WAEC Mathematics Past Questions & Answers - Page 150

746.

If x% of 240 equals 12, find x

A.

x = 1

B.

x = 3

C.

x = 5

D.

x = 7

Correct answer is C

x% of 240 = 12

\(\frac{x}{100} \times 240 = 12\)

x = \(\frac{12 \times 100}{240}\)

x = 5

747.

In the diagram, O is the centre of the circle and PQ is a diameter. Triangle RSO is an equilateral triangle of side 4cm. Find the area of the shaded region

A.

43.36cm2

B.

32.072

C.

18.212

D.

6.932

Correct answer is C

Area of shaded portion = Area of semicircle

Area of \(\bigtriangleup\) RSO

Area of semicircle = \(\frac {\pi r^2}{2} = \frac{22\times 4 \times 4}{7 \times 2}\)

= 25.14cm2; Area of \(\bigtriangleup\)RSO

=\(\sqrt{s(s - 1)(s - b)(s - c)}\); where

s = \(\frac{a + b + c}{2}\)

s = \(\frac{4 + 4 + 4}{2}\)

= 6cm

= \(\sqrt{6(6 - 4)(6 - 4) (6 - 4)}\)

= \(\sqrt{6(2) (2) (2)}\)

= \(\sqrt{18}\) = 6.93cm2

Area of shaded region

= 25.14 - 6.93

= 18.21cm2

748.

In the diagram, |XR| = 4cm

|RZ| = 12cm, |SR| = n, |XZ| = m and SR||YZ. Find m in terms of n

A.

m = 2n

B.

m = 3n

C.

m = 4n

D.

m = 5n

Correct answer is C

\(\frac{n}{4} = \frac{m}{4 + 12}\)

\(\frac{n}{4} = \frac{m}{16}\)

4m = 16n

\(\frac{4m}{4} = \frac{16n}{4}\)

m = 4n

749.

In the diagram, O is the centre of the circle and < PQR = 106º, find the value of y

A.

16

B.

37

C.

74

D.

127

Correct answer is C

2y + Reflex angle = 360

Reflex angle = 360 - 2y

360 - 2y = 2(106)(angles at a centre = 2ce < of circum.)

360 - 212 = 2y

148 = 2y

y \(\frac{148}{2}\)

y = 74o

750.

If the ratio x:y = 3:5 and y:z = 4:7, find the ratio x:y:z

A.

15 : 28 : 84

B.

12 : 20 : 35

C.

3 : 5 : 4

D.

5 : 4 : 7

Correct answer is B

x : y = 3.5, y : z = 4 : 7

x = 3 x 4 = 12

y = 5 x 4 = 20

y = 4 x 5 = 20

z = 7 x 5 = 35

then x : y : z = 12 : 20 : 35