Year : 
2010
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

11 - 20 of 49 Questions

# Question Ans
11.

Solve the inequality: 3(x + 1) \(\leq\) 5(x + 2) + 15

A. x \(\geq\) -14

B. x \(\leq\) - 14

C. x \(\geq\) -11

D. x \(\leq\) - 11

Detailed Solution

3(x + 1) \(\leq\) 5(x + 2) + 15

3x + 3 \(\leq\) 5x + 10 + 15

3x - 5x \(\geq\) 10 + 15 - 3

-2x \(\geq\) 22

x \(\leq\) \(\frac{-22}{2}\)

x \(\leq\) -11
12.

An empty rectangular tank is 250cm long and 120cm wide. If 180 litres of water is poured into the tank. Calculate the height of the water

A. 6.0cm

B. 5.5cm

C. 5.0cm

D. 4.5cm

Detailed Solution

Volume of water in tank = l x b x h = 180 litres

but 1 litre = 1000cm2

250 x 120 x hw = 180 x 1000

hw = \(\frac{180 \times 1000}{250 \times 120}\)

= 6cm
13.

Given that \(\frac{5^{n +3}}{25^{2n -2}}\) = 5o, find n

A. n = 1

B. n = 2

C. n = 3

D. n = 5

Detailed Solution

\(\frac{5^{n +3}}{25^{2n -2}}\) = 5o

\(\frac{5^{n + 3}}{5^{2(2n - 3)}}\) = 5o

n + 3 - 4n + 6 = 0

-3n + 9 = 0

-3n = -9

n = \(\frac{-9}{-3}\)

n = 3
14.

\(\begin{array}{c|c} \text{No. of pets} & 0 & 1 & 2 & 3 & 4 \\ \hline \text{No. of students} & 8 & 4 & 5 & 10 & 3\end{array}\) The table shows the number of pets kept by 30 students in a class. If a student is picked at random ftom the class. What is the probability that he/she kept more than one pet?

A. \(\frac{1}{5}\)

B. \(\frac{2}{5}\)

C. \(\frac{3}{5}\)

D. \(\frac{4}{5}\)

Detailed Solution

pr. (More than one pet)

= \(\frac{\text{No. of students with > 1 pet}}{\text{total no. of students}}\)

= \(\frac{5 + 10 + 3}{30}\)

= \(\frac{18}{30}\)

= \(\frac{3}{5}\)
15.

Simplify 2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)

A. 1

B. \(\frac{1}{3}\sqrt{3}\)

C. 2\(\sqrt{3} - 5\frac{2}{3}\)

D. 6\(\sqrt{3}\) - 17

Detailed Solution

2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)

= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{9 \times 3}}\)

= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} + \frac{3}{3\sqrt{3}}\)

= 2\(\sqrt{3} = 6 \frac{\sqrt{3}}{3} + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= 2\(\sqrt{3} - 2\sqrt{3} + \frac{\sqrt{3}}{3}\)

= \(\frac{\sqrt{3}}{3} = \frac{1}{3} \sqrt{3}\)
16.

If sin 3y = cos 2y and 0o \(\leq\) 90o, find the value of y

A. 18o

B. 36o

C. 54o

D. 90o

Detailed Solution

sin 3y = cos 2y, but sin \(\theta\) = cos(90 - \(\theta\))

sin 3y = cos(90 - 3y)

cos(90 - 3y) = cos 2y

90 - 3y = 2y

5y = 90

y = \(\frac{90}{5}\)

y = 18o
17.

A rectangle has length xcm and width (x - 1)cm. If the perimeter is 16cm. Find the value of x

A. 3\(\frac{1}{2}\)cm

B. 4cm

C. 4\(\frac{1}{2}\)cm

D. 5cm

Detailed Solution

l = x; b = x - 1

perimeter = 2(l + b) = 16

l + b = \(\frac{16}{2}\) = 8

l + b = 8

x + x - 1 = 8

2x = 8 + 1

2x = 9

x = \(\frac{9}{2}\)cm

x = 4\(\frac{1}{2}\)
18.

Given that tan x = 1, where 0o \(\geq\) x 90o, evaluate \(\frac{1 - \sin^2 x}{\cos x}\)

A. 2\(\sqrt{2}\)

B. \(\sqrt{2}\)

C. \(\frac{\sqrt{2}}{2}\)

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

Detailed Solution

Given tan x = 1

x = tan-1(1)

x = 45o

Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)

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

= \(\frac{1}{2} + \frac{1}{\sqrt{2}}\)

= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)

= \(\frac{\sqrt{2}}{2}\)
19.

What is the length of a rectangular garden whose perimeter is 32cm and area 39cm2?

A. 25cm

B. 18cm

C. 13cm

D. 9cm

Detailed Solution

perimeter = 2(l + b) = 32

l + b = \(\frac{32}{2}\)

l + b = 16

b = 16 - 1.......(1)

Area = l + b = 39

lb = 39 .....(2)

put (1) into (2)

l(16 - 1) = 39

16l - l2 = 39

l2 = 13l - 3l + 39 = 0

l(l - 13) - 3(1 - 13) = 0

(l - 3)(l + 130 = 0

l - 3 = 0 or l - 13 = 0

l = 3cm or l = 13cm; The length in 13cm
20.

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

A. \(\frac{4}{4 - x^3}\)

B. \(\frac{8}{4 - x^2}\)

C. \(\frac{4x}{4 - x^2}\)

D. \(\frac{8 - 4x}{4 - x^2}\)

Detailed Solution

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

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

= \(\frac{8}{4 - x62}\)
11.

Solve the inequality: 3(x + 1) \(\leq\) 5(x + 2) + 15

A. x \(\geq\) -14

B. x \(\leq\) - 14

C. x \(\geq\) -11

D. x \(\leq\) - 11

Detailed Solution

3(x + 1) \(\leq\) 5(x + 2) + 15

3x + 3 \(\leq\) 5x + 10 + 15

3x - 5x \(\geq\) 10 + 15 - 3

-2x \(\geq\) 22

x \(\leq\) \(\frac{-22}{2}\)

x \(\leq\) -11
12.

An empty rectangular tank is 250cm long and 120cm wide. If 180 litres of water is poured into the tank. Calculate the height of the water

A. 6.0cm

B. 5.5cm

C. 5.0cm

D. 4.5cm

Detailed Solution

Volume of water in tank = l x b x h = 180 litres

but 1 litre = 1000cm2

250 x 120 x hw = 180 x 1000

hw = \(\frac{180 \times 1000}{250 \times 120}\)

= 6cm
13.

Given that \(\frac{5^{n +3}}{25^{2n -2}}\) = 5o, find n

A. n = 1

B. n = 2

C. n = 3

D. n = 5

Detailed Solution

\(\frac{5^{n +3}}{25^{2n -2}}\) = 5o

\(\frac{5^{n + 3}}{5^{2(2n - 3)}}\) = 5o

n + 3 - 4n + 6 = 0

-3n + 9 = 0

-3n = -9

n = \(\frac{-9}{-3}\)

n = 3
14.

\(\begin{array}{c|c} \text{No. of pets} & 0 & 1 & 2 & 3 & 4 \\ \hline \text{No. of students} & 8 & 4 & 5 & 10 & 3\end{array}\) The table shows the number of pets kept by 30 students in a class. If a student is picked at random ftom the class. What is the probability that he/she kept more than one pet?

A. \(\frac{1}{5}\)

B. \(\frac{2}{5}\)

C. \(\frac{3}{5}\)

D. \(\frac{4}{5}\)

Detailed Solution

pr. (More than one pet)

= \(\frac{\text{No. of students with > 1 pet}}{\text{total no. of students}}\)

= \(\frac{5 + 10 + 3}{30}\)

= \(\frac{18}{30}\)

= \(\frac{3}{5}\)
15.

Simplify 2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)

A. 1

B. \(\frac{1}{3}\sqrt{3}\)

C. 2\(\sqrt{3} - 5\frac{2}{3}\)

D. 6\(\sqrt{3}\) - 17

Detailed Solution

2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)

= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{9 \times 3}}\)

= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} + \frac{3}{3\sqrt{3}}\)

= 2\(\sqrt{3} = 6 \frac{\sqrt{3}}{3} + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= 2\(\sqrt{3} - 2\sqrt{3} + \frac{\sqrt{3}}{3}\)

= \(\frac{\sqrt{3}}{3} = \frac{1}{3} \sqrt{3}\)
16.

If sin 3y = cos 2y and 0o \(\leq\) 90o, find the value of y

A. 18o

B. 36o

C. 54o

D. 90o

Detailed Solution

sin 3y = cos 2y, but sin \(\theta\) = cos(90 - \(\theta\))

sin 3y = cos(90 - 3y)

cos(90 - 3y) = cos 2y

90 - 3y = 2y

5y = 90

y = \(\frac{90}{5}\)

y = 18o
17.

A rectangle has length xcm and width (x - 1)cm. If the perimeter is 16cm. Find the value of x

A. 3\(\frac{1}{2}\)cm

B. 4cm

C. 4\(\frac{1}{2}\)cm

D. 5cm

Detailed Solution

l = x; b = x - 1

perimeter = 2(l + b) = 16

l + b = \(\frac{16}{2}\) = 8

l + b = 8

x + x - 1 = 8

2x = 8 + 1

2x = 9

x = \(\frac{9}{2}\)cm

x = 4\(\frac{1}{2}\)
18.

Given that tan x = 1, where 0o \(\geq\) x 90o, evaluate \(\frac{1 - \sin^2 x}{\cos x}\)

A. 2\(\sqrt{2}\)

B. \(\sqrt{2}\)

C. \(\frac{\sqrt{2}}{2}\)

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

Detailed Solution

Given tan x = 1

x = tan-1(1)

x = 45o

Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)

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

= \(\frac{1}{2} + \frac{1}{\sqrt{2}}\)

= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)

= \(\frac{\sqrt{2}}{2}\)
19.

What is the length of a rectangular garden whose perimeter is 32cm and area 39cm2?

A. 25cm

B. 18cm

C. 13cm

D. 9cm

Detailed Solution

perimeter = 2(l + b) = 32

l + b = \(\frac{32}{2}\)

l + b = 16

b = 16 - 1.......(1)

Area = l + b = 39

lb = 39 .....(2)

put (1) into (2)

l(16 - 1) = 39

16l - l2 = 39

l2 = 13l - 3l + 39 = 0

l(l - 13) - 3(1 - 13) = 0

(l - 3)(l + 130 = 0

l - 3 = 0 or l - 13 = 0

l = 3cm or l = 13cm; The length in 13cm
20.

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

A. \(\frac{4}{4 - x^3}\)

B. \(\frac{8}{4 - x^2}\)

C. \(\frac{4x}{4 - x^2}\)

D. \(\frac{8 - 4x}{4 - x^2}\)

Detailed Solution

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

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

= \(\frac{8}{4 - x62}\)