Quantitative Aptitude MCQs for UPSC Prelims

Correct answer: B - 48 km/hr

For equal distances at two speeds, the average speed is the harmonic mean: 2 x s1 x s2 / (s1 + s2) = 2 x 40 x 60 / (40 + 60) = 4800 / 100 = 48 km/hr. (Check: time A to B = 360/40 = 9 hr, time B to A = 360/60 = 6 hr; total distance 720 km in 15 hr = 48 km/hr.)

Correct answer: B - 6

Write the 11-digit number as N = 3798125 x 10000 + P x 1000 + 369 = 37981250369 + 1000P. Now reduce mod 7. 37981250369 = 7 x 5425892910 - 1, so 37981250369 leaves remainder 6 (that is, -1) mod 7. Also 1000 leaves remainder 6 mod 7. So N mod 7 = 6 + 6P (mod 7). For divisibility we need 6 + 6P congruent to 0 mod 7, i.e. 6P congruent to 1 mod 7. Since 6 is -1 mod 7, this is -P congruent to 1, giving P congruent to 6 mod 7, so P = 6. Testing the other options: P = 1, 7 or 9 each leaves a non-zero remainder, so only 6 works.

Correct answer: C - After buying the two tickets A will be left with 50 paise.

Rs. 10 is 80% of the total fare for two persons, so the full fare for two is 10 / 0.80 = Rs. 12.50. After B hands over Rs. 3, A has 10 + 3 = Rs. 13. Buying two tickets costs Rs. 12.50, leaving A with 13 - 12.50 = Rs. 0.50, which is 50 paise.

Correct answer: B - 4

35^n = 5^n x 7^n, so count factors of 5 and 7 in 7 x 343 x 385 x 1000 x 2401 x 77777. Factors of 5: 385 = 5 x 7 x 11 gives one 5, 1000 = 2^3 x 5^3 gives three 5s, total 4. Factors of 7: leading 7 (one), 343 = 7^3, 385 (one), 2401 = 7^4, 77777 = 7 x 11111 (one); total 1+3+1+4+1 = 10. The number of available 35s is limited by the scarcer prime, min(4, 10) = 4. So maximum n = 4. Answer: (b).

Correct answer: C - 12

Option (c) is correct. If x divides 7x + 96, then since x always divides 7x, x must divide the difference, which is 96. So x can be any positive divisor of 96 = 2^5 x 3, and the number of divisors is (5 + 1)(1 + 1) = 12. Options (a) 10 and (b) 11 miscount the divisors (1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96). Option (d) is wrong because the divisibility condition bounds x to divisors of 96, so the possibilities are finite.

Correct answer: D - 9

If 1186 leaves remainder 31 on division by n, then n divides 1186 - 31 = 1155 and n must exceed the remainder, so n > 31. Factorise: 1155 = 3 x 5 x 7 x 11, which has 16 divisors: 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155. Those greater than 31 are 33, 35, 55, 77, 105, 165, 231, 385 and 1155 - nine numbers, option (d). Options (a), (b) and (c) come from miscounting the divisor list or forgetting the divisor must be larger than 31.