Year : 
2008
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

31 - 40 of 45 Questions

# Question Ans
31.

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?

A. 1

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

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

D. zero

Detailed Solution

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

Note: A {horizontal}

B {vertical}

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

= 0
32.

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?

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

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

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

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

Detailed Solution

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

Note: A {horizontal}

B {vertical}

Pr(Difference of 6 or 7) = \(\frac{2}{16} = \frac{1}{8}\)
33.

Find the mean deviation of these numbers 10, 12, 14, 15, 17, 19.

A. 2.5

B. 2.6

C. 2.7

D. 2.8

Detailed Solution

\(\begin{array}{c|c} x & d = x - \bar{x} & |d| \\ \hline 10 & -4.5 & 4.5 \\ 12 & -2.5 & 2.5\\ 14 & -0.5 & 0.5\\ 15 & 0.5 & 0.5\\ 17 & 2.5 & 2.5\\ 19 & 4.5 & 4.5\\ \hline \sum x = 87 & & \sum|d| = 15\end{array}\)

\(\bar{x} = \frac{\sum x}{n}\)

= \(\frac{87}{6}\)

= 14.5
34.

In the diagram, < QPR = 90o. If q2 = 25 - r2. Find the value of p

A. 3

B. 4

C. 5

D. 6

Detailed Solution

Given q2 = 25 - r2.....(1)

from Pythagoras theorem

p2 = q2 + r2....(2)

put(1) into (2)

p2 = 25 - r2 + r2

p2 = 25

p = \(\sqrt{25}\)
35.

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

A. 150o

B. 120o

C. 90o

D. 60o

Detailed Solution

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

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

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

< PQR + 60 = 180o

< PR = 180 - 60 = 120o

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

< QPR +
36.

In the diagram, PQRS is a rhombus. /PR/ = 10cm and /QS/ = 24cm. Calculate the perimeter of the rhombus

A. 34cm

B. 52cm

C. 56cm

D. 96cm

Detailed Solution

For a rhombus, all the sides are equal and diagonals bisect each other at 90o. Hence, the triangles formed are congruent: (under RHS).

Thus: \(\bar{PS}^2\) = 52 + 122

\(\bar{PS} = \sqrt{25 + 144}\)

= \(\sqrt{169}\)

\(\bar{PS}\) = 13cm

perimeter = 4 x length of a side

= 4 x 13cm

= 52cm
37.

In the figure shown, PQs is a straight line. What is the value of < PRQ?

A. 128o

B. 108o

C. 98o

D. 78o

Detailed Solution

< QPR + < PRQ = < RQS

(Sum of two interior angles of a triangle = Opposite exterior angles)

70o + < PRQ = 148

< PRQ = 148o - 70o

= 78o
38.

In the diagram, PQ//RS, QU//PT and < PSR = 42o. Find angle x.

A. 84o

B. 48o

C. 42o

D. 32o

Detailed Solution

From the diagram, < QPS = xo (Corresponding angles)

Also, < QPS = < PSR(Alternate angles)

x = 42o
39.

In the figure /PX/ = /XQ/, PQ//YZ and XV//QR. What is the ratio of the area of XYZQ to te area of \(\bigtriangleup\)YZR?

A. 1:2

B. 2:1

C. 1:2

D. 3:1

Detailed Solution

From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A1, the area of \(\bigtriangleup\)PXY

= Area of \(\bigtriangleup\)YZR = A2

Area of \(\bigtriangleup\)PQR = A = A1 + 2A2

But from similarity of triangles

\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)

\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)

A = 4A2 But, A = A1 + 2A

A<
40.

In the diagram, O is the centre of the circle and PQRS is a cyclic quadrilateral. Find the value of x.

A. 25o

B. 65o

C. 115o

D. 130o

Detailed Solution

x = 65o (An interior angle of a cyclic quadrilateral = opposite exterior angle).
31.

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?

A. 1

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

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

D. zero

Detailed Solution

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

Note: A {horizontal}

B {vertical}

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

= 0
32.

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?

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

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

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

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

Detailed Solution

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

Note: A {horizontal}

B {vertical}

Pr(Difference of 6 or 7) = \(\frac{2}{16} = \frac{1}{8}\)
33.

Find the mean deviation of these numbers 10, 12, 14, 15, 17, 19.

A. 2.5

B. 2.6

C. 2.7

D. 2.8

Detailed Solution

\(\begin{array}{c|c} x & d = x - \bar{x} & |d| \\ \hline 10 & -4.5 & 4.5 \\ 12 & -2.5 & 2.5\\ 14 & -0.5 & 0.5\\ 15 & 0.5 & 0.5\\ 17 & 2.5 & 2.5\\ 19 & 4.5 & 4.5\\ \hline \sum x = 87 & & \sum|d| = 15\end{array}\)

\(\bar{x} = \frac{\sum x}{n}\)

= \(\frac{87}{6}\)

= 14.5
34.

In the diagram, < QPR = 90o. If q2 = 25 - r2. Find the value of p

A. 3

B. 4

C. 5

D. 6

Detailed Solution

Given q2 = 25 - r2.....(1)

from Pythagoras theorem

p2 = q2 + r2....(2)

put(1) into (2)

p2 = 25 - r2 + r2

p2 = 25

p = \(\sqrt{25}\)
35.

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

A. 150o

B. 120o

C. 90o

D. 60o

Detailed Solution

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

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

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

< PQR + 60 = 180o

< PR = 180 - 60 = 120o

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

< QPR +
36.

In the diagram, PQRS is a rhombus. /PR/ = 10cm and /QS/ = 24cm. Calculate the perimeter of the rhombus

A. 34cm

B. 52cm

C. 56cm

D. 96cm

Detailed Solution

For a rhombus, all the sides are equal and diagonals bisect each other at 90o. Hence, the triangles formed are congruent: (under RHS).

Thus: \(\bar{PS}^2\) = 52 + 122

\(\bar{PS} = \sqrt{25 + 144}\)

= \(\sqrt{169}\)

\(\bar{PS}\) = 13cm

perimeter = 4 x length of a side

= 4 x 13cm

= 52cm
37.

In the figure shown, PQs is a straight line. What is the value of < PRQ?

A. 128o

B. 108o

C. 98o

D. 78o

Detailed Solution

< QPR + < PRQ = < RQS

(Sum of two interior angles of a triangle = Opposite exterior angles)

70o + < PRQ = 148

< PRQ = 148o - 70o

= 78o
38.

In the diagram, PQ//RS, QU//PT and < PSR = 42o. Find angle x.

A. 84o

B. 48o

C. 42o

D. 32o

Detailed Solution

From the diagram, < QPS = xo (Corresponding angles)

Also, < QPS = < PSR(Alternate angles)

x = 42o
39.

In the figure /PX/ = /XQ/, PQ//YZ and XV//QR. What is the ratio of the area of XYZQ to te area of \(\bigtriangleup\)YZR?

A. 1:2

B. 2:1

C. 1:2

D. 3:1

Detailed Solution

From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A1, the area of \(\bigtriangleup\)PXY

= Area of \(\bigtriangleup\)YZR = A2

Area of \(\bigtriangleup\)PQR = A = A1 + 2A2

But from similarity of triangles

\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)

\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)

A = 4A2 But, A = A1 + 2A

A<
40.

In the diagram, O is the centre of the circle and PQRS is a cyclic quadrilateral. Find the value of x.

A. 25o

B. 65o

C. 115o

D. 130o

Detailed Solution

x = 65o (An interior angle of a cyclic quadrilateral = opposite exterior angle).