WAEC Mathematics Past Questions & Answers - Page 44

216.

If 3p = 4q and 9p = 8q - 12, find the value of pq.

A.

12

B.

7

C.

-7

D.

-12

Correct answer is A

9p = 8q - 12

9p = 2(4q) - 12

9p = 2(3q) - 12

9p = 6p - 12

3p = -12 

p = -4

\(\frac{3 \times -4}{4} = \frac{4q}{4}\) 

q = 13

pq = -3 x -4 

= 12

217.

Bala sold an article for #6,900.00 and made a profit of 15%. Calculate his percentage profit if he had sold it for N6,600.00. 

A.

5%

B.

10%

C.

12%

D.

13%

Correct answer is B

15 = (\(\frac{6,900 - C.P \times 100}{C.P}\))

15 C.P = 690000 - C.P 100

C.P = \(\frac{690000}{115}\)

C.P = N6,000

%profit = \(\frac{6,600 - 6,000}{6,000}\) x 100

= \(\frac{600}{6,000}\) x 100 

= 10%

218.

Simplify: \(\log_{10}\) 6 - 3 log\(_{10}\)  3 + \(\frac{2}{3} \log_{10} 27\)

A.

3 \(\log_{10}^2\)

B.

\(\log_{10}^2\)

C.

\(\log_{10}^3\)

D.

2 \(\log_{10}^3\)

Correct answer is B

log\(_{10}\) 6 - log\(_{10}\)3\(^3\) + log\(_{10}\) (\(\sqrt[3]{27}\))\(^2\)

= log \(_{10}\) 6 - log \(_{10}\) 27 + log\(_{10}\) 9

= log\(_{10}\) \(\frac{6  \times 9}{27}\)

= log\(_{10}\)2

219.

Solve 4x^{2}\) - 16x + 15 = 0.

A.

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

B.

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

C.

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

D.

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

Correct answer is B

4x\(^2\) - 16x + 15 = 0

(2x - 3)(2x - 5) = 0

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

220.

H varies directly as p and inversely as the square of y. If H = 1, p = 8 and y = 2, find H in terms of p and y.

A.

H = \(\frac{p}{4y^2}\)

B.

H = \(\frac{2p}{y^2}\)

C.

H = \(\frac{p}{2y^2}\)

D.

H = \(\frac{p}{y^2}\)

Correct answer is C

H \(\propto\) \(\frac{p}{y^2}\) 

H = \(\frac{pk}{y^2}\) 

1 = \(\frac{8k}{2^2}\)

k = \(\frac{4}{8}\)

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

H = \(\frac{p}{2y^2}\)