WAEC Past Questions and Answers - Page 3066

15,326.

The bearing of a point P from another point Q is 310o. If |PQ| = 200m, how far west of Q is P?

A.

128.6m

B.

153.2m

C.

167.8m

D.

187.9m

Correct answer is B

cos 40 = \(\frac{x}{200}\)

x - 200 x cosx

= 200 x 0.7660

x = 153.2m

15,327.

Q is 32 km away from P on a bearing 042o and R is 25km from P on a bearing of 132o. Calculate the bearing of R from Q.

A.

122o

B.

184o

C.

190o

D.

226o

Correct answer is B

No explanation has been provided for this answer.

15,328.

Divide the sum of 8, 6, 7, 2, 0, 4, 7, 2, 3, by their mean

A.

9

B.

8

C.

7

D.

6

Correct answer is A

Sum = 8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3 = 39

mean = \(\frac{39}{9}\) = 4.33

then \(\frac{sum}{mean}\) = \(\frac{39}{4.33}\)

= 9

15,329.

The table shows the ages(in years) of twenty children chosen at random from a community. What is the median of the distribution? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \\ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)

A.

3.5 years

B.

3.0 years

C.

2.5 years

D.

2.0 years

Correct answer is A

Median = (\(\frac{n + 1}{2})^{th}\)

= \(\frac{20 + 1}{2} = \frac{21}{2}\)

= 10.5

= \(\frac{{\text{10th term and 11th term}}}{2}\)

= \(\frac{3 + 4}{2} = \frac{7}{2}\)

= 3.5yrs

15,330.

The table shows the ages(in years) of twenty children chosen at random from a community. What is the mean age? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \\ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)

A.

4.46 years

B.

3.35 years

C.

3.30years

D.

3.00 years

Correct answer is B

Mean(age) = \(\frac{(1 \times 2) + (2 \times 3) + (3 \times 5) + (4 \times 6) + (5 \times 4)}{2 + 3 + 5 + 6 + 4}\)

= \(\frac{2 + 6 + 15 + 24 + 20}{20}\)

= \(\frac{67}{20}\)

= 3.35yrs