## Mathematics Study Mode

Try this quiz in CBT Mode

https://quizzerweb.com.ng/quiz?q=Mathematics&id=349

Question 1 Mathematics
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.

$$\sqrt{108} + \sqrt{125} - \sqrt{75}$$

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

= $$6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}$$

= $$\sqrt{3} + 5\sqrt{5}$$

Question 2 Mathematics
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




Question 3 Mathematics
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 = $$\frac{900}{x^2}$$

= $$yx^2 = 900$$

Question 4 Mathematics
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

Question 5 Mathematics
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 $$\frac{4}{5}$$

Question 6 Mathematics
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%

Question 7 Mathematics
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

Question 8 Mathematics
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

Question 9 Mathematics
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

Question 10 Mathematics
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$$\frac{1}{2}$$

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