Which one of the following is the average of first five multiples of each of the numbers from 11, 12, 13, ..., 20?
- (a)40.5
- (b)42.5
- (c)44.5
- (d)46.5
Correct — D, 46.5. The first five multiples of any number n are n, 2n, 3n, 4n and 5n, and they average (1 + 2 + 3 + 4 + 5)n ÷ 5 = 15n ÷ 5 = 3n. So the ten numbers 11 to 20 give averages 33, 36, 39 and so on up to 60. Averaging those is the same as taking 3 × (the average of 11 to 20). The numbers 11 to 20 are in arithmetic progression, so their average is the mean of the two ends, (11 + 20) ÷ 2 = 15.5. The answer is 3 × 15.5 = 46.5. Pooling all fifty multiples into one list gives the same value.
- (a)40.5 — 40.5 is 3 × 13.5, the answer if the numbers ran from 11 to 16. The set given runs to 20.
- (b)42.5 — 42.5 does not equal 3n for any average n of a run of whole numbers ending at 20; the multiplier 3 forces the answer to be a multiple of 1.5.
- (c)44.5 — This comes from averaging 11 to 20 as 14.83 or from mixing the mean of the first five natural numbers with the mean of their multiples. The correct chain is 3 × 15.5.
Two averaging facts do the whole job. First, the mean of the first five multiples of n is 3n, because the multipliers 1 to 5 average 3. Second, the mean of a set in arithmetic progression is the average of its first and last terms, so 11 to 20 averages 15.5. Averaging is linear: multiplying every member of a set by 3 multiplies the mean by 3.
The phrasing packs two averages into one sentence, and candidates lose time deciding whether to average within each number first or across all fifty multiples at once. Both routes agree, because the ten groups are the same size — that is worth knowing, since pooled means differ from means of means whenever group sizes differ. The arithmetic itself is short: 3 × 15.5.
- The mean of the first k multiples of n is n(k+1)/2; for k = 5 this is 3n.
- For numbers in arithmetic progression the mean equals the average of the first and last terms.
- The mean of 11, 12, …, 20 is (11 + 20) ÷ 2 = 15.5.
- Scaling every observation by a constant scales the mean by the same constant.
- Pooling all fifty multiples gives the same mean because each of the ten groups has equal size.
The ten groups are of equal size, so the mean of the group means equals the mean of all fifty multiples.
- Averaging the multipliers as 2.5 instead of 3 by counting 0 as a multiple.
- Stopping at 3n and forgetting to average across the ten numbers.
- Assuming a mean of means always equals the pooled mean — it does here only because the groups are equal in size.
A compressed averaging item: two averages in one sentence, both of them shortcuts a well-drilled candidate can do in the head.
The average of X1, X2 and X3 is 14. Twice the sum of X2 and X3 is 30. What is the value of X1?
- (a) 20
- (b) 27
- (c) 16
- (d) 12
Answer(b) 27
The same conversion between a mean and a sum, run backwards. There the mean is turned into a total to isolate one unknown; here the totals of ten small sets are turned into a single mean.
- practice — not a real PYQ
What is the average of the first five multiples of 14?
- (a)35
- (b)42
- (c)45
- (d)70
Answer(b) 42 — the first five multiples average 3n, and 3 × 14 = 42.
- practice — not a real PYQ
What is the average of the first ten multiples of 7?
- (a)35
- (b)38.5
- (c)42
- (d)45.5
Answer(b) 38.5 — the first ten multipliers average 5.5, so the mean is 5.5 × 7 = 38.5.