# Mathematics (2018) | Study Mode

Question 1
A
x < 8
B
x > -6
C
x < 4
D
x > -3

Question 2
A
x < 8
B
x > -6
C
x < 4
D
x > -3

Question 3
A
65
B
23
C
17
D
91
##### Explanation

Question 4
A
$$\frac{5}{8}$$
B
$$\frac{5}{16}$$
C
$$\frac{1}{2}$$
D
$$\frac{3}{8}$$
##### Explanation

Each multiplication of elements gives 16 results out of which the ones that are not greater than 6 are asterisked (*) and they are 8 in numbers

Pr(product of x and y NOT > 6 ) = $$\frac{8}{16}$$ = $$\frac{1}{2}$$

Question 5
A
1000
B
2000
C
3000
D
4000
##### Explanation

Question 6
A
8
B
$$\frac{1}{8}$$
C
1$$\frac{1}{8}$$
D
$$\frac{2}{5}$$

Question 7
A
7
B
6
C
5
D
4
##### Explanation

Profit (P) = 10$$_x$$ $$_x$$2

Maximum profit can be achieved when the differential of profit with respect to number of bags(x) is 0

i.e. $$\frac{dp}{dx}$$ = 0

$$\frac{dp}{dx}$$ = 10 - 2x = 0

10 = 2x

Then x = $$\frac{10}{2}$$ = 5

Question 8
A
10
B
28
C
36
D
40
##### Explanation

The diagram shows angles at a point, the total angle at a point is 360

x - 10 + 4x - 50 + 2x + 3x + 20 = 360

10x - 40 = 360

10x = 360 + 40

10x = 400

x = $$\frac{400}{10}$$

x = 40

Question 9
A
$$\frac{9}{13}$$
B
$$\frac{7}{13}$$
C
5
D
9$$\frac{3}{5}$$
##### Explanation

$$\frac{3}{4}$$ t + $$\frac{1}{3}$$ (21 - t) = 11

Multiply through by the LCM of 4 and 3 which is 12

12 x($$\frac{3}{4}$$ t) + 12 x ($$\frac{1}{3}$$ (21 - t)) = (11 x 12)

9t + 4(21 - t) = 132

9t + 84 - 4t = 132

5t + 84 = 132

5t = 132 - 84 = 48

t = $$\frac{48}{5}$$

t = 9 $$\frac{3}{5}$$

Question 10
A
$$\frac{5}{6}$$
B
$$\frac{7}{12}$$
C
$$\frac{5}{12}$$
D
$$\frac{1}{6}$$