WAEC Mathematics Past Questions & Answers - Page 129

641.

What is the length of a rectangular garden whose perimeter is 32cm and area 39cm2?

A.

25cm

B.

18cm

C.

13cm

D.

9cm

Correct answer is C

perimeter = 2(l + b) = 32

l + b = \(\frac{32}{2}\)

l + b = 16

b = 16 - 1.......(1)

Area = l + b = 39

lb = 39 .....(2)

put (1) into (2)

l(16 - 1) = 39

16l - l2 = 39

l2 = 13l - 3l + 39 = 0

l(l - 13) - 3(1 - 13) = 0

(l - 3)(l + 130 = 0

l - 3 = 0 or l - 13 = 0

l = 3cm or l = 13cm; The length in 13cm

642.

Given that tan x = 1, where 0o \(\geq\) x 90o, evaluate \(\frac{1 - \sin^2 x}{\cos x}\)

A.

2\(\sqrt{2}\)

B.

\(\sqrt{2}\)

C.

\(\frac{\sqrt{2}}{2}\)

D.

\(\frac{1}{2}\)

Correct answer is C

Given tan x = 1

x = tan-1(1)

x = 45o

Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)

= \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\)

= \(\frac{1}{2} + \frac{1}{\sqrt{2}}\)

= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)

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

643.

A rectangle has length xcm and width (x - 1)cm. If the perimeter is 16cm. Find the value of x

A.

3\(\frac{1}{2}\)cm

B.

4cm

C.

4\(\frac{1}{2}\)cm

D.

5cm

Correct answer is C

l = x; b = x - 1

perimeter = 2(l + b) = 16

l + b = \(\frac{16}{2}\) = 8

l + b = 8

x + x - 1 = 8

2x = 8 + 1

2x = 9

x = \(\frac{9}{2}\)cm

x = 4\(\frac{1}{2}\)

644.

If sin 3y = cos 2y and 0o \(\leq\) 90o, find the value of y

A.

18o

B.

36o

C.

54o

D.

90o

Correct answer is A

sin 3y = cos 2y, but sin \(\theta\) = cos(90 - \(\theta\))

sin 3y = cos(90 - 3y)

cos(90 - 3y) = cos 2y

90 - 3y = 2y

5y = 90

y = \(\frac{90}{5}\)

y = 18o

645.

Simplify 2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)

A.

1

B.

\(\frac{1}{3}\sqrt{3}\)

C.

2\(\sqrt{3} - 5\frac{2}{3}\)

D.

6\(\sqrt{3}\) - 17

Correct answer is B

2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)

= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{9 \times 3}}\)

= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} + \frac{3}{3\sqrt{3}}\)

= 2\(\sqrt{3} = 6 \frac{\sqrt{3}}{3} + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= 2\(\sqrt{3} - 2\sqrt{3} + \frac{\sqrt{3}}{3}\)

= \(\frac{\sqrt{3}}{3} = \frac{1}{3} \sqrt{3}\)