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**Question 1:Â **The value of $5\frac{5}{29}-\left[\frac{15}{4}\div\left\{\frac{3}{4}\times\left(\frac{3}{2}-\frac{1}{5}-\frac{1}{3}\right)\right\}\right]$ is:

a)Â 5

b)Â 1

c)Â 0

d)Â 10

**1)Â AnswerÂ (C)**

**Solution:**

According to the rule of BODMAS,

Parts of an equation enclosed in a ‘brackets’ must be solved first.

$5\frac{5}{29}-\left[\frac{15}{4}\div\left\{\frac{3}{4}\times\left(\frac{45-6-10}{30}\right)\right\}\right]$

$5\frac{5}{29}-\left[\frac{15}{4}\div\left\{\frac{3}{4}\times\frac{29}{30}\right\}\right]$

$5\frac{5}{29}-\left[\frac{15}{4}\div\frac{29}{40}\right]$

$5\frac{5}{29}-\frac{150}{29}$

$\frac{150}{29}-\frac{150}{29}$

= 0

**Question 2:Â **Find the greatest value of b so that 30a68b (a > b) is divisible by 11.

a)Â 4

b)Â 9

c)Â 3

d)Â 6

**2)Â AnswerÂ (C)**

**Solution:**

Given,Â 30a68b is divisible by 11.

$\Rightarrow$Â Â (3 + a + 8) – (0 + 6 + b) will be multiple of 11.

$\Rightarrow$Â (a – b + 5) will be multiple of 11.

$\Rightarrow$Â Â (a – b + 5) = 0 or (a – b + 5) = 11 or (a – b + 5) = 22 so on.

$\Rightarrow$Â a – b = -5Â orÂ a – b = 6Â orÂ a – b = 17Â so on.

Since a > b, a – b cannot be negative.

Both a and b are digits, so a – b cannot be a two digit number.

The only possibility is

a – b = 6

The pairs satisfying the above equation are (9,3), (8,2), (7,1), (6,0).

The greatest value of b can be 3.

Hence, the correct answer is Option C

**Question 3:Â **The sum of 3-digit numbers abc, cab and bca is not divisible by:

a)Â 37

b)Â 3

c)Â 31

d)Â a + b + c

**3)Â AnswerÂ (C)**

**Solution:**

Sum of the numbers =Â $\left(100a+10b+c\right)+\left(100c+10a+b\right)+\left(100b+10c+a\right)$

=Â $100\left(a+b+c\right)+10\left(a+b+c\right)+\left(a+b+c\right)$

=Â $111\left(a+b+c\right)$

=Â $37\times3\times\left(a+b+c\right)$

Sum is not divisible by 31.

Hence, the correct answer is Option C

**Question 4:Â **What is the sum of the digits of the largest five digit number which is divisible by 5, 35, 39 and 65?

a)Â 30

b)Â 33

c)Â 27

d)Â 35

**4)Â AnswerÂ (B)**

**Solution:**

LCM ofÂ 5, 35, 39 and 65 = 1365

When the largest five digit number 99999 is divided by 1365, the remainder will be 354.

So, 99999 – 354 = 99645 is the largest five digit number divisible by 5, 35, 39 and 65.

Sum of the digits = 9 + 9 + 6 + 4 + 5 = 33

Hence, the correct answer is Option B

**Question 5:Â **What is the value of k such that number 72k460k is divisible by 6?

a)Â 4

b)Â 7

c)Â 9

d)Â 8

**5)Â AnswerÂ (A)**

**Solution:**

Given,Â 72k460k is divisible by 6.

72k460k is divisible by both 2 and 3.

So, k is even andÂ sum of the digits is divisible by 3.

(7 + 2 + k + 4 + 6 + 0 + k =Â 19 + 2k) is divisible by 3.

since k is even, the only possibility is k = 4.

Hence, the correct answer is Option A

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**Question 6:Â **When a number is divided by 14, the remainder is 9. If the square of the same number is divided by 14, then the remainder will be:

a)Â 11

b)Â 9

c)Â 8

d)Â 10

**6)Â AnswerÂ (A)**

**Solution:**

Given,Â When the number is divided by 14, the remainder is 9

Let the number = 14k + 9

Square of the number = (14k + 9)$^2$

= 196k$^2$ + 252k + 81

= 196k$^2$ + 252k + 70 + 11

= 14(14k$^2$ + 18k + 5) + 11

= 14p + 11

$\therefore\ $When the square of the number is divided by 14, the remainder is 11

Hence, the correct answer is Option A

**Question 7:Â **If 4M37094267N is divisible by both 8 and 11, where M and N are single digit integers, then the values of M and N are:

a)Â M = 5, N = 6

b)Â M = 5, N = 4

c)Â M = 5, N = 2

d)Â M = 2, N = 5

**7)Â AnswerÂ (C)**

**Solution:**

Given,Â 4M37094267N is divisible by both 8 and 11

If the number is divisible by 8, then the three digits should be divisible by 8

$\Rightarrow$Â 67N is divisible by 8

$\Rightarrow$Â The only possible value for N is 2

If the number is divisible by 11, then

Sum of digits at odd place – Sum of digits at even place = 0 or multiple of 11

$\Rightarrow$Â (M+7+9+2+7) – (4+3+0+4+6+N) =Â 0 or multiple of 11

$\Rightarrow$Â M + 25 – 17 – N = 0 or multiple of 11

$\Rightarrow$Â M – N + 8 = 0Â or multiple of 11

$\Rightarrow$Â M – 2 + 8 = 0 or multiple of 11

$\Rightarrow$Â M + 6 = 0 orÂ multiple of 11

The possible value is M + 6 = 11

$\Rightarrow$Â M = 5

$\therefore\ $M = 5, N = 2

Hence, the correct answer is Option C

**Question 8:Â **Which number is divisible by both 9 and 11?

a)Â 10,098

b)Â 10,108

c)Â 10,089

d)Â 10,087

**8)Â AnswerÂ (A)**

**Solution:**

If a number is divisible by 9, then the sum of the digits of the number should be divisible by 9.

If a number is divisible by 11, then the difference of sum of the alternate numbers should be divisible by 11.

By Trial and Error method,

__Option A__

Sum of the digits of 10,098 = 1 + 0 + 0 + 9 + 8 = 18 which is divisible by 9. So 10,098 is divisible by 9.

Difference of sum of alternate digits = 1 + 0 + 8 – (0 + 9) = 0 which is divisible by 11. So 10,098 is divisible by 11.

Hence, the correct answer is Option A

**Question 9:Â **If the 8-digit number 43A5325B is divisible by 8 and 9, then the sum of A and B is equal to:

a)Â 15

b)Â 14

c)Â 12

d)Â 18

**9)Â AnswerÂ (B)**

**Solution:**

Given,Â 8-digit number 43A5325B is divisible by 8 and 9

If the number is divisible by 8, then the last three digits should be divisible by 8

$\Rightarrow$ 25B is divisible by 8

$\Rightarrow$Â B = 6

If the number is divisible by 9, then the sum of the digits of the number should be divisible by 9

$\Rightarrow$ 4 + 3 + A + 5 + 3 + 2 + 5 + B = multiple of 9

$\Rightarrow$ 22 + A + B = multiple of 9

$\Rightarrow$ 22 + A + 6 =Â multiple of 9

$\Rightarrow$ 28 + A = multiple of 9

The only possible value of A is 8

$\therefore\ $Sum of A and B = 8 + 6 = 14

Hence, the correct answer is Option B

**Question 10:Â **Find the value of $\sqrt {30 + \sqrt{ 30 +\sqrt {30 + \sqrt { 30 + … … … … … . \infty}}}}$

a)Â 4

b)Â 6

c)Â 5

d)Â 3

**10)Â AnswerÂ (B)**

**Solution:**

Let $\sqrt{30+\sqrt{30+\sqrt{30+\sqrt{30+…………….\infty}}}}=a$

$\Rightarrow$ $\sqrt{30+\sqrt{30+\sqrt{30+\sqrt{30+\sqrt{30+……….\infty}}}}}=a$

$\Rightarrow$ $\sqrt{30+a}=a$

$\Rightarrow$ $30+a=a^2$

$\Rightarrow$ $a^2-a-30=0$

$\Rightarrow$ $a^2-6a+5a-30=0$

$\Rightarrow$ $a\left(a-6\right)+5\left(a-6\right)=0$

$\Rightarrow$Â $\left(a-6\right)\left(a+5\right)=0$

$\Rightarrow$Â $a-6=0$Â Â orÂ Â $a+5=0$

$\Rightarrow$Â $a=6$ Â Â orÂ Â $a=-5$

$a$ cannot be negative

$\Rightarrow$ Â $a=6$

$\therefore\ $ $\sqrt{30+\sqrt{30+\sqrt{30+\sqrt{30+…………….\infty}}}}=6$

Hence, the correct answer is Option B

**Question 11:Â **If a positive integer n is divided by 7, the remainder is 2. Which of the numbers in the options yields a remainder of 0 when it is divided by 7?

a)Â n + 3

b)Â n + 5

c)Â n + 2

d)Â n + 1

**11)Â AnswerÂ (B)**

**Solution:**

Given, when ‘n’ is divided by 7 the remainder is 2.

Let n = 7k + 2 where k is an positive integer

By Trial and Error method,

__Option A__

n + 3 = 7k + 2 + 3 = 7k + 5

$\Rightarrow$Â When n + 3 is divided by 7, the remainder is 5.

__Option B__

n + 5 = 7k + 2 + 5 = 7k + 7 = 7(k+1)

$\Rightarrow$Â When n + 5 is divided by 7, the remainder is 0.

Hence, the correct answer is Option B

**Question 12:Â **The sum of the number of male and female students in an institute is 100. If the number of male students is $x$ , then the number of female students becomes $x$% of the total number of students. Find the number of male students.

a)Â 65

b)Â 50

c)Â 60

d)Â 45

**12)Â AnswerÂ (B)**

**Solution:**

Given, number of male students = $x$

Number of female students = $x$% of total students

Total number students = 100

$\Rightarrow$Â $x+\frac{x}{100}\times100=100$

$\Rightarrow$ Â $x+x=100$

$\Rightarrow$ Â $x=50$

$\therefore\ $Number of male students = 50

Hence, the correct answer is Option B

**Question 13:Â **Which of the following numbers is divisible by 2, 5 and 10?

a)Â 7,20,345

b)Â 149

c)Â 19,400

d)Â 1,25,372

**13)Â AnswerÂ (C)**

**Solution:**

From the options,

7,20,345 is not divisible by 2 because the units place is not an even number

149 is not divisible by 2 because the units place is not an even number

1,25,372 is not divisible by 10 because the units place is not zero

19,400 is divisible by 2,5,10 as it is even number and the units place is zero

Hence, the correct answer is Option C

**Question 14:Â **The sum of two numbers is 59 and their product is 840. Find the sum of their squares.

a)Â 2961

b)Â 1801

c)Â 1875

d)Â 1754

**14)Â AnswerÂ (B)**

**Solution:**

Let the two numbers are $a$ and $b$

Given,

Product of the numbers = 840

$=$>Â $ab = 840$

Sum of the numbers = 59

$=$>Â $a+b = 59$

$=$> Â $\left(a+b\right)^2=59^2$

$=$> Â $a^2+b^2+2ab=3481$

$=$> Â $a^2+b^2+2\left(840\right)=3481$

$=$> Â $a^2+b^2+1680=3481$

$=$> Â $a^2+b^2=3481-1680$

$=$> Â $a^2+b^2=1801$

$\therefore\ $Sum of their squares = 1801

Hence, the correct answer is Option B

**Question 15:Â **If the difference between two numbers is 6 and the difference between their squares is 60, what is the sum of their cubes ?

a)Â 894

b)Â 945

c)Â 678

d)Â 520

**15)Â AnswerÂ (D)**

**Solution:**

Let the two numbers are a,b

Difference between two numbers = 6

$=$>Â Â $a-b=6$ ………………………(1)

Difference between their squares = 60

$=$> Â $a^2-b^2=60$

$=$> Â $\left(a+b\right)\left(a-b\right)=60$

$=$> Â $\left(a+b\right)\left(6\right)=60$

$=$> Â $a+b=10$ …………………….(2)

Solving (1) and (2)

$2a=16$

$=$>Â $a=8$

Substituting $a=8$ in equation(2)

$=$>Â Â $8+b=10$

$=$> Â $b=2$

$\therefore\ $Sum of their cubes $=a^3+b^3=8^3+2^3=512+8=520$

Hence, the correct answer is Option D