Average Questions and Answers

Answer: A

The sixth month's sale must be Rs. 4991.

For an average of Rs. 6,500 over six months, the total must be 6 × 6,500 = Rs. 39,000. The first five months give:

\[ \begin{aligned} &6435 + 6927 + 6855 + 7230 + 6562 \\ &= 34009 \\ &39000 - 34009 = 4991 \end{aligned} \]

So the sixth month needs Rs. 4,991. The other options each give a total above or below Rs. 39,000; for example, 5,467 would make the average about Rs. 6,579.

Answer: B

The difference is 8.

Consecutive odd numbers go up in steps of 2, so five of them with an average of 61 are centred on 61: 57, 59, 61, 63 and 65. The difference between the highest and the lowest is

\[ 65 - 57 = 8 \]

In fact the average is irrelevant: any five consecutive odd numbers span four steps of 2, which is always 8. 4 is the span of three such numbers, and 12 and 16 are the spans of seven and nine.

Answer: C

The largest number is 30.

Four consecutive even numbers are n, n + 2, n + 4 and n + 6. Their average is the mean of the middle two, n + 3, so

\[ n + 3 = 27 \quad\Rightarrow\quad n = 24 \]

The numbers are 24, 26, 28 and 30, whose sum 108 divided by 4 is 27. The largest is 30. 28 is the third number, and 32 and 36 would push the average above 27.

Shortcut: for an even number of consecutive even numbers, the average lies midway between the two middle terms, so 26 and 28 surround 27, and the list ends two steps later at 30.

Answer: B

The number of boys is 8.

Let \(b\) be the number of boys. The total age of the class equals the boys' total plus the girls' total:

\[ \begin{aligned} 16(b + 10) &= 21b + 12 \times 10 \\ 16b + 160 &= 21b + 120 \\ 5b &= 40 \Rightarrow b = 8 \end{aligned} \]

Check: \(8 \times 21 + 10 \times 12 = 168 + 120 = 288\), and \(288/18 = 16\). The other values fail this check; for example 12 boys give an average of \((252 + 120)/22 \approx 16.9\).

Answer: B

The average is 48.

The first \(n\) natural numbers form an arithmetic progression, so their average is the average of the first and last terms:

\[ \text{average} = \frac{1 + 95}{2} = 48 \]

Equivalently, the sum is \(\frac{95 \times 96}{2} = 4560\), and \(4560 \div 95 = 48\).

50 is the average of the first 99 numbers, 49 of the first 97 and 51 of the first 101; each comes from using the wrong last term.

Answer: D

The average price is 300.

\[ \begin{aligned} \text{sum} &= 345 + 375 + 225 + 255 + 300 \\ &= 1500 \\ \text{average} &= \frac{1500}{5} = 300 \end{aligned} \]

A quick check: 345 and 255 average to 300, and so do 375 and 225, with 300 itself in the middle, so the average must be 300. 340 and 350 are near the highest prices, and 280 is below the middle of the set.

Answer: B

The actual average should have been 49.32.

The total of the recorded marks was 50 × 48 = 2400. The two mistakes understated the total by (67 - 37) + (73 - 37) = 30 + 36 = 66 marks.

\[ \begin{aligned} \text{correct total} &= 2400 + 66 = 2466 \\ \text{average} &= 2466 / 50 = 49.32 \end{aligned} \]

A quicker route: the average rises by 66/50 = 1.32, from 48 to 49.32. 46.68 is what you get by subtracting the correction instead of adding it, and 51.60 and 50.66 do not match the correction of 66 marks.

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,
\(\dfrac{x}{2}=20\\ \Rightarrow x = 20 \times 2 = 40\)