# Number System Questions for SSC CGL Tier 2 PDF

0
1134

## Number System Questions for SSC CGL Tier 2 PDF

Download SSC CGL Tier 2 Number System Questions PDF. Top 15 SSC CGL Tier 2 Number System questions based on asked questions in previous exam papers very important for the SSC exam.

Question 1: A number, when divided by 114, leaves remainder 21. If the same number is divided by 19, then the remainder will be

a) 1

b) 2

c) 7

d) 17

Question 2: Out of six consecutive natural numbers, if the sum of first three is 27, what is the sum of the other three ?

a) 36

b) 35

c) 25

d) 24

Question 3: The unit digit in the sum of (124)372 + (124)373 is

a) 5

b) 4

c) 20

d) 0

Question 4: The least number which must be added to 1728 to make it a perfect square is ……………..

a) 36

b) 32

c) 38

d) 30

Question 5: The square root of $(\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} – \sqrt{2}})$

a) $\sqrt{3} + \sqrt{2}$

b) $\sqrt{3} – \sqrt{2}$

c) $\sqrt{2} \pm \sqrt{3}$

d) $\sqrt{2} – \sqrt{3}$

Question 6: A positive integer when divided by 425 gives a remainder 45. When the same number is divided by 17, the remainder will be

a) 11

b) 8

c) 9

d) 10

Question 7: The ten’s digit of a 2-digit number is greater than the units digit by 7. If we subtract 63 from the number, the new number obtained is a number formed by interchange of the digits. Find the number.

a) 81

b) 70

c) 92

d) Cannot be determined

Question 8: Common factor of $12a^4b^6, 18a^6c^2, 36a^2b^2$ is

a) $36a^2$

b) $108b^2$

c) $6a^2b^2$

d) $6a^2$

Question 9: If $x^3 + 2x^2 – 5x + k$ is divisible by x + 1, then what is the value of k?

a) – 6

b) – 1

c) 0

d) 6

Question 10: The sum of a fraction and 7 times its reciprocal is 11/2. What is the fraction?

a) 7/2

b) 2/7

c) 3/4

d) 4/3

Question 11: What is the HCF of 6345 and 2160?

a) 45

b) 135

c) 270

d) 15

Question 12: $(2^{51}+2^{52}+2^{53}+2^{54}+2^{55})$ is divisible by

a) 23

b) 58

c) 124

d) 127

Question 13: Sum of the factors of $4b^2c^2 – (b^2 + c^2 – a^2)^2$ is

a) a + b + c

b) 2(a + b + c)

c) 0

d) 1

Question 14: If a number 657423547X46 is divisible by 11, then find the value of X.

a) 7

b) 9

c) 8

d) 6

Question 15: $4^{11}+4^{12}+4^{13}+4^{14}\$is divisible by_________.

a) 7

b) 14

c) 17

d) 9

Let the given number be x
Let a be the quotient when x is divided by 114
So $\frac{x}{114}$ = a$\frac{21}{114}$
so x = 114a + 21
when x is divided by 19 it can be written as
$\frac{x}{19} = \frac{114a + 21}{19}$
114 is divisible by 19 and 21 leaves a remainder of 2.

let’s say 6 consecutive numbers are (a-d), a, (a+d), (a+2d), (a+3d), (a+4d)
where d is the common difference i.e. 1 (given) and a is second term
summation of first three terms will be 3a = 27
hence second term a = 9
now sequence is 8,9,10,11,12,13,
so sum of last three terms 36

Both of numbers have unit digit as 4 and it has a repeating cycle of 2 with unit digits as 4 and 6
so in first number power is 372 which is exactly divisible by 2 hence unit digit of first number will be 6.
and in second number power is 373 which exceeds one with the reapeating cycle of 2 hence its unit digit will be 4.

now unit digit of the sum will be 6+4 = 10

We know that, 41 * 41 = 1681

and 42 * 42 = 1764

So, the least number to be added in 1728 to make it a perfect square = 1764-1728

= 36

we need to calculate square root of $(\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} – \sqrt{2}})$

let $(\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} – \sqrt{2}})$ be = y

on rationalizing y , we get

y = $(\surd2 + \surd3)^2$

hence square root of y => $\surd(y)$ = $\surd2 + \surd3$

Let the number be 45

So, when 45 is divided by 425 => remainder = 45

Now, when 45 is divided by 17

=> Remainder = 45%17 = 11

Let the unit’s digit of the number be $y$ and ten’s digit be $x$

=> Number = $10x + y$

According to ques, =>$x – y = 7$ ————–(i)

According to question, => $10x + y – 63 = 10y + x$

=> $9x – 9y = 63$

=> $x – y = \frac{63}{9} = 7$ ————–(ii)

Equations (i) and (ii) are same and thus we have two variables and one equation

Number can be = 92 , 81 and thus the solution cannot be determined.

=> Ans – (D)

Factors of :

$12a^4b^6$ = $(2 \times 6) \times (a^2 \times a^2) \times b^6$

$18a^6c^2$ = $(3 \times 6) \times (a^2 \times a^4) \times c^2$

$36a^2b^2$ = $(6 \times 6) \times (a^2) \times b^2$

The common factor in the 3 terms = $6a^2$

=> Ans – (D)

$f(x)=x^3 + 2x^2 – 5x + k$ is divisible by $(x+1)$

=> $x=-1$ is a factor of $f(x)$

=> $f(-1)=0$

=> $(-1)^3+2(-1)^2-5(-1)+k=0$

=> $-1+2+5+k=0$

=> $k=-6$

=> Ans – (A)

Let the fraction be $x$

According to ques,

=> $x+\frac{7}{x}=\frac{11}{2}$

=> $\frac{x^2+7}{x}=\frac{11}{2}$

=> $2x^2+14=11x$

=> $2x^2-11x+14=0$

=> $2x^2-4x-7x+14=0$

=> $2x(x-2)-7(x-2)=0$

=> $(x-2)(2x-7)=0$

=> $x=2,\frac{7}{2}$

=> Ans – (A)

6345 = 3*3*3*5*47

2160 = 3*3*3*5*16

HCF = product of common prime factors =  3*3*3*5 = 135

So the answer is option B.

Expression : $(2^{51}+2^{52}+2^{53}+2^{54}+2^{55})$

= $2^{51}(1+2+2^2+2^3+2^4)$

= $2^{51}(1+2+4+8+16)$

= $2^{51} \times 31$

Thus, the above expression is divisible by $31k$ and the only option that is divisible by 31 = 124

=> Ans – (C)

Expression : $4b^2c^2 – (b^2 + c^2 – a^2)^2$

= $(2bc)^2-(b^2+c^2-a^2)^2$

Using, $x^2-y^2=(x-y)(x+y)$, where $x=2bc$ and $y=b^2+c^2-a^2$

= $(2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2)$

= $[a^2-(-2bc+b^2+c^2)][(2bc+b^2+c^2)-a^2]$

= $[a^2-(b-c)^2][(b+c)^2-a^2]$

= $[(a-b+c)(a+b-c)][(b+c-a)(b+c+a)]$

Thus, sum of factors = $(a-b+c)+(a+b-c)+(b+c-a)+(b+c+a)$

= $2a+2b+2c=2(a+b+c)$

=> Ans – (B)

For a number $657423547X46$ to be divisible by 11, the difference between the sum of numbers at even position and odd position should be either ‘0’ or ’11’.

Sum of digits at odd position (starting from right) = $6+X+4+3+4+5=(22+X)$

Even positions = $4+7+5+2+7+6=31$

=> $22+X-31=0$

=> $X=9$

=> Ans – (B)

Expression : $4^{11}+4^{12}+4^{13}+4^{14}\$

= $4^{11}(1+4+4^2+4^3)$

= $4^{11}\times(1+4+16+64)$

= $4^{11}\times(85)$

$\because$ $85$ is divisible by 17, hence the above expression is also divisible by 17

=> Ans – (C)

We hope this Number System Questions pdf for SSC CGL Tier 2 exam will be highly useful for your Preparation.