1995 - WAEC Mathematics Past Questions and Answers - page 5
Without using tables, find the value of \(\frac{\sin 20°}{\cos 70°} + \frac{\cos 25°}{\sin 65°}\)
Note: sin 20° = cos 70° hence
\(\frac{\sin 20}{\cos 70} = 1\)
Also, cos 65° = sin 25°
\(\frac{\cos 65}{\sin 25} = 1\)
\(\therefore \frac{\sin 20}{\cos 70} + \frac{\cos 65}{\sin 25}\)
= 1 + 1
= 2
In the diagram above, ∠PRQ = 90°, ∠QPR = 30° and /PQ/ = 10 cm. Find y.
\(\sin 30° = \frac{y}{10}\)
\(y = 10 \sin 30° = 10 \times 0.5 = 5 cm\)
The bearing of two points Q and R from a point P are 030° and 120° respectively, lf /PQ/ = 12 m and /PR/ = 5 m, find the distance QR.
x\(^2\) = 12\(^2\) + 5\(^2\)
x\(^2\) = 144 + 25
x\(^2\) = 169
x = \(\sqrt{169}\) = 13m.
What is the mode of the numbers 8, 10, 9, 9, 10, 8, 11, 8, 10, 9, 8 and 14?
The mean of 20 observations in an experiment is 4, lf the observed largest value is 23, find the mean of the remaining observations.
Mean of 20 observations = 4
All the observations = 20 x 4 = 80
Largest observed value = 23
Remaining observation = 80 - 23 = 57
Mean of observation = 57/19 = 3
Find the median of the following numbers 2.64, 2.50, 2.72, 2.91 and 2.35.
Arranging the numbers in ascending order, we have
2.35, 2.50, 2.64, 2.72, 2.91.
The median = 2.64
Total balls = 2 + 6 + 5 = 13 balls p(black) = 5/13