Average Questions and Answers

Answer: A

M's salary is 50% of P's.

From the two averages,

\[ \begin{aligned} M + N + S &= 3 \times 4000 = 12000 \\ N + S + P &= 3 \times 5000 = 15000 \end{aligned} \]

With \(P = 6000\), \(N + S = 9000\), so \(M = 12000 - 9000 = 3000\). As a percentage of P's salary, \(3000/6000 = 50\%\).

Subtracting the two sums gives \(P - M = 3000\) directly, a useful check. 75%, 100% and 125% would need M to earn 4500, 6000 or 7500, none of which fits the averages.

Answer: B

The quantity of P sold is 50.

Let \(t\) items of T be sold. Then S, R and Q sold \(2t, 3t, 4t\), and P sold \(p\). Total revenue is \(250 \times 60 = 15000\).

\[ \begin{aligned} p + 10t &= 250 \\ 100p + 50(4t) + 40(3t) + 60(2t) + 60t &= 15000 \\ 100p + 500t &= 15000 \end{aligned} \]

Substituting \(p = 250 - 10t\): \(25000 - 1000t + 500t = 15000\), so \(t = 20\) and \(p = 50\). Check: \(5000 + 4000 + 2400 + 2400 + 1200 = 15000\). The values 40, 60 and 70 do not satisfy both the item count and the revenue.

Answer: C

A teacher is 32 years old.

Let a student be s years old and a teacher \(s + 24\). The total age of the 8 people is \(8 \times 14 = 112\).

\[ \begin{aligned} 2(s + 24) + 6s &= 112 \\ 8s + 48 &= 112 \Rightarrow s = 8 \\ \text{teacher} &= 8 + 24 = 32 \end{aligned} \]

Check: \(2 \times 32 + 6 \times 8 = 64 + 48 = 112\). Each of the other options fails this check: a teacher of 34, for example, makes a student 10 and the total 128.

Answer: C

His average in the last 30 innings is 110.4.

The new average after 120 innings is \(79.2 + 7.8 = 87\).

\[ \begin{aligned} 120 \times 87 &= 10440 \\ 90 \times 79.2 &= 7128 \\ \frac{10440 - 7128}{30} &= \frac{3312}{30} = 110.4 \end{aligned} \]

A quick check: the extra 30 innings raised the average of 120 by 7.8, so they exceeded the old average by \(7.8 \times 120 / 30 = 31.2\), and \(79.2 + 31.2 = 110.4\). 65.1 would have lowered the average, and 137.7 or 157.1 would have raised it by far more than 7.8.

Answer: D

The temperature on Sunday was 24°C.

Turn each average into a total:

\[ \begin{aligned} S + M + T &= 3 \times 27 = 81 \\ M + T + W &= 3 \times 29 = 87 \\ M + T &= 87 - 30 = 57 \\ S &= 81 - 57 = 24 \end{aligned} \]

A shortcut: the two totals differ by \(87 - 81 = 6\), which is \(W - S\), so \(S = 30 - 6 = 24\). The option 28 comes from subtracting the difference of the averages, 2, instead of the difference of the totals, 6.

Answer: C

The average age will be 56 years.

With no one leaving or joining, every one of the 16 professors is four years older, so the total age rises by \(16 \times 4 = 64\) years and the average rises by exactly 4 years: \(52 + 4 = 56\).

The size of the group does not matter: when every member ages by the same amount, the average ages by that amount too.

68 years adds 16 (the number of professors) instead of 4, and 58 years adds 6. 'None of these' is wrong because 56 is offered.

Answer: A

The average salary is ₹28,500.

Weight each salary by the share of employees earning it:

\[ \begin{aligned} &0.1 \times 20000 + 0.3 \times 25000 \\ &+ 0.5 \times 30000 + 0.1 \times 40000 \\ &= 2000 + 7500 + 15000 + 4000 = 28500 \end{aligned} \]

The simple average of the four salary levels, (20,000 + 25,000 + 30,000 + 40,000)/4 = ₹28,750, ignores the different numbers of employees. ₹36,500 and ₹19,000 lie outside any sensible weighting, and ₹24,000 is too low because half the staff earn ₹30,000.

Answer: D

The largest number is 72.

Let the second number be the common multiple of 2 and 3, say 6k. Then the first is half of it and the third a third of it:

\[ \begin{aligned} \text{first} &= 3k,\ \text{second} = 6k,\ \text{third} = 2k \\ 3k + 6k + 2k &= 3 \times 44 = 132 \\ 11k &= 132 \Rightarrow k = 12 \end{aligned} \]

The numbers are 36, 72 and 24, and the largest is the second, 72. 36 is the first number, 22 is not one of them, and 108 would need an average of 66.

Answer: C

C scored 67 runs.

The total is 5 × 41 = 205. Express everything in terms of E: D = E + 5, A = E + 8, and B = D + E - 5 = 2E. Since B + C = 117:

\[ \begin{aligned} A + D + E + (B + C) &= 205 \\ (E + 8) + (E + 5) + E + 117 &= 205 \\ 3E &= 75 \Rightarrow E = 25 \end{aligned} \]

So B = 50 and C = 117 - 50 = 67. Check: A = 33, D = 30, E = 25, B = 50, C = 67, total 205. 37, 85 and 53 do not satisfy all the conditions together.

Answer: C

The product is 315.

Four consecutive odd numbers are evenly spaced by 2, so their average lies midway between the second and third. With average 18, those two are 17 and 19, and the four numbers are 15, 17, 19 and 21.

\[ 15 \times 21 = 315 \]

Check: (15 + 17 + 19 + 21)/4 = 72/4 = 18. 187 = 11 × 17, 247 = 13 × 19 and 391 = 17 × 23 are products of other pairs of odd numbers that do not fit the average.