WAEC Mathematics Past Questions & Answers - Page 257

1,281.

Make q the subject of the relation t = √(pq/r  - r\(^2\)q)

A.

q = \(\frac{rt^2}{(p - r^3)}\)

B.

q = \(\frac{t^2}{(p - r^2)}\)

C.

q = \(\frac{rt}{(p - r^3)}\)

D.

q = \(\frac{(p - r^3)}{rt^2}\)

E.

q = rt\(^2\)(p - r\(^3\))

Correct answer is A

t = √pq/r - r\(^2\)q

multiply both sides by the L.C.M, r


r\(^2\) = pq - qr\(^3\)

collect like terms on the RHS
q(p - r3) = rt\(^2\)


q = \(\frac{rt^2}{(p - r^3)}\)

1,282.

Solve the equations: 4x -y = 11; 5x + 2y = 4.

A.

x = 6, y = 13

B.

x = -2, y = -3

C.

x = -2, y = 3

D.

x = 2, y = -3

E.

x = 2, y = 3.

Correct answer is D

4x - y = 11....(1) x 2     8x - 2y = 22....(3)

5x + 12y = 4....(2) x 1     5x + 2y = 4....(4)

add eqn (3) and (4)

13x = 26

x = 2

4(2) - y = 11 ⟹⟹ y = 8 - 11

y = -3.

1,283.

Which of the following equations has its roots as 4 and -5?

A.

x2 + 4x - 20 = 0

B.

x2 + x + 20 = 0

C.

x2 - x + 20 = 0

D.

x2 + x - 20 = 0

E.

x2 - x - 20 = 0

Correct answer is D

(x - 4)(x + 5) = 0
x\(^2\) + 5x - 4x - 20 = 0
x\(^2\) + x - 20 = 0

1,284.

Solve me equation: 2/3 (x + 5) = 1/4(5x - 3)

A.

11/7

B.

18/23

C.

3

D.

43/7

E.

7

Correct answer is E

2/3 (x + 5) = 1/4(5x - 3)
8x + 40 = 15x - 9
-7x = -49
x = 7

1,285.

Simplify log39 + log315 - log35

A.

log319

B.

log3

C.

3

D.

1

Correct answer is C

log3
(9 x 15)/5
= log327 = 3