# Expected Maths Questions For SSC MTS PDF

Download Top-20 SSC MTS Expected Maths Questions PDF. Most Expected Maths questions based on asked questions in previous year exam papers very important for the SSC MTS exam.

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**Question 1: **If X = 0.3 $\times$ 0.3, the value of X is

a) 0.009

b) 0.03

c) 0.09

d) 0.08

**Question 2: **An equation of the form ax + by + c = 0. Where, a ≠ 0, b ≠ 0 and c = 0 represents a straight line which passes through

a) (2, 4)

b) (0, 0)

c) (3, 2)

d) None of these

**Question 3: **The fifth term of the sequence for which $t_{1}=1$, $t_{2}=2$ and $t_{n+2}$ = $t_{n}+t_{n+1}$, is

a) 5

b) 10

c) 6

d) 8

**Question 4: **Reduce 3596 / 4292 to lowest terms.

a) 29/37

b) 17/43

c) 31/37

d) 19/23

**Question 5: **Reduce 2530/1430 to lowest terms.

a) 47/17

b) 23/13

c) 47/19

d) 29/17

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**Question 6: **The first and last terms of an arithmetic progression are -32 and 43. If the sum of the series is 88, then it has how many terms?

a) 16

b) 15

c) 17

d) 14

**Question 7: **29 is 0.8% of?

a) 3625

b) 1450

c) 7250

d) 10875

**Question 8: **5*[-0.6 (2.8 + 1.2)] of 0.3 is equal to

a) -1.44

b) -1.08

c) -1.2

d) -3.6

**Question 9: **Find the value of p if 3x + p, x – 10 and -x + 16 are in arithmetic progression.

a) 16

b) 36

c) -16

d) -36

**Question 10: **If 9/4th of 7/2 of a number is 126, then 7/2th of that number is …………..

a) 56

b) 284

c) 72

d) 26

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**Question 11: **Find equation of the perpendicular bisector of segment joining the points (2,-5) and (0,7)?

a) x – 6y = 5

b) x + 6y = -5

c) x – 6y = -5

d) x + 6y = 5

**Question 12: **Find equation of the perpendicular to segment joining the points A(0,4) and B(-5,9) and passing through the point P. Point P divides segment AB in the ratio 2:3.

a) x – y = 8

b) x – y = -8

c) x + y = -8

d) x + y = 8

**Question 13: **The co-ordinates of the centroid of a triangle ABC are (-1,-2) what are the co-ordinates of vertex C, if co-ordinates of A and B are (6,-4) and (-2,2) respectively?

a) (-7,-4)

b) (7,4)

c) (7,-4)

d) (-7,4)

**Question 14: **What is the slope of the line parallel to the line passing through the points (4,-2) and (-3,5)?

a) 3/7

b) 1

c) -3/7

d) -1

**Question 15: **The line passing through (-2,5) and (6,b) is perpendicular to the line 20x + 5y = 3. Find b?

a) -7

b) 4

c) 7

d) -4

**Question 16: **Find k, if the line 4x+y = 1 is perpendicular to the line 5x+ky = 2?

a) 20

b) -20

c) 4

d) -4

**Question 17: **The square root of 0.09 is

a) 0.30

b) 0.03

c) 0.81

d) 0.081

**Question 18: **How many perfect squares lie between 120 and 300 ?

a) 5

b) 6

c) 7

d) 8

**Question 19: **A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is:

a) 2.5

b) 3

c) 3.5

d) 4

**Question 20: **The product of two numbers is 45 and their difference is 4. The sum of squares of the two numbers is

a) 135

b) 240

c) 73

d) 106

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**Answers & Solutions:**

**1) Answer (C)**

Expression : $X=0.3\times0.3$

=> $X=0.09$

=> Ans – (C)

**2) Answer (B)**

As c=0, and substituting the point (0,0) in the equation, we get ax+by+c = 0 at the point (0,0).

Hence, the line passes through origin.

**3) Answer (D)**

$t_{1}=1$, $t_{2}=2$

$t_{n+2}$ = $t_{n}+t_{n+1}$

put n=3, then $t_{5}$ = $t_{3}+t_{4}$

$t_{3}$ = $t_{1}+t_{2}$ = 1+2 = 3

$t_{4}$ = $t_{2}+t_{3}$ = 2+3 = 5

$t_{5}$ = $t_{3}+t_{4}$ = 3+5 = 8

so the answer is option D.

**4) Answer (C)**

Expression : $\frac{3596}{4292}$

Dividing both numerator and denominator by 4, = $\frac{899}{1073}$

Similarly, dividing by 29, we get :

= $\frac{31}{37}$

=> Ans – (C)

**5) Answer (B)**

Expression : $\frac{2530}{1430}$

Dividing both numerator and denominator by 10, we get = $\frac{253}{143}$

Similarly, dividing by 11, we get :

= $\frac{23}{13}$

=> Ans – (B)

**6) Answer (A)**

First term of AP, $a=-32$ and last term, $l=43$

Let there be $n$ terms

Sum of AP = $\frac{n}{2}(a+l) = 88$

=> $\frac{n}{2}(-32+43)=88$

=> $\frac{11n}{2}=88$

=> $n=88 \times \frac{2}{11}$

=> $n=8 \times 2=16$

=> Ans – (A)

**7) Answer (A)**

Let the number be $x$

According to ques, 0.8% of $x$ = 29

=> $\frac{0.8}{100} \times x = 29$

=> $\frac{x}{125} = 29$

=> $x = 29 \times 125 = 3625$

=> Ans – (A)

**8) Answer (D)**

Expression : 5*[-0.6 (2.8 + 1.2)] of 0.3

= $5 [(-0.6) \times (4)] \times 0.3$

= $5 \times (-2.4) \times 0.3$

= $(-12) \times 0.3 = -3.6$

=> Ans – (D)

**9) Answer (D)**

Terms in arithmetic progression : $(3x + p) , (x – 10) , (-x + 16)$

=> Difference between first two terms is equal to the difference between last two terms

=> $(x – 10) – (3x + p) = (-x + 16) – (x – 10)$

=> $-2x -10 – p = -2x + 16 + 10$

=> $-p = 26 + 10 = 36$

=> $p = -36$

=> Ans – (D)

**10) Answer (A)**

Let the number be $x$

According to ques,

=> $\frac{9}{4} \times \frac{7}{2} \times x = 126$

=> $\frac{63}{8} x = 126$

=> $x = \frac{126}{63} \times 8$

=> $x = 2 \times 8 = 16$

$\therefore (\frac{7}{2})^{th}$ of the number = $\frac{7}{2} \times 16$

= $7 \times 8 = 56$

=> Ans – (A)

**11) Answer (C)**

Let line $l$ perpendicularly bisects line joining A(2,-5) and B(0,7) at C, thus C is the mid point of AB.

=> Coordinates of C = $(\frac{2 + 0}{2} , \frac{-5 + 7}{2})$

= $(\frac{2}{2} , \frac{2}{2}) = (1,1)$

Now, slope of AB = $\frac{y_2 – y_1}{x_2 – x_1} = \frac{(7 + 5)}{(0 – 2)}$

= $\frac{12}{-2} = -6$

Let slope of line $l = m$

Product of slopes of two perpendicular lines = -1

=> $m \times -6 = -1$

=> $m = \frac{1}{6}$

Equation of a line passing through point $(x_1,y_1)$ and having slope $m$ is $(y – y_1) = m(x – x_1)$

$\therefore$ Equation of line $l$

=> $(y – 1) = \frac{1}{6}(x – 1)$

=> $6y – 6 = x – 1$

=> $x – 6y = 1 – 6 = -5$

=> Ans – (C)

**12) Answer (B)**

Using section formula, the coordinates of point that divides line joining A = $(x_1 , y_1)$ and B = $(x_2 , y_2)$ in the ratio a : b

= $(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$

Coordinates of A(0,4) and B(-5,9). Let coordinates of P = (x,y) which divides AB in ratio = 2 : 3

=> $x = \frac{(2 \times -5) + (3 \times 0)}{2 + 3}$

=> $5x = -10$

=> $x = \frac{-10}{5} = -2$

Similarly, $y = \frac{(2 \times 9) + (3 \times 4)}{2 + 3}$

=> $5y = 18 + 12 = 30$

=> $y = \frac{30}{5} = 6$

=> Point P = (-2,6)

Slope of AB = $\frac{9 – 4}{-5 – 0} = \frac{5}{-5} = -1$

Let slope of line perpendicular to AB = $m$

Also, product of slopes of two perpendicular lines is -1

=> $m \times -1 = -1$

=> $m = 1$

Equation of lines having slope $m$ and passing through point P(-2,6) is

=> $(y – 6) = 1(x + 2)$

=> $y – 6 = x + 2$

=> $x – y = -8$

=> Ans – (B)

**13) Answer (A)**

Coordinates of centroid of triangle with vertices $(x_1 , y_1)$ , $(x_2 , y_2)$ and $(x_3 , y_3)$ is $(\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3})$

Let coordinates of vertex C = $(x , y)$

Vertex A(6,-4) and Vertex B(-2,2) and Centroid = (-1,-2)

=> $-1 = \frac{-2 + 6 + x}{3}$

=> $x + 4 = -1 \times 3 = -3$

=> $x = -3 – 4 = -7$

Similarly, => $-2 = \frac{-4 + 2 + y}{3}$

=> $y – 2 = -2 \times 3 = -6$

=> $y = -6 + 2 = -4$

$\therefore$ Coordinates of vertex C = (-7,-4)

=> Ans – (A)

**14) Answer (D)**

Slope of line passing through points (4,-2) and (-3,5)

= $\frac{5 + 2}{-3 – 4} = \frac{7}{-7} = -1$

Slope of two parallel lines is always equal.

=> Slope of the line parallel to the line having slope -1 = $-1$

=> Ans – (D)

**15) Answer (C)**

Slope of line having equation : $ax + by + c = 0$ is $\frac{-a}{b}$

=> Slope of line $20x + 5y = 3$ is $\frac{-20}{5} = -4$

Slope line passing through (-2,5) and (6,b) = $\frac{b – 5}{6 + 2} = \frac{(b – 5)}{8}$

Also, product of slopes of two perpendicular lines is -1

=> $\frac{(b – 5)}{8} \times -4 = -1$

=> $b – 5 = \frac{8}{4} = 2$

=> $b = 2 + 5 = 7$

=> Ans – (C)

**16) Answer (B)**

Slope of line having equation : $ax + by + c = 0$ is $\frac{-a}{b}$

Thus, slope of line $4x + y = 1$ is $\frac{-4}{1} = -4$

Similarly, slope of line $5x + ky = 2$ is $\frac{-5}{k}$

Also, product of slopes of two perpendicular lines is -1

=> $\frac{-5}{k} \times -4 = -1$

=> $\frac{20}{k} = -1$

=> $k = -20$

=> Ans – (B)

**17) Answer (A)**

0.09 = $\frac{9}{100}$

$\sqrt{\frac{9}{100}} = \frac{3}{10}$ = 0.3

**18) Answer (C)**

let’s say n be a number whose perfect square lie between 120 and 300

hence 120<$n^{2}$<300

or $121\leq n^{2} \leq289$

or $11\leq n^{2} \leq17$

**19) Answer (B)**

since we know volume will remain same while melting

$\pi r_{1}^{2}h= \frac{4}{3}\pi r_{2}^{3}$

where $r_{1}$ is radius of cylinderical wire and $r_{2}$ is radius of sphere and h is length of wire

putting values we will get $r_{2}$ = 3 cm.

**20) Answer (D)**

As we know $(a-b)^{2}$ = $a^{2} + b^{2} – 2ab$

We assume that first number is a and second number is b hence ab = 45

and a – b = 4

after putiing values we will get $a^{2} + b^{2}$ = 106