats clockwise, so 6 − 1 = **5** persons crossed before W). **(a)** in under a minute.

Key concepts:

Circular seating: 'between' = 3-chain
Chain merging via shared person
Opposite seats on n-ring = +n/2 mod n
Direction-test verification (clockwise from P meets R before T)

Sources: Direct logical deduction from adjacency and opposition constraints; verified by full clockwise traversal.

Q15. reasoning (standard)

The top of a table is rectangular and its dimensions are 6′ × 10′. Two rectangular portions of the table top are painted in blue colour; both these portions have dimensions 2·5′ × 8′ and each of them has exactly two sides common with two edges of the table top. If the table is fixed to the ground and the remaining portion of the table top is painted in white, how many different patterns are possible when observed from above?

(a) 2
(b) 4
(c) 6
(d) 8

Answer: (C) · Confidence: high

## Why the answer is (c) — Six distinct patterns ### Step 1 — Establish the geometry Table top: **6′ × 10′** (width × length). Each blue rectangle: **2.5′ × 8′** with exactly two sides flush with two edges of the table top. **Orientation:** the 8′ side cannot fit along the 6′ edge (8 > 6), so **the 8′ side must lie along the 10′ length edge** and the **2.5′ side along the 6′ width edge**. ### Step 2 — A 2-edge-flush blue rectangle sits in a corner "Exactly two sides common with two edges of the table top" means each blue rectangle occupies a **corner** — flush with one 10′ edge (top/bottom) AND one 6′ edge (left/right). There are **4 corners**: BL, BR, TL, TR. ### Step 3 — Place 2 blue rectangles (one per corner) — $\binom{4}{2}=6$ pairs | # | Pair | Visible union (above-view) | |---|---|---| | 1 | BL + BR | Full bottom strip $10 \times 2.5$ (rectangles overlap in centre $6 \times 2.5$) | | 2 | TL + TR | Full top strip $10 \times 2.5$ | | 3 | BL + TL | Two left strips, no overlap | | 4 | BR + TR | Two right strips, no overlap | | 5 | BL + TR | Diagonal pair, no overlap | | 6 | BR + TL | Anti-diagonal pair, no overlap | ### Step 4 — All six patterns are visually distinct Crucial: the problem says the table is **fixed to the ground** — so we do NOT collapse patterns related by rotation. Each of the 6 corner-pair placements yields a unique above-view image: - 1 and 2 are different solid strips (bottom vs top) - 3 and 4 are different left-strip vs right-strip layouts - 5 and 6 are mirror diagonals (still visually different on a fixed table) **Total distinct patterns: 6** → option **(c)**. ### ⚡ Quick trick (≈30 sec) **Corner-counting shortcut:** When two identical rectangles each occupy a corner of a fixed (non-rotatable) tabletop, **the count of distinct patterns = $\binom{N}{2}$ where N = number of corners**. Here N = 4 → $\binom{4}{2} = \mathbf{6}$. The fixed-to-ground clause is what stops you from collapsing rotational duplicates. **(c)** in under 30 seconds.

Key concepts:

Combinatorial placement on a fixed object (no rotation symmetry)
Corner-flush rectangle constraint
$\binom{n}{2}$ counting for unordered pairs
Overlap permitted (union is what's visible)

Sources: Direct geometric construction; all 6 corner pairs enumerated and verified distinct.

Passage: How is deflation done? Most countries use a method called 'double deflation', where input and output prices are deflated separately. Consider a manufacturer importing oil for use in production. If oil prices fall, output prices do not and quantities remain the same, real value added should not change. But if the same deflator is used for inputs and outputs, as in India, it would look as if the manufacturer had become more productive.

Q16. comprehension (standard)

Which of the following statements is/are correct?

Real value should not change in the instance of static output cost and unchanged quantities against falling oil prices
Deflators are to be used separately for inputs and outputs, and this is a practice universally adopted by all economies

Select the answer using the code given below?

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2

Answer: (A) · Confidence: high

## Why the answer is (a) — Only statement 1 is correct ### Statement 1 — Real value should not change when output cost is static and quantities unchanged despite falling oil prices **Passage:** "If oil prices fall, output prices do not and quantities remain the same, **real value added should not change**." Direct paraphrase. ✅ **VALID.** ### Statement 2 — Deflators are to be used separately for inputs and outputs, universally adopted by all economies **Passage:** "**Most** countries use a method called 'double deflation'… But if the same deflator is used for inputs and outputs, **as in India**, it would look as if the manufacturer had become more productive." Two passage facts directly contradict statement 2: - Passage says **"most"** countries — not **"all"** (universal-claim trap). - Passage explicitly names **India** as a counter-example (uses the same deflator). So the practice is NOT universal. ❌ **INVALID.** ### Match to options Valid: **1 only** → option **(a)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 only | ✓ | | (b) | 2 only | ✗ (universal claim wrong) | | (c) | Both | ✗ (2 invalid) | | (d) | Neither | ✗ (1 valid) | ### ⚡ Quick trick (≈20 sec) **Universal-quantifier trap:** Statement 2 uses **"universally adopted by all economies"** — but the passage explicitly says **"most countries"** and names **India** as the exception. **Any "all/every/universally" claim that the passage contradicts with even one counter-example** dies instantly. **(a)** in 20 seconds — eliminate any option containing 2, only (a) survives.

Key concepts:

Universal-quantifier trap (all vs most)
Counter-example invalidation
Passage-bound paraphrase
Double deflation methodology

Sources: Verified strictly against the deflation passage in CSAT 2026 paper.

Q17. comprehension (standard)

Which of the following assumptions is/are valid?

Deflation strategies can be used to make manufacturers appear to be doing better than they actually are
When input and output prices are both deflated against a single input price, it is referred to as 'double deflation'

Select the answer using the code given below?

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2

Answer: (A) · Confidence: high

## Why the answer is (a) — Only assumption 1 is valid ### Assumption 1 — Deflation strategies can make manufacturers appear to do better than they actually are **Passage:** "…if the **same deflator is used for inputs and outputs**, as in India, it would **look as if the manufacturer had become more productive**." The passage clearly states that a particular deflation choice (single deflator instead of separate ones) **creates the appearance of higher productivity** than what really occurred. This is the very definition of "appear better than actually are." ✅ **VALID.** ### Assumption 2 — When input and output prices are both deflated against a single input price, it is referred to as 'double deflation' **Passage definition:** "…a method called '**double deflation**', where input and output prices are deflated **separately**." Statement 2 reverses the definition — it describes the **single-deflator method** (the wrong one used in India) but mislabels it as "double deflation." Double deflation = **separate** deflators, not a single one. ❌ **INVALID** (definitional inversion). ### Match to options Valid: **1 only** → option **(a)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 only | ✓ | | (b) | 2 only | ✗ | | (c) | Both | ✗ | | (d) | Neither | ✗ | ### ⚡ Quick trick (≈25 sec) **Definitional-inversion trap:** Statement 2 sounds technical and uses the right keyword ("double deflation") but **inverts the passage's definition** (separate ↔ single). Whenever an option swaps the meaning of a defined term from the passage, it's an instant kill. **(a)** in under 30 seconds.

Key concepts:

Definitional-inversion trap (swapping the meaning of a passage-defined term)
Double deflation = separate deflators for input vs output
Productivity illusion from misapplied deflator
Passage-bound assumption testing

Sources: Verified strictly against the deflation passage in CSAT 2026 paper.

Q18. reasoning (standard)

A pattern formed by two characters $a$ and $b$ is repeated more than once in the following string : $$\times b \times a \times a \times \times a \times a \times bab$$ What is $\times \times$ in the 7th and 8th positions from the left in the above string?

(a) $aa$
(b) $ab$
(c) $ba$
(d) $bb$

Answer: (D) · Confidence: high

## Why the answer is (d) — bb ### Step 1 — Index the string The string has **15 characters**. Mark known (a/b) vs unknown (×): | Pos | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | |---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---| | Char | × | b | × | a | × | a | × | × | a | × | a | × | b | a | b | ### Step 2 — Find the repeating block length "A pattern formed by two characters $a$ and $b$ is repeated more than once" + total length 15 ⇒ block length must divide 15. Divisors > 1: **3, 5, 15**. Block length 15 = no repetition. Try **5** (gives 3 repetitions). ### Step 3 — Match positions modulo 5 If block length = 5, then positions $i$ and $i + 5$ must hold the same character: | Mod-5 class | Positions | Known values | Inferred | |---|---|---|---| | 1 | 1, 6, 11 | ×, a, a | **a** | | 2 | 2, 7, 12 | b, ×, × | **b** | | 3 | 3, 8, 13 | ×, ×, b | **b** | | 4 | 4, 9, 14 | a, a, a | a | | 5 | 5, 10, 15 | ×, ×, b | **b** | Repeating block: **a b b a b**. ### Step 4 — Reconstruct the full string and verify Full string = (a b b a b) × 3 = **a b b a b | a b b a b | a b b a b** (positions 1–15). Cross-check against every known character: pos 2 = b ✓, pos 4 = a ✓, pos 6 = a ✓, pos 9 = a ✓, pos 11 = a ✓, pos 13 = b ✓, pos 14 = a ✓, pos 15 = b ✓. **All matches.** ### Step 5 — Read off positions 7 and 8 Pos 7 = b, Pos 8 = b → **bb** → option **(d)**. ### ⚡ Quick trick (≈30 sec) **Period-detection from divisors:** Whenever a length-N string is said to be "a pattern repeated", the period **divides N**. For N = 15, try **5** (most useful divisor — 3 reps). Now align positions modulo 5 and inherit known letters across classes. Here position-2 class fixes letter at pos 7 = **b**, position-3 class fixes letter at pos 8 = **b**. **Answer (d)** in under 30 seconds.

Key concepts:

Period detection: period divides string length
Modular position alignment for repeating blocks
Inheritance of known characters across equivalent positions
Verification by full reconstruction

Sources: Direct construction by modular alignment of positions; verified by checking all 8 known characters.

Q19. numeracy (standard)

If $10^m \times 1000 \times n = 75^{25} \times 25^{32} \times 32^{75}$, where $n$ is not divisible by 10, then the value of $m$ is

(a) 101
(b) 111
(c) 121
(d) 131

Answer: (B) · Confidence: high

## Why the answer is (b) — $m = 111$ ### Step 1 — Prime-factorise the RHS $$75^{25} \times 25^{32} \times 32^{75}$$ - $75^{25} = (3 \cdot 5^2)^{25} = 3^{25} \cdot 5^{50}$ - $25^{32} = (5^2)^{32} = 5^{64}$ - $32^{75} = (2^5)^{75} = 2^{375}$ $$\text{RHS} = 2^{375} \cdot 3^{25} \cdot 5^{50+64} = 2^{375} \cdot 3^{25} \cdot 5^{114}.$$ ### Step 2 — Factorise the LHS $$10^m \times 1000 \times n = 2^m \cdot 5^m \cdot 2^3 \cdot 5^3 \cdot n = 2^{m+3} \cdot 5^{m+3} \cdot n.$$ ### Step 3 — Solve for $n$ $$n = \frac{2^{375} \cdot 3^{25} \cdot 5^{114}}{2^{m+3} \cdot 5^{m+3}} = 2^{372-m} \cdot 3^{25} \cdot 5^{111-m}.$$ ### Step 4 — Apply the constraint "$n$ is not divisible by 10" $n$ not divisible by 10 ⟺ $n$ does **not** contain both 2 and 5 as factors. Since 3 doesn't matter, we need either: - (i) exponent of 5 in $n$ = 0, i.e. $111 - m = 0 \Rightarrow \mathbf{m = 111}$, or - (ii) exponent of 2 in $n$ = 0, i.e. $372 - m = 0 \Rightarrow m = 372$ (rejected — exceeds available options). Also: for $n$ to remain a positive integer, both exponents must be **≥ 0**, giving $m \le 111$. So $m = 111$ is the **unique** valid value. ### Step 5 — Verify At $m = 111$: - $n = 2^{372-111} \cdot 3^{25} \cdot 5^{111-111} = 2^{261} \cdot 3^{25} \cdot 5^0 = 2^{261} \cdot 3^{25}$. - $n$ has no factor of 5 → not divisible by 10. ✓ **Answer: $m = 111$ → option (b).** ### ⚡ Quick trick (≈40 sec) **Cap-the-fives shortcut:** When the RHS has $5^k$ and the LHS pulls out $5^{m+3}$, the constraint "n has no factor 10" means **pull out ALL the 5's** — set $m + 3 = k$ (here $k = 114$ → $m = 111$). Bonus check: confirm 2's exponent stays non-negative ($372 - 111 = 261 \ge 0$ ✓). **(b)** in under a minute.

Key concepts:

Prime factorisation of composite powers
Divisibility by 10 = both 2 and 5 present
Cap the smaller prime exponent to extract maximum 10-factor
Non-negativity constraints on exponents

Sources: Direct prime factorisation; verified by reconstructing both sides at $m=111$.

Q20. numeracy (standard)

The speed of a train $T$ is 100 km per hour and the speed of a person $P$ is 4 km per hour. $T$ crosses $P$ in 15 seconds, if $P$ travels along the direction of motion of $T$. If $P$ travels along the opposite direction of $T$, then in how much time does $T$ cross $P$, in seconds, approximately?

(a) 13·51
(b) 13·65
(c) 13·85
(d) 14·05

Answer: (C) · Confidence: high

## Why the answer is (c) — Approximately 13·85 seconds ### Step 1 — Compute train length from the same-direction crossing When both move in the same direction, relative speed = $100 - 4 = 96$ km/h. Convert to m/s: $96 \times \frac{5}{18} = \frac{480}{18} = \frac{80}{3} \approx 26.67$ m/s. Length of train $L$ = relative speed × time = $\frac{80}{3} \times 15 = \mathbf{400}$ m. ### Step 2 — Compute opposite-direction crossing time Now relative speed = $100 + 4 = 104$ km/h. In m/s: $104 \times \frac{5}{18} = \frac{520}{18} = \frac{260}{9} \approx 28.89$ m/s. Time to cross = $\frac{L}{\text{rel speed}} = \frac{400}{260/9} = \frac{400 \times 9}{260} = \frac{3600}{260} = \frac{180}{13} \approx \mathbf{13.846}$ s. Rounded to 2 dp: **13·85 s** → option **(c)**. ### ⚡ Quick trick (≈30 sec) **Ratio shortcut — skip the absolute length:** $$\frac{t_{\text{opp}}}{t_{\text{same}}} = \frac{v_{\text{rel,same}}}{v_{\text{rel,opp}}} = \frac{96}{104} = \frac{12}{13}.$$ $$t_{\text{opp}} = 15 \times \frac{12}{13} = \frac{180}{13} \approx \mathbf{13.846} \approx 13.85 \text{ s}.$$ Skip the m/s conversion entirely — the ratio cancels. **(c)** in well under 30 seconds.

Key concepts:

Relative velocity (same vs opposite direction)
Length of train = relative speed × crossing time
Time ratio = inverse of speed ratio when distance is fixed
km/h ↔ m/s conversion factor 5/18

Sources: Direct kinematics with relative-velocity method; verified via length computation.

Q21. numeracy (standard)

For $\frac{1}{3} < x < y < 2$, which of the following statements is/are always correct? I. $x + \frac{1}{x} < y + \frac{1}{y}$ II. $\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}$

Select the answer using the code given below?

(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II

Answer: (B) · Confidence: high

## Why the answer is (b) — Only Statement II is always correct ### Statement I — $x + \frac{1}{x} < y + \frac{1}{y}$ for all $\frac{1}{3} < x < y < 2$ Consider $f(t) = t + \frac{1}{t}$. Its derivative is $f'(t) = 1 - \frac{1}{t^2}$, which is: - **negative for $t < 1$** (f is DECREASING), - **zero at $t = 1$** (minimum), - **positive for $t > 1$** (f is INCREASING). On the interval $(1/3, 2)$, f is not monotonic — it first falls (until $t=1$) then rises. **Counterexample (both< 1):** $x = 0.4$, $y = 0.5$: $f(x) = 0.4 + 2.5 = 2.9$ and $f(y) = 0.5 + 2 = 2.5$ → $f(x) > f(y)$. **Statement I FAILS.** ❌ ### Statement II — $\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}$ for all $\frac{1}{3} < x < y < 2$ Rewrite each side: $$\frac{\sqrt{1+t^2}}{t} = \sqrt{\frac{1+t^2}{t^2}} = \sqrt{\frac{1}{t^2} + 1}.$$ So Statement II becomes: $\sqrt{\frac{1}{y^2} + 1} < \sqrt{\frac{1}{x^2} + 1}$. Both sides positive — squaring preserves order: $$\frac{1}{y^2} + 1 < \frac{1}{x^2} + 1 \;\Longleftrightarrow\; \frac{1}{y^2} < \frac{1}{x^2} \;\Longleftrightarrow\; y^2 > x^2.$$ Since $x, y > 0$ and $x < y$, we always have $x^2 < y^2$. ✅ **Statement II is ALWAYS TRUE.** ### Match to options Valid: **II only** → option **(b)**. ### ⚡ Quick trick (≈40 sec) **Monotonicity check first:** $f(t) = t + 1/t$ has a U-shape with minimum at $t = 1$. The interval $(1/3, 2)$ straddles the minimum, so $f$ is **not monotonic** — kills Statement I instantly via the U-shape counterexample (try any pair both < 1 or both > 1 vs straddling 1). **Statement II is monotonic in $t$:** $\frac{\sqrt{1+t^2}}{t}$ is **strictly decreasing** on $t > 0$ (derivative of $\frac{1}{t^2} + 1$ is $-2/t^3 < 0$). So $y > x \Rightarrow \text{II RHS} > \text{II LHS}$. Always true. **(b)** in under a minute.

Key concepts:

Monotonicity check via derivative
$f(t) = t + 1/t$ has minimum at $t = 1$ for $t > 0$
Squaring preserves inequalities for positive sides
$\sqrt{1+t^2}/t$ is strictly decreasing on $t > 0$
Counterexample method for 'always correct' claims

Sources: Direct calculus + counterexample method; verified with x=0.4, y=0.5 for Statement I.

Q22. numeracy (standard)

What is the minimum number of times one needs to measure to get 298 litres of water from a tank, if the measuring cylinders have capacities 1 litre, 6 litres, 25 litres and 100 litres?

(a) 4
(b) 5
(c) 9
(d) 13

Answer: (B) · Confidence: high

## Why the answer is (b) — Minimum 5 measurements ### Step 1 — Recognise that 'measure' allows both pour-IN and remove-OUT Each measurement uses one cylinder, which can be **added to** or **removed from** the tank. We want a sequence of $\pm 1, \pm 6, \pm 25, \pm 100$ summing to **298 L** with minimum total operations. ### Step 2 — Greedy approach overshoots Pure addition: $298 = 2(100) + 3(25) + 3(6) + 5(1) = 200 + 75 + 18 + 5$ → $2+3+3+5 = \mathbf{13}$ measurements (option d). Too many. ### Step 3 — Use over-and-subtract **Key insight:** $298 = 300 - 2 = 3(100) - 2(1)$. - Pour 100 L three times → tank has 300 L (**3 ops**) - Remove 1 L twice → tank has 298 L (**2 ops**) **Total: 5 measurements.** ### Step 4 — Prove 4 measurements is impossible We need $\sum_{i=1}^{4} \epsilon_i \cdot c_i = 298$, where $\epsilon_i \in \{+1, -1\}$ and $c_i \in \{1, 6, 25, 100\}$. Check all 4-op combinations: - **All four 100s:** $\pm 100 \pm 100 \pm 100 \pm 100 \in \{0, \pm 200, \pm 400\}$ — no 298. - **Three 100s + one other:** $300 \pm c$, where $c \in \{1,6,25,100\}$. Values: $\{299, 301, 294, 306, 275, 325, 200, 400\}$ — none is 298. - **Two 100s + two others:** $200 + a + b$, where $a, b \in \{\pm 1, \pm 6, \pm 25\}$. Need $a + b = 98$. Max $|a + b| = 25 + 25 = 50$. Impossible. - **Fewer 100s:** Max achievable = $100 + 25 + 25 + 25 = 175$ or with one 100 only, ≤ $100 + 25 + 25 + 25 = 175$. Cannot reach 298. **No 4-op solution exists.** Minimum = 5. ✓ **Answer: (b) 5.** ### ⚡ Quick trick (≈30 sec) **Over-and-subtract heuristic:** When the target is close to a round multiple of the largest cylinder, **overshoot then subtract** the small difference: $$298 = 3 \times 100 - 2 \times 1 \;\Rightarrow\; 3 + 2 = \mathbf{5 \text{ ops}}.$$ This pattern (overshoot by k, correct with k smallest cylinders) almost always beats the greedy partition. **(b)** in under half a minute.

Key concepts:

Measurement = pour-in OR remove-out (signed contribution)
Over-and-subtract heuristic for near-round targets
Greedy partition often suboptimal
Exhaustive case-check for 4-op impossibility

Sources: Direct construction: 3×100 − 2×1 = 298; minimality proven by exhaustive 4-op case analysis.

Q23. numeracy (standard)

There are four types of weights, namely 1 kg, 2 kg, 5 kg and 10 kg. What is the maximum number of different ways one can measure 20 kg, if at least eight but not more than eleven weights of 1 kg are to be used while measuring?

(a) 7
(b) 8
(c) 9
(d) 10

Answer: (B) · Confidence: high

## Why the answer is (b) — 8 distinct ways ### Setup Let $a$ = # of 1-kg weights, $b$ = # of 2-kg, $c$ = # of 5-kg, $d$ = # of 10-kg. Find non-negative integer solutions to: $$a + 2b + 5c + 10d = 20, \quad 8 \le a \le 11.$$ Enumerate by fixed $a$. ### Case $a = 8$: $2b + 5c + 10d = 12$ | $d$ | $5c + 2b$ | $(c, b)$ | |---|---|---| | 0 | 12 | $(0, 6)$, $(2, 1)$ | | 1 | 2 | $(0, 1)$ | **Subtotal: 3.** ### Case $a = 9$: $2b + 5c + 10d = 11$ | $d$ | $5c + 2b$ | $(c, b)$ | |---|---|---| | 0 | 11 | $(1, 3)$ | (c = 0 gives $2b = 11$ which has no integer solution; $d = 1$ gives $5c + 2b = 1$, impossible.) **Subtotal: 1.** ### Case $a = 10$: $2b + 5c + 10d = 10$ | $d$ | $5c + 2b$ | $(c, b)$ | |---|---|---| | 0 | 10 | $(0, 5)$, $(2, 0)$ | | 1 | 0 | $(0, 0)$ | **Subtotal: 3.** ### Case $a = 11$: $2b + 5c + 10d = 9$ | $d$ | $5c + 2b$ | $(c, b)$ | |---|---|---| | 0 | 9 | $(1, 2)$ | **Subtotal: 1.** ### Total $$3 + 1 + 3 + 1 = \mathbf{8} \;\Rightarrow\; \text{option (b)}.$$ ### ⚡ Quick trick (≈45 sec) **Parity-shortcut:** After fixing $a$, the residual $R = 20 - a$ must equal $2b + 5c + 10d$. Note that $5c + 10d$ is always a multiple of 5, so $2b$ must equal $R \mod 5$. This rapidly narrows the search: - $a = 8 \Rightarrow R = 12$, $R \mod 5 = 2 \Rightarrow 2b \in \{2, 12\}$, i.e. $b \in \{1, 6\}$. 3 combos as enumerated. - $a = 9 \Rightarrow R = 11$, $R \mod 5 = 1$, need $2b \equiv 1 \mod 5$; smallest fit $b = 3$ ($2b = 6$). 1 combo. - $a = 10 \Rightarrow R = 10$, $R \mod 5 = 0 \Rightarrow b \in \{0, 5\}$. 3 combos. - $a = 11 \Rightarrow R = 9$, $R \mod 5 = 4 \Rightarrow b = 2$. 1 combo. **Total 8 → (b).**

Key concepts:

Non-negative integer partitions with a constrained component
Case enumeration by fixing the constrained variable
Modular arithmetic shortcut to narrow integer solutions
Diophantine equations: $a + 2b + 5c + 10d = N$

Sources: Direct enumeration across $a \in \{8, 9, 10, 11\}$; verified by tabulation.

Q24. reasoning (standard)

A cut on a solid object divides the object into two parts where the new surfaces thus produced are plane. On the other hand, one single cut can be used to cut more than one object at a time. In an experiment, the total number of pieces produced by applying $n$ cuts is denoted by $x_n$. The experiment is performed on a solid cube where pieces remain unmoved after each cut. In this experiment, if after the third cut, the pieces are identical, then which of the following is ***not*** a possible value for $x_4$?

(a) 16
(b) 12
(c) 8
(d) 5

Answer: (A) · Confidence: high

## Why the answer is (a) — $x_4 = 16$ is NOT achievable ### Step 1 — Enumerate configurations producing identical pieces after 3 cuts On a solid cube, three cuts producing **identical** pieces yield two principal configurations: - **Config A:** 3 parallel cuts equally spaced → **4 identical slabs** ($x_3 = 4$) - **Config B:** 3 mutually perpendicular cuts through the centre → **8 identical sub-cubes** ($x_3 = 8$) ### Step 2 — From Config A (4 slabs), the 4th cut yields $x_4 \in \{5, 6, 7, 8\}$ A planar cut on 4 slabs splits between 1 and 4 of them, so $x_4 = 4 + k$ for $k \in \{1, 2, 3, 4\}$: - Cut splits 1 slab → $x_4 = 5$ ✓ - Cut splits 4 slabs → $x_4 = 8$ ✓ ### Step 3 — From Config B (8 sub-cubes), the 4th cut yields $x_4 \in \{9, 10, \ldots, 15\}$ Same logic: $x_4 = 8 + k$ for the number $k$ of sub-cubes the planar cut splits. By careful tilting, a single plane can split **up to 7** of the 8 sub-cubes (the 8th remains on one side because it sits in a far corner). So $x_4 \le 8 + 7 = 15$. **Crucially, a single planar cut CANNOT split all 8 sub-cubes simultaneously** in a 2 × 2 × 2 arrangement — the opposite-corner sub-cubes lie on opposite sides of any plane that passes through the central region, leaving at least one sub-cube uncut. ### Step 4 — Match against options | Option | Value | Achievable? | Source | |---|---|---|---| | (a) | **16** | **✗ NO** | needs all 8 sub-cubes split — impossible | | (b) | 12 | ✓ Yes | Config B + cut through 4 sub-cubes (e.g. horizontal plane through middle of one layer) | | (c) | 8 | ✓ Yes | Config A + cut perpendicular through all 4 slabs | | (d) | 5 | ✓ Yes | Config A + cut splits only 1 slab | **Answer: (a) 16 — NOT a possible value.** ### ⚡ Quick trick (≈45 sec) **Maximum-pieces ceiling:** A single planar cut on a $2 \times 2 \times 2$ arrangement of sub-cubes splits **at most 7** of the 8 sub-cubes (the 8th always survives in a far corner). Therefore from Config B (the only config that can reach high $x_4$), $x_4 \le 15$. **16 exceeds the ceiling → (a)**. Other options achievable as tabulated.

Key concepts:

Identical-pieces configurations after 3 cuts: 4 slabs OR 8 sub-cubes
$x_4 = x_3 + (\text{number of pieces the new cut splits})$
Single-plane splitting ceiling for $2 \times 2 \times 2$ sub-cube arrangement = 7
16 = 8 + 8 requires all 8 sub-cubes split — geometrically impossible

Sources: Direct enumeration of identical-piece configurations and case analysis of 4th-cut outcomes; geometric impossibility of single-plane split of all 8 sub-cubes.

Q25. numeracy (standard)

The class average $x$ in a test increases by 4 when the score of a student is rectified, whose corrected score is 100 instead of 0. Later, the score of another student was found to have been recorded as 81 in place of 56. If there are no other corrections and the final corrected average is $y$, then $y - x$ is

(a) 2
(b) 3
(c) 5
(d) 6

Answer: (B) · Confidence: high

## Why the answer is (b) — $y - x = 3$ ### Step 1 — Find class size $N$ from the first correction First correction: a student's score changed from 0 to 100 → **total marks increase by 100** → average increases by $\frac{100}{N}$. Given that this increase equals 4: $$\frac{100}{N} = 4 \;\Rightarrow\; \boxed{N = 25}.$$ After the first correction, the average is $x + 4$ (where $x$ = original class average). ### Step 2 — Apply the second correction Second correction: a student's score changed from 81 to 56 → **total marks decrease by $81 - 56 = 25$** → average decreases by $\frac{25}{N} = \frac{25}{25} = 1$. After both corrections, the final average is: $$y = (x + 4) - 1 = x + 3.$$ $$y - x = \mathbf{3} \;\Rightarrow\; \text{option (b)}.$$ ### ⚡ Quick trick (≈25 sec) **Two-stage delta:** Each correction shifts the average by $\frac{\Delta \text{total}}{N}$. Find $N$ from the first delta, then apply both. - $\Delta_1 = +100$ shifts avg by $+4 \Rightarrow N = 25$ - $\Delta_2 = -25$ shifts avg by $-1$ - Net shift: $+4 - 1 = +3 \Rightarrow y - x = \mathbf{3}$. **(b)** in under 30 seconds.

Key concepts:

Average change = sum change / N
Inferring class size from a known delta
Net change = sum of individual deltas

Sources: Direct application of average-shift formula; verified by two-stage delta accounting.

Passage: When SARS-CoV-2 was first detected in 2019, it was a truly novel virus for the world. At that time, no one in the world had been exposed to SARS-CoV-2 or had specific immunity against it. In contrast, people across the world have been exposed to HMPV for decades and the virus is well-studied. HMPV and SARS-CoV-2 belong to two very different virus families with fundamentally different characteristics and epidemiology, with strong seasonality seen in HMPV, unlike SARS-CoV-2. Both viruses cause different severity of symptoms, particularly over the long term, and the affected population segments do not fully overlap. In general, HMPV causes milder illness with deaths being very rare and with no long-term post-viral symptoms.

Q26. comprehension (standard)

Which of the following conclusions is/are valid?

Though SARS-CoV-2 and HMPV are similar viruses with somewhat different epidemiology, the former became a pandemic because it was novel and people had not been exposed to it in the past
The two viruses have fundamentally different impacts on human populations and should not therefore be dealt with in a similar manner

Select the answer using the code given below?

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2

Answer: (B) · Confidence: high

## Why the answer is (b) — Only conclusion 2 is valid ### Conclusion 1 — "SARS-CoV-2 and HMPV are **similar viruses** with somewhat different epidemiology…" **Passage:** "HMPV and SARS-CoV-2 belong to **two very different virus families** with **fundamentally different characteristics and epidemiology**." The passage explicitly says the viruses belong to **different families** and have **fundamentally different** characteristics. Calling them "similar viruses" directly contradicts the passage. ❌ **INVALID.** ### Conclusion 2 — "The two viruses have fundamentally different impacts on human populations and should not therefore be dealt with in a similar manner" **Passage:** "Both viruses cause **different severity of symptoms**, particularly over the long term, and the **affected population segments do not fully overlap**… HMPV causes milder illness with deaths being very rare and with no long-term post-viral symptoms." The passage establishes: - Different severity of symptoms ✓ - Different long-term effects (no long-term post-viral symptoms in HMPV) ✓ - Different affected populations ✓ - HMPV milder, deaths rare ✓ All point to fundamentally different impacts. The inference "should not be dealt with in a similar manner" is a reasonable policy implication of these differences. ✅ **VALID.** ### Match to options Valid: **2 only** → option **(b)**. ### ⚡ Quick trick (≈20 sec) **Direct-contradiction kill:** Conclusion 1 calls the viruses "similar" while the passage uses the word **"fundamentally different"** — a literal contradiction (the strongest comprehension-trap signal). Kill conclusion 1, eliminate (a), (c). Conclusion 2 paraphrases the passage's differential-impact narrative directly → only (b) remains.

Key concepts:

Direct contradiction trap ("similar" vs passage's "fundamentally different")
Policy inference from passage-stated differences
Comprehension: passage-bound paraphrase

Sources: Verified strictly against the HMPV vs SARS-CoV-2 passage in CSAT 2026 paper.

Q27. comprehension (standard)

Which of the following reflect the intent of the writer in the above passage?

To evolve methodologies for objective analysis of the two viruses
To establish the epidemiological similarities and differences between the two viruses
To offer a better understanding of the remedies of HMPV when analysed in conjunction with SARS-CoV-2

Select the answer using the code given below?

(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) None of the above

Answer: (D) · Confidence: medium

## Why the answer is (d) — None of the listed options correctly captures the writer's intent ### Statement 1 — "To evolve methodologies for objective analysis of the two viruses" The passage **does not propose, evolve or describe any methodology** for analysing viruses. It simply states known facts about both. ❌ **NOT writer's intent.** ### Statement 2 — "To establish the epidemiological similarities and differences between the two viruses" The passage focuses overwhelmingly on **differences** — it explicitly uses the phrase "**fundamentally different** characteristics and epidemiology" and lists differences in family, seasonality, severity, long-term effects and affected populations. The passage **does NOT meaningfully establish similarities** — it mentions only that both are viruses (a trivial observation, not a substantive similarity). A claim to "establish similarities and differences" overstates what the passage does. ❌ **NOT a faithful description of intent** (intent is to contrast, not to balance similarities and differences). ### Statement 3 — "To offer a better understanding of the remedies of HMPV when analysed in conjunction with SARS-CoV-2" The passage contains **zero discussion of remedies, treatments, vaccines, or therapeutics** for either virus. The only outcome-related claim is that "HMPV causes milder illness with deaths being very rare" — not a remedy. ❌ **NOT writer's intent.** ### Writer's actual intent The passage's **real intent** is narrower: to **contrast** SARS-CoV-2 (novel, no prior immunity, pandemic potential) with HMPV (long-known, well-studied, milder). It is a **differentiating contrast**, not a balanced similarities-and-differences analysis, not a methodology paper, and not a remedies discussion. Since none of statements 1, 2, 3 captures this intent accurately, no combination in (a), (b) or (c) fits → **option (d)** is correct. ### Match to options | Option | Set | Verdict | |---|---|---| | (a) | 1 and 2 | ✗ (neither faithfully captures intent) | | (b) | 2 and 3 | ✗ (3 has no passage basis) | | (c) | 1 and 3 | ✗ (both irrelevant) | | (d) | None of the above | ✓ | ### ⚡ Quick trick (≈25 sec) **Out-of-scope statement kill:** - "Methodologies" — not in passage → kill statement 1 - "Remedies" — not in passage → kill statement 3 Both kills eliminate options (a), (b), (c). Only **(d)** remains. Statement 2's borderline status ("similarities AND differences" overstates a contrast-focused passage) is the secondary check, but the elimination by 1 and 3 alone forces (d).

Key concepts:

Writer's intent vs incidental content
Out-of-scope-statement kill technique
Contrast-focused passage vs balanced comparison
Strict elimination logic in CSAT comprehension

Sources: Verified strictly against the HMPV vs SARS-CoV-2 passage in CSAT 2026 paper; no remedies, methodologies, or balanced-similarity content present.

Q28. comprehension (standard)

Which of the following statements reflect the logical and rational inferences that can be drawn from the passage?

HMPV has, historically, had longer documentation of studies when compared with SARS-CoV-2
The two viruses are different from each other and this results in markedly different outcomes amongst those affected
Long-term impacts of the two viruses are dissimilar and this is an important differentiator between them
One of the common factors between the two viruses is seasonal specificity

Select the answer using the code given below?

(a) 1 and 2 only
(b) 1, 2 and 3
(c) 3 and 4
(d) 1 and 3 only

Answer: (B) · Confidence: high

## Why the answer is (b) — Inferences 1, 2 and 3 ### Inference 1 — HMPV has historically had longer documentation of studies than SARS-CoV-2 **Passage:** SARS-CoV-2 was "**first detected in 2019**" and was "a **truly novel** virus for the world". HMPV: "people across the world have been exposed to HMPV for **decades** and the virus is **well-studied**." Decades-old exposure + well-studied vs novel-since-2019 ⇒ HMPV has longer documentation. ✅ **VALID.** ### Inference 2 — The viruses are different and this results in markedly different outcomes among those affected **Passage:** "Both viruses cause **different severity of symptoms**… HMPV causes **milder illness** with **deaths being very rare**". Different impacts on those affected. ✅ **VALID.** ### Inference 3 — Long-term impacts of the two viruses are dissimilar and this is an important differentiator **Passage:** "Both viruses cause different severity of symptoms, **particularly over the long term**" + "HMPV causes milder illness with… **no long-term post-viral symptoms**". Long-term impacts explicitly differ. ✅ **VALID.** ### Inference 4 — One of the common factors between the two viruses is seasonal specificity **Passage:** "strong seasonality seen in HMPV, **unlike** SARS-CoV-2". Passage flags seasonality as a **DIFFERENCE**, not a similarity. Inference 4 inverts the relationship. ❌ **INVALID.** ### Match to options Valid: **1, 2 and 3** → option **(b)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 2 only | ✗ (drops valid 3) | | (b) | 1, 2 and 3 | ✓ | | (c) | 3 and 4 | ✗ (4 invalid) | | (d) | 1 and 3 only | ✗ (drops valid 2) | ### ⚡ Quick trick (≈25 sec) **Inversion trap:** Inference 4 claims seasonality is a **common factor**, but the passage uses the word **"unlike"** for seasonality — flipping a passage-stated difference into a similarity is a classic UPSC trap. Kill 4 → eliminate (c). Then verify 1, 2, 3 all have direct anchors → **(b)** in 25 seconds.

Key concepts:

Inversion trap (turning a passage-stated difference into a similarity)
Direct comprehension anchors via keyword matching
"Unlike" as a difference-marker

Sources: Verified strictly against the HMPV vs SARS-CoV-2 passage in CSAT 2026 paper.

Q29. numeracy (standard)

Three variables $x$, $y$ and $z$ take values 2, 3, 4 or 5 such that their values are always distinct. If $M$ and $N$ denote the largest possible value and the smallest possible value, respectively, for the expression $\{(x \times y) + z\}$; then $M - N$ is

(a) 11
(b) 12
(c) 13
(d) 14

Answer: (C) · Confidence: high

## Why the answer is (c) — $M - N = 13$ ### Setup $x, y, z$ take distinct values from $\{2, 3, 4, 5\}$. Expression $= xy + z$. Maximise to get $M$, minimise to get $N$. ### Maximise: $M$ Strategy: make the multiplicative term $xy$ as large as possible (since it dominates), then maximise $z$ with what's left. | $(x, y)$ | $xy$ | best $z$ remaining | $xy + z$ | |---|---|---|---| | $\{4, 5\}$ | 20 | 3 | **23** ✓ | | $\{3, 5\}$ | 15 | 4 | 19 | | $\{3, 4\}$ | 12 | 5 | 17 | | $\{2, 5\}$ | 10 | 4 | 14 | $M = \mathbf{23}$ (with $\{x, y\} = \{4, 5\}$ and $z = 3$). ### Minimise: $N$ Strategy: minimise $xy$, then minimise $z$ with what's left. | $(x, y)$ | $xy$ | best $z$ remaining | $xy + z$ | |---|---|---|---| | $\{2, 3\}$ | 6 | 4 | **10** ✓ | | $\{2, 4\}$ | 8 | 3 | 11 | | $\{2, 5\}$ | 10 | 3 | 13 | | $\{3, 4\}$ | 12 | 2 | 14 | $N = \mathbf{10}$ (with $\{x, y\} = \{2, 3\}$ and $z = 4$). ### Compute $M - N$ $$M - N = 23 - 10 = \mathbf{13} \;\Rightarrow\; \text{option (c)}.$$ ### ⚡ Quick trick (≈20 sec) **Product-dominates-sum heuristic:** When the expression is (product + small term), the product dominates. So: - Max: put two LARGEST values into product → $4 \times 5 = 20$, add next-largest available → $20 + 3 = 23$. - Min: put two SMALLEST values into product → $2 \times 3 = 6$, add next-smallest available → $6 + 4 = 10$. $M - N = 23 - 10 = \mathbf{13}$. **(c)** in 20 seconds.

Key concepts:

Optimisation over distinct selections
Product-dominates-sum heuristic
Constraint: distinct values from a fixed set

Sources: Direct exhaustive enumeration; verified by tabulation.

Q30. numeracy (standard)

Suppose $x$, $y$ and $z$ are variables taking positive real numbers as their possible values. It is given that $y$ is directly proportional to $x^2$ and $x$ is inversely proportional to $z$. For $z = \frac{7}{25}$, the values of $x$ and $y$ are 5 and 50, respectively. If $y = 98$, what is $z$ equal to?

(a) $\frac{1}{7}$
(b) $\frac{1}{5}$
(c) $\frac{5}{7}$
(d) 1

Answer: (B) · Confidence: high

## Why the answer is (b) — $z = \frac{1}{5}$ ### Step 1 — Write the proportionality relations - $y \propto x^2 \Rightarrow y = a \cdot x^2$ for some constant $a$. - $x \propto \frac{1}{z} \Rightarrow x \cdot z = b$ for some constant $b$ (i.e., $x = b/z$). ### Step 2 — Find $a$ and $b$ from the given pair At $z = 7/25$, $x = 5$, $y = 50$: - $a = \frac{y}{x^2} = \frac{50}{25} = \mathbf{2}$ → $y = 2x^2$. - $b = x \cdot z = 5 \cdot \frac{7}{25} = \frac{7}{5}$ → $x = \frac{7}{5z}$. ### Step 3 — Use $y = 98$ to find $x$ $98 = 2x^2 \Rightarrow x^2 = 49 \Rightarrow x = 7$ (positive real). ### Step 4 — Use $x = 7$ to find $z$ $x = \frac{7}{5z} \Rightarrow 7 = \frac{7}{5z} \Rightarrow 5z = 1 \Rightarrow \mathbf{z = \frac{1}{5}}$. **Answer: (b) $\frac{1}{5}$.** ### Verification At $z = 1/5$: $x = \frac{7/5}{1/5} = 7$. Then $y = 2 \cdot 49 = 98$. ✓ ### ⚡ Quick trick (≈25 sec) **Ratio shortcut — skip the constants:** $$\frac{y_2}{y_1} = \left(\frac{x_2}{x_1}\right)^2 \;\Rightarrow\; \frac{98}{50} = \left(\frac{x_2}{5}\right)^2 \;\Rightarrow\; \frac{x_2}{5} = \sqrt{\frac{49}{25}} = \frac{7}{5} \;\Rightarrow\; x_2 = 7.$$ $$\frac{z_2}{z_1} = \frac{x_1}{x_2} \;\Rightarrow\; z_2 = \frac{7}{25} \cdot \frac{5}{7} = \frac{1}{5}.$$ **(b)** in under 30 seconds — no need to compute the proportionality constants explicitly.

Key concepts:

Direct & inverse proportionality
Ratio method skips proportionality constants
Chain of proportionalities: $y \propto x^2$, $x \propto 1/z$

Sources: Direct algebra; verified by substitution back into both proportionalities.

Passage: India is starting to deploy AI for critical use cases such as weather forecasting, pest detection and control, and crop yield optimisation. However, penetration is limited to a small subset of tech-savvy farmers. In the US and in Europe, generative AI tools have started offering precision farming at scale, integrating large datasets to provide real-time agronomic insights. For at-scale integration and accessibility of AI tools in India, it would be helpful to develop Indian languages-based AI tools for smallholder farmers, partner with AgTechs to create affordable AI solutions, and disseminate AI-based advisory services through government programmes.

Q31. comprehension (standard)

Which of the following assumptions is/are valid?

Agricultural productivity has marched ahead in the West because of the economies of scale facilitated by the adoption of AI tools
Affordable AI tools rendered available in local languages can help AI-based solutions reach more and more small farmers
Though penetration is as yet low, critical application areas deploying AI tools are already in use in India

Select the answer using the code given below?

(a) 1, 2 and 3
(b) 2 only
(c) 1 and 3 only
(d) 2 and 3 only

Answer: (D) · Confidence: high

## Why the answer is (d) — Assumptions 2 and 3 only ### Assumption 1 — Agricultural productivity has marched ahead in the West **because of the economies of scale facilitated by the adoption of AI tools** **Passage:** "In the US and in Europe, generative AI tools have **started offering precision farming at scale**." Two problems with assumption 1: - Passage says AI tools have only **"started offering"** precision farming — does not claim productivity has "marched ahead" already as a result. - "Economies of scale" and "at scale" are different concepts. The passage uses "at scale" (= deployment scale of the tool itself), not "economies of scale" (= cost-savings from large operations). Assumption 1 over-attributes Western productivity to AI economies of scale, a claim the passage does not make. ❌ **INVALID.** ### Assumption 2 — Affordable AI tools available in local languages can help reach more small farmers **Passage:** "For at-scale integration and accessibility of AI tools in India, it would be helpful to develop **Indian languages-based AI tools for smallholder farmers**, partner with AgTechs to create **affordable AI solutions**." Direct paraphrase of the passage's policy prescription. ✅ **VALID.** ### Assumption 3 — Though penetration is low, critical application areas deploying AI tools are already in use in India **Passage:** "India is **starting to deploy AI** for **critical use cases** such as weather forecasting, pest detection and control, and crop yield optimisation. However, **penetration is limited** to a small subset of tech-savvy farmers." Direct paraphrase — both clauses (critical use cases in use + limited penetration) are explicitly stated. ✅ **VALID.** ### Match to options Valid: **2 and 3 only** → option **(d)**. | Option | Set | Verdict | |---|---|---| | (a) | 1, 2 and 3 | ✗ (1 invalid) | | (b) | 2 only | ✗ (drops valid 3) | | (c) | 1 and 3 only | ✗ (1 invalid) | | (d) | 2 and 3 only | ✓ | ### ⚡ Quick trick (≈25 sec) **Causal-over-link trap:** Assumption 1 forges a cause-effect link the passage doesn't make: "productivity marched ahead BECAUSE OF economies of scale FACILITATED BY AI tools." The passage only says AI tools have **started** offering precision farming at scale — no productivity-marched-ahead claim, no economies-of-scale claim. Kill 1 → eliminate (a), (c). Assumptions 2 and 3 are direct paraphrases → **(d)** in 25 seconds.

Key concepts:

Causal over-link trap (cause-effect chains the passage doesn't draw)
"At scale" ≠ "economies of scale"
Passage's policy prescription as direct support
Started doing ≠ has resulted in

Sources: Verified strictly against the AI-in-agriculture passage in CSAT 2026 paper.

Q32. comprehension (standard)

Which of the following statements is/are ***not*** correct?

Tech-savvy farmers will drive the AgTech companies of the future
The development of advisory services by the government programmes for the use of AI tools in agriculture would be helpful
In the US and Europe, AI tools have replaced traditional agricultural practices
The integration of large datasets for use in real-time agronomic analysis is already a reality

Select the answer using the code given below?

(a) 1 and 3 only
(b) 1, 3 and 4
(c) 2 and 4
(d) 3 only

Answer: (A) · Confidence: high

## Why the answer is (a) — Statements 1 and 3 are NOT correct ### Statement 1 — Tech-savvy farmers will **drive** the AgTech companies of the future **Passage:** "penetration is limited to a small subset of **tech-savvy farmers**" + "partner with **AgTechs** to create affordable AI solutions". Passage states tech-savvy farmers are current limited users; it does NOT claim they will **drive** future AgTech companies. This is an over-extrapolation. ❌ **NOT correct.** ### Statement 2 — Government programmes' development of advisory services for AI tools would be helpful **Passage:** "disseminate **AI-based advisory services through government programmes**". Direct paraphrase. ✅ **Correct.** ### Statement 3 — In the US and Europe, AI tools have **replaced** traditional agricultural practices **Passage:** "In the US and in Europe, generative AI tools have **started offering** precision farming at scale". "Started offering" ≠ "replaced". The passage indicates AI tools are an EMERGING addition, not a wholesale replacement of traditional practices. ❌ **NOT correct.** ### Statement 4 — The integration of large datasets for real-time agronomic analysis is already a reality **Passage:** "generative AI tools have started offering precision farming at scale, **integrating large datasets to provide real-time agronomic insights**." Direct paraphrase. ✅ **Correct.** ### Match to options NOT correct: **1 and 3** → option **(a)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 3 only | ✓ | | (b) | 1, 3 and 4 | ✗ (4 IS correct) | | (c) | 2 and 4 | ✗ (both 2 and 4 are correct) | | (d) | 3 only | ✗ (drops invalid 1) | ### ⚡ Quick trick (≈20 sec) **Verb-overreach trap-spotting:** - Statement 1: "will **drive**" (a future-leadership claim absent from passage) → kill. - Statement 3: "have **replaced**" (a completed-substitution claim contradicting passage's "started offering") → kill. Both kills land in option (a). **(a)** in 20 seconds.

Key concepts:

Verb-overreach trap ("will drive", "have replaced")
Started doing ≠ replaced
Current users ≠ future drivers
Direct-paraphrase verification

Sources: Verified strictly against the AI-in-agriculture passage in CSAT 2026 paper.

Q33. reasoning (standard)

$P$, $Q$, $R$, $S$ and $T$ are ranked 1 to 5 (not necessarily in that order). The rank of $P$ is 4, the rank of $Q$ is not 5, the rank of $R$ is 1, the rank of $S$ is not 2, the rank of $T$ is not 3. Then which of the following is/are correct? I. If the rank of $S$ is 3, then that of $T$ is 2. II. If the rank of $Q$ is 3, then that of $T$ is 5

Select the answer using the code given below?

(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II

Answer: (D) · Confidence: high

## Why the answer is (d) — Neither I nor II is correct ### Setup Persons P, Q, R, S, T get distinct ranks from $\{1, 2, 3, 4, 5\}$. - **P = 4** (fixed) - **R = 1** (fixed) - Q, S, T share ranks $\{2, 3, 5\}$. - Constraints: $Q \ne 5$, $S \ne 2$, $T \ne 3$. So: - $Q \in \{2, 3\}$ - $S \in \{3, 5\}$ - $T \in \{2, 5\}$ ### Enumerate the valid combinations | Q | S | T | Valid? | |---|---|---|---| | 2 | 3 | 5 | ✓ | | 2 | 5 | (T∈{2,5}\{2,5}=∅) | ✗ | | 3 | 5 | 2 | ✓ | | 3 | (S∈{3,5}\{3}={5}) → 5 | 2 | (same as above) | Only **2 valid assignments**: - **Case A:** Q = 2, S = 3, T = 5 - **Case B:** Q = 3, S = 5, T = 2 ### Test Statement I — "If S = 3, then T = 2" S = 3 only in **Case A**: Q = 2, S = 3, T = **5** (not 2). Statement I claims T = 2 in this case, but actually T = 5. ❌ **Statement I FALSE.** ### Test Statement II — "If Q = 3, then T = 5" Q = 3 only in **Case B**: Q = 3, S = 5, T = **2** (not 5). Statement II claims T = 5 in this case, but actually T = 2. ❌ **Statement II FALSE.** ### Match to options Both statements false → option **(d)** Neither I nor II. ### ⚡ Quick trick (≈40 sec) **Only-two-cases insight:** After fixing P = 4 and R = 1, only 2 valid (Q, S, T) assignments exist: - **(2, 3, 5)** — Q = 2, S = 3, T = 5 - **(3, 5, 2)** — Q = 3, S = 5, T = 2 Notice that **whenever S = 3, T = 5**, and **whenever Q = 3, T = 2**. Both statements invert the actual relationship → both false → **(d)**.

Key concepts:

Constraint-based rank assignment
Exhaustive enumeration of valid permutations
Conditional statements: 'if X then Y' tested by counterexample
Distinct-value constraint forces narrow case-set

Sources: Direct enumeration of all valid (Q, S, T) assignments; verified by elimination.

Q34. reasoning (standard)

Two identical straight rods are painted in five distinct colours so that each of them gets divided into five equal parts along the length. In one of them, the portions are marked $P1$, $P2$, $P3$, $P4$ and $P5$ (not necessarily in that order) whereas in the other, they are marked $Q1$, $Q2$, $Q3$, $Q4$ and $Q5$ (not necessarily in that order). When the rods are kept parallel to each other side by side, $P1$ and $Q3$ match, $P4$ matches $Q1$ or $Q2$, and $Q4$ matches $P3$ or $P5$. If $Q3$ and $Q5$ are adjacent, which of the following is/are possible? I. $Q3$ is marked at the middle portion of the straight rod. II. $P2$ is marked at one of the extreme portions of the straight rod

Select the answer using the code given below?

(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II

Answer: (C) · Confidence: high

## Why the answer is (c) — Both I and II are possible (a single configuration realises both) ### Setup Two identical rods, each divided into 5 equal portions (positions 1–5). Rod 1 has portions P1-P5; rod 2 has Q1-Q5, both in some order. "Match" = same position on both rods (side-by-side). **Constraints:** - P1 matches Q3 (same position) - P4 matches Q1 or Q2 - Q4 matches P3 or P5 - **Q3 and Q5 are adjacent** (consecutive positions on rod 2) ### Test Statement I — Q3 at middle (position 3) If Q3 at position 3, then P1 also at position 3 (since P1 matches Q3). Q5 adjacent to Q3 ⇒ Q5 at position 2 or 4. Try Q5 at position 2. The remaining Q's (Q1, Q2, Q4) go to positions 1, 4, 5. Try: Q1 at pos 1, Q2 at pos 4, Q4 at pos 5. Then: - P4 must match Q1 or Q2 ⇒ P4 at pos 1 or 4. - Q4 (at pos 5) must match P3 or P5 ⇒ P3 or P5 at pos 5. **Try:** P4 at pos 4 (matches Q2), P5 at pos 5 (matches Q4). Remaining P2, P3 go to pos 1, 2. ### A configuration that realises BOTH I and II | Position | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---| | **Rod 1 (P)** | **P2** | P3 | **P1** | **P4** | **P5** | | **Rod 2 (Q)** | Q1 | **Q5** | **Q3** | **Q2** | **Q4** | **Verification:** - P1 at pos 3 = Q3 at pos 3 ✓ (P1 matches Q3) - P4 at pos 4 = Q2 at pos 4 ✓ (P4 matches Q2 — one of {Q1, Q2}) - Q4 at pos 5 = P5 at pos 5 ✓ (Q4 matches P5 — one of {P3, P5}) - Q3 at pos 3, Q5 at pos 2 ✓ (adjacent) - **Q3 at middle position 3** ✓ (Statement I) - **P2 at position 1 (extreme)** ✓ (Statement II) Both statements hold simultaneously in this valid configuration. ### Match to options Both I and II are achievable → option **(c)**. ### ⚡ Quick trick (≈60 sec) **Constructive existence:** "Is/are possible?" only requires that you exhibit ONE valid configuration realising each statement. Set Q3 at middle (forces P1 also at middle), then assign Q5 next to Q3, populate remaining Q's so P4-Q1/Q2 and Q4-P3/P5 align, and check if the leftover P2 can land at an extreme. The configuration above (P2 P3 P1 P4 P5 vs Q1 Q5 Q3 Q2 Q4) shows both I and II true → **(c)** in about a minute.

Key concepts:

Position-matching constraint between two rods
Adjacency on a linear arrangement
Existence proof by construction
Constraint satisfaction across two interacting permutations

Sources: Direct construction of valid configuration; verified by checking all 4 constraints + both statements simultaneously.

Q35. reasoning (standard)

Seven persons $A$, $B$, $C$, $D$, $E$, $F$ and $G$ travel by three cars $X$, $Y$, $Z$. $A$ and another two of them travel by $X$. Only $E$ travels with $G$. $C$ travels by $Z$, but $B$ does not travel by $Y$. Besides, $A$ and $B$ do not travel by the same car. Then which of the following are correct? I. No one travels alone. II. Only $D$ travels with $F$. III. Only $C$ travels with $B$

Select the answer using the code given below?

(a) I and II only
(b) I and III only
(c) II and III only
(d) All the three

Answer: (B) · Confidence: high

## Why the answer is (b) — Statements I and III only ### Step 1 — Decode the constraints - A in X, with **2 others** in X (so $|X| = 3$). - **Only E travels with G** ⇒ E and G are in a car by themselves ($|\text{their car}| = 2$). - C in Z, B not in Y, A and B not in the same car. ### Step 2 — Locate the E-G car E and G cannot be in X (would mean A and a third person travel with G, contradicting "only E with G"). E and G cannot be in Z (would put 3+ people with G, since C is also in Z). ⇒ **E and G are in Y**, and Y has exactly 2 people: E, G. ### Step 3 — Count remaining seats Total 7 = 3 (X) + 2 (Y) + $|Z|$ ⇒ $|Z| = 2$. So Z has C plus exactly one other person. ### Step 4 — Place B - B not in Y (given). - A in X and A ≠ B's car ⇒ B not in X. - ⇒ **B in Z**, together with C. ### Step 5 — Place D and F Remaining: D and F must fill the 2 open seats in X (alongside A). ### Final allocation | Car | Passengers | |---|---| | **X** | A, D, F | | **Y** | E, G | | **Z** | B, C | ### Step 6 — Test each statement **I. No one travels alone.** Each car has $\ge 2$ passengers. ✅ **TRUE.** **II. Only D travels with F.** F is in X with **both A and D**. So F has TWO companions — D is not F's only companion. ❌ **FALSE.** **III. Only C travels with B.** B is in Z with only C. ✅ **TRUE.** ### Match to options Valid: **I and III** → option **(b)**. | Option | Set | Verdict | |---|---|---| | (a) | I and II only | ✗ (II false) | | (b) | I and III only | ✓ | | (c) | II and III only | ✗ (II false) | | (d) | All the three | ✗ (II false) | ### ⚡ Quick trick (≈45 sec) **Chain deduction:** 1. "Only E with G" + capacities ⇒ E and G are alone in Y. 2. C in Z + 7−2−3=2 seats in Z ⇒ Z has exactly 2: C + one other. 3. B can't be in Y or X ⇒ **B is in Z with C**. 4. Remaining D, F join A in X. Statement II fails because F has TWO companions (A and D) — kills II → option must be **(b)**.

Key concepts:

"Only X travels with Y" = X is Y's sole companion
Constraint-driven seat counting
Sequential placement by elimination
Distinguishing 'with' from 'only with'

Sources: Direct logical deduction; verified by checking all constraints against the unique final allocation.

Q36. reasoning (standard)

There are four statements $X$, $Y$, $Z$ and $W$. Their relations are as follows : If $X$ is incorrect, then so is $Z$; if $Y$ is incorrect, then $W$ is correct; if $W$ is correct, then $X$ is incorrect. Which of the following is/are correct? I. If $X$ is correct, then so is $Y$. II. If $Z$ is correct, then it is not necessary that $Y$ is correct

Select the answer using the code given below?

(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II

Answer: (A) · Confidence: high

## Why the answer is (a) — Only Statement I is correct ### Setup: the three given implications Let ¬ denote "incorrect" and unaccented denote "correct". - **R1:** ¬X → ¬Z (i.e., X incorrect ⇒ Z incorrect) - **R2:** ¬Y → W (Y incorrect ⇒ W correct) - **R3:** W → ¬X (W correct ⇒ X incorrect) **Their contrapositives** (logically equivalent): - **R1':** Z → X (Z correct ⇒ X correct) - **R2':** ¬W → Y (W incorrect ⇒ Y correct) - **R3':** X → ¬W (X correct ⇒ W incorrect) ### Test Statement I — If X correct, then Y correct $$X \overset{R3'}{\Longrightarrow} \lnot W \overset{R2'}{\Longrightarrow} Y.$$ Direct chain. ✅ **Statement I is CORRECT.** ### Test Statement II — If Z correct, then it is **not necessary** that Y is correct $$Z \overset{R1'}{\Longrightarrow} X \overset{R3'}{\Longrightarrow} \lnot W \overset{R2'}{\Longrightarrow} Y.$$ So Z correct **forces** Y correct. Statement II claims it is NOT necessary that Y is correct — this is FALSE, since Y is in fact necessary. ❌ **Statement II is FALSE.** ### Match to options | Option | Set | Verdict | |---|---|---| | (a) | I only | ✓ | | (b) | II only | ✗ | | (c) | Both | ✗ | | (d) | Neither | ✗ | ### ⚡ Quick trick (≈40 sec) **Contrapositive-chain shortcut:** Convert each "¬A → ¬B" form to "B → A" (the contrapositive). Then chain: $X \to \lnot W \to Y$ via R3', R2'. So X correct forces Y correct (Statement I ✓). For Statement II, the same chain prepended by R1' ($Z \to X$) shows $Z \to X \to \lnot W \to Y$, so Z correct forces Y correct — statement II's "not necessary" claim is wrong. **(a)** in 40 seconds.

Key concepts:

Contrapositive: (P → Q) ≡ (¬Q → ¬P)
Implication chaining
Distinguishing 'possible' from 'necessary'
De Morgan's laws on conditionals

Sources: Direct propositional logic; verified by chained contrapositives.

Q37. numeracy (standard)

$X$ and $Y$ are two runners who run for the same duration of time on the same circular track. They started running at the same time in the same direction with uniform speeds. When $X$ completed 7 rounds, $Y$ did exactly 5. After completing 5 rounds, $Y$ changed his direction and started running in the opposite direction with speed which is double of his earlier speed. On the other hand, $X$ continued to run with the same speed. They stopped running when $X$ completed exactly 21 rounds. How many times did $X$ and $Y$ meet after they had started and before they finally stopped?

(a) 35
(b) 34
(c) 31
(d) 29

Answer: (A) · Confidence: high

## Why the answer is (a) — 35 meetings ### Step 1 — Set speeds from the given ratio When X completed 7 rounds, Y completed 5 → **speed ratio X : Y = 7 : 5**. Let X's speed = 7 (in rounds/unit time) and Y's initial speed = 5. ### Step 2 — Find the duration of each phase **Phase 1 (same direction):** Y completes 5 rounds at speed 5 → $t_1 = 1$ unit. In this time, X completes 7 rounds. Both back at start position at $t = 1$. **Phase 2 (Y reverses, doubles speed):** X needs to complete a total of 21 rounds. Remaining for X: $21 - 7 = 14$ rounds at speed 7 → $t_2 = 14/7 = 2$ units. So Phase 2 runs from $t = 1$ to $t = 3$. Y's new speed = $2 \times 5 = 10$, opposite direction. ### Step 3 — Count Phase 1 meetings Same direction → meetings happen each time X **laps** Y. Relative speed = $7 - 5 = 2$. Over $t_1 = 1$, X gains $2 \times 1 = 2$ laps on Y. Meetings (excluding $t=0$): at $t = 0.5$ (first lap) and $t = 1$ (second lap, both back at start). **2 meetings.** ### Step 4 — Count Phase 2 meetings Opposite directions → relative speed = $7 + 10 = 17$. Over $t_2 = 2$, relative distance covered = $17 \times 2 = 34$ laps. Meeting condition (mod 1): $7t \equiv 5 - 10(t-1) \pmod{1} \Rightarrow 17t \equiv 15 \pmod{1} \Rightarrow t = \frac{15 + k}{17}$ for integer $k$. Valid $t \in [1, 3] \Rightarrow k \in [2, 36]$ → 35 values total. But $k = 2$ gives $t = 1$ (already counted as the Phase 1 end-meeting) and $k = 36$ gives $t = 3$ (when they STOP — excluded). Strictly $1 < t < 3 \Rightarrow k \in \{3, 4, \ldots, 35\}$ → **33 meetings**. ### Step 5 — Total meetings strictly between start and stop $$\underbrace{2}_{\text{Phase 1}} + \underbrace{33}_{\text{Phase 2, strict}} = \mathbf{35} \;\Rightarrow\; \text{option (a)}.$$ ### ⚡ Quick trick (≈90 sec) **Relative-lap counter:** - **Same direction**: meetings = $|v_X - v_Y| \times t$ (number of laps gained). - **Opposite direction**: meetings = $(v_X + v_Y) \times t$ (combined relative distance). Phase 1: $2 \times 1 = 2$. Phase 2: $17 \times 2 = 34$. Total (before exclusion) = 36. Exclude the start ($t=0$) and final stop ($t=3$) → $36 - 1 = 35$ (the $t=1$ boundary is counted once since it's a genuine meeting, the $t=3$ endpoint is excluded). **(a)** in about 90 seconds.

Key concepts:

Circular-track meeting count: same vs opposite direction
Relative speed in modular position arithmetic
Phase decomposition for changing speeds
Endpoint inclusion/exclusion at meeting boundaries

Sources: Direct kinematic analysis with modular position equations; verified by case-by-case meeting enumeration.

Q38. numeracy (standard)

In an objective type question paper, 5 marks are awarded for a correct answer and 2 marks are deducted for a wrong answer. A student attempted all the questions and got a score of 69. Had he been awarded 4 marks for a correct answer and 1 mark deducted for a wrong answer, he would have scored 84. How many questions were there in the question paper?

(a) 99
(b) 81
(c) 84
(d) 79

Answer: (B) · Confidence: high

## Why the answer is (b) — 81 questions in total ### Step 1 — Variables Let $c$ = number of correct answers, $w$ = number of wrong answers. Total questions $n = c + w$. ### Step 2 — Write the two scoring equations - **Scheme 1** (+5, −2): $5c - 2w = 69$ … (1) - **Scheme 2** (+4, −1): $4c - w = 84$ … (2) ### Step 3 — Solve From (2): $w = 4c - 84$. Substitute in (1): $$5c - 2(4c - 84) = 69 \;\Rightarrow\; 5c - 8c + 168 = 69 \;\Rightarrow\; -3c = -99 \;\Rightarrow\; c = 33.$$ Then $w = 4 \cdot 33 - 84 = 132 - 84 = 48$. ### Step 4 — Total questions $$n = c + w = 33 + 48 = \mathbf{81} \;\Rightarrow\; \text{option (b)}.$$ ### Verification Scheme 1 check: $5 \cdot 33 - 2 \cdot 48 = 165 - 96 = 69$ ✓ Scheme 2 check: $4 \cdot 33 - 48 = 132 - 48 = 84$ ✓ ### ⚡ Quick trick (≈25 sec) **Subtract the equations cleverly:** Multiply (2) by 2 and subtract (1): $2(4c - w) - (5c - 2w) = 168 - 69 \Rightarrow 8c - 2w - 5c + 2w = 99 \Rightarrow 3c = 99 \Rightarrow c = 33$. The $w$-terms cancel cleanly. Plug back to get $w = 48$, total $= 81$. **(b)** in 25 seconds.

Key concepts:

Linear system from two scoring schemes
Elimination by coefficient-matching (the 2:1 ratio between w-coeffs)
Total = correct + wrong (no unattempted, per problem statement)

Sources: Direct linear algebra; verified by back-substitution into both scoring schemes.

Passage: The Juvenile Justice (Care and Protection of Children) Act, or the JJ Act, 2015 allows for the possibility for trying adolescents above 16 as adults if they are accused of committing a heinous offence. A heinous offence is one with a minimum punishment of seven years. Offences such as culpable homicide and causing death by negligence, which are common in drunken driving cases, are not heinous offences because they do not have a prescribed minimum punishment. The JJ Act, amended in 2021, now categorises an offence that has no minimum sentence, but has a maximum sentence of seven years or more as a serious offence which nonetheless, in the opinion of activists, does not merit the transfer of a case to the adult criminal justice system.

Q39. comprehension (standard)

Which of the following conclusions is/are valid?

Only a serious offence as categorised by the revised JJ Act, justifies the transfer of a case to the adult judicial system
The JJ Act, 2021, categorises an offence as a serious offence based on the maximum sentence it carries, rather than on the minimum sentence

Select the answer using the code given below?

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2

Answer: (B) · Confidence: high

## Why the answer is (b) — Only conclusion 2 is valid ### Conclusion 1 — Only a 'serious offence' as categorised by the revised JJ Act justifies transfer to the adult judicial system **Passage:** "The JJ Act, 2015 allows for the possibility for trying adolescents above 16 as adults if they are accused of committing a **heinous offence**." Then: "The JJ Act, amended in 2021, now categorises an offence that has no minimum sentence, but has a maximum sentence of seven years or more as a **serious offence** which nonetheless, **in the opinion of activists, does not merit the transfer** of a case to the adult criminal justice system." Two passage facts contradict Conclusion 1: - The trigger for adult trial is **heinous offence**, not 'serious offence'. - Activists explicitly say 'serious offence' DOES NOT merit transfer to adult court. ❌ **INVALID** — confuses 'serious' with 'heinous'. ### Conclusion 2 — The JJ Act 2021 categorises an offence as 'serious' based on the maximum sentence, rather than the minimum sentence **Passage:** The 2021 amendment defines a 'serious offence' as one with "**no minimum sentence, but has a maximum sentence of seven years or more**". Direct paraphrase — the 2021 amendment uses **maximum sentence** (not minimum) as the categorisation basis. ✅ **VALID.** ### Match to options Valid: **2 only** → option **(b)**. ### ⚡ Quick trick (≈25 sec) **Term-confusion trap-spotting:** The passage uses TWO related but distinct terms — **heinous** (triggers adult trial) and **serious** (does NOT trigger adult trial). Conclusion 1 swaps them. Whenever a comprehension option uses one term where the passage uses another, kill instantly. Eliminate (a), (c). Conclusion 2 is a direct paraphrase. **(b)** in 25 seconds.

Key concepts:

Term-confusion trap (heinous vs serious)
JJ Act 2015 vs 2021 amendment distinction
Min-sentence (heinous) vs max-sentence (serious) basis for categorisation
Direct paraphrase verification

Sources: Verified strictly against the JJ Act passage in CSAT 2026 paper.

Q40. comprehension (standard)

Which of the following statements is/are correct?

If an offence has no minimum prescribed punishment, it cannot be considered heinous as per the JJ Act, 2015
As per the JJ Act, 2021, an offence for which there is a provision for a maximum sentence of seven years or more, but no minimum sentence, is to be considered a serious offence

Select the answer using the code given below?

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2

Answer: (C) · Confidence: high

## Why the answer is (c) — Both statements are correct ### Statement 1 — If an offence has no minimum prescribed punishment, it cannot be considered heinous per the JJ Act 2015 **Passage (direct quote):** "A heinous offence is one with a **minimum punishment of seven years**. Offences such as culpable homicide and causing death by negligence, which are common in drunken driving cases, are **not heinous offences because they do not have a prescribed minimum punishment**." The passage explicitly states: no-min-punishment ⇒ not heinous. Statement 1 paraphrases this exactly. ✅ **VALID.** ### Statement 2 — Per the JJ Act 2021, an offence with max sentence ≥ 7 years but no minimum is considered a serious offence **Passage (direct quote):** "The JJ Act, amended in 2021, now categorises an offence that has **no minimum sentence, but has a maximum sentence of seven years or more as a serious offence**." Direct paraphrase. ✅ **VALID.** ### Match to options Both valid → option **(c)** Both 1 and 2. | Option | Set | Verdict | |---|---|---| | (a) | 1 only | ✗ (drops valid 2) | | (b) | 2 only | ✗ (drops valid 1) | | (c) | Both 1 and 2 | ✓ | | (d) | Neither | ✗ | ### ⚡ Quick trick (≈20 sec) **Both-direct-paraphrase recognition:** Both statements use the exact passage logic — no inferences, no over-extensions, no term swaps. When both numbered statements are clean paraphrases of the passage, **(c) Both** is almost always the answer.

Key concepts:

Direct paraphrase verification
Heinous offence definition (min ≥ 7 yrs)
Serious offence per 2021 amendment (no min, max ≥ 7 yrs)

Sources: Verified strictly against the JJ Act passage in CSAT 2026 paper — both statements are word-for-word paraphrases.

Q41. numeracy (standard)

An explosion takes place at a certain distance from an army camp. As soon as the sensor in the camp receives the sound of the explosion, a drone starts flying towards the spot of explosion. The drone clicks a picture from the spot and the camp receives it at the same time. Immediately another drone starts flying to the spot and it also sends a picture as soon as it reaches the spot. The two pictures were received at 5:02 PM and 5:05 PM, respectively. If the speed of the drones is 30 m/s, at what time did the explosion take place? Assume that the speed of sound is 300 m/s.

(a) 4:59:00 PM
(b) 4:59:02 PM
(c) 4:58:42 PM
(d) 4:56:32 PM

Answer: (C) · Confidence: high

## Why the answer is (c) — 4:58:42 PM ### Step 1 — Label the events Let $d$ = distance from camp to explosion site (in metres) and $t_0$ = time of explosion. Sequence: - **$t_0$:** explosion occurs. - **$t_1 = t_0 + d/300$:** sensor at camp hears the explosion (sound travels at 300 m/s). - Drone 1 launches immediately at $t_1$; reaches spot at $t_1 + d/30$ (drone at 30 m/s); picture received at camp instantly → **$t_2 = t_1 + d/30 = $ 5:02 PM**. - Drone 2 launches immediately at $t_2$; reaches spot at $t_2 + d/30$; picture received instantly → **$t_3 = t_2 + d/30 = $ 5:05 PM**. ### Step 2 — Find $d$ from the 3-minute gap between pictures $t_3 - t_2 = d/30 = 5{:}05 - 5{:}02 = 3$ minutes $= 180$ seconds. $$d = 30 \times 180 = \mathbf{5400 \text{ m}}.$$ ### Step 3 — Find $t_1$ $t_1 = t_2 - d/30 = 5{:}02 - 180\text{ s} = \mathbf{4{:}59{:}00}$ PM. ### Step 4 — Find $t_0$ (time of explosion) Sound travel time: $d/300 = 5400/300 = 18$ seconds. $$t_0 = t_1 - 18 = 4{:}59{:}00 - 18\text{ s} = \mathbf{4{:}58{:}42}\text{ PM}.$$ **Answer: (c) 4:58:42 PM.** ### ⚡ Quick trick (≈45 sec) **Pipeline shortcut:** 1. Picture gap = drone travel time = $d/30 = 180$ s → $d = 5400$ m. 2. First picture arrives at camp **180 s after** sensor heard explosion → sensor heard at $5{:}02 - 3{:}00 = 4{:}59{:}00$. 3. Sound takes $d/300 = 18$ s to reach the camp → explosion was at $4{:}59{:}00 - 0{:}18 = \mathbf{4{:}58{:}42}$. **(c)** in under a minute. The key insight: the picture gap IS the drone-travel time, which back-solves the distance.

Key concepts:

Sound vs drone speed ratio (300:30 = 10:1)
Picture-arrival gap = drone one-way travel time
Backward chronology: events occurred in reverse temporal order
Speed-distance-time fundamentals

Sources: Direct kinematics with reverse-chronology event chain; verified by forward simulation.

Q42. numeracy (standard)

The digit in the unit place of the number $6^{129} \times 7^{307}$ is

(a) 2
(b) 4
(c) 8
(d) 6

Answer: (C) · Confidence: high

## Why the answer is (c) — Unit digit is 8 ### Step 1 — Unit digit of $6^{129}$ $6^n$ always ends in **6** for any positive integer $n$ (since $6 \times 6 = 36$, $6 \times 6 \times 6 = 216$, etc.). So unit digit of $6^{129}$ = **6**. ### Step 2 — Unit digit of $7^{307}$ Powers of 7 cycle through unit digits every **4 powers**: | $n$ | $7^n$ | Unit digit | |---|---|---| | 1 | 7 | **7** | | 2 | 49 | **9** | | 3 | 343 | **3** | | 4 | 2401 | **1** | | 5 | 16807 | **7** (cycle restarts) | Cycle: $7, 9, 3, 1$ (period 4). $307 \div 4 = 76$ remainder $3 \Rightarrow 307 \equiv 3 \pmod 4 \Rightarrow$ unit digit of $7^{307}$ = unit digit of $7^3$ = **3**. ### Step 3 — Multiply unit digits $$\text{unit of } (6^{129} \times 7^{307}) = \text{unit of }(6 \times 3) = \text{unit of }18 = \mathbf{8}.$$ **Answer: (c) 8.** ### ⚡ Quick trick (≈20 sec) **Two facts to memorise:** - $6^n$ always ends in 6. - $7^n$ cycles $7, 9, 3, 1$ with period 4; reduce exponent mod 4. $307 \bmod 4 = 3$ → $7^{307}$ ends in 3. $6 \times 3 = 18$ → unit = **8**. **(c)** in 20 seconds.

Key concepts:

Cyclicity of unit digits: $7^n$ has period 4
$6^n$ unit digit is invariant (always 6)
Modular reduction of exponent
Product unit digit = product of unit digits (mod 10)

Sources: Direct application of cyclicity rules for unit digits; verified by tabulation.

Q43. numeracy (standard)

A person saves 10% of his salary every month. If his salary increases by 12% and the expenditure increases by 10%, then what will be the change in his saving per month?

(a) 20% increase
(b) 30% increase
(c) 03% decrease
(d) 02% decrease

Answer: (B) · Confidence: high

## Why the answer is (b) — 30% increase in savings ### Step 1 — Set baseline values Let salary $= S$. Saves 10% → **Savings $= 0.10 \, S$**, **Expenditure $= 0.90 \, S$**. ### Step 2 — Apply the changes - New salary $= 1.12 \, S$ (12% increase). - New expenditure $= 1.10 \times 0.90 \, S = 0.99 \, S$ (10% increase on the original expenditure). ### Step 3 — New savings $$\text{New savings} = 1.12 \, S - 0.99 \, S = 0.13 \, S.$$ ### Step 4 — Percentage change in savings $$\frac{\text{New} - \text{Old}}{\text{Old}} \times 100 = \frac{0.13S - 0.10S}{0.10S} \times 100 = \frac{0.03}{0.10} \times 100 = \mathbf{30\%} \text{ increase}.$$ **Answer: (b) 30% increase.** ### ⚡ Quick trick (≈30 sec) **Plug-in shortcut:** Set $S = 100$ to make arithmetic mental. - Old savings = 10, old expenditure = 90. - New salary = 112, new expenditure = 90 × 1.10 = 99. - New savings = 112 − 99 = 13. - Change = (13 − 10) / 10 = 30%. **(b)** in 30 seconds.

Key concepts:

Savings = Income − Expenditure
Multiplicative percentage compounding
Plug-in S = 100 for fast mental arithmetic

Sources: Direct percentage arithmetic; verified with S = 100 base case.

Q44. reasoning (standard)

$P$ has a son and a daughter. $S$ is the mother of $T$. $S$ is $R$'s spouse. $Q$ and $R$ are children of $P$. Then how is $Q$ related to $S$?

(a) $Q$ is a sister of $S$
(b) $Q$ is a daughter of $S$
(c) $Q$ is the mother of $S$
(d) $Q$ is a sister of the husband of $S$

Answer: (D) · Confidence: high

## Why the answer is (d) — Q is a sister of the husband of S ### Step 1 — Identify each person's gender - P has a son and a daughter → P has exactly 2 children of opposite gender. - S is the **mother** of T → S is **female**. - S is R's spouse → since S is female, **R is male**. - Q and R are P's children → {Q, R} = {son, daughter} of P. Since R is male (son), **Q is female (daughter)**. ### Step 2 — Trace the relationship Q → S - Q is a daughter of P (female sibling of R). - R is the husband of S. - So Q is **R's sister** = **sister of S's husband**. ### Step 3 — Match to options | Option | Claim | Verdict | |---|---|---| | (a) | Q is a sister of S | ✗ (Q and S are not siblings) | | (b) | Q is a daughter of S | ✗ (Q is P's daughter, not S's) | | (c) | Q is the mother of S | ✗ (Q is younger generation than S/R, in fact same generation as R) | | (d) | Q is a sister of the husband of S | ✓ | **Answer: (d).** ### ⚡ Quick trick (≈30 sec) **Gender-pinning shortcut:** Use "mother of T" to fix S as female, which fixes R as male (S's husband). With R fixed as P's son, **the other child Q must be P's daughter**. Then Q is sister of R, and R is S's husband → Q is sister-in-law of S, i.e. **sister of S's husband**. **(d)** in 30 seconds.

Key concepts:

Gender inference from kinship labels (mother → female; spouse → opposite gender)
Sibling relations on a generation
Sister-in-law decomposition (sister of spouse)

Sources: Direct family-tree deduction; verified by gender chain.

Q45. numeracy (standard)

How many three-digit numbers can be expressed as an integral power of 2?

(a) 1
(b) 2
(c) 3
(d) 4

Answer: (C) · Confidence: high

## Why the answer is (c) — 3 such numbers ### Step 1 — Identify the 3-digit range 3-digit numbers: $100 \le n \le 999$. ### Step 2 — List powers of 2 in this range | Exponent | $2^k$ | 3-digit? | |---|---|---| | 6 | 64 | ✗ (2-digit) | | 7 | **128** | ✓ | | 8 | **256** | ✓ | | 9 | **512** | ✓ | | 10 | 1024 | ✗ (4-digit) | Exactly **3** powers of 2 are three-digit: $2^7, 2^8, 2^9$. **Answer: (c) 3.** ### ⚡ Quick trick (≈15 sec) **Boundary memory:** $2^{10} = 1024$ (the famous "kilobyte" boundary). So $2^7, 2^8, 2^9 = 128, 256, 512$ are the only 3-digit powers, and $2^6 = 64$ falls just short on the low end. **3 numbers → (c)** in 15 seconds.

Key concepts:

Powers of 2: 64, 128, 256, 512, 1024 (memory anchor)
Range-counting in a sequence
$2^{10} = 1024$ as boundary marker

Sources: Direct enumeration of powers of 2 in the 3-digit range.

Passage: The key source of the battle for clean skies and clear lungs is the fuel we burn—from household Chulhas to the thermal power plants. In most cases, it is biomass or coal. The Supreme Court banned the use of pet coke—the dirtiest of such fuels. The Delhi Government banned the use of coal, which was later extended to the entire National Capital Region. It was also agreed that the thermal power plants would clean up or shut down. Action on this has been patchy to say the least. The lesson from the transition to CNG is that people need alternatives for a ban to be effective. When diesel buses were stopped, CNG supply had to be assured. It also had to be feasible in terms of cost. The Supreme Court agreed that fiscal measures were needed to keep clean fuel cheaper than dirty fuel. Now even as coal is banned, the price of natural gas makes industry uncompetitive.

Q46. comprehension (standard)

Which of the following inferences is/are correct?

The source of the energy we consume is the key to the battle for cleaner air
Bans are effective where the will is strong and the people are convinced that such bans are for the greater good of society
There is judicial approval for a policy that intervenes fiscally to facilitate benevolent pricing for cleaner fuel

Select the answer using the code given below?

(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) 1 only

Answer: (C) · Confidence: high

## Why the answer is (c) — Inferences 1 and 3 only ### Inference 1 — The source of energy we consume is the key to the battle for cleaner air **Passage:** "The key source of the battle for clean skies and clear lungs is the **fuel we burn**—from household Chulhas to the thermal power plants." Direct paraphrase (fuel = energy source). ✅ **VALID.** ### Inference 2 — Bans are effective where the **will is strong** and people are convinced of **greater good of society** **Passage:** "The lesson from the transition to CNG is that **people need alternatives** for a ban to be effective." Passage cites the condition for ban-effectiveness as **availability of alternatives** (CNG supply for diesel buses, with feasibility in cost). It does NOT speak of "will" or "conviction of greater good". Inference 2 substitutes the actual condition with a different (un-supported) one. ❌ **INVALID.** ### Inference 3 — Judicial approval exists for a policy that intervenes fiscally for benevolent pricing of cleaner fuel **Passage:** "The **Supreme Court agreed** that **fiscal measures** were needed to keep clean fuel cheaper than dirty fuel." Direct paraphrase — judicial (SC) approval + fiscal intervention + cheaper clean fuel = benevolent pricing. ✅ **VALID.** ### Match to options Valid: **1 and 3** → option **(c)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 2 | ✗ (2 invalid) | | (b) | 2 and 3 | ✗ (2 invalid) | | (c) | 1 and 3 | ✓ | | (d) | 1 only | ✗ (drops valid 3) | ### ⚡ Quick trick (≈25 sec) **Substitution-trap spotting:** Inference 2 swaps the passage's reason ("need alternatives") with a different reason ("will + greater-good conviction"). Whenever a comprehension option **changes the cited cause**, kill it. Inferences 1 and 3 are direct paraphrases → **(c)** in 25 seconds.

Key concepts:

Substitution trap (replacing passage's stated cause with another)
Direct paraphrase verification
Judicial-approval keywords ("Supreme Court agreed")
Fiscal vs willpower-based policy levers

Sources: Verified strictly against the clean-air passage in CSAT 2026 paper.

Q47. comprehension (standard)

Which of the following statements is/are correct?

Thermal power stations in Delhi were required to summarily shut down
CNG supplies had to be assured once diesel vehicles were prohibited from plying
The Supreme Court banned the use of coal across the National Capital Region

Select the answer using the code given below?

(a) 1 and 2
(b) 3 only
(c) 2 only
(d) 2 and 3

Answer: (C) · Confidence: high

## Why the answer is (c) — Only statement 2 is correct ### Statement 1 — Thermal power stations in Delhi were required to **summarily shut down** **Passage:** "It was also agreed that the thermal power plants would **clean up OR shut down**. Action on this has been **patchy** to say the least." Two issues with statement 1: - The plants were given a **choice** (clean up OR shut down) — not a summary shutdown directive. - Action was "patchy", which contradicts "summarily" (= immediately and without exception). ❌ **INVALID.** ### Statement 2 — CNG supplies had to be assured once diesel vehicles were prohibited from plying **Passage:** "When diesel buses were stopped, **CNG supply had to be assured**." Direct paraphrase. ✅ **VALID.** ### Statement 3 — The **Supreme Court** banned the use of coal across the National Capital Region **Passage:** "The Supreme Court banned the use of pet coke—the dirtiest of such fuels. **The Delhi Government** banned the use of coal, which was later extended to the entire National Capital Region." The **Delhi Government** (not the Supreme Court) banned coal. The SC banned **pet coke**, a different fuel. Misattribution of the banning authority. ❌ **INVALID.** ### Match to options Valid: **2 only** → option **(c)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 2 | ✗ (1 misstates "summarily") | | (b) | 3 only | ✗ (SC banned pet coke, not coal) | | (c) | 2 only | ✓ | | (d) | 2 and 3 | ✗ (3 misattributes to SC) | ### ⚡ Quick trick (≈25 sec) **Authority-attribution trap-spotting:** Statement 3 confuses **Supreme Court** with **Delhi Government** as the coal-ban authority. Whenever a comprehension option swaps the agent (judiciary ↔ executive ↔ legislature ↔ private body), kill instantly. Statement 1's "summarily" overstates the soft policy. Only statement 2 survives → **(c)** in 25 seconds.

Key concepts:

Authority-attribution trap (SC vs Delhi Government)
Pet coke (SC ban) vs coal (Delhi Govt ban)
"Summarily" overstates a policy with patchy enforcement
Direct paraphrase verification

Sources: Verified strictly against the clean-air passage in CSAT 2026 paper.

Q48. reasoning (standard)

Consider the following statements : Every red is blue. Every blue is green. Every green is yellow. Which of the following statements denoted by P, Q and R are correct? P. Every blue is yellow. Q. Every red is green. R. Every red is yellow

Select the answer using the code given below?

(a) P and Q only
(b) Q and R only
(c) P and R only
(d) P, Q and R

Answer: (D) · Confidence: high

## Why the answer is (d) — All three (P, Q, R) are correct ### Step 1 — Set-inclusion chain from the premises - Every red is blue → Red ⊆ Blue. - Every blue is green → Blue ⊆ Green. - Every green is yellow → Green ⊆ Yellow. Chaining: **Red ⊆ Blue ⊆ Green ⊆ Yellow.** ### Step 2 — Test each conclusion | Statement | Set claim | Valid? | |---|---|---| | **P.** Every blue is yellow | Blue ⊆ Yellow | ✓ (via Blue ⊆ Green ⊆ Yellow) | | **Q.** Every red is green | Red ⊆ Green | ✓ (via Red ⊆ Blue ⊆ Green) | | **R.** Every red is yellow | Red ⊆ Yellow | ✓ (full chain transitivity) | All three follow from transitivity of subset inclusion. **Answer: (d) P, Q and R.** ### ⚡ Quick trick (≈15 sec) **Transitivity chain:** "Every X is Y" + "Every Y is Z" → "Every X is Z". With three nested inclusions (Red ⊆ Blue ⊆ Green ⊆ Yellow), every pair of categories in the chain inherits the inclusion → **all 3 conclusions true** → **(d)** in 15 seconds.

Key concepts:

Transitive set inclusion (A ⊆ B ⊆ C ⇒ A ⊆ C)
Categorical syllogisms with universal-affirmative premises
Chained 'all X are Y' deductions

Sources: Direct application of transitive subset inclusion (basic set theory).

Q49. numeracy (standard)

How many times does 5 appear in all two-digit positive integers?

(a) 18
(b) 19
(c) 20
(d) 21

Answer: (B) · Confidence: high

## Why the answer is (b) — Digit 5 appears 19 times ### Step 1 — Count 5's in the TENS place Numbers with 5 in tens place: **50, 51, 52, 53, 54, 55, 56, 57, 58, 59** → 10 numbers. Each contributes one 5 in tens place → **10 fives**. ### Step 2 — Count 5's in the UNITS place Numbers with 5 in units place (two-digit only, so tens place 1-9): **15, 25, 35, 45, 55, 65, 75, 85, 95** → 9 numbers. Each contributes one 5 in units place → **9 fives**. ### Step 3 — Double-counting check **55** appears in both lists — but it contains TWO fives (one in each place), so counting it in both lists correctly counts both occurrences. No subtraction needed. ### Step 4 — Total $$10 + 9 = \mathbf{19} \;\Rightarrow\; \text{option (b)}.$$ ### ⚡ Quick trick (≈20 sec) **Place-value separation:** Count digit-d occurrences in tens place (10 per fixed tens-digit) + units place (9 per fixed units-digit, since tens-digit can't be 0 for 2-digit numbers). For d ∈ {1, …, 9}: total = 10 + 9 = **19**. **(b)** in 20 seconds.

Key concepts:

Place-value counting separates occurrences correctly
Two-digit range: 10–99 (tens-digit ≥ 1)
55 has two 5's — no double-count adjustment needed
Formula generalises: digit-d count in 2-digit range = 19 for d ∈ {1,…,9}

Sources: Direct enumeration by place value; verified by listing.

Q50. numeracy (standard)

$X$ travels 6 km on a bicycle with average speeds of 5 km per hour, 10 km per hour and 4 km per hour during the first 1 km, the next 2 km and the remaining 3 km, respectively. $Y$ travels the same distances with average speeds of 4 km per hour, 10 km per hour and 5 km per hour, respectively. How many minutes early will $Y$ complete the journey if both $X$ and $Y$ start at the same time?

(a) 3
(b) 4
(c) 5
(d) 6

Answer: (D) · Confidence: high

## Why the answer is (d) — Y arrives 6 minutes earlier than X ### Step 1 — Compute X's total time | Segment | Distance | Speed | Time | |---|---|---|---| | 1 | 1 km | 5 km/h | $\frac{1}{5}$ h = **12 min** | | 2 | 2 km | 10 km/h | $\frac{2}{10}$ h = **12 min** | | 3 | 3 km | 4 km/h | $\frac{3}{4}$ h = **45 min** | **X total: $12 + 12 + 45 = 69$ minutes.** ### Step 2 — Compute Y's total time | Segment | Distance | Speed | Time | |---|---|---|---| | 1 | 1 km | 4 km/h | $\frac{1}{4}$ h = **15 min** | | 2 | 2 km | 10 km/h | $\frac{2}{10}$ h = **12 min** | | 3 | 3 km | 5 km/h | $\frac{3}{5}$ h = **36 min** | **Y total: $15 + 12 + 36 = 63$ minutes.** ### Step 3 — Time difference $$X - Y = 69 - 63 = \mathbf{6} \text{ minutes.}$$ **Y arrives 6 minutes earlier → option (d).** ### ⚡ Quick trick (≈30 sec) **Only segments 1 and 3 differ** (segment 2 is identical 12 min for both). So compute only the difference: - Seg 1: X is faster (12 min) vs Y slower (15 min) → X saves 3 min here. - Seg 3: Y faster (36 min) vs X slower (45 min) → Y saves 9 min here. Net: Y saves $9 - 3 = \mathbf{6}$ minutes overall. **(d)** in 30 seconds — no need to add segment 2 since it cancels.

Key concepts:

Time = distance / speed
Multi-segment journey time = sum of segment times
Cancellation of identical segments to speed up computation

Sources: Direct kinematic computation; verified segment-by-segment.

Q51. reasoning (standard)

Seven cubes are identical in shape. Out of these, the weight of each of the six cubes is equal and the weight of the remaining cube is less than the weight of any other cube. A balance is used to identify the lightest cube. What is the minimum number of attempts required to distinguish the odd cube with certainty?

(a) 2
(b) 3
(c) 4
(d) 1

Answer: (A) · Confidence: high

## Why the answer is (a) — Minimum 2 attempts ### Setup 7 cubes, 1 is lighter, 6 are equal. Use a 2-pan balance. Find minimum **worst-case** number of weighings. ### Step 1 — Optimal split: 3 vs 3 (1 left aside) Divide 7 cubes into three groups: **(3, 3, 1)**. Weigh the two 3-cube groups against each other. **Outcomes:** - **Balanced** → lighter cube is the 1 set aside. **Found in 1 attempt** (best case). - **Unbalanced** → lighter cube is in the lighter pan's group of 3. Proceed to attempt 2. ### Step 2 — From the 3 suspects, split into (1, 1, 1) Weigh any 2 of the 3 against each other. **Outcomes:** - **Balanced** → lighter is the 1 left aside → **found in 2 attempts**. - **Unbalanced** → lighter is the cube on the lighter pan → **found in 2 attempts**. ### Worst case = 2 attempts Why not 1? A single weighing yields 3 outcomes (left lighter / balanced / right lighter), distinguishing only 3 possibilities. With 7 cubes, 1 attempt is insufficient. $$\lceil \log_3 7 \rceil = \lceil 1.77 \rceil = \mathbf{2}.$$ **Answer: (a) 2.** ### ⚡ Quick trick (≈25 sec) **Ternary logic shortcut:** A balance gives 3 outcomes per weighing, so $k$ weighings distinguish up to $3^k$ items. For $n = 7$, need $3^k \ge 7$ ⇒ $k \ge 2$. The (3, 3, 1) split realises this bound. **(a) 2** in 25 seconds.

Key concepts:

Balance scale gives ternary information per weighing
Lower bound: $\lceil \log_3 n \rceil$
Optimal split for n=7 is (3, 3, 1)
Worst-case vs best-case analysis

Sources: Classical balance puzzle; verified by exhaustive case analysis of 7-cube partition.

Passage: Previous waves of customer-service technology, including email and those pesky voice menus, stoked concerns of job losses, only for them to fail to materialise. AI could yet prove different. And if it does, its effects may be salutary. Human agents could be freed up to spend more time on creative and rewarding tasks, like using feedback to make products and services better—and thereby spend less time listening to irate customers!

Q52. comprehension (standard)

Which one among the following statements most appropriately reflects the point of view of the given passage?

(a) If AI were to take over customer service, there would be no work left for human subjects to do.
(b) Irritating voice menus and email could not achieve human redundancy to the extent that AI might.
(c) The value of human intervention in the workplace affected by AI might be enhanced through redirection towards more fulfilling tasks.
(d) Unlike previous waves in customer-service technology, AI has raised the alarm of worker replacement.

Answer: (C) · Confidence: high

## Why the answer is (c) — Redirection to fulfilling tasks reflects the writer's optimistic POV ### Passage's core point The passage argues that even if AI does displace customer-service workers (unlike past tech waves), the **effects may be salutary** — humans freed up to do more creative and rewarding work. The writer takes an **optimistic reframe** of AI's labour impact. Key clause: "Human agents could be **freed up to spend more time on creative and rewarding tasks**, like using feedback to make products and services better." ### Evaluate each option **(a) "If AI takes over, no work left for humans."** — Passage says exactly the OPPOSITE: humans are freed up for other (better) work. ❌ Contradicts passage. **(b) "Voice menus and email could not achieve human redundancy to the extent AI might."** — This is a sub-observation in the passage ("AI could yet prove different"), but it is NOT the writer's main point. The writer's POV is the **salutary reframe**, not just the comparative redundancy claim. ⚠️ Partially correct as fact, but not the POV. **(c) "Value of human intervention might be enhanced through redirection to more fulfilling tasks."** — Direct paraphrase of the passage's optimistic conclusion: "Human agents could be freed up to spend more time on creative and rewarding tasks…". ✅ **CAPTURES POV.** **(d) "Unlike previous waves, AI has raised the alarm of worker replacement."** — WRONG: passage says previous waves ALSO "stoked concerns of job losses" (only those concerns failed to materialise). The alarm is NOT unique to AI. ❌ Misstates the contrast. ### Match **Answer: (c).** The writer's POV is the silver lining — humans redirected from drudgery ("listening to irate customers") to fulfilling work ("creative and rewarding tasks"). ### ⚡ Quick trick (≈30 sec) **Last-sentence-is-the-POV rule:** The writer's POV usually lives in the **concluding sentence**. Here it's "Human agents could be freed up to spend more time on creative and rewarding tasks…" — option (c) is its near-verbatim restatement. Options (a) and (d) contradict the passage; (b) captures a sub-point but not the POV. **(c)** in 30 seconds.

Key concepts:

Writer's POV = the optimistic/silver-lining conclusion
Concluding-sentence test for identifying POV
Distinguishing main point from supporting observations
Salutary reframe of technological displacement

Sources: Verified strictly against the AI-customer-service passage in CSAT 2026 paper.

Q53. comprehension (standard)

Which of the following conclusions, made on the basis of the given passage, is/are correct?

The advent of new customer-service technology had invariably sparked fears about job losses
Often it is found that instead of job losses, alternative channels for employee engagement are discovered while certain tasks are replaced by technology
The advent of technology inevitably leads to stressful outcomes

Select the answer using the code given below?

(a) 1 and 3
(b) 2 only
(c) 3 only
(d) 1 and 2

Answer: (D) · Confidence: high

## Why the answer is (d) — Conclusions 1 and 2 only ### Conclusion 1 — New customer-service technology had **invariably** sparked fears about job losses **Passage:** "**Previous waves** of customer-service technology, **including email and those pesky voice menus**, **stoked concerns of job losses**…" The passage's examples (email, voice menus, now AI) all stoked concerns — supporting the invariability claim. ✅ **VALID.** ### Conclusion 2 — Often, instead of job losses, alternative channels for employee engagement are discovered **Passage:** "…only for them to **fail to materialise**." + "Human agents could be freed up to spend more time on **creative and rewarding tasks**". The fears failed to materialise (no actual losses), and humans are freed up for alternative engagement. Direct support. ✅ **VALID.** ### Conclusion 3 — The advent of technology **inevitably** leads to **stressful outcomes** **Passage:** "…its effects may be **salutary**." + humans get "creative and rewarding" work. Passage explicitly says effects may be POSITIVE (salutary). "Inevitably stressful" directly contradicts the passage's optimistic framing. ❌ **INVALID.** ### Match to options Valid: **1 and 2** → option **(d)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 3 | ✗ (3 contradicts passage) | | (b) | 2 only | ✗ (drops valid 1) | | (c) | 3 only | ✗ (3 contradicts passage) | | (d) | 1 and 2 | ✓ | ### ⚡ Quick trick (≈20 sec) **Sentiment-contradiction kill:** Conclusion 3 calls outcomes "stressful" while the passage uses the word **"salutary"** — direct sentiment contradiction. Kill 3 → eliminate (a), (c). Conclusions 1 and 2 both have direct passage anchors → **(d)** in 20 seconds.

Key concepts:

Sentiment-contradiction kill ("stressful" vs passage's "salutary")
"Invariably" supported by enumerated examples
Alternative engagement = creative-rewarding redirection
Passage-bound paraphrase check

Sources: Verified strictly against the AI-customer-service passage in CSAT 2026 paper.

Passage: Cattle from the nearby villages came to the common ground to graze, and there was still a cool freshness in the air. Hori took several deep breaths and thought of sitting down for a while, since he'd be dying of heat in the scorching 'loo' wind the rest of the day. A number of farmers were eager to lease this bit of land and had offered a good price, but Rai Sahib—God bless him—had plainly told them it was reserved for grazing and would not be relinquished for any price. If he'd been one of those selfish Zamindars, he'd have said the cattle could go to hell, that there was no reason for him to miss the chance to make a little money. But the Rai Sahib still held to the old values, feeling that any landlord who didn't look after his tenants was less than human.

Q54. comprehension (standard)

Which of the following conclusions is/are correct?

All landlords essentially have some goodness trapped within them
The common grazing grounds of a village are intended for use by the cattle of that village
Landlords who believe in tradition tend to be more concerned about their tenants
Winds later in the day tend to be cooler post the hot winds of the morning

Select the answer using the code given below?

(a) 1 and 3
(b) 2 and 4
(c) 3 only
(d) 2 only

Answer: (C) · Confidence: high

## Why the answer is (c) — Only conclusion 3 is correct ### Conclusion 1 — **All** landlords essentially have some goodness trapped within them **Passage:** Rai Sahib is contrasted with hypothetical **"selfish Zamindars"** who would have leased the land for money. The passage distinguishes good vs selfish landlords. "All landlords have goodness" is an over-generalisation — passage acknowledges some are selfish. ❌ **INVALID.** ### Conclusion 2 — The common grazing grounds are intended for use by the cattle of **that** village **Passage:** "Cattle from the **nearby villages** came to the common ground to graze." Passage indicates cattle from MULTIPLE nearby villages used the ground — not just one village. The conclusion's "that village" (singular) is too restrictive. ❌ **INVALID.** ### Conclusion 3 — Landlords who believe in tradition tend to be more concerned about their tenants **Passage:** "Rai Sahib still held to the **old values**, feeling that any landlord who didn't look after his tenants was less than human." Direct linkage: old values (tradition) ⇒ concern for tenants. The passage's normative claim (traditional landlords care more) is explicit. ✅ **VALID.** ### Conclusion 4 — Winds later in the day tend to be cooler post the hot winds of the morning **Passage:** "there was still a **cool freshness** in the air" (morning) + "he'd be **dying of heat** in the scorching 'loo' wind the **rest of the day**". Morning = cool, rest of day = hot — **reverse of the conclusion's claim**. ❌ **INVALID.** ### Match to options Valid: **3 only** → option **(c)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 3 | ✗ (1 over-generalises) | | (b) | 2 and 4 | ✗ (both invalid) | | (c) | 3 only | ✓ | | (d) | 2 only | ✗ (2 invalid; drops valid 3) | ### ⚡ Quick trick (≈25 sec) **Two quick kills:** - Conclusion 4 reverses temporal order (passage says morning cool, day hot — conclusion claims opposite). - Conclusion 1 uses **"All"** when the passage contrasts good vs selfish landlords. Two kills + conclusion 2's over-restrictive "that village" → only **3** survives → **(c)** in 25 seconds.

Key concepts:

Over-generalisation kill ("All" without exhaustive support)
Temporal-reversal kill (morning cool, day hot ≠ conclusion's reverse)
Singular vs plural village specification
Traditional values → tenant-care linkage

Sources: Verified strictly against the Hori/Rai Sahib passage in CSAT 2026 paper.

Q55. comprehension (standard)

Which of the following statements are ***not*** correct?

The landholdings of Rai Sahib were currently not being used for farming
Temperamentally, Rai Sahib was as greedy as other landlords
It cannot be ascertained that Rai Sahib could have made some money by leasing out the grazing land
It may be asserted that Rai Sahib valued his tenants and wanted to protect their livelihood

Select the answer using the code given below?

(a) 1 and 2 only
(b) 1 and 3 only
(c) 3 and 4
(d) 1, 2 and 3

Answer: (D) · Confidence: high

## Why the answer is (d) — Statements 1, 2 and 3 are NOT correct ### Statement 1 — Rai Sahib's landholdings were currently not being used for farming **Passage:** "A number of farmers were eager to lease **this bit of land** and had offered a good price, but Rai Sahib… plainly told them it was **reserved for grazing**." The passage discusses ONLY one specific plot (the grazing common). It says nothing about Rai Sahib's OTHER landholdings, which presumably are farmed (he is a Zamindar). Statement 1's plural "landholdings" over-generalises. ❌ **NOT correct.** ### Statement 2 — Temperamentally, Rai Sahib was as greedy as other landlords **Passage:** "If he'd been one of those **selfish** Zamindars, he'd have said the cattle could go to hell, that there was no reason for him to miss the chance to make a little money." Passage explicitly **contrasts** Rai Sahib with selfish/greedy landlords. Statement 2 directly contradicts this. ❌ **NOT correct.** ### Statement 3 — It **cannot be ascertained** that Rai Sahib could have made money by leasing **Passage:** "A number of farmers were eager to lease this bit of land and **had offered a good price**." The passage ascertains that he COULD have made money (good price was offered). Statement 3's "cannot be ascertained" directly contradicts this. ❌ **NOT correct.** ### Statement 4 — It may be asserted that Rai Sahib valued his tenants and wanted to protect their livelihood **Passage:** "the Rai Sahib still held to the old values, feeling that **any landlord who didn't look after his tenants was less than human**." Direct support — Rai Sahib's tenant-care motive is explicit. ✅ **Correct.** ### Match to options NOT correct: **1, 2 and 3** → option **(d)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 2 only | ✗ (drops invalid 3) | | (b) | 1 and 3 only | ✗ (drops invalid 2) | | (c) | 3 and 4 | ✗ (4 is correct) | | (d) | 1, 2 and 3 | ✓ | ### ⚡ Quick trick (≈30 sec) **Triple-kill via direct contradictions:** - Statement 2 contradicts passage's explicit "if he'd been selfish" contrast → kill. - Statement 3 contradicts passage's "good price was offered" → kill. - Statement 1 over-generalises beyond the single plot discussed → kill. Three kills converge on **(d)** in 30 seconds.

Key concepts:

Singular plot vs plural landholdings (scope over-generalisation)
Direct contradiction of passage's contrast (selfish vs Rai Sahib)
"Cannot be ascertained" contradicted by explicit passage information
Tenant-care motive as direct paraphrase

Sources: Verified strictly against the Hori/Rai Sahib passage in CSAT 2026 paper.

Q56. reasoning (standard)

A person facing the East travels 4 km straight and then turns right and travels 3 km, then further turns left and travels 2 km and finally turns left and travels 3 km. The minimum distance between the final point and the initial point, and the direction in which the person is facing at the final point are, respectively

(a) 12 km, East
(b) 6 km, East
(c) 8 km, North
(d) 6 km, North

Answer: (D) · Confidence: high

## Why the answer is (d) — 6 km East-of-start, facing North ### Step 1 — Use coordinates (E = +x, N = +y); start at origin facing East | Move | Heading | Δposition | New position | |---|---|---|---| | Start | East | (0, 0) | (0, 0) | | 4 km straight (E) | East | (+4, 0) | (4, 0) | | Turn **right** → South; 3 km | South | (0, −3) | (4, −3) | | Turn **left** → East; 2 km | East | (+2, 0) | (6, −3) | | Turn **left** → North; 3 km | North | (0, +3) | **(6, 0)** | ### Step 2 — Compute distance from start (0,0) to end (6,0) $$d = \sqrt{(6-0)^2 + (0-0)^2} = \sqrt{36} = \mathbf{6}\text{ km}.$$ Final position is **6 km due East** of start; the displacement vector points East. ### Step 3 — Direction person is facing at the final point After the last left turn from East, the person now faces **North**. **Answer: (d) 6 km, North.** ### ⚡ Quick trick (≈30 sec) **Vector-sum shortcut:** Track only net E-displacement and net N-displacement: - Net East: $4 + 2 = 6$ km (from steps 1 and 3) - Net North: $-3 + 3 = 0$ km (steps 2 cancels step 4) Distance = $\sqrt{6^2 + 0^2} = 6$ km. Heading after R, L, L turns from East: $E \to S \to E \to N$ → facing **North**. **(d)** in 30 seconds.

Key concepts:

Coordinate-based displacement tracking
Right/left turn = 90° clockwise/anticlockwise
Net E-W and N-S vector decomposition
Pythagorean distance formula

Sources: Direct coordinate-tracking; verified by step-by-step turn analysis.

Q57. numeracy (standard)

In a sequence of numbers, each number other than the first two is the sum of the two immediately preceding numbers from it. If the first two numbers in the sequence are 4 and 7, then the sixth number is

(a) 29
(b) 37
(c) 43
(d) 47

Answer: (D) · Confidence: high

## Why the answer is (d) — Sixth number is 47 ### Sequence rule Fibonacci-like: $a_n = a_{n-1} + a_{n-2}$ for $n \ge 3$, with $a_1 = 4$, $a_2 = 7$. ### Compute term-by-term | $n$ | $a_n$ | Calculation | |---|---|---| | 1 | 4 | (given) | | 2 | 7 | (given) | | 3 | 11 | $4 + 7$ | | 4 | 18 | $7 + 11$ | | 5 | 29 | $11 + 18$ | | **6** | **47** | $18 + 29$ | **Answer: (d) 47.** ### ⚡ Quick trick (≈15 sec) **Mental fibonacci-roll:** 4, 7 → 11 → 18 → 29 → **47**. Six additions, each a single-digit sum. **(d)** in 15 seconds.

Key concepts:

Fibonacci-like recurrence: $a_n = a_{n-1} + a_{n-2}$
Term-by-term computation

Sources: Direct sequence computation; verified by tabulation.

Q58. numeracy (standard)

The ratio of male to female workers in two companies $A$ and $B$ is 13 : 10 and 7 : 5, respectively. If both the companies have the same number of female workers, then what is the ratio of the total number of workers in $A$ to those in $B$?

(a) 24 : 23
(b) 23 : 24
(c) 18 : 17
(d) 27 : 18

Answer: (B) · Confidence: high

## Why the answer is (b) — A : B = 23 : 24 ### Step 1 — Express each company in its own variable - **Company A:** M : F = 13 : 10 → M = 13k, F = 10k → Total A = 23k. - **Company B:** M : F = 7 : 5 → M = 7m, F = 5m → Total B = 12m. ### Step 2 — Equate female counts $$F_A = F_B \Rightarrow 10k = 5m \Rightarrow m = 2k.$$ ### Step 3 — Compute total ratio - Total A = 23k. - Total B = 12m = 12(2k) = 24k. $$\frac{A}{B} = \frac{23k}{24k} = \mathbf{\frac{23}{24}} \;\Rightarrow\; \text{ratio } 23:24.$$ **Answer: (b) 23 : 24.** ### ⚡ Quick trick (≈30 sec) **LCM of female counts:** LCM(10, 5) = 10. So set F = 10 in both: - A: 13 males + 10 females = 23. - B: females = 10, but ratio is 7:5, so multiply by 2: males = 14, females = 10 → total = 24. Ratio A : B = **23 : 24**. **(b)** in 30 seconds.

Key concepts:

Ratio normalisation across two systems by equating common quantity
LCM method for matching ratios
Total = (numerator + denominator) × scaling factor

Sources: Direct ratio algebra; verified by LCM shortcut.

Q59. numeracy (standard)

If the product of the HCF and LCM of two distinct numbers is the cube of one of the numbers, then which of the following statements is/are correct? I. The difference of the numbers is an even number. II. One of the numbers is a perfect square

Select the answer using the code given below?

(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II

Answer: (C) · Confidence: high

## Why the answer is (c) — Both I and II are correct ### Step 1 — Use the identity $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ Given: $\text{HCF} \times \text{LCM} = $ cube of one number, say $a^3$. $$ab = a^3 \Rightarrow b = a^2.$$ (If the cube is $b^3$ instead, then $a = b^2$ — same structure with a and b swapped.) ### Step 2 — Test Statement I: difference is even $$|a - b| = |a - a^2| = a \cdot |1 - a| = a(a-1)$$ (for $a \ge 2$; $a = 1$ would give $b = 1$, but numbers are distinct). $a(a-1)$ is the **product of two consecutive integers** → always contains an even factor → always **even**. ✅ **Example:** $a = 2, b = 4$ → diff = 2 (even). $a = 3, b = 9$ → diff = 6 (even). ### Step 3 — Test Statement II: one number is a perfect square $b = a^2$ is by definition a **perfect square**. ✅ (If $a = b^2$ instead, then $a$ is the perfect square. Either way, one of the two numbers is a perfect square.) ### Verification | $a$ | $b = a^2$ | HCF × LCM | Diff (even?) | $b$ perfect sq? | |---|---|---|---|---| | 2 | 4 | 2 × 4 = 8 = $2^3$ ✓ | 2 ✓ | 4 ✓ | | 3 | 9 | 3 × 9 = 27 = $3^3$ ✓ | 6 ✓ | 9 ✓ | | 5 | 25 | 5 × 25 = 125 = $5^3$ ✓ | 20 ✓ | 25 ✓ | Both statements hold in all valid cases → **option (c)**. ### ⚡ Quick trick (≈40 sec) **One identity, two conclusions:** From $\text{HCF} \times \text{LCM} = ab = a^3$, derive $b = a^2$ immediately. Both Statement I (consecutive-integer product = even) and Statement II (square of a) fall out directly. **(c)** in 40 seconds.

Key concepts:

Identity: HCF(a, b) × LCM(a, b) = a × b
Consecutive-integer product is always even
Perfect square: $a^2$ for integer $a$
$b = a^2$ structural relationship

Sources: Direct number-theory derivation from HCF×LCM identity; verified with multiple examples.

Q60. numeracy (standard)

If $x$ and $y$ are two digits and the number $4x5y790$ is divisible by 11, then what is the remainder, if $x + y$ is divided by 11?

(a) 1
(b) 3
(c) 5
(d) 7

Answer: (D) · Confidence: high

## Why the answer is (d) — Remainder is 7 ### Step 1 — Apply the 11-divisibility rule A number is divisible by 11 iff the **alternating sum** of its digits (odd-position − even-position, from the right or left consistently) is divisible by 11. Digits of $4x5y790$ (positions 1 → 7 from the left): | Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 | |---|---|---|---|---|---|---|---| | Digit | 4 | x | 5 | y | 7 | 9 | 0 | Alternating sum: $$\underbrace{(4 + 5 + 7 + 0)}_{\text{odd positions}} - \underbrace{(x + y + 9)}_{\text{even positions}} = 16 - (x + y + 9) = 7 - (x + y).$$ ### Step 2 — Set the alternating sum divisible by 11 $$7 - (x + y) \equiv 0 \pmod{11} \;\Rightarrow\; x + y \equiv 7 \pmod{11}.$$ ### Step 3 — Compute (x + y) mod 11 Since $x, y \in \{0, 1, \ldots, 9\}$, $x + y \in [0, 18]$. Values satisfying $x + y \equiv 7 \pmod{11}$ in this range: $x + y = 7$ or $x + y = 18$. Both give the **same remainder** when divided by 11: - $7 \div 11 = 0$ remainder **7** - $18 \div 11 = 1$ remainder **7** **Answer: (d) 7.** ### ⚡ Quick trick (≈30 sec) **Direct from the rule:** Alternating sum of fixed digits = $16 - 9 = 7$; variable part = $x + y$ (in the even-position group, subtract). So the equation reduces to $x + y \equiv 7 \pmod{11}$, and the answer to the question ("remainder of $x + y$ divided by 11") is **exactly 7**. **(d)** in 30 seconds.

Key concepts:

11-divisibility rule: alternating digit sum
Modular arithmetic on digit sums
Two valid $(x+y)$ values (7 and 18) give same mod-11 remainder

Sources: Direct application of 11-divisibility rule; verified by modular reduction.

Passage: Was it the sun-dappled ambience, the strawberries and cream, the frustration of Flavio Cobolli's unforced errors against Serbian Novak Djokovic on Centre Court or simply the crushing weight of being a 64-year-old man in the third act of a very public life? Whatever the reason, Hugh Grant, the actor, deserves empathy. There he was, in the Royal Box at Wimbledon, flanked by Britain's well-dressed and well-rested spectators, watching the men's singles quarterfinals, when the actor did something quietly radical : head at a tilt, eyes closed, utterly unbothered, he took a nap. So praise be to Grant for serving up an unexpected ace. In that small, delicious moment, he didn't merely catch forty winks, he made an elegant case for surrender. Not to laziness, but to limits. To the body's quiet wisdom over society's relentless performance metrics. Wimbledon had its tennis. The perpetually sleep-deprived discovered a leading man, not of action, but of rest.

Q61. comprehension (standard)

Which of the following statements is/are correct?

Radical action can also be attributed to mild surrender where one acts against societal expectations
Submitting to one's limitations, given the effect of age and other factors, ought not to be conflated with laziness
'Leading man' usually refers to one who plays the lead role in a movie; in this instance the implication is that Hugh Grant is performing the role of not an action hero, but that of a resting one!

Select the answer using the code given below?

(a) 1 and 2 only
(b) 2 and 3 only
(c) 1, 2 and 3
(d) 3 only

Answer: (C) · Confidence: high

## Why the answer is (c) — All three statements are correct ### Statement 1 — Radical action can be attributed to mild surrender against societal expectations **Passage:** Hugh Grant napping at Wimbledon was "**quietly radical**" and represented "surrender… to the body's quiet wisdom over **society's relentless performance metrics**". Direct match: napping (mild act of surrender) was radical against societal performance expectations. ✅ **VALID.** ### Statement 2 — Submitting to one's limitations should not be conflated with laziness **Passage:** "an elegant case for surrender. **Not to laziness, but to limits**." Direct paraphrase. ✅ **VALID.** ### Statement 3 — 'Leading man' usually means lead role in a movie; here Hugh Grant is the 'leading man of rest', not of action **Passage:** "The perpetually sleep-deprived discovered a **leading man, not of action, but of rest**." Direct paraphrase of the passage's playful word-twist. ✅ **VALID.** ### Match to options All three valid → option **(c)** 1, 2 and 3. ### ⚡ Quick trick (≈25 sec) **Triple direct-paraphrase check:** Each statement maps onto a specific passage line: - Statement 1 ↔ "quietly radical" + "society's relentless performance metrics" - Statement 2 ↔ "Not to laziness, but to limits" - Statement 3 ↔ "leading man, not of action, but of rest" When all three pair with passage lines, **(c) All three** is the answer.

Key concepts:

Direct paraphrase verification
Word-play recognition ("leading man of rest")
Soft surrender as radical action
Passage-bound inference

Sources: Verified strictly against the Hugh Grant / Wimbledon passage in CSAT 2026 paper.

Q62. comprehension (standard)

Which of the following statements is/are correct?

Hugh Grant was watching, from the Royal Box, the men's semifinal match on Centre Court between Flavio Cobolli and Novak Djokovic
The phrase 'unexpected ace' in the context uses a term from the game of tennis to highlight Hugh Grant's somewhat uncharacteristic act of catching 'forty winks'; an act that is viewed with opprobrium
Grant subjects the demands of society to the wisdom of his body

Select the answer using the code given below?

(a) 1 and 2
(b) 3 only
(c) 2 and 3
(d) 2 only

Answer: (B) · Confidence: high

## Why the answer is (b) — Only statement 3 is correct ### Statement 1 — Hugh Grant was watching the men's **semifinal** between Cobolli and Djokovic **Passage:** "watching the men's **singles quarterfinals**". Round mismatch: quarterfinals, NOT semifinals. ❌ **INVALID.** ### Statement 2 — 'unexpected ace' highlights Grant's uncharacteristic act of catching forty winks, viewed with **opprobrium** **Passage:** "**Praise be** to Grant for serving up an unexpected ace." "Opprobrium" means harsh disapproval / disgrace. The passage's tone is the **opposite** — it praises Grant. Calling the act "viewed with opprobrium" inverts the passage's celebratory framing. ❌ **INVALID.** ### Statement 3 — Grant subjects the demands of society to the wisdom of his body **Passage:** "surrender… to the body's quiet wisdom **over** society's relentless performance metrics." Body's wisdom prevails over society's demands. Statement 3 phrases this as "subjects the demands of society to the wisdom of his body" — i.e. subordinates society to body. Direct match. ✅ **VALID.** ### Match to options Valid: **3 only** → option **(b)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 2 | ✗ (both invalid) | | (b) | 3 only | ✓ | | (c) | 2 and 3 | ✗ (2 inverts tone) | | (d) | 2 only | ✗ (2 inverts tone) | ### ⚡ Quick trick (≈25 sec) **Tone-flip + factual-detail kills:** - Statement 1 changes 'quarterfinals' to 'semifinals' — factual error. - Statement 2 adds 'opprobrium' (disapproval) to a passage that explicitly **praises** Grant — tone inversion. Both kills → only statement 3 remains → **(b)** in 25 seconds.

Key concepts:

Factual-detail trap (quarterfinals vs semifinals)
Tone-inversion trap (opprobrium vs praise)
Body wisdom prevailing over societal demands

Sources: Verified strictly against the Hugh Grant / Wimbledon passage in CSAT 2026 paper.

Passage: The process by which countries close their labour-productivity gap with the technology leader is based on convergence theory. The convergence model divides economic eras into three phases : the breakaway, the catch-up, and the fine-tuning phase. It also divides economic entities into two categories : the technology leaders and the technology followers. The process begins with the development of a new technology, such as scavenging three million years ago (MYA), hunting—one MYA, farming—12 thousand years ago, and industrial technology—a little more than 200 years ago. During the breakaway phase, the per capita income of the technology leaders (e.g., Western Europe and North America in the industrial era) rises, but is unchanged for the technology followers. In the catch-up phase, the followers adopt the new technology and close their per capita income gap with the technology leaders. In the fine-tuning phase, where participants try to extract the remaining benefits from an increasingly exhausted technology, leaders and followers have similar per capita incomes.

Q63. comprehension (standard)

Which of the following conclusions are correct?

In the breakaway phase, economic progress is slow for the technology followers
In the catch-up phase, leaders stagnate and followers, therefore, close the gap between them and the leaders
In the fine-tuning phase, technology is exhausted, as it were, and both leaders and followers attempt to extract leftover benefits, leading to more or less similar per capita income levels
Industrial technology followed scavenging, which preceded hunting, which itself was followed by farming

Select the answer using the code given below?

(a) 3 and 4 only
(b) 1, 2, 3 and 4
(c) 1 and 2 only
(d) 2, 3 and 4 only

Answer: (A) · Confidence: high

## Why the answer is (a) — Only conclusions 3 and 4 are correct ### Conclusion 1 — In the breakaway phase, economic progress is **slow** for the technology followers **Passage:** "During the breakaway phase, the per capita income of the technology leaders… rises, but is **unchanged** for the technology followers." Passage says followers' income is **UNCHANGED** (zero progress), not merely "slow". "Slow" implies some forward movement; "unchanged" means none. The statement understates by mischaracterising — a UPSC trap. ❌ **INVALID.** ### Conclusion 2 — In the catch-up phase, leaders **stagnate** and followers therefore close the gap **Passage:** "In the catch-up phase, the followers **adopt the new technology** and close their per capita income gap with the technology leaders." Two problems: - Passage does NOT say leaders stagnate during catch-up. It is silent on leaders' growth here. - The cause of catch-up is **followers adopting the technology**, not leaders stagnating. The conclusion's "therefore" forges an unstated causal link. ❌ **INVALID** (false cause attribution). ### Conclusion 3 — In the fine-tuning phase, technology is exhausted; both leaders and followers extract leftover benefits, leading to similar per capita incomes **Passage:** "In the fine-tuning phase, where participants try to extract the **remaining benefits** from an **increasingly exhausted** technology, leaders and followers have **similar per capita incomes**." Direct paraphrase. ✅ **VALID.** ### Conclusion 4 — Industrial technology followed scavenging, which preceded hunting, which itself was followed by farming **Passage timeline:** scavenging (3 MYA) → hunting (1 MYA) → farming (12 KYA) → industrial (~200 years ago). Statement order: scavenging → hunting → farming → industrial. Matches the passage chronologically. ✅ **VALID.** ### Match to options Valid: **3 and 4** → option **(a)**. | Option | Set | Verdict | |---|---|---| | (a) | 3 and 4 only | ✓ | | (b) | 1, 2, 3 and 4 | ✗ (1 and 2 invalid) | | (c) | 1 and 2 only | ✗ (1 and 2 invalid) | | (d) | 2, 3 and 4 only | ✗ (2 invalid) | ### ⚡ Quick trick (≈30 sec) **Two key traps:** - Statement 1 swaps **"unchanged"** (zero growth) with **"slow"** (some growth) — soft mischaracterisation. - Statement 2 invents a cause ("leaders stagnate") not stated in the passage. Both kills + statements 3 and 4 as clean paraphrases → **(a)** in 30 seconds.

Key concepts:

Convergence theory: breakaway → catch-up → fine-tuning
Unchanged vs slow (zero vs positive growth)
False cause attribution trap ("therefore" + unsupported cause)
Chronological-order verification

Sources: Verified strictly against the convergence theory passage in CSAT 2026 paper.

Q64. comprehension (standard)

Which of the following statements is/are correct?

The convergence model divides nations into three phases of economic progress
At the heart of the convergence theory is the closing of the gap between labour and productivity
Technology leaders typically have arrived earlier at different economic eras
The time period covered by the convergence theory presented herein encompasses, as mentioned, 4012200 years

Select the answer using the code given below?

(a) 2 and 4
(b) 2 and 3
(c) 3 only
(d) 1, 3 and 4

Answer: (C) · Confidence: high

## Why the answer is (c) — Only statement 3 is correct ### Statement 1 — The convergence model divides **nations** into three phases **Passage:** "The convergence model divides **economic eras** into three phases : the breakaway, the catch-up, and the fine-tuning phase. It also divides **economic entities into two categories** : the technology leaders and the technology followers." Two categorical errors: - It's **economic eras** (not nations) that are divided into three phases. - Nations/entities are divided into **TWO categories**, not three phases. ❌ **INVALID.** ### Statement 2 — At the heart of convergence theory is the closing of the gap between **labour and productivity** **Passage:** "countries close their **labour-productivity gap** with the technology leader". The hyphenated compound "labour-productivity" is **one concept** (= productivity of labour) — not two separate quantities (labour AND productivity). Statement 2 misreads the compound, fabricating a gap between two distinct concepts. ❌ **INVALID** (compound-word misreading). ### Statement 3 — Technology leaders typically have arrived earlier at different economic eras **Passage:** "Western Europe and North America in the industrial era" — named as the technology leaders of that era — implicitly indicating leaders arrive first at each new technology era. ✅ **VALID** (reasonable inference from named examples). ### Statement 4 — The time period encompasses 4,012,200 years (as mentioned) **Passage timeline:** scavenging (3 MYA), hunting (1 MYA), farming (12 KYA), industrial (~200 years ago). $3{,}000{,}000 + 1{,}000{,}000 + 12{,}000 + 200 = 4{,}012{,}200$ — but this is a **sum of era-ages**, NOT the total time period covered. The time period is the **earliest era** = **3 million years** (since scavenging began). The number 4,012,200 is a mathematical artifact, not the actual time period. ❌ **INVALID.** ### Match to options Valid: **3 only** → option **(c)**. | Option | Set | Verdict | |---|---|---| | (a) | 2 and 4 | ✗ (both invalid) | | (b) | 2 and 3 | ✗ (2 invalid) | | (c) | 3 only | ✓ | | (d) | 1, 3 and 4 | ✗ (1 and 4 invalid) | ### ⚡ Quick trick (≈40 sec) **Compound-word and category traps:** - Statement 1 confuses what is divided (eras vs nations). - Statement 2 splits "labour-productivity" into two concepts — classic misread. - Statement 4 adds era-ages as if they were time-period — meaningless sum. Only statement 3 (named-example inference) survives → **(c)** in 40 seconds.

Key concepts:

Category confusion (eras vs nations)
Compound-word misreading ("labour-productivity" as one concept)
Time period vs sum of era ages (the 4,012,200 trap)
Named-example inference

Sources: Verified strictly against the convergence theory passage in CSAT 2026 paper.

Q65. numeracy (standard)

An alloy $P$ contains 20% copper and 80% zinc by weight. Another alloy $Q$ contains 60% copper and 40% zinc by weight. A third alloy $R$ is to be prepared from $P$ and $Q$ so that it contains equal amount of copper and zinc. In what ratio, amounts of $P$ and $Q$ be mixed in order to get $R$?

(a) 1 : 3
(b) 3 : 1
(c) 2 : 3
(d) 3 : 2

Answer: (A) · Confidence: high

## Why the answer is (a) — Mix P and Q in ratio 1 : 3 ### Step 1 — Set up the copper-balance equation Let amounts mixed be $a$ grams of P and $b$ grams of Q. Target copper percentage = **50%**. $$\frac{0.20 \, a + 0.60 \, b}{a + b} = 0.50.$$ ### Step 2 — Solve for $a : b$ $$0.20a + 0.60b = 0.50(a + b) = 0.50a + 0.50b.$$ $$0.10b = 0.30a \;\Rightarrow\; \frac{b}{a} = 3 \;\Rightarrow\; a : b = \mathbf{1 : 3}.$$ **Answer: (a) 1 : 3.** ### Verification Mix 1 g of P + 3 g of Q (total 4 g): - Cu: $0.20 \times 1 + 0.60 \times 3 = 0.20 + 1.80 = 2.00$ g → 2/4 = 50% ✓ - Zn: $0.80 \times 1 + 0.40 \times 3 = 0.80 + 1.20 = 2.00$ g → 2/4 = 50% ✓ ### ⚡ Quick trick (≈20 sec) **Alligation method:** ``` P (20% Cu) Q (60% Cu) \ / 50% (target) / \ 60-50=10 50-20=30 ``` Ratio P : Q = 10 : 30 = **1 : 3**. **(a)** in 20 seconds.

Key concepts:

Weighted-average / alligation rule
Mixture ratio inversely proportional to distance from target
Conservation of solute mass in mixing

Sources: Direct alligation; verified by component-wise mass conservation.

Q66. numeracy (standard)

$A$ is a 2-digit number with different digits. $B$ is also a 2-digit number and is obtained by reversing the digits of $A$. If $A - B$ is a multiple of 27, where $A > B$, how many such different $A$'s are possible?

(a) 6
(b) 9
(c) 12
(d) 18

Answer: (B) · Confidence: high

## Why the answer is (b) — 9 such values of A ### Step 1 — Set up Let A = $10x + y$ where $x$ = tens digit, $y$ = units digit, $x \ne y$, $x \ge 1$. Then B = $10y + x$. **Both A and B are 2-digit** ⇒ $x \ge 1$ AND $y \ge 1$. ### Step 2 — Express A − B and apply the divisibility condition $$A - B = (10x + y) - (10y + x) = 9(x - y).$$ A > B ⇒ $x > y$. So $A - B = 9(x - y) > 0$. $A - B$ is a multiple of 27 ⇒ $9(x - y) \equiv 0 \pmod{27}$ ⇒ $x - y \equiv 0 \pmod 3$. Since digits give $1 \le x - y \le 8$ (with $y \ge 1$, $x \le 9$): $$x - y \in \{3, 6\}.$$ (Note: $x - y = 9$ would require $y = 0$, violating B being 2-digit.) ### Step 3 — Enumerate the (x, y) pairs **$x - y = 3$:** $(4,1), (5,2), (6,3), (7,4), (8,5), (9,6)$ → **6 pairs** **$x - y = 6$:** $(7,1), (8,2), (9,3)$ → **3 pairs** ### Step 4 — Total $$6 + 3 = \mathbf{9} \;\Rightarrow\; \text{option (b)}.$$ ### Verification Example: A = 41, B = 14, A − B = 27 ✓ (27 × 1). Example: A = 96, B = 69, A − B = 27 ✓ (27 × 1). Example: A = 71, B = 17, A − B = 54 = 27 × 2 ✓. ### ⚡ Quick trick (≈30 sec) **Algebra shortcut:** $A - B = 9(x - y)$, so divisibility by 27 reduces to divisibility of $(x - y)$ by 3. With $1 \le x - y \le 8$, valid differences are 3 and 6. Count $(x, y)$ pairs: 6 + 3 = **9**. **(b)** in 30 seconds.

Key concepts:

Reversed two-digit subtraction: $A - B = 9(x - y)$
Divisibility transfer: $9(x-y)$ ÷ 27 ⟺ $(x-y)$ ÷ 3
Digit constraint: both digits 1–9 (no leading zero)
Enumeration by difference class

Sources: Direct algebraic factorisation of A − B; verified by exhaustive enumeration.

Q67. reasoning (standard)

If ZERO is encoded as ADSN, then how do you encode STOP?

(a) SPOT
(b) TSPO
(c) TSOP
(d) POST

Answer: (B) · Confidence: high

## Why the answer is (b) — TSPO ### Step 1 — Decode the ZERO → ADSN pattern | Position | Source | Target | Shift | |---|---|---|---| | 1 | Z (26) | A (1) | **+1** (Z wraps to A) | | 2 | E (5) | D (4) | **−1** | | 3 | R (18) | S (19) | **+1** | | 4 | O (15) | N (14) | **−1** | **Pattern: +1, −1, +1, −1** (alternating shifts). ### Step 2 — Apply the same pattern to STOP | Position | Source | Shift | Target | |---|---|---|---| | 1 | S (19) | +1 | **T (20)** | | 2 | T (20) | −1 | **S (19)** | | 3 | O (15) | +1 | **P (16)** | | 4 | P (16) | −1 | **O (15)** | Encoded STOP = **TSPO** → option **(b)**. ### ⚡ Quick trick (≈25 sec) **Alternating shift pattern:** Note Z+1 wraps to A. Pattern is clearly +1/−1 alternating. Apply to STOP: S→T, T→S, O→P, P→O → **TSPO**. **(b)** in 25 seconds.

Key concepts:

Alphabetic letter-shift coding
Alternating-shift pattern recognition
Cyclic wraparound (Z + 1 = A)

Sources: Direct coding analysis; verified by digit-position shift application.

Q68. numeracy (standard)

There are three types of rectangular tiles : 3′ × 3′, 3′ × 7′ and 3′ × 11′. An area of rectangular shape of dimensions 3′ × 100′ is to be covered using these tiles without breaking them. If $x$ and $y$ are the maximum and minimum numbers of tiles of various sizes, respectively, that can be used to cover the area exactly, then $x - y$ is

(a) 20
(b) 12
(c) 10
(d) 7

Answer: (A) · Confidence: high

## Why the answer is (a) — $x - y = 20$ ### Setup Tile a $3 \times 100$ rectangle with $3 \times 3$, $3 \times 7$, and $3 \times 11$ tiles. All tiles are 3-wide, so only **lengths must sum to 100**: $$3a + 7b + 11c = 100, \quad a, b, c \ge 0.$$ Find $x$ = max(a + b + c) and $y$ = min(a + b + c). ### Maximum tiles ($x$): favour small (length-3) tiles 100 isn't divisible by 3 (100 mod 3 = 1), so we need at least one non-3 tile. Try one tile of length 7: $100 - 7 = 93 = 3 \cdot 31$. So $a = 31, b = 1, c = 0$ → total = **32**. Check via residue: $3a + 7b + 11c \equiv 0 + b + 2c \pmod 3 \equiv 100 \equiv 1$. So $b + 2c \equiv 1 \pmod 3$. With $c = 0$, $b = 1$ (smallest) maximises $a$. **$x = 32$.** ### Minimum tiles ($y$): favour large (length-11) tiles Total = $a + b + c$ with $3a + 7b + 11c = 100$. Subtract: $4b + 8c = 100 - 3(\text{total})$. For total = 10: $4b + 8c = 70$ → $2b + 4c = 35$, no integer solution (parity mismatch). For total = 11: $4b + 8c = 67$, no integer solution (parity). For total = 12: $4b + 8c = 64$ → $b + 2c = 16$. Multiple solutions: $(a, b, c) = (4, 0, 8), (3, 2, 7), (2, 4, 6), \ldots$ ✓ **$y = 12$.** ### Compute $x - y$ $$x - y = 32 - 12 = \mathbf{20} \;\Rightarrow\; \text{option (a)}.$$ ### ⚡ Quick trick (≈60 sec) **Max (small tiles):** Use as many 3s as possible. Since 100 mod 3 = 1, swap one 3 for a 7 (lengths differ by 4 ≡ 1 mod 3). So $a = 31, b = 1$ → 32 tiles. **Min (large tiles):** Use as many 11s as possible. $100 / 11 \approx 9$, but 9·11 = 99 leaves 1 (unfittable). Try 8·11 = 88, leaves 12 = 4·3 → total 12 tiles. $32 - 12 = \mathbf{20}$. **(a)** in about a minute.

Key concepts:

Linear Diophantine equation: $3a + 7b + 11c = 100$
Residue arithmetic mod 3 for feasibility
Parity-based elimination of impossible totals
Greedy small-tile (max) vs greedy large-tile (min)

Sources: Direct enumeration with modular feasibility checks; verified by constructing both maximum and minimum tilings.

Q69. numeracy (standard)

A train has to complete a journey of 800 km. If it meets a minor accident, its speed becomes half of the existing speed. If there is a mechanical defect, the speed becomes one-fourth of the existing speed. On its way, the train meets with a minor accident after 200 km; and 400 km thereafter, it develops a mechanical defect. Had the train developed the mechanical defect after 200 km and met the minor accident 400 km thereafter, it would have taken 4 more hours to reach its destination. What was the original speed of the train in km per hour?

(a) 200
(b) 190
(c) 150
(d) 100

Answer: (A) · Confidence: high

## Why the answer is (a) — Original speed = 200 km/h ### Key insight: "speed becomes X of the existing speed" Each event modifies the **current** speed, not the original. So the events COMPOUND multiplicatively. Let original speed = $v$ km/h. ### Scenario A (actual): minor accident first, then defect | Segment | Distance | Speed | Time | |---|---|---|---| | 0 → 200 km | 200 | $v$ | $200/v$ | | 200 → 600 km | 400 | $v/2$ (after accident) | $800/v$ | | 600 → 800 km | 200 | $(v/2)/4 = v/8$ (after defect on current speed) | $1600/v$ | $$T_A = \frac{200}{v} + \frac{800}{v} + \frac{1600}{v} = \frac{2600}{v}.$$ ### Scenario B (swapped): defect first, then minor accident | Segment | Distance | Speed | Time | |---|---|---|---| | 0 → 200 km | 200 | $v$ | $200/v$ | | 200 → 600 km | 400 | $v/4$ (after defect) | $1600/v$ | | 600 → 800 km | 200 | $(v/4)/2 = v/8$ (after accident on current speed) | $1600/v$ | $$T_B = \frac{200}{v} + \frac{1600}{v} + \frac{1600}{v} = \frac{3400}{v}.$$ ### Solve for $v$ using the 4-hour difference $$T_B - T_A = \frac{3400 - 2600}{v} = \frac{800}{v} = 4 \text{ hours}.$$ $$v = \frac{800}{4} = \mathbf{200 \text{ km/h}}.$$ **Answer: (a) 200.** ### ⚡ Quick trick (≈60 sec) **Compound-speed insight:** In Scenario A, the second segment (400 km) runs at $v/2$ (slower than original, faster than $v/4$); in B, it runs at $v/4$ (much slower). The third segment is the SAME speed in both ($v/8$). Difference comes entirely from middle segment: $$\Delta T = \frac{400}{v/4} - \frac{400}{v/2} = \frac{1600}{v} - \frac{800}{v} = \frac{800}{v} = 4 \Rightarrow v = 200.$$ **(a)** in about a minute. The mechanical-defect-applied-on-already-halved-speed (v/8 in both scenarios) is the key cancellation.

Key concepts:

"Existing speed" modifier compounds multiplicatively
Segment-time = distance / speed
Difference-of-scenarios isolates the variable segment
Order matters: half-then-quarter = 1/8, quarter-then-half = 1/8 (same)

Sources: Direct kinematic computation; verified by segment-wise time accounting in both scenarios.

Q70. numeracy (standard)

In a recruitment process, the selection of candidates is based on their performance in three components. The weightages of the components 1, 2 and 3 are 0·2, 0·3 and 0·5, respectively. Use the data given below and find the cutoff score if exactly three candidates are to be selected:

(a) 5·1
(b) 5·2
(c) 5·3
(d) 5·4

Answer: (A) · Confidence: high

## Why the answer is (a) — Cutoff = 5·1 ### Step 1 — Compute weighted score for each candidate Weights: 0.2 (C1), 0.3 (C2), 0.5 (C3). Weighted score = $0.2 s_1 + 0.3 s_2 + 0.5 s_3$. | Candidate | s₁ | s₂ | s₃ | Weighted score | |---|---|---|---|---| | 1 | 5 | 4 | 6 | $1.0 + 1.2 + 3.0 = $ **5.2** | | 2 | 4 | 6 | 5 | $0.8 + 1.8 + 2.5 = $ **5.1** | | 3 | 3 | 2 | 8 | $0.6 + 0.6 + 4.0 = $ **5.2** | | 4 | 9 | 4 | 3 | $1.8 + 1.2 + 1.5 = $ 4.5 | | 5 | 8 | 8 | 2 | $1.6 + 2.4 + 1.0 = $ 5.0 | ### Step 2 — Rank descending and pick top 3 Sorted: **5.2 (C1), 5.2 (C3), 5.1 (C2), 5.0 (C5), 4.5 (C4)**. Top 3 selected: C1, C3, C2 with scores 5.2, 5.2, **5.1**. ### Step 3 — Cutoff is the lowest-selected score $$\text{Cutoff} = 5.1 \;\Rightarrow\; \text{option (a)}.$$ ### ⚡ Quick trick (≈45 sec) **Component-3-dominates heuristic:** Weight 0.5 on C3 is the largest, so candidates with high $s_3$ score well. C1 ($s_3 = 6$), C3 ($s_3 = 8$) lead. Then C2 ($s_3 = 5$) and C5 ($s_3 = 2$) compete for third. Compute precisely: C2 = 5.1, C5 = 5.0. C2 wins → cutoff = **5.1**. **(a)** in 45 seconds.

Key concepts:

Weighted average calculation
Rank-and-select for cutoff determination
Cutoff = lowest score of selected candidates
Dominant-weight heuristic for quick estimation

Sources: Direct arithmetic of weighted averages; verified by ranking and selection.

Q71. reasoning (standard)

Consider the following three statements, namely S1, S2 and S3:

S3 is a counterpoint to S1
S3 is unconnected to S1 and S2
S2 could be the reason for S3

Select the answer using the code given below?

(a) 2 and 3 only
(b) 1, 2 and 3
(c) 1 and 2 only
(d) 3 only

Answer: (D) · Confidence: medium

## Why the answer is (d) — Only relationship 3 is correct ### The three premises - **S1:** Protecting the environment is existential for humans (due to climate change). - **S2:** Scientific consensus on extent of human contribution to climate change has NOT been achieved. - **S3:** Environmental activism includes climate alarmism and extremist views that climate change deniers focus on. ### Relationship 1 — S3 is a counterpoint to S1 S1 makes a strong urgency claim; S3 talks about pitfalls of extreme activism that deniers exploit. S3 doesn't **directly oppose** S1's urgency — both can be true simultaneously. S3 is a tangential observation about the FORM of activism, not a denial of S1's substance. ❌ **Not a proper counterpoint.** ### Relationship 2 — S3 is unconnected to S1 and S2 S3 explicitly mentions **"climate change deniers"** — the topic of climate change directly ties S3 to both S1 (climate change impact) and S2 (climate change consensus). S3 is clearly **connected** to both. ❌ **INVALID.** ### Relationship 3 — S2 could be the reason for S3 S2 (lack of scientific consensus on human contribution) provides the **rhetorical ammunition** that climate change deniers exploit. Without S2's uncertainty, deniers would have less ground to focus on activist excesses. So S2 ENABLES the dynamic described in S3 (alarmism + deniers' counter-attack). ✅ **Plausible cause → effect.** ### Match to options Valid: **3 only** → option **(d)**. | Option | Set | Verdict | |---|---|---| | (a) | 2 and 3 only | ✗ (2 invalid — S3 IS connected) | | (b) | 1, 2 and 3 | ✗ | | (c) | 1 and 2 only | ✗ | | (d) | 3 only | ✓ | ### ⚡ Quick trick (≈30 sec) **Quick kill on 2:** S3 literally mentions "climate change deniers" — explicitly connected to climate change theme of S1 and S2. Saying "unconnected" is a direct contradiction. Eliminate (a), (b), (c). Verify (d): S2's uncertainty plausibly fuels denialist arguments (S3). **(d)** in 30 seconds.

Key concepts:

Relationship analysis between premises (counterpoint, cause, independence)
Thematic connection through shared topic (climate change)
Strict definition of "counterpoint" (direct contradiction, not tangent)
Scientific uncertainty as enabler of denial

Sources: Logical relationship analysis between the three statements; verified by elimination.

Q72. reasoning (standard)

Match List–I with List–II and select the answer using the code given below the Lists:

(a) A→1, B→3, C→4, D→2
(b) A→1, B→4, C→3, D→2
(c) A→2, B→4, C→3, D→1
(d) A→2, B→3, C→4, D→1

Answer: (B) · Confidence: high

## Why the answer is (b) — A=1, B=4, C=3, D=2 ### Match each relationship by **formality** and **decisiveness** **Two axes** to evaluate communication type: - **Formal vs Informal** = level of protocol / institutional rigidity. - **Firm vs Open-ended** = decisiveness vs collaborative/exploratory. | Relationship | Formality | Decisiveness | Match | |---|---|---|---| | **A.** Cricket captain ↔ team members | Informal (sports team) | Firm (captain's call is final) | **1. Informal and firm** | | **B.** Judge ↔ lawyers in court | Formal (judicial protocol) | Firm (judicial decisions) | **4. Formal and firm** | | **C.** Vice Chancellor ↔ Deputy Registrar | Formal (administrative hierarchy) | Open-ended (planning, discussion) | **3. Formal and open-ended** | | **D.** Peers and coworkers | Informal (no hierarchy) | Open-ended (collaborative) | **2. Informal and open-ended** | ### Final mapping **A=1, B=4, C=3, D=2** → option **(b)**. ### ⚡ Quick trick (≈25 sec) **Two-axis classification:** - Courtroom = formal + firm (judge's authority) → B = 4 - Peers = informal + open (no hierarchy) → D = 2 - VC-Registrar = formal admin + planning = formal + open-ended → C = 3 - Captain-team = sports informal + tactical decisions = informal + firm → A = 1 **(b)** in 25 seconds — start with the two extremes (judicial court is most formal-firm, peers are most informal-open) and infer the middle two.

Key concepts:

Communication classification by formality and decisiveness
Two-axis matching
Hierarchy-implies-formality
Open-ended = collaborative / exploratory

Sources: Direct conceptual matching; verified by axis-based classification.

Q73. reasoning (standard)

Match List–I with List–II and select the answer using the code given below the Lists:

(a) A→1, B→4, C→2, D→3
(b) A→1, B→2, C→4, D→3
(c) A→3, B→4, C→2, D→1
(d) A→3, B→2, C→4, D→1

Answer: (C) · Confidence: high

## Why the answer is (c) — A=3, B=4, C=2, D=1 ### Direct matching by tool purpose | Tool | Primary purpose | List–II match | |---|---|---| | **A. Memorandum** | An internal official document used to issue directives, instructions or policy notes. | **3. To intimate a directive** | | **B. Flyer** | A printed (or digital) leaflet for mass distribution to unspecified audiences (ads, public notices). | **4. To disseminate non-targeted information** | | **C. Bcc (blind carbon copy)** | Email header field where recipients listed are hidden from each other — used to send a copy privately without others seeing. | **2. To inform confidentially** | | **D. Minutes** | The official record of decisions, discussions and actions taken at a meeting. | **1. To record decisions** | ### Final mapping **A=3, B=4, C=2, D=1** → option **(c)**. ### ⚡ Quick trick (≈20 sec) **Definitional anchors:** - **Minutes** are by definition a written record of a meeting's decisions → D = 1. - **Bcc** is by definition a confidentiality mechanism → C = 2. These two are unambiguous; A and B follow by elimination (Memorandum → directive, Flyer → mass distribution). **(c)** in 20 seconds.

Key concepts:

Communication tool definitions: memo, flyer, Bcc, minutes
Bcc as confidentiality mechanism in email
Minutes = meeting decisions record
Memo = internal directive

Sources: Standard business-communication definitions; verified by direct tool-purpose mapping.

Q74. decision-making (standard)

Which among the following actions would constitute the most appropriate directive(s) in resolving interpersonal conflict in an office with culturally diverse personnel?

Direct personnel to practise activities that are the cultural markers of diverse groups
Allow conflicts to resolve naturally over time to set an appropriate precedent of leadership
Encourage personnel to seek each other's perspectives

Select the answer using the code given below?

(a) 1 and 3
(b) 2 only
(c) 3 only
(d) 2 and 3

Answer: (C) · Confidence: high

## Why the answer is (c) — Only directive 3 is appropriate ### Directive 1 — Direct personnel to practise activities that are the cultural markers of diverse groups Forcing employees to **practise** others' cultural markers (rituals, customs, dress codes, etc.) can be: - **Culturally insensitive** (risks superficial or stereotyped enactments). - **Coercive** (people from minority cultures may feel objectified). - A form of **forced participation** that violates personal autonomy. Cultural appreciation is best built through **awareness and respect**, not mandated practice. ❌ **NOT APPROPRIATE.** ### Directive 2 — Allow conflicts to resolve naturally over time Passive leadership lets interpersonal conflicts **fester and escalate**. Effective management requires timely, structured intervention — not abdication framed as "setting a precedent". ❌ **NOT APPROPRIATE.** ### Directive 3 — Encourage personnel to seek each other's perspectives Encouraging perspective-taking: - Builds **empathy** and **mutual understanding**. - Surfaces hidden assumptions / cultural lenses. - Is a well-established conflict-resolution best practice (active listening, perspective-taking). ✅ **APPROPRIATE.** ### Match to options Valid: **3 only** → option **(c)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 3 | ✗ (1 is coercive) | | (b) | 2 only | ✗ (passive leadership) | | (c) | 3 only | ✓ | | (d) | 2 and 3 | ✗ (2 is passive) | ### ⚡ Quick trick (≈20 sec) **Two quick eliminations:** - Directive 1 = forced cultural practice → coercive → kill. - Directive 2 = passive/laissez-faire → fails leadership duty → kill. Only directive 3 (perspective-taking) survives as ethical management → **(c)** in 20 seconds.

Key concepts:

Conflict resolution: perspective-taking promotes empathy
Coercive cultural practice ≠ cultural respect
Passive leadership fails to manage interpersonal disputes
Active, structured intervention is best practice

Sources: Standard organisational-behaviour / conflict-resolution principles.

Q75. reasoning (standard)

Match List–I with List–II and select the answer using the code given below the Lists:

(a) A→3, B→1, C→4, D→2
(b) A→3, B→4, C→1, D→2
(c) A→2, B→1, C→4, D→3
(d) A→2, B→4, C→1, D→3

Answer: (D) · Confidence: high

## Why the answer is (d) — A=2, B=4, C=1, D=3 ### Match each barrier to its example | Barrier (List–I) | Definition | Example (List–II) | |---|---|---| | **A. Semantic** | Concerning meaning of words/symbols | **2. Misunderstanding the meaning of a word** | | **B. Cognitive** | Concerning mental processing capacity | **4. Information overload** | | **C. Organisational** | Arising from organisational structure / hierarchy | **1. Lack of feedback** | | **D. Affective** | Emotional / psychological | **3. Fear of social stigma** | ### Final mapping **A=2, B=4, C=1, D=3** → option **(d)**. ### ⚡ Quick trick (≈25 sec) **Definitional anchors:** - **Semantic** literally means "about meaning" → A = 2 (word meaning). - **Affective** literally means "about emotion" → D = 3 (fear/stigma). These two are unambiguous. Remaining: Cognitive (mental load = overload) → B = 4, and Organisational (structural lack of feedback loop) → C = 1. **(d)** in 25 seconds.

Key concepts:

Barriers to communication: semantic, cognitive, organisational, affective
Semantic = meaning of words
Affective = emotion/feeling
Cognitive = mental processing capacity
Organisational = structural / hierarchical

Sources: Standard communication-theory categorisation; verified by definitional matching.

Q76. decision-making (standard)

You are required to design a 'questionnaire' to be filled on-location by visitors, based on the following objective while writing a report : "To determine the feasibility of setting up a family-oriented vacation resort in the vicinity of a lake destination in the mountains" Which of the following heads would you include in the questionnaire to make it most appropriate for your purpose?

Size of family
Budget
Number of earners in the family
Food allergies and dietary restrictions

Select the answer using the code given below?

(a) 1 and 3
(b) 1, 2 and 4
(c) 1 and 2 only
(d) 2 and 3

Answer: (C) · Confidence: high

## Why the answer is (c) — Heads 1 and 2 only ### Objective recap Assess **feasibility** of a **family-oriented vacation resort** near a lake destination. The questionnaire collects market-validation data from visitors. ### Head 1 — Size of family Directly determines the **target room/cottage capacities** (4-person, 6-person, etc.) and overall accommodation mix. ✅ **Essential** for feasibility. ### Head 2 — Budget Determines the **price band** that families are willing to pay → directly informs revenue projections and feasibility. ✅ **Essential.** ### Head 3 — Number of earners in the family **Intrusive personal/financial detail** that goes beyond what a feasibility questionnaire needs. Budget (head 2) already captures **spending capacity** directly; asking number of earners adds privacy concerns without new actionable information. ❌ **NOT APPROPRIATE.** ### Head 4 — Food allergies and dietary restrictions Useful for **menu design** AFTER resort is built — relevant to operational planning, not to the upstream **feasibility** decision (whether to build). For a feasibility study, knowing dietary specifics is premature. ❌ **NOT for this stage.** ### Match to options Valid: **1 and 2 only** → option **(c)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 3 | ✗ (3 intrusive) | | (b) | 1, 2 and 4 | ✗ (4 is operational, not feasibility) | | (c) | 1 and 2 only | ✓ | | (d) | 2 and 3 | ✗ (drops valid 1; includes intrusive 3) | ### ⚡ Quick trick (≈25 sec) **Feasibility-question filter:** Ask only what's needed to validate the BUSINESS DECISION (build or not). Family size (capacity) and budget (price point) are core feasibility inputs. Number of earners is intrusive; food allergies are operational (post-build). **(c)** in 25 seconds.

Key concepts:

Feasibility-questionnaire scope: market validation, not operational planning
Privacy-respect principle (avoid intrusive financial questions when proxy data suffices)
Stage-appropriate data collection
Family size + budget = core feasibility inputs

Sources: Standard market-research / feasibility-study methodology.

Q77. decision-making (standard)

With reference to 'circular letters', which of the following statements is/are correct?

Circular letters are usually addressed to a group of people
Non-standard and customized content is typical of circular letters
Circular letters are used to intimate appraisals and increments of employees within organisations
Circular letters are less cost-effective than personalised and specific recipient-directed letters

Select the answer using the code given below?

(a) 1 and 2
(b) 2 and 4
(c) 1, 3 and 4
(d) 1 only

Answer: (D) · Confidence: high

## Why the answer is (d) — Only statement 1 is correct ### Statement 1 — Circular letters are usually addressed to a group of people By definition, a **circular** is a single document broadcast to many recipients (a group). ✅ **CORRECT.** ### Statement 2 — Non-standard and customized content is typical of circular letters Circulars are characterised by **STANDARDISED**, identical content sent to all recipients — the OPPOSITE of customised. ❌ **INCORRECT.** ### Statement 3 — Circular letters are used to intimate appraisals and increments of employees Appraisals and salary increments are **personal, confidential** communications between employer and individual employee — NOT broadcast circulars. ❌ **INCORRECT** (privacy violation if circulated). ### Statement 4 — Circular letters are LESS cost-effective than personalised letters Circulars are far MORE cost-effective (one draft, mass distribution, no per-recipient customisation cost) than personalised letters. Statement inverts the truth. ❌ **INCORRECT.** ### Match to options Valid: **1 only** → option **(d)**. | Option | Set | Verdict | |---|---|---| | (a) | 1 and 2 | ✗ (2 inverts definition) | | (b) | 2 and 4 | ✗ (both inverted) | | (c) | 1, 3 and 4 | ✗ (3 and 4 wrong) | | (d) | 1 only | ✓ | ### ⚡ Quick trick (≈20 sec) **Definitional kill chain:** - A circular is by definition **standard** and **broadcast** → statement 2 ("customized") is inverted, statement 4 ("less cost-effective") is inverted. - Appraisals are **private**, never broadcast → statement 3 wrong. Three inverted/wrong statements + statement 1 as direct definition → **(d)** in 20 seconds.

Key concepts:

Circular = same message to many (mass-broadcast)
Standardisation, not customisation, is the circular's defining feature
Appraisals are confidential (private letters)
Cost-effectiveness of broadcast > personalised

Sources: Standard business-communication definitions; verified by definitional matching.

Q78. numeracy (standard)

Three partners $A$, $B$ and $C$ entered into a business. $A$ invested one-third of the capital for one-third duration. $B$ invested one-fourth of the capital for one-fourth duration. $C$ invested the remaining capital for the whole duration. Out of a profit of ₹17,000, how much profit will $C$ get?

(a) ₹12,000
(b) ₹10,000
(c) ₹12,500
(d) ₹10,750

Answer: (A) · Confidence: high

## Why the answer is (a) — C gets ₹12,000 ### Step 1 — Normalise capital and time using LCM Total capital = 12 units (LCM of 3 and 4); total duration = 12 units (same LCM). Compute each partner's capital × time contribution: | Partner | Capital share | Capital (units) | Duration share | Duration (units) | Capital × Time | |---|---|---|---|---|---| | A | 1/3 | 4 | 1/3 | 4 | $4 \times 4 = 16$ | | B | 1/4 | 3 | 1/4 | 3 | $3 \times 3 = 9$ | | C | remaining | $12 - 4 - 3 = 5$ | whole | 12 | $5 \times 12 = 60$ | ### Step 2 — Ratio of profit shares Profit-share ratio = ratio of capital × time = **A : B : C = 16 : 9 : 60**. Total parts = 85. ### Step 3 — C's profit $$\text{C's share} = \frac{60}{85} \times 17{,}000 = 60 \times 200 = \mathbf{₹12{,}000}.$$ **Answer: (a) ₹12,000.** ### ⚡ Quick trick (≈45 sec) **LCM normalisation:** Use 12 for both capital and time to avoid fractions. Then capital-time products are: A=16, B=9, C=60 → ratio 16:9:60, total 85. **Profit-per-part shortcut:** ₹17,000 / 85 = ₹200 per part. C has 60 parts → 60 × 200 = ₹12,000. **(a)** in 45 seconds.

Key concepts:

Profit-sharing rule: share ∝ capital × time
LCM normalisation to clear fractions
Per-part profit shortcut for fast ratio division

Sources: Direct partnership accounting; verified by ratio computation.

Q79. numeracy (standard)

There are two chemicals which do not react with each other. A container contains 10 litres of the chemical $A$. One litre of this chemical is removed from it and one litre of the chemical $B$ is poured. Then one litre of the mixture is removed from the container and one litre of $B$ is poured. If this process of replacing one litre of the mixture by one litre of $B$ is performed once more, then what is the volume of $B$ that is present in the container approximately (in percentage)?

(a) 25
(b) 27
(c) 29
(d) 31

Answer: (B) · Confidence: high

## Why the answer is (b) — Approximately 27% B in the container ### Setup Container starts with 10 L of A. Each operation: remove 1 L of mixture, then add 1 L of B. Volume stays constant at 10 L. Repeat 3 times. ### Use the standard replacement formula After $n$ operations, fraction of original substance (A) remaining: $$\left(1 - \frac{r}{V}\right)^n$$ where $r = 1$ L removed each time, $V = 10$ L total, $n = 3$. $$\text{Fraction A} = \left(1 - \frac{1}{10}\right)^3 = \left(\frac{9}{10}\right)^3 = \frac{729}{1000} = 0.729.$$ ### Compute B's fraction and percentage $$\text{Fraction B} = 1 - 0.729 = 0.271 = 27.1\%.$$ **Rounded: ≈ 27%** → option **(b)**. ### Verification by step-by-step | Step | Action | A in container | B in container | |---|---|---|---| | Start | — | 10 L | 0 L | | After op 1 | Remove 1 L A, add 1 L B | 9 L | 1 L | | After op 2 | Remove 0.9 L A + 0.1 L B, add 1 L B | 8.1 L | 1.9 L | | After op 3 | Remove 0.81 L A + 0.19 L B, add 1 L B | 7.29 L | 2.71 L | B = 2.71 L out of 10 L = **27.1%** ✓. ### ⚡ Quick trick (≈25 sec) **Replacement formula:** $(1 - r/V)^n$ gives the fraction of original substance left. Here $(0.9)^3 = 0.729$, so 72.9% A → 27.1% B → **27%**. **(b)** in 25 seconds — memorise the formula and skip the step-by-step.

Key concepts:

Replacement formula: $(1 - r/V)^n$
Substance conservation in repeated dilution
Geometric decay of original substance

Sources: Standard dilution/replacement formula; verified by step-by-step simulation.

Q80. decision-making (standard)

A shopkeeper employs a delivery boy and gives him a motorcycle for home delivery. For every delivery, the boy is given ₹5. At the end of the day, he also gets ₹2 for every kilometre of the distance covered in the day. The boy wants to earn more than ₹500 a day, but does not want to travel more than 100 km. Which of the following numbers of deliveries would definitely meet his target?

(a) 80
(b) 85
(c) 90
(d) The question cannot be answered due to insufficient data

Answer: (D) · Confidence: high

## Why the answer is (d) — Cannot be definitively answered due to insufficient data ### Setup Earnings = $5D + 2K$ where $D$ = deliveries, $K$ = kilometres travelled. Constraints from the problem: - Boy wants earnings > ₹500. - Boy doesn't want to travel more than 100 km, so $K \le 100$. But **no relationship is given between $D$ and $K$** — i.e., we don't know how much travel each delivery requires. $K$ could in principle range from 0 to 100 for any given $D$. ### Test each option for "definitely meets" (= meets for ALL valid K) "Definitely meets the target" means the inequality $5D + 2K > 500$ must hold for every valid $K \in [0, 100]$ — in particular even at the **worst case** $K = 0$. Worst case ($K = 0$): earnings = $5D$. Need $5D > 500 \Rightarrow D > 100$. | Option | D | Worst-case earnings (K=0) | > 500? | |---|---|---|---| | (a) | 80 | $5 \times 80 = 400$ | ✗ | | (b) | 85 | $5 \times 85 = 425$ | ✗ | | (c) | 90 | $5 \times 90 = 450$ | ✗ | **None of D = 80, 85, 90 guarantees > ₹500** because all fall short when $K$ is small. ### Why "insufficient data" Even though best-case earnings (with $K = 100$) would exceed ₹500 for all three options, we cannot guarantee the boy will travel a particular distance. **The boy's earnings depend on a parameter ($K$) for which only an upper bound is given, not a fixed value or a $D$-to-$K$ relationship.** Hence the question **cannot be definitively answered** with the data provided → option **(d)**. ### ⚡ Quick trick (≈30 sec) **Worst-case-K test:** For "definitely meets", set $K$ to its minimum ($K = 0$ — no link given between D and K). Need $5D > 500 \Rightarrow D > 100$. None of (a), (b), (c) satisfy. → **(d) insufficient data**. **30 seconds.**

Key concepts:

"Definitely" = guaranteed under all valid parameter values
Worst-case analysis with unspecified parameter
Insufficient data when relationship between variables is missing
Distinguishing best-case from definite outcome

Sources: Direct logical analysis of guarantee condition; verified by worst-case parameter test.

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