WAEC Mathematics Past Questions & Answers - Page 144

716.

In the diagram, /PQ/ = /QR/ and /PR/ = /RS/ = /SP/, calculate the side of < QRS

150^o

120^o

90^o

60^o

View answer & explanation

Correct answer is C

Since |PR| = |RS| = |SP|

\(\bigtriangleup\) PRS is equilateral and so < RPS = < PRS = < PSR = 70^o

But < PQR + < PSR = 180^o (Opposite interior angles of a cyclic quadrilateral)

< PQR + 60 = 180^o

< PR = 180 - 60 = 120^o

But in \(\bigtriangleup\) PQR, PQ = PR, hence < QPR = < PRQ(Base angles of an Isosceles triangle)

< QPR + < PRQ + < PQR = 180^o (Angles in a triangle)

2 < QPR + 120 = 18-

2 < QPR = 180 - 120

QPR = \(\frac{60}{2}\) = 30^o

From the diagram, < QRS = < PRQ + < PRS

30 + 60 = 90^o

717.

Given the sets A = [2, 4, 6, 8] and B = [2, 3, 5, 9]. If a number is picked at random from each of the two sets, what is the probability that their difference is 6 or 7?

\(\frac{1}{256}\)

\(\frac{1}{16}\)

\(\frac{1}{8}\)

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

View answer & explanation

Correct answer is C

\(\begin{array}{c|c} - & 2 & 3 & 5 & 9 \\ \hline 2 & 0 & 1 & 3 & 7 \\ \hline 4 & 2 & 1 & 1 & 5\\ \hline 6 & 4 & 3 & 1 & 3 \\ \hline 8 & 6 &5 & 3 & 1 \end{array}\)

Note: A {horizontal}

B {vertical}

Pr(Difference of 6 or 7) = \(\frac{2}{16} = \frac{1}{8}\)

718.

Given the sets A = [2, 4, 6, 8] and B = [2, 3, 5, 9]. If a number is picked at random from each of the two sets, what is the probability that their difference is odd?

\(\frac{3}{4}\)

\(\frac{1}{4}\)

zero

View answer & explanation

Correct answer is D

\(\begin{array}{c|c} x & 2 & 3 & 5 & 9 \\ \hline 2 & 4 & 6 & 10 & 18 \\ \hline 4 & 8 & 12 & 20 & 36 \\ \hline 6 & 12 & 18 & 30 & 54 \\ \hline 8 & 16 & 24 & 40 & 72 \end{array}\)

Note: A {horizontal}

B {vertical}

Pr (Odd Product) = \(\frac{0}{16}\)

= 0

719.

Given the sets A = [2, 4, 6, 8] and B = [2, 3, 5, 9]. If a number is selected at random from set B, what is the probability that the number is prime?

\(\frac{3}{4}\)

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

\(\frac{1}{4}\)

View answer & explanation

Correct answer is B

A = [2, 4, 6, 8}

B = {2, 3, 5, 9}

Pr = (Prime in B) = \(\frac{3}{4}\)

720.

Solve the inequality 1 - 2x < - \(\frac{1}{3}\)

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

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

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

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

View answer & explanation

Correct answer is C

1 - 2x < - \(\frac{1}{3}\); -2x < -\(\frac{1}{3}\) - 1

-2x < - \(\frac{1- 3}{3}\)

-2x < - \(\frac{4}{-6}\)

3x -2x < -4; -8x < -4

x > -\(\frac{4}{-6}\) = x > \(\frac{2}{3}\)