WAEC Mathematics Past Questions & Answers - Page 75

371.

The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45cm in 3 seconds, how far will it fall in 6 seconds?

A.

90cm

B.

135cm

C.

180cm

D.

225cm

Correct answer is C

d \(\alpha\) t2

d = t2 k

where k is a constant. d = 45cm, when t = 3s; thus 45 = 32 x t

k = \(\frac{45}{9}\) = 5

thus equation connecting d and t is d = 5t2

when t = 6s, d = 5 x 62

= 5 x 36

= 180cm

372.

Simplify:(\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))2

A.

75.00

B.

15.00

C.

8.66

D.

3.87

Correct answer is B

Note that \(\frac{10\sqrt{3}}{\sqrt{5}} = \frac{10\sqrt{3}}{\sqrt{5}} \times - \frac{\sqrt{5}}{\sqrt{5}}\)

= \(\frac{10\sqrt{15}}{\sqrt{5}} = 2\sqrt{15}\)

hence, (\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))2 = (\(2\sqrt{15} - \sqrt{15}\))2

= (\(2\sqrt{15} - \sqrt{15}\))(\(2\sqrt{15} - \sqrt{15}\))

= 4\(\sqrt{15 \times 15} - 2\sqrt{15 \times 15} - 2\sqrt{15 x 15} + \sqrt{15 \times 15}\)

= 4 x 15 - 2 x 15 - 2 x 15 + 15

= 60 - 30 - 30 + 15

= 15

373.

Simplify: (\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\)

A.

\(\frac{1}{60}\)

B.

\(\frac{5}{72}\)

C.

\(\frac{1}{10}\)

D.

1\(\frac{7}{10}\)

Correct answer is C

(\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\)

= (\(\frac{9 - 8}{12} \times \frac{6}{5}\))

= \(\frac{1}{12} \times \frac{6}{5}\)

= \(\frac{1}{10}\)

374.

If 23x + 101x = 130x, find the value of x

A.

7

B.

6

C.

5

D.

4

Correct answer is D

23x + 101x = 130x

2 x X1 + 3 x Xo + 1 x X2 + 0 x X1 + 1 x Xo

= 1 x Xo = 1 x X2 + 3 x X1 + 0 x Xo

= X2 + 3x + 0

2x + 3 = x2 + 0 + 1 + x2 + 3x

2x - 3x + x2 - x2 = -3 - 1

- x = -4

x = 4

375.

In the diagram, \(\bar{OX}\) bisects < YXZ and \(\bar{OZ}\) bisects < YZX. If < XYZ = 68o, calculate the value of < XOZ

A.

68o

B.

72o

C.

112o

D.

124o

Correct answer is D

In \(\Delta\) XYZ, 2m + 2n + 68o = 180o

2(m + n) + 68o = 180o...(1)

in \(\Delta\) XOZ, m + n + q = 180o ...(2)

(m + n) = 180o - q...(3)

substituting 180o - q for (m + n) in (1) gives

2(180o - q) + 68o = 180o

360o - 2q = 180o - 68o

360o - 2q = 112o

360o - 112o = 2q

248o = 2q

q = \(\frac{248^o}{2}\)

= 124o

hence, < XOZ = 124o