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: B

Let the sale in the sixth month \(=x\)
Then, \(\dfrac{6435+6927+6855+7230+6562+x}{6}=6500\)
\(\Rightarrow 6435+6927+6855+7230+6562+x=6 \times 6500\\ \Rightarrow 34009 + x = 39000\\ \Rightarrow x = 39000 - 34009 = 4991\)

Answer: D

Since the month begins with a Sunday, to there will be five Sundays in the month.
Required average
\(\left( \dfrac {510 \times 5 + 240 \times 25}{30} \right)\\ =\dfrac{8550}{30}\\ =285\)

Answer: B

Let the total number of students \(=x\)
The average marks increased by \(\dfrac12\) due to an increase of \(83 - 63 = 20\) marks.
But total increase in the marks
\(\dfrac{1}{2}\times x = \dfrac{x}{2}\)
Hence we can write as,
\(\dfrac{x}{2}=20\\ \Rightarrow x = 20 \times 2 = 40\)

Answer: B

Equal distances each way call for the harmonic mean.

\[ \bar{v} = \frac{2 \times 50 \times 30}{50 + 30} = \frac{3000}{80} = 37.5 \text{ km/hr} \]

The 150 km is not needed - it cancels. The arithmetic mean of 40 would be too high, since more time is spent on the slower return.