Data interpretation & sufficiency

Answer: (c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

m + n > mn with m > n forces n = 1, but leaves m open; mn = 24 alone allows several pairs. Together, n = 1 and m = 24.

1. m + n > mn is the same as mn − m − n + 1 < 1, i.e. (m − 1)(n − 1) < 1.
2. For natural numbers this means (m − 1)(n − 1) = 0; since m > n ≥ 1, n = 1. Statement I fixes n = 1 but any m ≥ 2 works, so it is not sufficient.
3. Statement II alone: mn = 24 allows (24, 1), (12, 2), (8, 3), (6, 4) and their reversals, so it is not sufficient.
4. Together: n = 1 and mn = 24 give m = 24. Check: 24 + 1 = 25 > 24.

Remember · Rewrite m + n > mn as (m − 1)(n − 1) < 1; with natural numbers, one of them must be 1.

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Answer: (c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

Download time = size ÷ transfer rate. Statement I gives only the size and Statement II only the rate; together they fix the time.

1. Time = size of the software ÷ transfer rate.
2. Statement I gives the size (12 megabytes) but no rate, so it is not sufficient.
3. Statement II gives the rate (2.4 kilobytes per second) but no size, so it is not sufficient.
4. Together: 12 megabytes = 12,000 kilobytes (taking 1 MB = 1000 KB), so time = 12,000 ÷ 2.4 = 5,000 seconds; with 1 MB = 1024 KB it is 5,120 seconds. Either way the time is fixed.

Remember · In data sufficiency, write the formula first and tick off which statement supplies each quantity.

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Answer: (c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

‘x/y is odd’ alone allows endless pairs, and xy = 12 alone allows six ordered pairs. Together, of the pairs with product 12, only x = 6, y = 2 gives an odd quotient (3).

1. Statement I alone: x = 3y, 5y, 7y, …, for example (3, 1), (5, 1), (6, 2); not unique.
2. Statement II alone: (1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1); not unique.
3. Together, test x/y for each pair: 12/1 = 12 (even), 6/2 = 3 (odd); 4/3, 3/4, 2/6 and 1/12 are not whole numbers.
4. Only x = 6, y = 2 works, so both statements together answer the question.

Remember · When one statement gives a short list, test each item on it against the other statement.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

Statement I makes X’s share 4/9 of the total and Statement II makes Y’s share 2/9; together Z gets 3/9, so Y received the least. Either statement alone leaves the split between the other two open.

1. Statement I: X = (4/5)(Y + Z), so X : (Y + Z) = 4 : 5 and X = 4/9 of the total. Y and Z share 5/9 in an unknown way, so it is not sufficient.
2. Statement II: Y = (2/7)(X + Z), so Y = 2/9 of the total. X and Z share 7/9 in an unknown way (X could even be less than Y), so it is not sufficient.
3. Together: X = 4/9 and Y = 2/9, so Z = 1 − 6/9 = 3/9.
4. Y received the least.

Remember · ‘A gets p/q of what the others get’ means A’s share is p/(p + q) of the total.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) The Question cannot be answered even by using both the Statements together

Statement II only tells us that the highest and lowest marks add up to 120 (twice the average). Statement I confirms 70 + 50 = 120, but neither statement says anything about how many students there are.

1. Let there be n students, with total marks 60n.
2. Statement II: (60n − H − L) ÷ (n − 2) = 60 gives H + L = 120, which is true for any n.
3. Statement I: H = 70 and L = 50, consistent with H + L = 120, but n is still free.
4. Example: marks 50, 60, 70 (n = 3) and 50, 60, 60, 70 (n = 4) both fit everything. So the question cannot be answered.

Remember · If a statement holds for every value of the unknown, it cannot fix that unknown; test two different cases.

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Answer: (a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone

Three distinct primes with a prime sum must all be odd, because including 2 makes the sum even. With the sum below 23, the only possibility is 3 + 5 + 11 = 19, so Statement I alone suffices. Statement II allows 3, 5, 11 and 5, 7, 11 and more.

1. If 2 is one of them, the sum is 2 + odd + odd = even (and more than 2), so not prime. All three must be odd primes.
2. Statement I: odd-prime triples with sum below 23 are 3 + 5 + 7 = 15, 3 + 5 + 11 = 19, 3 + 5 + 13 = 21 and 3 + 7 + 11 = 21; only 19 is prime. So the numbers are 3, 5 and 11: sufficient.
3. Statement II: triples containing 5 with a prime sum include {3, 5, 11} (19), {5, 7, 11} (23) and {3, 5, 23} (31), so it is not sufficient.
4. So the question can be answered by one statement (I) alone but not by the other.

Remember · A prime sum of three distinct primes cannot include 2 (the sum turns even). Start listing from the smallest odd primes.

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Answer: (d) The Question cannot be answered even by using both the Statements together

Neither statement alone fixes x + y. Even together they only make 3(x + y) an integer, so x + y can be a fraction such as 2/3 (x = y = 1/3) or an integer (x = y = 1).

1. Statement I alone: x = 0.5, y = 0 gives 2x + y = 1 but x + y = 0.5; x = y = 1 gives x + y = 2. Not sufficient.
2. Statement II alone: the same examples with x and y swapped show it is not sufficient.
3. Together: adding them, 3x + 3y is an integer, so x + y is only known to be an integer divided by 3.
4. x = y = 1/3 gives 2x + y = 1 and x + 2y = 1, yet x + y = 2/3; x = y = 1 gives x + y = 2. Both fit, so the answer is still open.

Remember · In ‘is it an integer?’ questions, try fractions like 1/3 or 1/2; they often break an apparent ‘yes’.

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Answer: (c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone

p + r = 50 − 16 = 34, with both more than 16. Statement I gives p ≤ 17 and Statement II gives p ≥ 17, so together p = ₹17; either alone leaves a range, since prices need not be whole rupees.

1. p + r = 50 − 16 = 34, and p and r are both more than 16 because q is the least.
2. Statement I (p ≤ r): p ≤ 17, so 16 < p ≤ 17; p could be ₹16.50 or ₹17. Not sufficient.
3. Statement II (r ≤ p): p ≥ 17, so 17 ≤ p < 18. Not sufficient.
4. Together, p ≤ r and r ≤ p give p = r = 34 ÷ 2 = ₹17.
5. Note: only if prices had to be whole rupees would each statement alone give ₹17; the question does not say so.

Remember · In data sufficiency, don’t assume whole numbers unless told. ‘Not more than’ in both directions means equality.

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Answer: (d) The Question cannot be answered even by using both the Statements together

Combining the statements gives only P > R and S > Q; nothing compares P with Q directly. Examples with P above Q and with P below Q both satisfy the data.

1. Statement I alone: P + Q = R + S says nothing about P versus Q.
2. Statement II alone: P + S > Q + R also depends on S and R, so it is not sufficient.
3. Together: putting R = P + Q − S into Statement II gives 2S > 2Q, i.e. S > Q; similarly P > R. Neither compares P with Q.
4. Example: P = 5, Q = 1, R = 4, S = 2 (P > Q) and P = 5, Q = 6, R = 4, S = 7 (P < Q) both satisfy I and II. So the question cannot be answered.

Remember · Combine the equations to see what they really fix, then build two examples that give opposite answers.

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Answer: (a) The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone

With ages 10a + b and 10b + a, Statement II gives 11(a + b) = (11/6) × 9|a − b|, i.e. 2(a + b) = 3|a − b|, whose only solution is 51 and 15. The difference is 36 whichever is older, so Statement II alone suffices; Statement I alone does not.

1. Let the ages be 10a + b and 10b + a, with a and b non-zero digits. Sum = 11(a + b); difference = 9|a − b|.
2. Statement I alone only says which one is older, so it is not sufficient.
3. Statement II: 11(a + b) = (11/6) × 9|a − b|, which simplifies to 2(a + b) = 3|a − b|.
4. If a > b: 2a + 2b = 3a − 3b, so a = 5b; b = 1 gives a = 5 (b = 2 would need a = 10). The ages are 51 and 15, in some order.
5. Difference = 36 either way, so Statement II alone answers the question.
6. Check: 51 + 15 = 66 = (11/6) × 36.

Remember · For a number and its reverse: sum = 11(a + b), difference = 9(a − b). Ask only what the question actually needs.

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