# SSC > Factorisation

Explore popular questions from Factorisation for SSC. This collection covers Factorisation previous year SSC questions hand picked by experienced teachers.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.

Correct2

Incorrect-0.5

If {tex} a^{2}+b^{2}=24,\ ab=4 {/tex} find the value of {tex} \left(a-b\right) {/tex}

4

B

6

C

3

D

2

##### Explanation

{tex} \left(a-b\right)^{2} = a^{2}+b^{2}-2ab {/tex}

={tex} 24 - 2 \times 4 = 24 - 8 =16 {/tex}

Therefore, {tex} \left(a-b\right)^{2} = 16 {/tex}

{tex} a-b = \sqrt{16} = 4 {/tex}

Q 2.

Correct2

Incorrect-0.5

If {tex} x^{2}+\frac{1}{x^{2}}=38, {/tex} find the value of {tex} x-\frac{1}{x} {/tex}

A

8

6

C

4

D

2

##### Explanation

{tex} \left(x-\frac{1}{x}\right)^{2} = x^{2}+\frac{1}{x^{2}}-2 {/tex}

= 38 - 2 = 36

Therefore, {tex} x-\frac{1}{x} = \sqrt{36} = 6 {/tex}

Q 3.

Correct2

Incorrect-0.5

Simpllify: {tex} \frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}} {/tex}

A

{tex} \left(a + b\right) {/tex}

B

{tex} \left(a - b\right)^{2} {/tex}

{tex} \left(a - b\right) {/tex}

D

{tex} \left(a + b\right)^{2} {/tex}

##### Explanation

{tex} \frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}} = \frac{ \left(a-b\right) \left(a^{2}+ab+b^{2}\right) }{ \left(a^{2}+ab+b^{2}\right) } {/tex}

= {tex} \left(a - b\right) {/tex}

Q 4.

Correct2

Incorrect-0.5

Find the factors of {tex} x^{2}+10x+24 {/tex}

A

{tex} \left(x-6\right) \left(x-4\right) {/tex}

B

{tex} \left(x-6\right) \left(x+4\right) {/tex}

C

{tex} \left(x+6\right) \left(x-4\right) {/tex}

{tex} \left(x+6\right) \left(x+4\right) {/tex}

##### Explanation

{tex} x^{2}+10x+24 = x^{2}+6x+4x+24 {/tex}

={tex} x \left(x+6\right)+4 \left(x+6\right) {/tex}

={tex} \left(x+6\right) \left(x+4\right) {/tex}

Therefore, the factors of {tex} x^{2}+10x+24 {/tex} are ={tex} \left(x+6\right) \left(x+4\right) {/tex}

Q 5.

Correct2

Incorrect-0.5

Simplify {tex} \frac{ \left(a^{3}+b^{3}\right) \left(a^{2}-b^{2}\right) }{ \left(a^{2}-ab+b^{2}\right) \left(a-b\right) } {/tex}

{tex} \left(a+b\right)^{2} {/tex}

B

{tex} \left(a-b\right)^{2} {/tex}

C

{tex} \left(a+b\right) \left(a-b\right) {/tex}

D

None of these

##### Explanation

{tex} \frac{ \left(a^{3}+b^{3}\right) \left(a^{2}+b^{2}\right) }{ \left(a^{2}-ab+b^{2}\right) \left(a-b\right) } = \frac{ \left(a+b\right) \left(a^{2}-ab+b^{2}\right) \left(a+b\right) \left(a-b\right) }{ \left(a^{2}-ab+b^{2}\right) \left(a-b\right) } {/tex}

={tex} \left(a+b\right) \left(a+b\right) = \left(a+b\right)^{2} {/tex}

Q 6.

Correct2

Incorrect-0.5

Find the common factor of {tex} x^{2}+\frac{1}{x^{2}}=102 {/tex}, then the value of {tex} \left(x-\frac{1}{x}\right) {/tex} is

10

B

12

C

6

D

4

##### Explanation

{tex} x^{2}+\frac{1}{x^{2}} = 102 {/tex}

{tex} \left(x-\frac{1}{x}\right)^{2} = x^{2}+\frac{1}{x^{2}}-2 {/tex}

= 102 -2 = 100

{tex} x-\frac{1}{x} = \sqrt{100} = 10 {/tex}

Q 7.

Correct2

Incorrect-0.5

Factorise: {tex} x^{2}+18x+81 {/tex}

A

{tex} \left(x+9\right) \left(x-9\right) {/tex}

{tex} \left(x+9\right)^{2} {/tex}

C

{tex} \left(x+3\right)^{2} {/tex}

D

{tex} \left(x-9\right)^{2} {/tex}

##### Explanation

{tex} x^{2}+18x+81 = x^{2}+2 \times 9x+9^{2} {/tex}

={tex} \left(x+9\right)^{2} {/tex}

Q 8.

Correct2

Incorrect-0.5

Simplify {tex} \frac{ \left(x^{2}-4\right) \left(x^{3}-8\right) }{ \left(x-2\right)^{2} \left(x^{2}+2x+4\right) }=? {/tex}

A

{tex} \left(x-2\right)^{2} {/tex}

B

{tex} \left(x+2\right)^{2} {/tex}

{tex} \left(x+2\right) {/tex}

D

{tex} \left(x-2\right) {/tex}

##### Explanation

The factors of {tex} x^{2}-4 = \left(x+2\right) \left(x-2\right) {/tex}

The factors of {tex} x^{3}-8 = \left(x-2\right) \left(x^{2}+2x+4\right) {/tex}

{tex} \frac{ \left(x^{2}-4\right) \left(x^{3}-8\right) }{ \left(x-2\right)^{2} \left(x^{2}+2x+4\right) } = \frac{ \left(x+2\right) \left(x-2\right) \left(x-2\right) \left(x^{2}+2x+4\right) }{ \left(x-2\right)^{2} \left(x^{2}+2x+4\right) } {/tex}

= {tex} \left(x+2\right) {/tex}

Q 9.

Correct2

Incorrect-0.5

If a+b=5, ab=4 find the value of {tex} a^{3}+b^{3} {/tex}

A

60

B

50

C

56

65

##### Explanation

{tex} a+b=5, ab=4, a^{3}+b^{3}=? {/tex}

Therefore, {tex} \left(a+b\right)^{3} = a^{3}+b^{3}+3ab \left(a+b\right) {/tex}

{tex} 5^{3} = a^{3}+b^{3}+3 \times 4 \times 5 {/tex}

{tex} a^{3}+b^{3}=125 - 60 = 65 {/tex}

Q 10.

Correct2

Incorrect-0.5

Simplify {tex} \frac{x^{2}+11x+30}{ \left(x^{2}+2x-15\right) } {/tex}

{tex} \frac{x+6}{x-3} {/tex}

B

{tex} \frac{x+3}{x-3} {/tex}

C

{tex} \frac{x+6}{x-5} {/tex}

D

{tex} \frac{x+5}{x+6} {/tex}

##### Explanation

Factors of {tex} x^{2}+11x+30= \left(x+6\right) \left(x+5\right) {/tex}

Factors of {tex} x^{2}+2x -15 = \left(x+5\right) \left(x-3\right) {/tex}

{tex} \frac{x^{2}+11x+30}{x^{2}+2x-15} - \frac{ \left(x+6\right) \left(x+5\right) }{ \left(x+5\right) \left(x-3\right) } = \frac{x+6}{x-3} {/tex}

Q 11.

Correct2

Incorrect-0.5

If a=5, d=3, find 40th term.

122

B

160

C

142

D

124

##### Explanation

a=5, d=3, n=40

Formula: {tex} t_{n}=a+ \left(n-1\right)d {/tex}

{tex} 5+ \left(40-1\right) \times 3 = 5 + 39 \times 3 = 5 + 117 = 122 {/tex}

Q 12.

Correct2

Incorrect-0.5

Find the sum 9+15+27+....+195.

A

3244

B

3344

3264

D

3364

##### Explanation

Here a=9, d=6, t=195

Therefore, {tex} n=\frac{l-a}{d}+1 {/tex}

={tex} \frac{195-9}{6}+1 = \frac{186}{6}+1 = 32 {/tex}

{tex} S_{n} =\frac{n}{2} \left(2a \left(n-1\right)d \right) {/tex}

{tex} \frac{32}{2} \{ 18+ \left(32-1\right) \times 6 \} {/tex}

={tex} 16 \left(18+31 \times 6\right) {/tex}

={tex} 16 \left(18+186\right) {/tex}

={tex} 16 \times 204 = 3264 {/tex}

Q 13.

Correct2

Incorrect-0.5

In a GP 96, 48, 24 ...... find the value of the 6th term.

3

B

6

C

{tex} \frac{3}{4} {/tex}

D

{tex} \frac{3}{16} {/tex}

##### Explanation

Given GP is 96, 48, 24.....

{tex} t_{n}=ar^{n-1} {/tex} her a=96, r={tex} \frac{1}{2} {/tex}, n=6

{tex} t_{6} = 96 \times \left(\frac{1}{2}\right)^{6-1} {/tex}

= {tex} 96 \left(\frac{1}{2}\right)^{5} {/tex}

={tex} \frac{96}{32} = 3 {/tex}

Q 14.

Correct2

Incorrect-0.5

In a GP 2, 6, 18, 54 .... find how many terms are there, if their sum is 728.

A

5

6

C

12

D

7

##### Explanation

Given GP is 2, 6, 18, 54

{tex} S_{n} {/tex} = 728 Here a = 2, r = 3

{tex} S_{n} {/tex} = {tex} \frac{a \left(r^{n}-1\right) }{r-1} {/tex}

728 = {tex} \frac{2 \left(3^{n}-1\right) }{3-1} = 3^{n}-1 {/tex}

{tex} 3^{n} = 728+1 = 729 {/tex}

{tex} 3^{n} = 3^{6} {/tex}

n = 6

Q 15.

Correct2

Incorrect-0.5

Find the sum 24+25+....+100

4774

B

4674

C

4764

D

4467

##### Explanation

{tex} \sum_{n=1}^{100} n = \frac{100 \left(100+1\right) }{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050 {/tex}

{tex} \sum_{n=1}^{23} \frac{23 \times 24}{2}=23 \times 12 = 276 {/tex}

Therefore, 24+25+....+100 = 5050-276=4774

Q 16.

Correct2

Incorrect-0.5

Find the sum of 1+3+5+....+45

A

540

B

539

529

D

579

##### Explanation

1+3+5+....+45

Here a=1, d=2, l=45

{tex} n=\frac{l-a}{d}+1 = \frac{45-1}{2}+1 = 22+1 = 23{/tex}

Therefore, 1+3+5+....+45 = {tex} n^{2} {/tex}

={tex} 23^{2} = 529 {/tex}

Q 17.

Correct2

Incorrect-0.5

Find the sum of {tex} 21^{2}+22^{2}+....+60^{2} {/tex}

70940

B

2870

C

12810

D

9946

##### Explanation

{tex} 21^{2}+22^{2}+....+60^{2}= \left(1^{2}+2^{2}+....+60^{2}\right) {/tex}- {tex} \left(a^{2}+2^{2}+....+20^{2}\right) {/tex}

={tex} \frac{60 \left(60+1\right) \left(120+1\right) }{6} - \frac{20 \times 21 \times 41}{6} {/tex}

= {tex} \frac{60 \times 61 \times 121}{6} - \frac{20 \times 21 \times 41}{6} {/tex}

= {tex} 610 \times 121-70 \times 41=73810-2870 = 70940 {/tex}

Q 18.

Correct2

Incorrect-0.5

Find the sum of {tex} 1^{3}+2^{3}+....+10^{3} {/tex}

A

163675

3025

C

163765

D

cannot be determined

##### Explanation

{tex} 1^{3}+2^{3}+....+10^{3} {/tex}

Formula: {tex} \sum n^{3}=\left[ \frac{n \left(n+1\right) }{2} \right]^{2} {/tex}

At sum {tex} 10^{3}= \left(\frac{10 \times 11}{2}\right)^{2} = 55^{2} = 3025 {/tex}

Q 19.

Correct2

Incorrect-0.5

Find the sum of {tex} 15^{3}+16^{3}+....+50^{3} {/tex}

1614600

B

1614660

C

1614670

D

1615676

##### Explanation

{tex} 15^{3}+16^{3}+....+50^{3}= \left(1^{3}+2^{3}+....+50^{3}\right) - {/tex} {tex} \left(1^{3}+2^{3}+.....+14^{3}\right) {/tex}

{tex} 1^{3}+2^{3}+3^{3}+....+50^{3} = \left(\frac{50 \times 51}{2}\right)^{2} = 1275^{2} {/tex}

= 1625625

{tex} 1^{3}+2^{3}+....+14^{3} = \left(\frac{14 \times 15}{2}\right)^{2}=105^{2}=11025 {/tex}

{tex} 15^{3}+16^{3}+.....+50^{3}=1625625 - 11025 = 1614600 {/tex}

Q 20.

Correct2

Incorrect-0.5

Find the square root of {tex} x^{4}+2x^{2}y^{2}+y^{4} {/tex}

{tex} x^{2}+y^{2} {/tex}

B

{tex} \left(x+y\right)^{2} {/tex}

C

{tex} x^{2}-y^{2} {/tex}

D

None of these

##### Explanation

{tex} x^{4}+2x^{2}y^{2}+y^{4} = \left(x^{2}+y^{2}\right)^{2} {/tex}

{tex} \sqrt{x^{4}+2x^{2}y^{2}+y^{4}}=\sqrt{ \left(x^{2}+y^{2}\right)^{2} } = x^{2}+y^{2} {/tex}

Q 21.

Correct2

Incorrect-0.5

Find the square root of {tex} \left(a^{2}-1\right) \left(a^{2}-2a-3\right) \left(a^{2}-4a+3\right) {/tex}

A

{tex} \left(a+1\right) \left(a-1\right) \left(a-3\right) {/tex}

{tex} \left(a-1\right) \left(a+3\right) \left(a+1\right) {/tex}

C

{tex} \left(a-1\right) \left(a+1\right) {/tex}

D

None of these

##### Explanation

{tex} a^{2}-1 = \left(a+1\right) \left(a-1\right) {/tex}

{tex} a^{2}-2a-3= \left(a-3\right) \left(a+1\right) {/tex}

{tex} a^{2}-4a+3 = \left(a-3\right) \left(a-1\right) {/tex}

Therefore, {tex} \sqrt{ \left(a^{2}-1\right) \left(a^{2}-2a-3\right) \left(a^{2}-4a+3\right) } {/tex}

= {tex} \sqrt{ \left(a+1\right) \left(a-1\right) \left(a-3\right) \left(a+1\right) \left(a-3\right) \left(a-1\right) } {/tex}

{tex} = \left(a+1\right) \left(a-1\right) \left(a-3\right) {/tex}

Q 22.

Correct2

Incorrect-0.5

Which of the following is correct. The squares root of {tex} 25x^{4}+30x^{3}+19x^{2}+6x+1 {/tex}

A

{tex} 5x^{2}-3x-1 {/tex}

{tex} 5x^{2}+3x+1 {/tex}

C

{tex} 5x^{2}+6x+1 {/tex}

D

{tex} 5x^{2}-6x+1 {/tex}

##### Explanation

{tex} 5x^{2}+3x+1 {/tex}

{tex} 25x^{4}+30x^{3}+19x^{2}+6x+1 \div 5x^{2} = 25x^{4} {/tex}

{tex} 30x^{3}+19x^{2} \div 10x^{2}+3x=10x^{2}+6x+1 {/tex}

{tex} 10x^{2}+6x+1 \div 10x^{2}+6x+1 = 0 {/tex}

The square root of {tex} 25x^{4}+30x^{3}+19x^{2}+6x+1 {/tex} is

{tex} 5x^{2}+3x+1 {/tex}

Q 23.

Correct2

Incorrect-0.5

If {tex} 25x^{4}+20x^{3}-26x^{2}+ax+b {/tex} is a perfect square, find the value of a and b

A

a = 12, c = 9

B

a = 12, b = 9

a = -12, b = 9

D

a = 6, b = 9

##### Explanation

{tex} 5x^{2}+2x-3 {/tex}

{tex} 25x^{4}+20x^{3}-26x^{2}+ax+b \div 5x^{2}=25x^{4} {/tex}

{tex} 20x^{3}-26x^{2} \div 10x^{2}+2x = 20x^{3+4x^{2}} {/tex}

{tex} -30x^{2}+ax+b \div 10x^{2}+4x-3 = -30x^{2}-12x+9 = 0 {/tex}

Therefore, a = -12, b = 9

Q 24.

Correct2

Incorrect-0.5

In a rectangle, the length of the rectangle is 10 mts more than the breadth. If the area of the rectangle is {tex} 144 m^{2} {/tex}, then its perimeter is

A

52 mts.

54 mts.

C

18 mts.

D

36 mts.

##### Explanation

Let x be the breadth of the rectangle.

length = x+10

{tex} x \left(x+10\right) = 144 {/tex} (Given)

{tex} x^{2}+10x-144 = 0 {/tex}

{tex} x^{2}+18x-8x-144 = 0 {/tex}

{tex} x \left(x+18\right)-8 \left(x+18\right) = 0 {/tex}

{tex} \left(x+18\right) \left(x-8\right)=0 {/tex}

x-8 = 0, x=8

x+18 =0, x = -18

breadth of the rectangle = 8mts.length of the rectangle = 10+8 = 18 mts

perimeter = {tex} 2 \left(l+b\right)=2 \left(18+8\right) = 52 mts. {/tex}

Q 25.

Correct2

Incorrect-0.5

If {tex} \alpha {/tex} and {tex} \beta {/tex} are the roots of the equation {tex} x^{2}-7x+12=0 {/tex}, then find the value of {tex} \alpha ^{2}+ \beta ^{2} {/tex}

25

B

24

C

49

D

12

##### Explanation

If {tex} \alpha and \beta {/tex} are the roots of the eqn.

{tex} ax^{2}+bx+c=0 {/tex}

then {tex} \alpha + \beta = -\frac{b}{a} {/tex}

and {tex} \alpha \beta =\frac{c}{a} {/tex}

Here the given equation is {tex} x^{2}-7x+12=0 {/tex}

Here, {tex} \alpha + \beta = 7, \alpha \beta = 12 {/tex}

{tex} \alpha ^{2}+ \beta ^{2}= \left( \alpha + \beta \right)^{2}-2 \alpha \beta {/tex}

= {tex} 7^{2}-2 \times 12 {/tex}

= 49 - 24 = 25