Mathematics 2018 | Study Mode

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Question 41
A
90\(^o\)
B
60\(^o\)
C
45\(^o\)
D
30\(^o\)
Explanation

In the diagram, < WOZ = 180\(^o\) (angle on a straight line)

< WOX = < XOY = < YOZ

(|WX| = |XY| = |YZ|)

\(\frac{180^o}{3}\) = 60\(^o\)

= 60\(^o\)

M + m =2m (base angles of isosceles \(\bigtriangleup\), |OY| and |OZ| are radii)

< YOZ + 2m (base angles of a \(\bigtriangleup\))

60\(^o\) + 2m = 180\(^o\) (sum of a \(\bigtriangleup\))

60\(^o\) + 2m = 180\(^o\)

2m = 180\(^o\) - 60\(^o\)

2m = 120\(^o\)

m = \(\frac{120^o}{2}\)

= 60\(^o\)


Correct option: B
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Question 42
A
x = 240\(^o\) - y - z
B
x = 180\(^o\) - y - z
C
x = 360\(^o\) + y -z
D
x = 360\(^o\) - y - z
Explanation

In the diagram,

a = z (alternate angles)

b = 180\(^o\) - a (angles on a straight line)

b = 180\(^o\) - z

c = 180\(^o\) - x (angles on a straight line)

y = b + c (sum of oposite interior angles)

y = 180\(^o\) - z + 180\(^o\) - x

y = 360\(^o\) - z - x

x = 360\(^o\) - y - z

Maths-explained-image

Correct option: D
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Question 43
A
2 : 3 : 4
B
3 : 4 : 5
C
4 : 5 : 6
D
5 : 6 : 7
Explanation

m + n = 110\(^o\), (n + r) = 130\(^o\)

(m + n) = 120\(^o\)

then, r = 130\(^o\) - n

and;

m + (130^o - n) = 120\(^o\)

m - n = -10\(^o\)

2m + (n + r) = 110 + 120 = 230

2m + 130 = 230

2m = 230 - 130

m = \(\frac{100}{2}\) = 50\(^o\)

n = 110\(^o\) - 50\(^o\)

= 60\(^o\)

r = 130\(^o\) - 60\(^o\) = 70\(^o\)

Hence, the ratio m : n : r

= 50 : 60 : 70

= 5 : 6 : 7


Correct option: D
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Question 44
A

N26,792.00

B

N26,972.00

C

N62,792.00

D

N62,972.00

Explanation

Total donation = 4 x 500 + 7 x 2000 + 20 x 1000 + 9 x 700 + 4 x 500 + 5 x 100 + 3 x 50 + 1 x 2 + 2 x 10

= 20000 + 14000 + 20000 + 6300 + 2000 + 500 + 150 + 2 + 20

= N62,972


Correct option: D
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Question 45
A
170\(^o\)
B
177\(^o\)
C
182\(^o\)
D
192\(^o\)
Explanation

Length of arc, L = 21.4 - 2 x 4.2cm

= 21.4 - 8.4

= 13cm

But L = \(\frac{\theta}{360^o}\) x 2\(\pi r\)

i.e 13 = \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 4.2

= 13 x 360\(^o\) x 7

= \(\theta\) x 2 x 22 x 4.2

\(\theta\) = \(\frac{13 \times 360^o \times 7}{44 \times 4.2}\)

= \(\approx\) 177.27\(^o\)

\(\approx\) 177\(^o\) (to the nearest degree)

Maths-explained-image

Correct option: B
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Question 46
A
\(\frac{1}{10}\)
B
\(\frac{1}{5}\)
C
\(\frac{5}{12}\)
D
1\(\frac{2}{5}\)
Explanation

From the diagram,

h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras')

h\(^2\) = 16 + 9 = 25

h = \(\sqrt{25}\) = 5

Hence, sin x - cos x

= \(\frac{4}{5} - \frac{3}{5}\)

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

Maths-explained-image

Correct option: B
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Question 47
A
10cm
B
15cm
C
25cm
D
30cm
Explanation

In \(\bigtriangleup\)YSC, sin 30\(^o\) = \(\frac{YS}{20}\)

|YS| = 20 sin 30\(^o\)

= 20 x 0.5

10m

Maths-explained-image

Correct option: A
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Question 48
A
42cm
B
48cm
C
52cm
D
60cm
Explanation

Let the length of a side of the rhombus be n

Then, n\(^2\) = 5\(^2\) + 12\(^2\)

= 25 + 144 = 169

n = \(\sqrt{169}\)

= 13cm

Hence, perimeter of rhombus = 4n = 4 x 13

= 52cm

Maths-explained-image

Correct option: C
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Question 49
A
18
B
20
C
30
D
38
Explanation

Let n(M \(\cup\) N \) = x

Then 20 - x + x + 30

- x = n(M \(\cup\) N)

50 - x = 40

50 - 40 = x

10 = x

x = 10

Hence, n(M \(\cup\N)' = 8 + (20 - 10) + (30 + 10)

= 8 + 10 + 20

= 38

Maths-explained-image

Correct option: D
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Question 50
A
(0,0), (1,1)
B
(0,0), (0,1)
C
(1, 0), (0, 0)
D
(0, 0) (0, 0)
Explanation

y = x\(^2\) ....(1)

y = x ......(2)

y = y

x\(^2\) - x

x\(^2\) - x = 0

x(x - 1) = 0

x = 0 or x - 1 = 0

x = 0 or x = 1

when x = 0, y = 0\(^2\) = 0

when x = 1, y = 1\(^2\) = 1

Hence; the two graphs interest at (0, 0) and (1, 1)


Correct option: A

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