WAEC Mathematics Past Questions & Answers - Page 112

556.

The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation

A.

12n - 6 = 0

B.

3n - 12 = 0

C.

2n - 35 = 0

D.

5n - 33 = 0

Correct answer is D

12 = \(\frac{n}{3} - 2n = 1\), multiply through by 3

36 + n - 6n = 3

-5n = 3 - 36

-5n = -33

-5n + 33 = 0

5n - 33 = 0

557.

If x + y = 2y - x + 1 = 5, find the value of x

A.

3

B.

2

C.

1

D.

-1

Correct answer is B

x + y = 2y - x + 1 = 5

x + y = 2y - x + 1

x + x + y - 2y = 1

2x - y = 1....(i)

2y - x + 1 = 5

-x + 2y = 5 + 1

-x = 2y = 4

x - 2y = -4 .....(ii)

solve simultaneously (i) x 2x - y = 1

(ii) x x - 2y = -4

2x - y = 1

=2x - 4y = -8

3y = 9

y = \(\frac{9}{3}\)

y = 3

substitute y = 3 into equation (i)

2x - y = 1

2x - 3 = 1

2x = 1 + 3

2x = 4

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

= 2

558.

Make p the subject of the relation: q = \(\frac{3p}{r} + \frac{s}{2}\)

A.

p = \(\frac{2q - rs}{6}\)

B.

p = 2qr - sr - 3

C.

p = \(\frac{2qr - s}{6}\)

D.

p = \(\frac{2qr - rs}{6}\)

Correct answer is D

q = \(\frac{3p}{r} + \frac{s}{2}\)

q = \(\frac{6p + rs}{2r}\)

6p + rs = 2qr

6p = 2qr - rs

p = \(\frac{2qr - rs}{6}\)

559.

Simplify: \(\frac{54k^2 - 6}{3k + 1}\)

A.

6(1 - 3k2)

B.

6(3k2 - 1)

C.

6(3k - 1)

D.

6(1 - 3k)

Correct answer is C

\(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\)

= \(\frac{6(3k + 1) - (3k - 1)}{3k + 1}\)

= 6(3k - 1)

560.

Solve for x in the equation; \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\)

A.

5

B.

4

C.

3

D.

1

Correct answer is A

\(\frac{1}{8} + \frac{2}{3x} = \frac{1}{3}\)

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

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

3x = 15

x = \(\frac{15}{3}\)

= 5