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SSC > Factorisation

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

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