# 2016 - WAEC Mathematics Past Questions and Answers - page 2

11
Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10?
A
1
B
3
C
7
D
17
correct option: c
2x + y = 7...(1)

3x - 2y = 3...(2)

From (1), y = 7 - 2x for y in (2)

3x - 2(7 - 2x) = 3

3x - 14 + 4x = 3

7x + 3 + 14 = 17

x = $$\frac{17}{7}$$

Hence, 7 x $$\frac{17}{7}$$

= 17 - 10

= 7
12
Simplify; $$\frac{2}{1 - x} - \frac{1}{x}$$
A
$$\frac{x + 1}{x(1 - x)}$$
B
$$\frac{3x - 1}{ x(1 - x)}$$
C
$$\frac{3x + 1}{ x(1 - x)}$$
D
$$\frac{x + 1}{ x(1 - x)}$$
correct option: b
$$\frac{2}{1 - x} - \frac{1}{x}$$ = $$\frac{2x - 1(1 - x)}{x(1 - x)}$$

= $$\frac{2x - 1(1 + x)}{x(1 - x)}$$

= $$\frac{3x - 1}{x(1 - x)}$$
13
Make s the subject of the relation: P = S + $$\frac{sm^2}{nr}$$
A
s = $$\frac{mrp}{nr + m^2}$$
B
s = $$\frac{nr + m^2}{mrp}$$
C
s = $$\frac{nrp}{mr + m^2}$$
D
s = $$\frac{nrp}{nr + m^2}$$
correct option: d
P = S + $$\frac{sm^2}{nr}$$

P = S(1 + $$\frac{m^2}{nr}$$)

P = S(1 + $$\frac{nr + m^2}{nr}$$)

nrp = S(nr + m2)

S = $$\frac{nrp}{nr + m^2}$$
14
Factorize; (2x + 3y)2 - (x - 4y)2
A
(3x - y)(x + 7y)
B
(3x + y)(2x - 7y)
C
(3x + y)(x - 7y)
D
(3x - y)(2x + 7y)
correct option: a
(2x + 3y)2 - (x - 4y)2

= (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y)

= 4x2 + 12xy + 9y2 - (x2 - 8xy+ 16y2)

= 4x2 + 12xy + 19y2 - x2 + 8xy - 16y2

= 3x2 + 20xy - 7y2

= 3x2 + 21xy - xy - 7y2

= 3x(x + 7y) - y(x + 7y)

= (3x - y)(x + 7y)
15
The curve surface area of a cylinder, 5cm high is 110cm 2. Find the radius of its base. [Take $$\pi = \frac{22}{7}$$]
A
2.6cm
B
3.5cm
C
3.6cm
D
7.0cm
correct option: b
Curved surface area of cylinder = 2$$\pi$$rh

110 = 2 x $$\frac{22}{7}$$ x r x 5

r = $$\frac{110 \times 7}{44 \times 5}$$

= 3.5cm
16
The volume of a pyramid with height 15cm is 90cm3. If its base is a rectangle with dimension xcm by 6cm, find the value of x
A
3
B
5
C
6
D
8
correct option: a
Volume of pyramid = $$\frac{1}{3}$$lbh

90 = $$\frac{1}{3} \times x \times 6 \times 15$$

x = $$\frac{90 \times 33}{6 \times 15}$$

= 3
17

A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ

A
$$\frac{3}{5}$$
B
$$\frac{2}{3}$$
C
$$\frac{3}{2}$$
D
$$\frac{5}{3}$$
correct option: c

Let: (x1, y1) = (1, 2)

(x2, y2) = (5, 8)

The gradient m of $$\bar{PQ}$$ is given by

m = $$\frac{y_2 y_1}{x_2 - x_1}$$

= $$\frac{8 - 2}{5 - 1}$$

= $$\frac{6}{4}$$

= $$\frac{3}{2}$$

18
A straight line passes through the point P(1,2) and Q
(5,8). Calculate the length PQ
A
$$4\sqrt{11}$$
B
$$4\sqrt{10}$$
C
$$2\sqrt{17}$$
D
$$2\sqrt{13}$$
correct option: d
|PQ| = $$\sqrt{(x_2 - X- 1) + (y_2 - y_1)^2}$$

= $$\sqrt{(5 - 1)^2 + (8 - 2)^2}$$

= $$\sqrt{4^2 + 6^2}$$

= $$\sqrt{16 + 36}$$

= $$\sqrt{52}$$

= 2$$\sqrt{13}$$
19
If cos $$\theta$$ = x and sin 60o = x + 0.5 0o < $$\theta$$ < 90o, find, correct to the nearest degree, the value of $$\theta$$
A
32o
B
40o
C
60o
D
69o
correct option: d
sin 60o = x + 0.5 0o(given)

0.8660 = x + 0.5

0.8660 - 0.5 = x

x = 0.3660

cos$$\theta$$ = x(given)

cos$$\theta$$ = 0.3660

Hence, $$\theta$$ = cos-1(0.3660)

= 68.53o

= 69o (nearest degree)
20
$$\begin{array}{c|c} Age(years) & 13 & 14 & 15 & 16 & 17 \ \hline Frequency & 10 & 24 & 8 & 5 & 3 \end{array}$$

The table shows the ages of students in a club. How many students are in the club?
A
50
B
55
C
60
D
65
correct option: a
Number of students in the club is

10 + 24 + 8 + 5 + 3 = 50