WAEC Mathematics Past Questions & Answers - Page 148

736.

Given that x = 2 and y = -\(\frac{1}{4}\), evaluate \(\frac{x^2y - 2xy}{5}\)

A.

zero

B.

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

C.

1

D.

2

Correct answer is A

Given; x = 2; y = \(\frac{-1}{4}\)

= \(\frac{x^2y - 2xy}{5}\)

= \(\frac{2^2(\frac{-1}{4}) - 2(2)(\frac{-1}{4})}{5}\)

= \(\frac{4(\frac{-1}{4}) + 4(\frac{-1}{4})}{5}\)

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

= \(\frac{0}{5}\)

= 0

737.

Simplify 3\(\sqrt{27x^3y^9}\)

A.

9xy3

B.

3xy6

C.

3xy3

D.

9y3

Correct answer is C

3\(\sqrt{27x^3y^9}\) = 3\(\sqrt{27} \times 3\sqrt{3^3} \times 3\sqrt{y^9}\)

= 3 \(\times x \times y^3\)

= 3xy3

738.

Simplify \(\frac{\frac{1}{x} + \frac{1}{y}}{x + y}\)

A.

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

B.

\(\frac{1}{xy}\)

C.

x + y

D.

xy

Correct answer is B

\(\frac{\frac{1}{x} + \frac{1}{y}}{x + y}\) = \(\frac{\frac{y + x}{xy}}{x + y}\)

= \(\frac{x + y}{xy}\)

= \(\frac{x + y}{xy} \times \frac{1}{x + y}\)

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

739.

If p-2g + 1 = g + 3p and p - 2 = 0, find g

A.

-2

B.

-1

C.

1

D.

2

Correct answer is B

p - 2g + 1 = g + 3p.........(1)

p - 2 = 0 .........(2)

From (2), p = 2; put p = 2 into (1);

2 - 2g + 1 = g + 3(2)

3 - 2g = g + 6

-2g - g = 6 - 3

-3g = 3

g = \(\frac{3}{-3}\)

g = -1

740.

A trader bought 100 tubers at 5 for N350.00. She sold them in sets of 4 for N290.00. Find her gain percent.

A.

3.6%

B.

3.5%

C.

3.4%

D.

2.55

Correct answer is A

Cost price, c.p = \(\frac{100}{5}\) x N350 = N7000

Selling price, s.p = \(\frac{100}{5}\) x N290 = N7250

%Gain = \(\frac{S.p - C.p}{C.p}\) x 100%

= \(\frac{7250 - 7000}{7000}\) x 100% = \(\frac{250 \times 100}{7000}\)

= 3.6% (approx.)