# Mathematics (2018) | Study Mode

Question 31
A
20
B
30
C
48
D
60

Question 32
A
5x + 7y = 20
B
7x + 5y = 70
C
xy = 7
D
15x + 17y = 90

Question 33
A
N828.00
B
N756.00
C
N720.00
D
N684.00
##### Explanation

Since the salary increases by 36 after every 6 months

Every 6 months that can be counted on the eight month of the third year is 5

(i.e. 2 times in the first year, 2 times in the second year and once in the third year)

His salary then = initial salary + increment

= 540 + 5(36)

= 540 + 180

= ?720.00

It can also be solves using a sequence in form of an AP

Question 34
A
52cm
B
43cm
C
40cm
D
15cm
##### Explanation

Tan 74 = 150/x

x = 150/tan 74

= 43.01cm

Question 35
A
3x$$^3$$ - 2x$$^2$$ + x + c
B
2x$$^3$$ - x$$^2$$ + x + c
C
2x$$^3$$ 3x$$^2$$ + c
D
x$$^3$$ + x$$^2$$ x + c

Question 36
A
72
B
144
C
360
D
720
##### Explanation

The word LEADER has 1L 2E 1A 1D and 1R making total of 6! $$\frac{6}{1!2!1!1!1!}$$ = $$\frac{6!}{2!}$$

= $$\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}$$

= 360

Question 37
A
32cm
B
24 cm
C
16 cm
D
12 cm
##### Explanation

From the figure,

XMN = XZY

Angle X is common

So, XNM = XYZ

Then from the angle relationship

$$\frac{XM}{XZ}$$ = $$\frac{XN}{XY}$$ = $$\frac{MN}{ZY}$$

XM = 8, XZ = 12 + 4 = 16,

XN = 12, XY = 8 + YM

$$\frac{8}{16}$$ = $$\frac{12}{(8 +YM) }$$

Cross multiply

8(8 + YM) = 192

64 + 8YM = 192

8YM = 128

YM = $$\frac{128}{8}$$

= 16cm

Question 38
A
3%
B
3 $$\frac{1}{3}$$%
C
15%
D
30%
##### Explanation

The two numbers must be expressed in the same unit. To convert 495g to kg, it will be divided by 1000

495g = $$\frac{495}{1000}$$

= 0.495kg

To express in percentage, 0.495 will be divided by 16.5 and then multiplied by 100

% will be added to the answer $$\frac{0.4950}{16.5}$$ x 100

= 3%

Question 39
A
-4
B
-2
C
2
D
4
##### Explanation

23 - 4) ( 23 + 4)

= 12 + 83 - 83 16

= 12 16

= -4

The two expressions in the bracket are conjugate of each other

Question 40
A
y = 2x - 4
B
y = 2x + 4
C
y = 2x - 2
D
y = 2x + 2
##### Explanation

The gradient to the curve is found by differentiating the curve equation with respect to x

So $$\frac{dy}{dx}$$ 2x - 2

The gradient of the curve is the same with that of the tangent.

At point (2, 0) $$\frac{dy}{dx}$$ = 2(2) - 2

= 4 2 = 2

The equation of the tangent is given by (y - y1) $$\frac{dy}{dx}$$ (x x1)

At point (x1, y1) = (2, 0)

y - 0 = 2(x - 2)

y = 2x - 4

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