WAEC Mathematics Past Questions & Answers - Page 73

361.

Factorize; (2x + 3y)2 - (x - 4y)2

A.

(3x - y)(x + 7y)

B.

(3x + y)(2x - 7y)

C.

(3x + y)(x - 7y)

D.

(3x - y)(2x + 7y)

Correct answer is A

(2x + 3y)2 - (x - 4y)2

= (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y)

= 4x2 + 12xy + 9y2 - (x2 - 8xy+ 16y2)

= 4x2 + 12xy + 19y2 - x2 + 8xy - 16y2

= 3x2 + 20xy - 7y2

= 3x2 + 21xy - xy - 7y2

= 3x(x + 7y) - y(x + 7y)

= (3x - y)(x + 7y)

362.

Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\)

A.

s = \(\frac{mrp}{nr + m^2}\)

B.

s = \(\frac{nr + m^2}{mrp}\)

C.

s = \(\frac{nrp}{mr + m^2}\)

D.

s = \(\frac{nrp}{nr + m^2}\)

Correct answer is D

P = S + \(\frac{sm^2}{nr}\)

P = S(1 + \(\frac{m^2}{nr}\))

P = S(1 + \(\frac{nr + m^2}{nr}\))

nrp = S(nr + m2)

S = \(\frac{nrp}{nr + m^2}\)

363.

Simplify; \(\frac{2}{1 - x} - \frac{1}{x}\)

A.

\(\frac{x + 1}{x(1 - x)}\)

B.

\(\frac{3x - 1}{ x(1 - x)}\)

C.

\(\frac{3x + 1}{ x(1 - x)}\)

D.

\(\frac{x + 1}{ x(1 - x)}\)

Correct answer is D

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

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

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

364.

Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10?

A.

1

B.

3

C.

7

D.

17

Correct answer is C

2x + y = 7...(1)

3x - 2y = 3...(2)

From (1), y = 7 - 2x for y in (2)

3x - 2(7 - 2x) = 3

3x - 14 + 4x = 3

7x + 3 + 14 = 17

x = \(\frac{17}{7}\)

Hence, 7 x \(\frac{17}{7}\)

= 17 - 10

= 7

365.

Find the values of y for which the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined

A.

6, -7

B.

3, -6

C.

3, -7

D.

-3, -7

Correct answer is C

\(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\)

Factorize the denominator;

Y2 + 7y - 3y - 21

= y(y + 7) -3 (y + 7)

= (y - 3)(y + 7)

Hence the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined

when y2 + 4y - 21 = 0

ie. y = 3 or -7