# Mathematics (2017) | Study Mode

Question 11
A

3$$^{2n}$$

B

9

C

3n

D

3 $$^{n + 1}$$

##### Explanation
3$$^{n + 1}$$ $$\times$$ 27$$\frac{n + 1}{81^n}$$

= 3$$^{n + 1}$$ $$\times$$ 3 $$\frac{3^{(n + 1)}}{3^{4n}}$$

= 3$$^{n - 1 + 3n + 3 - 4n}$$

= 3$$^{4n - 4n - 1 + 3}$$

= 32

= 9

Question 12
A
circle centre P
B
pair of parallel lines each opposite to PQ
C
circle centre Q
D
perpendicular line to PQ
##### Explanation
The locus of points at a fixed distance from the point P is a circle with the given P at its centre.

The locus of points at a fixed distance from the point Q is a circle with the given point Q at its centre

The locus of points equidistant from two points P and Q i.e line PQ is the perpendicular bisector of the segment determined by the points

Hence, The locus of a point which is equidistant from the line PQ forms a perpendicular line to PQ

Question 13
A
{2, 3}
B
{2, 3, 4}
C
{3, 4, 6}
D
{2}
##### Explanation
Given T = {even numbers from 1 to 12}
= { 2, 4, 6, 8,10, 12}

N = {common factors of 6, 8 and 12}

= {2} Find T n N = {2}

Question 14
A

100o

B

140o

C

120o

D

10o

##### Explanation
If RST = 60o

RXT = 2 $$\times$$ RST

(angle at the centre twice angle at the circumference)

RXT = 2 $$\times$$ 60

= 120o

Question 15
A

16

B

14

C

12

D

10

##### Explanation

Range = Highest Number - Lowest Number

Mode is the number with highest occurrence
10, 9, 10, 9, 8, 7, 7, 10, 8, 4, 6,, 9, 10, 9, 7, 10, 6, 5

Range = 10 - 4 = 6

Mode = 10

Sum of range and mode = range + mode = 6 + 10

= 16

Question 16
A
$$\frac{1}{2}$$
B
$$\frac{3}{5}$$
C
$$\frac{-1}{5}$$
D
$$\frac{73}{12}$$
##### Explanation
Sum to infinity

? = arn ? 1

= $$\frac{a}{1}$$ ? r

a = $$\frac{1}{4}$$

r = $$\frac{1}{8}$$ $$\frac{1}{4}$$

r = $$\frac{1}{s}$$ $$\times$$ $$\frac{4}{1}$$

= $$\frac{1}{2}$$

S = $$\frac{1 \div 4}{1}$$ ? $$\frac{1}{2}$$

= $$\frac{1}{4}$$ $$\frac{1}{2}$$

= $$\frac{1}{4}$$ $$\times$$ $$\frac{2}{1}$$

= $$\frac{1}{2}$$

Question 17
A

6

B

2

C

4

D

5

Question 18
A
16
B
10
C
18
D
24

Question 19
A
8
B
6
C
4
D
3
##### Explanation
2x + x = 180o

3x = 180o

x = 60o (exterior angle of the polygon)

angle = $$\frac{\text{total angle}}{\text{number of sides}}$$

60 = $$\frac{360}{n}$$

n = $$\frac{360}{60}$$

n = 6 sides

Question 20
A

23m

B

25m

C

20m

D

22m

##### Explanation
Capacity = Volume = 3080m3

base diameter = 14m

radius = $$\frac{\text{diameter}}{2}$$

= 7m

Volume of Cylidner = Capacity of cylinder

r2h = 3080

$$\frac{22}{7}$$ $$\times$$ 7 $$\times$$ 7 $$\times$$ h = 3080

h = $$\frac{3080}{22 \times 7}$$

h = 20m

Try this quiz in in E-test/CBT Mode
switch to

Question Map