Year : 
2018
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

1 - 10 of 50 Questions

# Question Ans
1.

Simplify: \(\sqrt{108} + \sqrt{125} - \sqrt{75}\)

A. \(\sqrt{3} + 5\sqrt{5}\)

B. \(6 \sqrt{3} - 5 \sqrt{5}\)

C. \(6 \sqrt{3} + \sqrt{2}\)

D. \(6\sqrt{3} - \sqrt{2}\)

Detailed Solution

\(\sqrt{108} + \sqrt{125} - \sqrt{75}\)
= \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\)
= \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\)
= \(\sqrt{3} + 5\sqrt{5}\)
2.

Evaluate: \((64^{\frac{1}{2}} + 125^{\frac{1}{3}})^2\)

A. 121

B. 144

C. 169

D. 196

Detailed Solution

\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\)
\([8 + 5]^2\) = \([13]^2\)
= 169
3.

Given that y varies inversely as the square of x. If x = 3 when y = 100, find the equation connecting x and y.

A. \(yx^2 = 300\)

B. \(yx^2 = 900\)

C. y = \(\frac{100x}{9}\)

D. \(y = 900x^2\)

Detailed Solution

Y \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\)
If x = 3 and y = 100,
then, \(\frac{100}{1} = \frac{k}{3^2}\)
\(\frac{100}{1} = \frac{k}{9}\)
k = 100 x 9 = 900
Substitute 900 for k in
y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\)
= \(yx^2 = 900\)
4.

Find the value of x for which \(32_{four} = 22_x\)

A. three

B. five

C. six

D. seven

Detailed Solution

\(32_4 = 22_x\)
\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\)
12 + 2 x 1 = 2x + 2 x 1
14 = 2x + 2
14 - 2 = 2x
12 = 2x
x = \(\frac{12}{2}\)
x = 6
5.

Simplify; 2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\)

A. \(\frac{5}{9}\)

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

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

D. 1\(\frac{4}{5}\)

6.

There are 250 boys and 150 girls in a school, if 60% of the boys and 40% of the girls play football, what percentage of the school play football?

A. 40.0%

B. 42.2%

C. 50.0%

D. 52.5%

Detailed Solution

Population of school = 250 + 150 = 400
60% of 250 = \(\frac{\text{60%}}{\text{100%}}\) x 250 = 150
40% of 150 = \(\frac{\text{40%}}{\text{100%}}\) x 150 = 60
Total number of students who plays football;
150 + 60 = 210
Percentage of school that play football;
\(\frac{210}{400}\) x 100% = 52.5%
7.

If \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1, solve for x.

A. 2

B. 3

C. 4

D. 5

Detailed Solution

\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1
\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10
\(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10
\(\frac{6x - 4}{2}\) = 10
6x - 4 = 2 x 10
= 20
6x = 20 + 4
6x = 20
x = \(\frac{24}{6}\)
x = 4
8.

If F = \(\frac{9}{5}\)C + 32, find C when F = 98.6

A. 30

B. 37

C. 39

D. 41

Detailed Solution

F = \(\frac{9}{5}\)C + 32
When F = 98.6
98.6 = \(\frac{9}{5}\)C + 32
98.6 - 32 = \(\frac{9}{5}\)C
66.6= \(\frac{9}{5}\)C
66.6 x 5 = 9C
C = \(\frac{66.6 \times 5}{9}\)
= 37
9.

If y + 2x = 4 and y - 3x = -1, find the value of (x + y)

A. 3

B. 2

C. 1

D. -1

Detailed Solution

y + 2x = 4 .....(1)
y - 3x = -1 ......(2)
Subtract (2) from (1)
2x - (-3x) = 4 - (-1)
2x + 3x = 4 + 1
5x = 5
X = \(\frac{5}{5}\)
= 1
Substitute 1 for x in (1);
y + 2(1) = 4
y + 2 = 4
y = 4 - 2 = 2
Hence, (x + y) = (1 + 2)
= 3
10.

If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6z - 2y}\)

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

B. 2

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

D. 3

Detailed Solution

If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\)
\(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\)
Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\)
y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\)
Using y = y
\(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\)
\(\frac{T_1}{T_2}\) = \(\frac{15}{21}\)
\(T_1\) = 15 and \(T_2\) = 21
Therefore;
x = \(\frac{2}{5}\) x 15 = 6
y = \(\frac{3}{5}\) x 15 = 9
y = \(\frac{3}{7}\) x 21 = 9 (again)
z = \(\frac{4}{7}\) x 21 = 12
Hence;
\(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) -
1.

Simplify: \(\sqrt{108} + \sqrt{125} - \sqrt{75}\)

A. \(\sqrt{3} + 5\sqrt{5}\)

B. \(6 \sqrt{3} - 5 \sqrt{5}\)

C. \(6 \sqrt{3} + \sqrt{2}\)

D. \(6\sqrt{3} - \sqrt{2}\)

Detailed Solution

\(\sqrt{108} + \sqrt{125} - \sqrt{75}\)
= \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\)
= \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\)
= \(\sqrt{3} + 5\sqrt{5}\)
2.

Evaluate: \((64^{\frac{1}{2}} + 125^{\frac{1}{3}})^2\)

A. 121

B. 144

C. 169

D. 196

Detailed Solution

\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\)
\([8 + 5]^2\) = \([13]^2\)
= 169
3.

Given that y varies inversely as the square of x. If x = 3 when y = 100, find the equation connecting x and y.

A. \(yx^2 = 300\)

B. \(yx^2 = 900\)

C. y = \(\frac{100x}{9}\)

D. \(y = 900x^2\)

Detailed Solution

Y \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\)
If x = 3 and y = 100,
then, \(\frac{100}{1} = \frac{k}{3^2}\)
\(\frac{100}{1} = \frac{k}{9}\)
k = 100 x 9 = 900
Substitute 900 for k in
y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\)
= \(yx^2 = 900\)
4.

Find the value of x for which \(32_{four} = 22_x\)

A. three

B. five

C. six

D. seven

Detailed Solution

\(32_4 = 22_x\)
\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\)
12 + 2 x 1 = 2x + 2 x 1
14 = 2x + 2
14 - 2 = 2x
12 = 2x
x = \(\frac{12}{2}\)
x = 6
5.

Simplify; 2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\)

A. \(\frac{5}{9}\)

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

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

D. 1\(\frac{4}{5}\)

6.

There are 250 boys and 150 girls in a school, if 60% of the boys and 40% of the girls play football, what percentage of the school play football?

A. 40.0%

B. 42.2%

C. 50.0%

D. 52.5%

Detailed Solution

Population of school = 250 + 150 = 400
60% of 250 = \(\frac{\text{60%}}{\text{100%}}\) x 250 = 150
40% of 150 = \(\frac{\text{40%}}{\text{100%}}\) x 150 = 60
Total number of students who plays football;
150 + 60 = 210
Percentage of school that play football;
\(\frac{210}{400}\) x 100% = 52.5%
7.

If \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1, solve for x.

A. 2

B. 3

C. 4

D. 5

Detailed Solution

\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1
\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10
\(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10
\(\frac{6x - 4}{2}\) = 10
6x - 4 = 2 x 10
= 20
6x = 20 + 4
6x = 20
x = \(\frac{24}{6}\)
x = 4
8.

If F = \(\frac{9}{5}\)C + 32, find C when F = 98.6

A. 30

B. 37

C. 39

D. 41

Detailed Solution

F = \(\frac{9}{5}\)C + 32
When F = 98.6
98.6 = \(\frac{9}{5}\)C + 32
98.6 - 32 = \(\frac{9}{5}\)C
66.6= \(\frac{9}{5}\)C
66.6 x 5 = 9C
C = \(\frac{66.6 \times 5}{9}\)
= 37
9.

If y + 2x = 4 and y - 3x = -1, find the value of (x + y)

A. 3

B. 2

C. 1

D. -1

Detailed Solution

y + 2x = 4 .....(1)
y - 3x = -1 ......(2)
Subtract (2) from (1)
2x - (-3x) = 4 - (-1)
2x + 3x = 4 + 1
5x = 5
X = \(\frac{5}{5}\)
= 1
Substitute 1 for x in (1);
y + 2(1) = 4
y + 2 = 4
y = 4 - 2 = 2
Hence, (x + y) = (1 + 2)
= 3
10.

If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6z - 2y}\)

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

B. 2

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

D. 3

Detailed Solution

If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\)
\(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\)
Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\)
y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\)
Using y = y
\(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\)
\(\frac{T_1}{T_2}\) = \(\frac{15}{21}\)
\(T_1\) = 15 and \(T_2\) = 21
Therefore;
x = \(\frac{2}{5}\) x 15 = 6
y = \(\frac{3}{5}\) x 15 = 9
y = \(\frac{3}{7}\) x 21 = 9 (again)
z = \(\frac{4}{7}\) x 21 = 12
Hence;
\(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) -