WAEC 2016 · Paper 2 · Q6

(a)
Find the sum of the Arithmetic Progression (A.P.) $1, 3, 5, \ldots, 101$ .
(b)
Out of 95 travellers interviewed, 7 travelled by bus and train only, 3 by train and car only and 8 travelled by all three means of transport. The number, $x$ , of travellers who travelled by bus only was equal to the number who travelled by bus and car only. If 47 people travelled by bus and 30 by train: (i) represent this information in a Venn diagram; (ii) calculate the: I. value of $x$ ; II. number who travelled by at least two means.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

The first term is $a = 1$ and the common difference is $d = 2$ .
Find how many terms: $1 + 2(n - 1) = 101$ , so $n - 1 = 50$ and $n = 51$ .
$S_n = \frac{n}{2}(\text{first} + \text{last})$
$= \frac{51}{2}(1 + 101)$
$= 51 \times 51$
$= 2601$ .

Draw three overlapping circles B (bus), T (train) and C (car) in a rectangle of 95.
Put 8 in the centre, 7 in B and T only, 3 in T and C only, and $x$ in both B only and B and C only.

I.** The bus circle holds 47: $x + 7 + 8 + x = 47$ , so $2x + 15 = 47$ , $2x = 32$ and $x = 16$ .
II. At least two means: B and T only, T and C only, B and C only, and all three: $7 + 3 + 16 + 8 = 34$ .