WAEC Mathematics Past Questions & Answers - Page 54

266.

Solve: - \(\frac{1}{4}\) < \(\frac{3}{4}\) (3x - 2) < \(\frac{1}{2}\)

A.

-\(\frac{5}{9}\) < x <\(\frac{8}{9}\)

B.

-\(\frac{8}{9}\) < x <\(\frac{7}{9}\)

C.

-\(\frac{8}{9}\) < x <\(\frac{5}{9}\)

D.

-\(\frac{7}{9}\) < x <\(\frac{8}{9}\)

Correct answer is A

\(\frac{3}{4}\) (3x - 2) < \(\frac{1}{2}\); \(\frac{3}{4}\) (3x - 2) > - \(\frac{1}{4}\)

3(3x - 2) < 2; 3(3x - 2) > -1

9x - 6 < 2; 9x - 6 > -1

9x < 8; 9x > 5

x < \(\frac{5}{9}\); x > \(\frac{8}{9}\)

267.

One factor of \(7x^2 + 33x - 10\) is

A.

7x + 5

B.

x - 2

C.

7x - 2

D.

x - 5

Correct answer is C

\(7x^2 + 33x - 10\)

\(7x^2 + 35x - 2x - 10\)

7x (x + 5) - 2 (x + 5)

(7x - 2) (x + 5)

268.

A sum of N18,100 was shared among 5 boys and 4 girls with each boy taking N20.00 more than each girl. Find a boy's share.

A.

N1,820.00

B.

2,000.00

C.

N2,020.00

D.

N2,040.00

Correct answer is C

Let a girl's share = x + 20

4x + 5(x + 20) = 18,100

4x + 5x + 100 = 18,100

9x + 100 = 18,100

9x = 18,000

x = \(\frac{18,000}{9}\)

x = 2,000

\(\therefore\) Each boy gets N(2,000 + 20)

 = N2,020.

269.

Solve the equation: \(\frac{1}{5x} + \frac{1}{x}\)= 3

A.

\(\frac{1}{5}\)

B.

\(\frac{2}{5}\)

C.

\(\frac{3}{5}\)

D.

\(\frac{4}{5}\)

Correct answer is B

\(\frac{1}{5x} + \frac{1}{x}\)= 3

\(\frac{1 + 5}{5x}\) = 3

6 = 15x

x = \(\frac{6}{15}\)

= \(\frac{2}{5}\)

270.

If x = \(\frac{2}{3}\) and y = - 6, evaluate xy - \(\frac{y}{x}\)

A.

0

B.

5

C.

8

D.

9

Correct answer is B

x = \(\frac{2}{3}\) and y = - 6

xy - \(\frac{y}{x}\)

\(\frac{2}{3} - (6)^2 - (-6) \div \frac{2}{3}\)

= -4 - (6) x \(\frac{3}{2}\)

= -4 - (-6) x \(\frac{3}{2}\)

= -4 - (-9)

= -4 + 9

= 5