# Mathematics 2014 | Study Mode

Question 41
A
$$\frac{1}{5}$$
B
$$\frac{1}{2}$$
C
$$\frac{2}{5}$$
D
$$\frac{3}{4}$$
##### Explanation
Let E demote the event of obtaining at least a 4
Then n(E) = 16 + 10 + 14 = 40

Hence, prob (E) = $$\frac{n(E)}{n(S)}$$

$$= \frac{40}{100}$$

$$= \frac{2}{5}$$

Question 42
A
$$\frac{2}{15}$$
B
$$\frac{1}{10}$$
C
$$\frac{1}{3}$$
D
$$\frac{2}{5}$$
##### Explanation
Sample space S = {10, 11, 12, ... 30}

Let E denote the event of choosing a number divisible by 3

Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7

Prob (E) = $$\frac{n(E)}{n(E)}$$

Prob (E) = $$\frac{7}{21}$$

Prob (E) = $$\frac{1}{3}$$

Question 43
A
(P$$\cap$$Q)$$\cup$$(P$$\cap$$R)
B
(P$$\cup$$Q)$$\cap$$(P$$\cap$$R)
C
(P$$\cup$$Q)$$\cup$$(P$$\cup$$R)
D
(P$$\cap$$Q)$$\cup$$(P$$\cup$$R)

Question 44
A
91o
B
89o
C
37o
D
19o
##### Explanation
In the diagram above, $$\alpha$$ = 54o(alternate angles; KL||MN) < KNM = 2$$\alpha$$ (LN is bisector of < KNM) = 108o

35o + < KMN + 108o = 180o(sum of angles of $$\bigtriangleup$$)

< KMN + 143o = 180o

< KMN = 180o - 143o

= 37o

Question 45
A
135o
B
90o
C
60o
D
45o
##### Explanation
In the figure above, qo = 30o (vertically opposite angles)

(P + 2q)o + 30o = 180o(angles on a straight line)

p + 2 x 30o + 30o = 180o

p + 60o + 30o = 180o

p + 90o = 180o

p = 180o - 90o

= 90o

Question 46
A
20$$\sqrt{3}$$cm
B
10$$\sqrt{3}$$cm
C
5$$\sqrt{3}$$cm
D
4$$\sqrt{3}$$cm
##### Explanation
In the figure above, $$\frac{x}{\sin 60^o} = \frac{10}{\sin 30^o}$$ (Sine rule)

x = $$\frac{10 \sin 60^o}{\sin 30^o}$$

= 10 x $$\frac{\sqrt{3}}{2} \times \frac{1}{2}$$

= 10 x $$\frac{\sqrt{3}}{2} \times \frac{2}{1}$$

= 10$$\sqrt{3}$$cm

Question 47
A
y = x + 5
B
y = -x + 5
C
y = x - 5
D
y = -x - 5
##### Explanation
(x1, y1) = (0,5)

(x2, y2) = (5, 0)

Using $$\frac{y - y_1}{y_1 - y_1} = \frac{x - x_1}{x_1 - x_1}$$

$$\frac{y - 5}{0 - 5} = \frac{x - 0}{5 - 0}$$

$$\frac{y - 5}{-5} = \frac{x}{5}$$

5(y - 5) = -5x

y - 5 = -x

x + y = 5

y = -x + 5

Question 48
A
N42,000
B
N18,000
C
N16,000
D
N12,000
##### Explanation
Angle of sector subtended by yam

= 360o - (70 + 80 + 50)o

= 360o - 200o

= 160o

But $$\frac{80^o}{360^o}$$ x T = 8000

T = $$\frac{8000 \times 360^o}{80^o}$$

= N36,000

Hence the amount spent on yam = $$\frac{160^o}{260} \times N36,000$$

= N16,000

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