Data interpretation & sufficiency

Answer: (a) इस विकल्प का चयन कीजिए, यदि प्रश्न का उत्तर, इन कथनों में से केवल एक का उपयोग कर दिया जा सकता है, किन्तु दूसरे कथन का उपयोग कर नहीं दिया जा सकता है

There are only ten ways to pick three different coins, so list their totals. Every total whose digits add up to more than 5 includes the 5-coin, so Statement II settles it; Statement I leaves 32 (2 + 10 + 20) as an exception.

1. Three different coins from {1, 2, 5, 10, 20} can be chosen in 10 ways.
2. With the 5-coin: 1+2+5 = 8, 1+5+10 = 16, 1+5+20 = 26, 2+5+10 = 17, 2+5+20 = 27, 5+10+20 = 35.
3. Without the 5-coin: 1+2+10 = 13, 1+2+20 = 23, 1+10+20 = 31, 2+10+20 = 32.
4. Statement I (m not prime): m can be 8, 16, 26, 27, 35 (5-coin present) or 32 (5-coin absent). Not sufficient.
5. Statement II (digit sum > 5): 8, 16, 26, 17, 27, 35 have digit sums 8, 7, 8, 8, 9, 8; the totals without the 5-coin have digit sums 4, 5, 4, 5. So m always includes the 5-coin. Sufficient.
6. One statement alone works and the other does not.

• ✗ Statement I Not sufficient alone: non-prime totals include 32 = 2 + 10 + 20, which has no 5-coin, as well as 8, 16, 26, 27 and 35, which do.
• ✓ Statement II Sufficient alone: every total with digit sum above 5 (8, 16, 17, 26, 27, 35) uses the 5-coin; totals without it (13, 23, 31, 32) have digit sums of only 4 or 5.

Remember · In data sufficiency with a small finite set, list every case first, then test each statement against the list.

Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) इस विकल्प का चयन कीजिए, यदि प्रश्न का उत्तर दोनों में से किसी भी कथन का उपयोग कर नहीं दिया जा सकता है

Both statements together only trap y between x² and √x, and for 0 < x < 1 the number x itself lies inside that gap. So y can sit on either side of x and the question stays open.

1. Statement I: x = 0.9, y = 0.85 fits (0.81 < 0.85 < 1) and gives x > y; x = 0.1, y = 0.5 also fits and gives y > x. Not sufficient.
2. Statement II: √x < 1 means 0 ≤ x < 1. x = 0.25, y = 0.4 fits (0.4 < 0.5) with y > x; x = 0.25, y = 0.1 fits with y < x. Not sufficient.
3. Together: x² < y < √x with 0 < x < 1. Take x = 0.25: y must lie between 0.0625 and 0.5.
4. y = 0.4 gives y > x, y = 0.1 gives y < x. Still not decided.
5. Check: for 0 < x < 1, x² < x < √x, so x always lies inside the allowed range of y.

• ✗ Statement I Not sufficient alone: x = 0.9, y = 0.85 and x = 0.1, y = 0.5 both satisfy it but order x and y differently.
• ✗ Statement II Not sufficient alone: with x = 0.25, both y = 0.4 and y = 0.1 are below √x = 0.5, yet one is above x and one below.

Remember · For numbers between 0 and 1, x² < x < √x. A range that straddles x never decides which is bigger.

Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) इस विकल्प का चयन कीजिए, यदि प्रश्न का उत्तर दोनों कथनों का एक साथ उपयोग कर दिया जा सकता है, किन्तु दोनों में से केवल किसी भी एक कथन का उपयोग कर नहीं दिया जा सकता है

Statement I says at least one of x and y is even; Statement II says x and y have the same parity. Only together do they force both to be even, which answers the question.

1. Statement I: x²y² = (xy)² is even, so xy is even, so at least one of x, y is even. x could be odd (x = 1, y = 2). Not sufficient.
2. Statement II: 1 + x² + y² is odd, so x² + y² is even, so x and y are both even or both odd. Not sufficient.
3. Together: they share parity and at least one is even, so both are even. x is even — answered.
4. Check: x = 1, y = 1 satisfies II but not I; x = 2, y = 2 satisfies both.

• ✗ Statement I Not sufficient alone: x = 1, y = 2 gives x²y² = 4 (even) with x odd, while x = 2, y = 1 gives x even.
• ✗ Statement II Not sufficient alone: it only says x and y have the same parity; x = y = 1 and x = y = 2 both fit.

Remember · Parity items: turn each statement into a plain rule (at least one even / same parity), then combine the rules.

Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) इस विकल्प का चयन कीजिए, यदि प्रश्न का उत्तर दोनों में से किसी भी कथन का उपयोग कर नहीं दिया जा सकता है

Each statement only tells us that T cannot mix odd and even numbers. An all-odd T and an all-even T satisfy both statements, so whether T comes wholly from Y cannot be decided.

1. Statement I: the sum of two numbers is even only when both have the same parity. So T is all odd or all even. Not sufficient.
2. Statement II: if T is all odd, p − 1 is even, so (p − 1)q is even. If T is all even, q is even, so (p − 1)q is even. Both cases fit.
3. A mixed T fails II: an even p and an odd q give (odd) × (odd) = odd. So II also means 'all odd or all even'. Not sufficient.
4. Together: an all-odd T (from X) and an all-even T (from Y) satisfy both statements. Still not answered.

• ✗ Statement I Not sufficient alone: T = {1, 3} and T = {2, 4} both have even pairwise sums.
• ✗ Statement II Not sufficient alone: for T = {1, 3}, p − 1 is always even; for T = {2, 4}, q is always even. Both give (p − 1)q even.

Remember · When every statement is satisfied by two opposite cases, even combined, mark 'cannot be answered'.

Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) इस विकल्प का चयन कीजिए, यदि प्रश्न का उत्तर, इन कथनों में से केवल एक का उपयोग कर दिया जा सकता है, किन्तु दूसरे कथन का उपयोग कर नहीं दिया जा सकता है

Factorise both numbers: 116 = 2² × 29 and 261 = 3² × 29. The first leaves only y = 2 and x = 29, a prime; the second allows x = 3, 9, 29 or 87, so it cannot decide.

1. 116 = 2² × 29. With y > 1, the only square factor is y² = 4, so y = 2 and x = 29.
2. 29 is prime, so Statement I alone answers the question (yes).
3. 261 = 3² × 29. With x, z > 1, x can be 3, 9, 29 or 87.
4. 3 and 29 are prime but 9 and 87 are not, so Statement II alone cannot answer.
5. One statement alone works and the other does not.

• ✓ Statement I Sufficient alone: 116 = 4 × 29 forces y = 2 and x = 29, which is prime.
• ✗ Statement II Not sufficient alone: 261 = 9 × 29 allows x = 3 or 29 (prime) and x = 9 or 87 (not prime).

Remember · Prime factorisation first; then ask which factor splits are allowed by the given conditions (here, every variable > 1).

Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 5·1

Weighted scores are 5·2, 5·1, 5·2, 4·5 and 5·0 for candidates 1 to 5. The top three are 5·2, 5·2 and 5·1, so a cutoff of 5·1 selects exactly three.

1. Candidate 1: 0·2 × 5 + 0·3 × 4 + 0·5 × 6 = 1 + 1·2 + 3 = 5·2.
2. Candidate 2: 0·8 + 1·8 + 2·5 = 5·1.
3. Candidate 3: 0·6 + 0·6 + 4 = 5·2.
4. Candidate 4: 1·8 + 1·2 + 1·5 = 4·5.
5. Candidate 5: 1·6 + 2·4 + 1 = 5·0.
6. Ranked: 5·2, 5·2, 5·1, 5·0, 4·5. For exactly three selections the cutoff is 5·1 (a cutoff of 5·2 would select only two).

Remember · Weighted score = Σ(weight × score). The cutoff is the lowest score among those who must be selected.

Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 1 Oct 2026 (how we verify). Permalink · Report an error