Mathematics 2018 | Study Mode

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Question 11
A

y = \(x^2 - 5x - 6\)

B

y = \(x^2 + 5x - 6\)

C

y = \(x^2 + x - 6\)

D

y = \(x^2 - x - 6\)

Explanation

Since the curve cuts the x-axis at x = -2 and x = 3,

(x + 2)(x - 3) = 0

\(x^2 - 3x + 2x - 6\) = 0

\(x^2 - x - 6\) = 0

Hence, the equation of the curve is

y = \(x^2 - x - 6\)


Correct option: D
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Question 12
A
14m
B
10m
C
7m
D
5m
Explanation

Volume of a cylinder = \( \pi r^2\)h

385 = \(\frac{22}{7}\) x \(r^2\) x 10

385 x 7 = 22 x \(r^2\) x 10

\(r^2\) = \(\frac{385 \times 7}{22 \times 10}\)

= 12.25

r = \(\sqrt{12.25}\)

= 3.5m

Hence, diameter of tank = 2r

= 2 x 3.5 = 7m


Correct option: C
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Question 13
A
\(45^o\)
B
\(84^o\)
C
\(85^o\)
D
\(95^o\)
Explanation

x + 2x + 2x + (x + \(30^o\)) + (x + \(20^o\)) + (x - \(10^o\)) = (2n - 4) x \(90^o\)

8x + 50 \(^o\) - 10\(^o\) = (2 x 6 -4) x 90\(^o\)

8x + 40\(^o\) = 8 x 90\(^o\) = 720\(^o\)

8x = 720\(^o\) - 40\(^o\) = 680\(^o\)

x = \(\frac{680^o}{8}\)

= 85\(^o\)


Correct option: C
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Question 14
A
8 units
B
11 units
C
15 units
D
17 units
Explanation

|MN| = \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\)

= \(\sqrt{(-3 -5)^2 + (8 - 7)^2}\)

= \(\sqrt{(-8)^2 + (8 + 7)^2}\)

= \(\sqrt{64 + (15)^2}\)

= \(\sqrt{64 + 225}\)

= \(\sqrt{289}\)

= 17 units


Correct option: D
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Question 15
A
1
B
2
C
3
D
9
Explanation

Using the two - point from

\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)

\(\frac{y - 2}{-2 - 2} = \frac{x - 4}{-8 - 4}\)

\(\frac{y - 2}{-4} = \frac{x - 4}{-12}\)

\(\frac{-12(y -2)}{-4}\) = x - 4

3(y -2) = x -4

3y - 6 = x - 4

3y = x - 4 + 6

3y = x + 2...

By comparing the equations;

3y = px + , p = 1


Correct option: A
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Question 16
A
2
B
4
C
6
D
8
Explanation

Mean = \(\frac{\sum x}{n}\)

4 = \(\frac{1 + 3 + 5 + 7 + x}{5}\)

4 x 5 = 16 + x

20 - 16 = x

4 = x

x = 4


Correct option: B
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Question 17
A
\(\frac{9}{25}\)
B
\(\frac{1}{5}\)
C
\(\frac{6}{25}\)
D
\(\frac{2}{5}\)
Explanation

Prob. (team scored 4 goals) = Prob. (team scored 7 goals) = \(\frac{3}{25}\)

Hence, probability that a team selected at random scored either 4 or 7 goals;

= \(\frac{6}{25} + \frac{3}{25}\)

= \(\frac{9}{25}\)


Correct option: A
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Question 18
A
\(\frac{3}{25}\)
B
\(\frac{1}{5}\)
C
\(\frac{6}{25}\)
D
\(\frac{2}{5}\)
Explanation

No. of teams that scored at most 3 goals = 3 + 1 + 6 = 10

Hence, probability that a team selected at random scored at most 3 goals

= \(\frac{10}{25}\) = \(\frac{2}{5}\)


Correct option: D
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Question 19
A
5.0 cm
B
7.0 cm
C
8.5 cm
D
12.0 cm
Explanation

Total surface area of hemisphere is

3\(\pi r^2\) = 75\(\pi cm^2\)

\(r^2\) = \(\frac{75 \pi}{3 \pi}\)

\(r^2\) = 25

r = \(\sqrt{25}\)

r = 5cm


Correct option: A
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Question 20
A
1\(\frac{1}{2}\)
B
2
C
2\(\frac{1}{2}\)
D
3
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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