If 19a + 19b + 19c = 437, then what is the mean of a, b and c?
- (a)6·33
- (b)7·66
- (c)9·33
- (d)11·55
Correct — B, 7·66. The left side has a common factor: 19a + 19b + 19c = 19(a + b + c). So 19(a + b + c) = 437, and a + b + c = 437 ÷ 19 = 23. The mean of three numbers is their sum divided by three, that is 23 ÷ 3 = 7.666…, printed here as 7·66. The individual values of a, b and c are never needed, and the question supplies no way of finding them.
- (a)6·33 — 6.33 is 19 ÷ 3. It follows from dividing the common factor by three instead of the sum.
- (c)9·33 — 9.33 is 28 ÷ 3. It needs a sum of 28, which does not come from 437 ÷ 19.
- (d)11·55 — 11.55 is roughly 23 ÷ 2, the value if the mean were taken over two numbers instead of three.
The arithmetic mean is the sum divided by the count, so any equation that yields the sum yields the mean. Factoring the common multiplier out of 19a + 19b + 19c is the whole of the algebra: it converts a three-variable equation into a single statement about the total.
Three unknowns and one equation look under-determined, and that is the point — the question asks only for a quantity that depends on the sum. 437 = 19 × 23 is worth spotting quickly; 19 × 20 = 380 and 19 × 3 = 57 add to 437. The printed value 7·66 is the truncation of 7.666…, so a candidate who rounds to 7.67 should still read option (b) as the intended match rather than hunting for a closer figure.
- 19a + 19b + 19c = 19(a + b + c), so the equation fixes only the sum.
- 437 ÷ 19 = 23, so a + b + c = 23.
- Mean = sum ÷ number of terms = 23 ÷ 3 = 7.666…
- The individual values of a, b and c are not determined and are not needed.
- 19 × 23 = 437 is a useful factorisation to recognise, since 437 is not divisible by 2, 3, 5, 7, 11 or 13.
Three unknowns, one equation, and a question that asks only about their total.
- Dividing 437 by 3 first and then by 19, or dividing 19 by 3.
- Assuming a, b and c must be found individually.
- Rounding 7.666 to 7.7 and hunting for a closer printed option.
A one-line algebra item testing whether a candidate spots that only the sum of the variables is required.
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 movement between a mean and a total, taken one step further. That item converts a mean into a sum and then isolates a single term; this one converts an equation into a sum and stops at the mean.
- practice — not a real PYQ
If 7x + 7y = 91, what is the mean of x and y?
- (a)6.5
- (b)7
- (c)13
- (d)26
Answer(a) 6.5 — 7(x + y) = 91 gives x + y = 13, and the mean of two numbers is 13 ÷ 2 = 6.5.
- practice — not a real PYQ
The mean of five numbers is 18. If one number is removed the mean of the rest is 20. Which number was removed?
- (a)8
- (b)10
- (c)12
- (d)14
Answer(b) 10 — the total was 90 and the remaining four total 80, so the removed number is 10.