Mathematics | Study Mode

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Question 1
A
\(\sqrt{3} + 5\sqrt{5}\) 
B
\(6 \sqrt{3} - 5 \sqrt{5}\) 
C
\(6 \sqrt{3} + \sqrt{2}\) 
D
\(6\sqrt{3} - \sqrt{2}\) 
Explanation.

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

= \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\)

= \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\)

= \(\sqrt{3} + 5\sqrt{5}\)


Correct Option:
A
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Question 2
A
121 
B
144 
C
169 
D
196 
Explanation.

\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\)

\([8 + 5]^2\) = \([13]^2\)

= 169




Correct Option:
C
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Question 3
A
\(yx^2 = 300\) 
B
\(yx^2 = 900\) 
C
y = \(\frac{100x}{9}\) 
D
\(y = 900x^2\) 
Explanation.

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 = \(\frac{900}{x^2}\)

= \(yx^2 = 900\)


Correct Option:
B
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Question 4
A
three 
B
five 
C
six 
D
seven 
Explanation.

\(32_4 = 22_x\)

\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\)

12 + 2 x 1 = 2x + 2 x 1

14 = 2x + 2

14 - 2 = 2x

12 = 2x

x = \(\frac{12}{2}\)

x = 6


Correct Option:
C
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Question 5
A

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

 
B

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

 
C

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

 
D

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

 
Explanation.

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

= \(\frac{9}{4} \times \frac{7}{2} \div \frac{35}{8}\)

= \(\frac{9}{4} \times \frac{7}{2} \div \frac{8}{35}\)

= \(\frac{9}{5}\)

= 1 \(\frac{4}{5}\)


Correct Option:
D
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Question 6
A
40.0% 
B
42.2% 
C
50.0% 
D
52.5% 
Explanation.

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%


Correct Option:
C
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Question 7
A
B
C
D
Explanation.

\(\log_{10}\)(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


Correct Option:
C
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Question 8
A
30 
B
37 
C
39 
D
41 
Explanation.

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


Correct Option:
B
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Question 9
A
B
C
D
-1 
Explanation.

y + 2x = 4 .....(1)

9 - 3x = -1 ......(2)

Substract (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


Correct Option:
A
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Question 10
A
1\(\frac{1}{2}\) 
B
C
2\(\frac{1}{2}\) 
D
Explanation.

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;

\(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) - 2(9)}\)

\(\frac{54 + 27}{72 - 18}\) = \(\frac{81}{54}\) = \(\frac{3}{2}\)

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


Correct Option:
A

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