# Mathematics (2015) | Study Mode

Question 61
A

4

B

5

C

6

D

3

##### Explanation

Each exterior angle = 180 -120=60

Exterior angle =360/n

60 = 360/n

n =360/6

= 6

Question 62
A
N100
B
N20
C
N80
D
N150
##### Explanation
25% of N80 = 25/100 $$\times$$ 80

= N20.00

Question 63
A
7.51
B
7.71
C
5.5
D
7.55
2.04 x 3.7
=
204
x 37
-----
7548 = 7.548

= 7.55(2dp)

Question 64
A

3%

B

2%

C

4%

D

1%

##### Explanation

Let assume the cost price is 100%

Marked up price + cost price = 20 + 100 = 120%

Discount at 20% = 20/100 $$\times$$ 120 % of cost price

Selling price = cost price - gain

= (120 - 24)% of cost price

= 96% of cost price

Loss = (100- 96)% of cost price

= 4% of cost price

- He will wake 4% loss

Question 65
A

3(2 + $$\sqrt{5}$$)

B

2(3 + $$\sqrt{5}$$)

C

5(2 + $$\sqrt{3}$$)

D

$$3\sqrt{3}$$ + 1

##### Explanation

5 (2 - $$\sqrt{3}$$)

Using conjugate surds

[5 (2 - $$\sqrt{3}$$)]$$\times$$(2 + $$\sqrt{3}$$) (2 + $$\sqrt{3}$$)]

[5(2+$$\sqrt{3}$$)((2- $$\sqrt{3}$$))2]

[5(2 +$$\sqrt{3}$$) (2)2- ($$\sqrt{3}$$)2]

[5(2 +$$\sqrt{3}$$) (4- 3)]

= 5(2 + $$\sqrt{3}$$)

Question 66
A

(a + b)(a - b)

B

(a - 2 + b)

C

(a + 1)(a - 2 + b)

D

(a + b) 2

##### Explanation

The trinomial = $$a^2 - 4a + 4 \\ a^2 - b^2- 4a + 4 = (a^2 - 4a + 4) - b^2 \\ (a^2 - 2a - 2a + 4) - b^2 \\ [a(a - 2)- 2(a - 2)] - b^2 \\ (a - 2)2 - b^2 \\ (a - 2 + b)(a - 2 - b )$$

Question 67
A
$$\frac{2\sqrt{13}}{13}$$
B
$$\frac{3\sqrt{13}}{13}$$
C
$$\frac{4\sqrt{13}}{13}$$
D
$$\frac{6\sqrt{13}}{13}$$
##### Explanation
tan x = $$\frac{2}{3}$$(given), is illustrated in a right-angled $$\Delta$$

thus m2 = 22 + 32

= 4 + 9 = 13

m = $$\sqrt{13}$$

Hence, 2sin x = 2 x $$\frac{2}{m}$$

2 x$$\frac{2}{\sqrt{13}}$$

= $$\frac{4}{\sqrt{13}}$$

= $$\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}$$

= $$\frac{4\sqrt{13}}{13}$$

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