Year : 
2007
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

51 - 60 of 67 Questions

# Question Ans
51.

Find the smallest value of k such that 2\(^2\) x 3\(^3\) x 5 x k is a perfect square.

A. 3

B. 5

C. 15

D. 30

Detailed Solution

2\(^2\) x 3\(^3\) x 5\(^1\) x k;
2\(^2\) x 3\(^2\) x 3 x 5 x k
2\(^2\) x 3\(^2\) x 15 x k
smallest value for k
2\(^2\) x 3\(^2\) x 15 = 2\(^2\) x 3\(^2\) x 15\(^2\)
k = 15
52.

Expand the expression(3a - xy)(3a + xy)

A. 9a2 - x2y2

B. 9a2 + x2y2

C. 9a2 - xy

D. 9a2 + x2y

Detailed Solution

(3a - xy)(3a + xy); (3a)2 - (xy)2

difference of two sqs; 9a2 - x2y2
53.

Simplify \(\frac{4}{2x} - \frac{2x + x}{x}\)

A. -1

B. -2x

C. 2x

D. \(\frac{2 - x}{2x}\)

Detailed Solution

\(\frac{4}{2x} - \frac{2 + x}{x} = \frac{4 - 2(2 + x)}{2x}\)

= \(\frac{4 - 4 - 2x}{2x} = \frac{-2x}{2x}\)

= 1
54.

Two circles have radii 16cm and 23cm. What is the difference between their circumference? take \(\pi = \frac{22}{7}\)

A. 422.92cm

B. 149.92cm

C. 44.00cm

D. 43.96cm

Detailed Solution

r = 16, R = 23; 2\(\pi R - 2 \pi r\)

= 2\(\pi(R - r)\)

= 2 x \(\frac{22}{7} (23 - 16)\)

= 2 x \(\frac{22}{7} \times (7)\)

= 44cm
55.

The volume of a cylinder is 1200cm3 and the area of its base is 150cm2. Find the height of the cylinder.

A. 80.00cm

B. 8.00cm

C. 0.80cm

D. 0.08cm

Detailed Solution

Vol. of cylinder = \(\pi r^2h\) = 1200cm2

Area of base = \(\pi^2\) = 150cm2

h = \(\frac{\pi r^2}{\pi r^2} = \frac{1200}{150}\)

= 8.00cm
56.

In the diagram |LN| = 4cm, LNM = 90o and tan y = \(\frac{2}{3}\). What is the area of the \(\bigtriangleup\)LMN?

A. 24cm2

B. 12cm2

C. 10cm2

D. 6cm2

Detailed Solution

Since tan y = \(\frac{2}{3}\) and LN = 4

tan y = \(\frac{2 \times 4}{3 \times 4} = \frac{8}{12} = \frac{4}{6}\)

then tan y = \(\frac{opp}{adj}\)

MN = 6cm

Area of angle LMN = \(\frac{1}{2}\)bh

= \(\frac{1}{2} \times 6 \times 4\)

= 12cm3
57.

From the diagram, find the value of x in the diagram.

A. 80o

B. 70o

C. 55o

D. 35o

Detailed Solution

y + 110o = 180o(angles on a straight line)

y = 180 - 110

= 70o

x = y(corresponding angles)

x = 70o
58.

In the diagram, |SP| = |SR| and < PRS = 50o. Calculate < PQR

A. 120o

B. 110o

C. 100o

D. 80o

Detailed Solution

< PRS = RPS = 50o (base of isosceles)

RPS = 50o; < RSP + < PRS + < RPS = 180(sum of < s in a triangle)

< RPS + 50 + 50 = 180

< RSP = 180 - 100 = 80

then < PQR + < RSp = 180 (opp. < S of cyclic quad.)

< PQR + 80 = 180o

< PQR = 180 - 80

= 100
59.

Find the value of x in the diagram

A. 281o

B. 269o

C. 201o

D. 179o

Detailed Solution

a = 34o (alternate angles)

b = 45o (alternate ngles)

a + b = 34 + 45 = 79o

x + 79 = 360(sum of angles at a point)

x = 360 - 79

x = 281o
60.

The venn diagram shows the choice of food of a number of visitors to a canteen. If there were 35 visitors in all, find the value of x

A. 5

B. 4

C. 3

D. 2

Detailed Solution

5 + 6 + 4 + 8 + 5 + 2 + x = 35

30 + x = 35

x = 35 - 30

x = 5
51.

Find the smallest value of k such that 2\(^2\) x 3\(^3\) x 5 x k is a perfect square.

A. 3

B. 5

C. 15

D. 30

Detailed Solution

2\(^2\) x 3\(^3\) x 5\(^1\) x k;
2\(^2\) x 3\(^2\) x 3 x 5 x k
2\(^2\) x 3\(^2\) x 15 x k
smallest value for k
2\(^2\) x 3\(^2\) x 15 = 2\(^2\) x 3\(^2\) x 15\(^2\)
k = 15
52.

Expand the expression(3a - xy)(3a + xy)

A. 9a2 - x2y2

B. 9a2 + x2y2

C. 9a2 - xy

D. 9a2 + x2y

Detailed Solution

(3a - xy)(3a + xy); (3a)2 - (xy)2

difference of two sqs; 9a2 - x2y2
53.

Simplify \(\frac{4}{2x} - \frac{2x + x}{x}\)

A. -1

B. -2x

C. 2x

D. \(\frac{2 - x}{2x}\)

Detailed Solution

\(\frac{4}{2x} - \frac{2 + x}{x} = \frac{4 - 2(2 + x)}{2x}\)

= \(\frac{4 - 4 - 2x}{2x} = \frac{-2x}{2x}\)

= 1
54.

Two circles have radii 16cm and 23cm. What is the difference between their circumference? take \(\pi = \frac{22}{7}\)

A. 422.92cm

B. 149.92cm

C. 44.00cm

D. 43.96cm

Detailed Solution

r = 16, R = 23; 2\(\pi R - 2 \pi r\)

= 2\(\pi(R - r)\)

= 2 x \(\frac{22}{7} (23 - 16)\)

= 2 x \(\frac{22}{7} \times (7)\)

= 44cm
55.

The volume of a cylinder is 1200cm3 and the area of its base is 150cm2. Find the height of the cylinder.

A. 80.00cm

B. 8.00cm

C. 0.80cm

D. 0.08cm

Detailed Solution

Vol. of cylinder = \(\pi r^2h\) = 1200cm2

Area of base = \(\pi^2\) = 150cm2

h = \(\frac{\pi r^2}{\pi r^2} = \frac{1200}{150}\)

= 8.00cm
56.

In the diagram |LN| = 4cm, LNM = 90o and tan y = \(\frac{2}{3}\). What is the area of the \(\bigtriangleup\)LMN?

A. 24cm2

B. 12cm2

C. 10cm2

D. 6cm2

Detailed Solution

Since tan y = \(\frac{2}{3}\) and LN = 4

tan y = \(\frac{2 \times 4}{3 \times 4} = \frac{8}{12} = \frac{4}{6}\)

then tan y = \(\frac{opp}{adj}\)

MN = 6cm

Area of angle LMN = \(\frac{1}{2}\)bh

= \(\frac{1}{2} \times 6 \times 4\)

= 12cm3
57.

From the diagram, find the value of x in the diagram.

A. 80o

B. 70o

C. 55o

D. 35o

Detailed Solution

y + 110o = 180o(angles on a straight line)

y = 180 - 110

= 70o

x = y(corresponding angles)

x = 70o
58.

In the diagram, |SP| = |SR| and < PRS = 50o. Calculate < PQR

A. 120o

B. 110o

C. 100o

D. 80o

Detailed Solution

< PRS = RPS = 50o (base of isosceles)

RPS = 50o; < RSP + < PRS + < RPS = 180(sum of < s in a triangle)

< RPS + 50 + 50 = 180

< RSP = 180 - 100 = 80

then < PQR + < RSp = 180 (opp. < S of cyclic quad.)

< PQR + 80 = 180o

< PQR = 180 - 80

= 100
59.

Find the value of x in the diagram

A. 281o

B. 269o

C. 201o

D. 179o

Detailed Solution

a = 34o (alternate angles)

b = 45o (alternate ngles)

a + b = 34 + 45 = 79o

x + 79 = 360(sum of angles at a point)

x = 360 - 79

x = 281o
60.

The venn diagram shows the choice of food of a number of visitors to a canteen. If there were 35 visitors in all, find the value of x

A. 5

B. 4

C. 3

D. 2

Detailed Solution

5 + 6 + 4 + 8 + 5 + 2 + x = 35

30 + x = 35

x = 35 - 30

x = 5