Average Questions and Answers

Answer: B

The attendance on the fifth day is 40.

Convert each average into a total:

\[ \begin{aligned} \text{first 4 days} &= 4 \times 30 = 120 \\ \text{first 5 days} &= 5 \times 32 = 160 \\ \text{fifth day} &= 160 - 120 = 40 \end{aligned} \]

Shortcut: the fifth day must lift the average of five days by 2, so it is the old average plus \(5 \times 2\): \(30 + 10 = 40\). 32 is the new average itself, a common slip, and 36 and 38 do not give a five-day total of 160.

Answer: A

The largest number is 46.

For an odd count of consecutive even numbers, the average is the middle number. So the five numbers are centred on 42, going up and down in steps of 2:

\[ 38,\ 40,\ 42,\ 44,\ 46 \]

Check: \(38 + 40 + 42 + 44 + 46 = 210 = 5 \times 42\). 44 is the next number after the middle, not the largest. 43 and 45 are odd, so they cannot belong to a set of even numbers at all.

Answer: A

The average is \(\frac{2n+5}{2}\).

Five consecutive integers starting at m are m to m + 4, and their average is the middle one: \(n = m + 2\). Six consecutive integers starting at m + 2 run from m + 2 to m + 7, and their average is the midpoint of the ends:

\[ \begin{aligned} \text{average} &= \frac{(m+2) + (m+7)}{2} = m + 4.5 \\ &= n + 2.5 = \frac{2n + 5}{2} \end{aligned} \]

n + 2 and n + 3 are near misses on either side of n + 2.5, and \(\frac{n+5}{2}\) halves n by mistake.

Answer: C

80% of the second number is 48.

Let the numbers be \(a\) and \(b\). From \(0.45a = 0.6b\), \(a = \frac{4}{3}b\). Their average is 10 less than \(a\).

\[ \begin{aligned} \frac{a + b}{2} &= a - 10 \\ b &= a - 20 = \frac{4}{3}b - 20 \\ b &= 60,\ a = 80 \end{aligned} \]

So 80% of 60 is 48. Check: 45% of 80 = 36 = 60% of 60, and the average 70 is 10 less than 80. 64 is 80% of the first number.

Answer: B

The seventh member is 81 years old.

Work with totals rather than averages:

\[ \begin{aligned} \text{total of 7} &= 7 \times 75 = 525 \\ \text{total of 6} &= 6 \times 74 = 444 \\ \text{7th member} &= 525 - 444 = 81 \end{aligned} \]

75 is just the overall average, and 80 or 70 do not come from the totals.

A shortcut: adding the seventh member raises the average of the other six from 74 to 75, so he must be \(74 + 7 \times 1 = 81\) years old.

Answer: D

The first number is 132.

Let the third number be \(t\). Then the second is \(2t\) and the first is \(4t\), and their total is three times the average:

\[ \begin{aligned} t + 2t + 4t &= 3 \times 77 = 231 \\ 7t &= 231, \ t = 33 \end{aligned} \]

So the numbers are 132, 66 and 33, and the first is 132. 33 and 66 are the third and second numbers, and 77 is the average.

Check: \((132 + 66 + 33)/3 = 231/3 = 77\), the given average.

Answer: D

The mean of the remaining observations is 8.9.

\[ \begin{aligned} \text{total of 12} &= 12 \times 7.5 = 90 \\ \text{total of 7} &= 7 \times 6.5 = 45.5 \\ \text{mean of 5} &= \frac{90 - 45.5}{5} = 8.9 \end{aligned} \]

The remaining observations number \(12 - 7 = 5\); dividing by 7 instead would give about 6.4, which is not offered. 5.5, 8.5 and 9.2 do not come from the totals.

Working with totals rather than averages is the standard method for questions in which part of a group is removed.

Answer: A

The teacher is 32 years old.

Work with totals. The 20 students' ages add to \(20 \times 11 = 220\). With the teacher there are 21 people averaging 12 years.

\[ \begin{aligned} \text{new total} &= 21 \times 12 = 252 \\ \text{teacher} &= 252 - 220 = 32 \end{aligned} \]

Shortcut: the teacher brings his own share (12) plus 1 year for each of the 20 students, \(12 + 20 = 32\). 36 and 42 fit no step, so 'None of the above' is wrong too.

Answer: D

The teacher is 57 years old.

Compare the total ages before and after the teacher is included:

\[ \begin{aligned} \text{Students} &= 40 \times 16 = 640 \\ \text{With teacher} &= 41 \times 17 = 697 \\ \text{Teacher} &= 697 - 640 = 57 \end{aligned} \]

Shortcut: the teacher must supply his own share of the new average (17) plus one extra year for each of the 40 students, which is 17 + 40 = 57.

A teacher of 56 would make the new total 696, an average just under 17, and 48 or 60 miss it by more. Only 57 makes the average of all 41 people exactly 17.

Answer: B

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