Averages & Ages: previous-year questions

Answer: (C) 80%

Use alligation on the averages: boys 52, girls 42, combined 50. Boys : girls $= (50 - 42) : (52 - 50) = 8 : 2 = 4 : 1$, so boys are $\frac{4}{5} = 80\%$ of the class. Algebra gives the same result: if $b$ is the fraction of boys, $52b + 42(1 - b) = 50$, so $b = 0.8$. Option D is the trap, since the actual numbers are not needed; a ratio is enough to give a percentage. Common-sense check: the combined average (50) lies much closer to the boys' average, so boys must be the large majority.

Answer: (C) 47

Total of 11 results = 11 × 32 = 352; first five total 5 × 27 = 135 and last five total 5 × 34 = 170. The sixth result is 352 − 135 − 170 = 47. The other values would make the overall total wrong.

Answer: (B) 64

The wrong total was 9 × 65 = 585, which is 76 − 67 = 9 too high. The correct total is 576, and 576 ÷ 9 = 64. 66 would follow from adding the error instead of subtracting it.

Answer: (D) 0·5

With eight values the median is the mean of the 4th and 5th, (36 + 37) ÷ 2 = 36·5. After removing 35 there are seven values (30, 34, 36, 37, 38, 39, 40) and the median is the 4th, 37. The increase is 0·5, not 1 or more.

Answer: (A) 37·5

Total for 28 students = 1400; the remaining 20 students average 55, totalling 1100. The 8 who left scored 1400 − 1100 = 300, an average of 300 ÷ 8 = 37·5. The other options would leave the new average different from 55.

Answer: (C) 20

Total weight = 30 × 45 = 1350 kg. With b boys, 49b + 37(30 − b) = 1350, so 12b = 240 and b = 20. 10 is the number of girls, the most tempting wrong answer.

Answer: (B) 7 years

The son reaches the father's present age after (F − S) years, when both have aged by that amount: 36 + 2(F − S) = 80, so F − S = 22. With F + S = 36, the son is (36 − 22) ÷ 2 = 7 years old. The other ages do not give a 22-year gap.

Answer: (A) 76

He has 10 × 32 = 320 runs; an average of 36 over 11 innings needs 396 runs. So the next innings must bring 396 − 320 = 76 runs. Option C, 4, is only the rise wanted in the average, not the runs he must score.

Answer: (D) 85 kg

An increase of 1 kg in the average of 25 persons means the total rose by 25 kg. So the new person weighs 60 + 25 = 85 kg. 61 kg adds only the 1 kg rise in the average.

Answer: (C) 43

The group's total is 10 × 41 = 410; adding 63 makes 473 for 11 students. The new average is 473 ÷ 11 = 43. An average must rise when a higher score is added, so 39 and 40 are impossible.

Answer: (D) 14 years

Four years ago the sum was 72 − 12 = 60, shared 1 : 2 : 3 as 10, 20 and 30. A's present age is 10 + 4 = 14 years. 12 years wrongly divides the present total of 72 in the ratio.

Answer: (A) 14 years

After t years, 50 + t = 2(18 + t), so 50 + t = 36 + 2t and t = 14. Check: A will be 64 and B 32. 16 years would make them 66 and 34.

Answer: (B) ₹ 17,600

Let the 6th man give x; then x = $\frac{100000 + x}{6}$ − 2000. So 6x = 100000 + x − 12000, 5x = 88000 and x = ₹ 17,600. Check: the average is 117600 ÷ 6 = ₹ 19,600, which is ₹ 2,000 more than x; ₹ 19,600 is the average, not his share.

Answer: (C) 20

Total runs $=5\times35=175$. Let E's score be $e$. Then $D=e+5$, $A=e+20$ (E scored 20 fewer than A) and $B=D+E=2e+5$. Since $B+C=75$, the total is $A+(B+C)+D+E=(e+20)+75+(e+5)+e=3e+100=175$, so $e=25$. Hence $D=30$, $A=45$, $B=55$ and $C=75-55=20$. Check: $45+55+20+30+25=175$. Options (A) 45, (B) 55 and (D) 30 are the scores of A, B and D, so make sure you answer for the batsman actually asked.

Answer: (A) 50 years

Let the younger son be $y$ years old; then the elder son is $y + 15$ and the father is $2(y + 15)$. After ten years: $2(y + 15) + 10 = 3(y + 10)$, i.e. $2y + 40 = 3y + 30$, so $y = 10$. Hence the elder son is 25 and the father is 50. Check: after 10 years the father is 60 and the younger son 20, and $60 = 3 \times 20$.

Answer: (D) 21

The original mean is $\dfrac{9+12+15+18+20+22}{6} = \dfrac{96}{6} = 16$. The new mean must be 17, so the new total is $17 \times 6 = 102$, an increase of $102-96=6$ over the original total — so 15 must be replaced by $15+6=21$.

Answer: (D) 30 years

Let the ages be $3x$ and $5x$. The difference is $5x-3x = 2x = 12$, so $x = 6$ and the ages are 18 and 30 years. The key takes Axar as the elder (30 years); the question does not actually say who is older, but 18 is not among the options, so 30 is the only possible answer.