WAEC Mathematics Past Questions & Answers - Page 126

626.

In an athletic composition, the probability that an athlete wins a 100m race is \(\frac{1}{8}\) and the probability that he wins in high jump is \(\frac{1}{4}\). What is the probability that he wins only one of the events?

A.

\(\frac{3}{32}\)

B.

\(\frac{7}{3}\)

C.

\(\frac{5}{3}\)

D.

\(\frac{5}{16}\)

Correct answer is D

Pr. (winning 100m race) = \(\frac{1}{8}\)

Pr. (losing 100m race) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)

Pr. (winning high jump) = \(\frac{1}{4}\)

Pr. (losing high jump ) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)

Pr. (winning only one) = Pr. (Winning 100m race and losing high jump) or Pr.(Losing 100m race and winning high jump)

= (\(\frac{1}{8} \times \frac{3}{4}\)) + (\(\frac{7}{8} \times \frac{1}{4}\))

= \(\frac{3}{32} + \frac{7}{32}\)

= \(\frac{10}{32}\)

= \(\frac{5}{16}\)

627.

The mean of the numbers 2, 5, 2x and 7 is less than or equal to 5. Find the range of the values of x

A.

x \(\leq\) 3

B.

x \(\geq\) 3

C.

x < 3

D.

x > 3

Correct answer is A

mean \(\leq\) 5; \(\frac{2 + 5 + 2x + 7}{4}\) \(\leq\) 5

= \(\frac{14 + 2x}{4} \leq 5\)

= 14 + 2x \(\leq\) 5 x 4

14 + 2x \(\leq\) 20 ; 2x \(\leq\) 20 - 14

2x \(\leq\) 20 - 14

2x \(\leq\) 6

x \(\leq\) \(\frac{6}{2}\)

x \(\leq\) 3

628.

If x km/h = y m/s, then y =

A.

\(\frac{7}{18}\)x

B.

\(\frac{11}{20}\)x

C.

\(\frac{4}{15}\)x

D.

\(\frac{5}{18}\)x

Correct answer is D

x kmh-1 = y ms-1

\(\frac{x km}{1 hr}\) = y ms-1

\(x \times \frac{1km}{1hr}\) = y ms-1

\(x \times \frac{1000m}{60 \times 60s}\) = y ms-1

\(x \times \frac{1000}{3600} \frac{m}{s}\) = y ms-1

\(x \times \frac{5}{18} ms^{-1}\)

\(x \times \frac{5}{18} ms^{-1}\) = y ms-1

y = \(\frac{5}{18}\)x

629.

If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k

A.

\(\frac{8}{3}\)

B.

\(\frac{7}{3}\)

C.

\(\frac{5}{3}\)

D.

\(\frac{2}{3}\)

Correct answer is A

x2 + kx + \(\frac{16}{9}\); Perfect square

But, b2 - 4ac = 0, for a perfect square

where a - 1; b = k; c = \(\frac{16}{9}\)

k2 - 4(1) x \(\frac{16}{9}\) = 0

k2 - \(\frac{64}{9}\) = 0

k2 = \(\frac{64}{9}\)

k = \(\sqrt{\frac{64}{9}}\)

k = \(\frac{8}{3}\)

630.

If the sum of the roots of the equation (x - p)(2x + 1) = 0 is 1, find the value of x

A.

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

B.

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

C.

-\(\frac{3}{2}\)

D.

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

Correct answer is A

(x - p)(2x + 1) = 0

2x2 + x - 2px - p = 0

2x2 + x (1 - 2p) - p = 0

2x2 - (2p - 1)x - p = 0

divide through by 2

x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0

compare to x2 - (sum of roots)x + product of roots = 0

sum of roots = \(\frac{2p - 1}{2}\)

But sum of roots = 1

Given; \(\frac{2p - 1}{2}\) = 1

2p - 1 = 2 x 1

2p - 1 = 2

2p = 2 + 1 = 3

p = \(\frac{3}{2}\)

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