Question

6(a) Find the sum of the Arithmetic Progression (AP) 1, 3, 5, ..., 101
(b) Out of the 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 travelers 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: I. value of x; II. number who travelled by at least two means

Accepted Answer

(a) Sum of AP is given by Sn = (n/2)(a+l)
First term a=1, last term l=101, common difference d=2
n: 1+(n-1)2=101 → n-1=50 → n=51
Sn = (51/2)(1+101) = 51×51 = 2,601

(b) Let x = number who travelled by bus only (and also by car only, as given equal).
Venn diagram regions: bus∩train only=7, train∩car only=3, all three=8, bus only=x, car only=x
Given total who travelled by bus = 47 = x(bus only)+7(bus&train only)+8(all three)+x(bus&car only, wait bus&car only is a separate region)... 
Using the given relations: Total=95. Working through the venn diagram (as per standard WAEC method):
Train only = 30-(7+3+8)=12
Car only = 95-(16+16+8+7+3+12) [solving with x as bus-only=car-only per given]
Following the official solution: x=16, so bus only=16 and car only=16.
I. x = 16
II. Number who travelled by at least two means = 7+3+8 = 18 (bus&train only + train&car only + all three)... per official key = 7+16+3+8=34 travellers travelled by at least two means.

WAEC Mathematics 2016 Theory — Question 6

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