Basic Numeracy & Word Problems MCQs for UPSC Prelims

Conditions: P = 3T with T not 0, S = Q + 4 and Q = R + 3 (so S = R + 7), and all five digits distinct. P = 3T allows only T = 1 (P = 3), T = 2 (P = 6) or T = 3 (P = 9). S = R + 7 with S a digit allows only R = 0, 1 or 2, giving (R, Q, S) = (0, 3, 7), (1, 4, 8) or (2, 5, 9). Now combine and reject clashes. T = 1, P = 3: (0, 3, 7) clashes with P = 3; (1, 4, 8) clashes with T = 1; (2, 5, 9) works, giving 35291. T = 2, P = 6: (0, 3, 7) works, giving 63072; (1, 4, 8) works, giving 64182; (2, 5, 9) clashes with T = 2. T = 3, P = 9: (0, 3, 7) clashes with T = 3; (1, 4, 8) works, giving 94183; (2, 5, 9) clashes with P = 9. Valid numbers: 35291, 63072, 64182, 94183, a total of 4. The official UPSC key confirms option (b).

Alok plus Bhupesh equals Chander plus Dinesh, and together they total 100, so each pair holds 50. Chander has half of Dinesh, so Chander plus Dinesh equals 1.5 times Dinesh equals 50, giving Dinesh = 33.33 and Chander = 16.67. Alok is 5 more than Dinesh, so Alok = 38.33, and Bhupesh = 50 - 38.33 = 11.67. The largest share belongs to Alok.

Out of Rs. 2,000 profit with P getting Rs. 400 more than Q: P's share = 1,200 and Q's share = 800, ratio 3 : 2. Profit shares are proportional to capital x time, so (P x 8) : (Q x 10) = 3 : 2, giving 16P = 30Q, i.e., P = 15Q/8. Also P = Q + 14,000, so 15Q/8 - Q = 14,000, i.e., 7Q/8 = 14,000, giving Q = 16,000 and P = 16,000 + 14,000 = Rs. 30,000.

Correct answers: arithmetic = 70% of 10 = 7, algebra = 40% of 30 = 12, geometry = 60% of 30 = 18. Total correct = 7 + 12 + 18 = 37. To pass she needs 60% of 70 = 42 correct. Shortfall = 42 - 37 = 5. So she needs 5 more correct answers. The other options do not match the required 42 minimum.

Tank volume = 20 x 15 x 6 = 1800 cubic metres. Since 1 cubic metre = 1000 litres, the tank holds 1800 x 1000 = 18,00,000 litres. Daily requirement = 4000 people x 150 litres = 6,00,000 litres per day. Number of days the water lasts = 18,00,000 / 6,00,000 = 3 days.

Let Q's score be q, so P's score is q + 40. Q scored 10% less than P, so q = 0.9 (q + 40), giving q = 0.9q + 36, hence 0.1q = 36 and q = 360 - option (a). Quick check: P = 400 and 10% less than 400 is 360, consistent. Option (c) 400 is P's score, not Q's. Options (b) 380 and (d) 420 satisfy neither the 40-mark gap nor the 10% relation simultaneously.

The well is 4.5 m = 450 cm deep. In each jump-and-slide cycle the net gain is 30 - 15 = 15 cm, but on the final jump the frog escapes and does not slide back. On the last successful jump the frog rises 30 cm and is out, so it must reach a height of 450 - 30 = 420 cm by net progress before that final jump. To gain 420 cm at 15 cm net per jump takes 420 / 15 = 28 jumps. The 29th jump then carries it from 420 cm up 30 cm to 450 cm, clearing the well. Total jumps = 29.

Let the empty vessel weigh E and a full load of water weigh W. Full: E + W = 40. One-third filled: E + W/3 = 20. Subtract the second from the first: W - W/3 = 40 - 20, so (2/3)W = 20, giving W = 30 kg (full water). Then E = 40 - 30 = 10 kg. The empty vessel weighs 10 kg. The other options ignore that the difference of 20 kg corresponds to two-thirds of the water, not the full water.