SSC CHSL ALGEBRA QUESTIONS WITH SOLUTIONS

SSC CHSL Algebra Questions with solutions from all shifts of August 2021. Solved Algebra question answer with short tricks are very useful for upcoming SSC CGL, CHSL and other Competitive Exams.

SSC CHSL ALGEBRA QUESTIONS

Q.1: If a + b = p, ab = q, then (a4 + b4 ) is equal to:
(A) p4 – 4p2q + q2
(B) p4 – 4p2q2 + 2q2
(C) p4 – 2p2q2 + q2
(D) p4 – 4p2q + 2q2

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Ans : (D) p4 – 4p2q + 2q2
a + b = p , ab = q
(a + b)2 = p2
a2 + 2ab + b2 = p2
a2 + b2 = p2 – 2q
(a2 + b2)2 = (p2 – 2q)2
a4 + b4 + 2a2b2 = p4 – 4p2q + 4q2 [ab = q, a2 b2 = q2]
a4 + b4 = p4 – 4p2q + 2q2

Q.2: If (x +\frac 1x)^3 = 27 , then what is the value of (x^2 +\frac {1}{x^2}) ? Given the x is real.
(A) 9
(B) 25
(C) 7
(D) 11

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Ans : (C) 7
(x +\frac 1x)^3 = 27
x +\frac 1x = (27)^{1/3} = 3
x^2 +\frac {1}{x^2} + 2 = 9
x^2 +\frac {1}{x^2} = 9 - 2 = 7

Q.3: If x -\frac 2x = 4 , then what will be the value of x^2 + \frac {4}{x^2} ?
(A) 18
(B) 8
(C) 12
(D) 20

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Ans : (D) 20
({x -\frac 2x)^2} = 4
x^2 +\frac {4}{x^2} - 2\times2 = 16
x^2 +\frac {4}{x^2} = 16 + 4 = 20

Q.4: If \sqrt{x} + \frac {1} {\sqrt x} = \sqrt 6 , then the value of x^6 + \frac {1}{x^6} will be:
(A) 2712
(B) 2502
(C) 2270
(D) 2702

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Ans : (D) 2702
\sqrt{x} + \frac {1} {\sqrt x} = \sqrt6
(\sqrt{x} + \frac {1} {\sqrt x})^2 = (\sqrt6)^2
x +\frac 1x +2 = 6
x +\frac 1x = 4
(x +\frac 1 x)^2 = (4)^2
x^2 +\frac {1}{x^2} = 16 - 2 = 14
(x^2 +\frac {1}{x^2})^3 = (14)^3
x^6 +\frac {1}{x^6} + 3x^2\times\frac {1}{x^2}\times (x^2 + (\frac {1}{x^2}) = 2744
x^6 +\frac {1}{x^6} = 2744 - 42 = 2702

Q.5: If x2 + 1 – 2x = 0, x > 0 , then x2(x2 – 2) =_______.
(A) 0
(B) -1
(C) 1
(D) \sqrt2

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Ans : (B) -1
x2 + 1 – 2x = 0
x > 0
x = 1, so 1 + 1 – 2 = 0
Put x = 1 in x2(x2 – 2)
1 (1 – 2)
= -1

Q.6: If x^2 -3\sqrt2x + 1 =0 , then what is the value of x^3 + (\frac {1}{x^3}) ?
(A) 15\sqrt6
(B) 30\sqrt6
(C) 45\sqrt2
(D) 30\sqrt2

Show Answer
Ans : (C) 45\sqrt2
x^2 - 3\sqrt2x + 1 = 0
x - 3\sqrt2 +\frac1x = 0
x +\frac 1x =3\sqrt2
(x +\frac 1x)^3 = (3\sqrt2)^3
x^3 +\frac {1}{x^3} + 3x\times\frac1x\times (x +\frac1x) = 54\sqrt2
x^3 +\frac{1}{x^3} = 54\sqrt2 - 9\sqrt2
= 45\sqrt2

Q.7: If x – y = 4 and xy = 3 , then what is the value of x3 – y3 ?
(A) 88
(B) 100
(C) 64
(D) 28

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Ans : (B) 100
x – y = 4
(x – y)3 = 43
x3 – y3 – 3xy (x – y) = 64 [Put xy = 3, x – y =4]
x3 – y3 – 3 x 3 x 4 = 64
x3 – y3 = 64 + 36 = 100

Q.8: If x -\frac 1x = 2\sqrt2 , then what will be the value of x^3 -\frac {1}{x^3} ?
(A) 12\sqrt2
(B) 10\sqrt2
(C) 20\sqrt2
(D) 22\sqrt2

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Ans : (D) 22\sqrt2
x -\frac 1x =2\sqrt2
(x -\frac 1x)^3 = (2\sqrt2)^3
x^3 -\frac {1}{x^3} - 3\times x\times\frac 1x (x -\frac1x) = 16\sqrt2
x^3 -\frac {1}{x^3} - 3\times 2\sqrt2 = 16\sqrt2
x^3 -\frac {1}{x^3} = 16\sqrt2 + 6\sqrt2
= 22\sqrt2

Q.9: If x + 2y = 19 and x3 + 8y3 = 361 , then xy is equal to:
(A) 58
(B) 56
(C) 55
(D) 57

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Ans : (D) 57
x + 2y = 19
(x + 2y)3 = (19)3
x3 + 8y3 + 6xy (x + 2y) = 6859
x3 + 8y3 + 6xy x 19 = 6859
6xy x 19 = 6859 – 361 = 6498
xy =\frac {6498}{19\times6}
xy = 57

Q.10: If (x^2 + \frac{1}{49x^2}) = 15\frac 57 , then what is the value of (x +\frac{1}{7x}) ?
(A) 4
(B) \pm 7
(C) \pm 4
(D) 7

Show Answer
Ans : (C) \pm 4
(x^2 + \frac{1}{49x^2}) = 15\frac57
(x^2 + \frac{1}{7x})^2 + 2\times x\times \frac {1}{7x} =\frac{110}{7} + \frac27
(x + \frac{1}{7x})^2 = \frac{112}{7} = 16
x + \frac{1}{7x} =\pm 4

SSC CHSL Algebra Questions with solutions for Competitive Exams

Q.11: If x + y = 27 and x2 + y2 = 425 , then the value of (x – y)2 will be:
(A) 121
(B) 225
(C) 169
(D) 144

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Ans : (A) 121
x + y = 27
x2 + y2 = 425
(x + y)2 = (27)2
x2 + y2 + 2xy = 729
2xy = 729 – 425
2xy = 304
(x – y)2 = x2 + y2 – 2xy
= 425 – 304
(x – y)2 = 121

Q.12: If 3x + y = 12 and xy = 9 , then the value of (3x – y) is:
(A) 6
(B) 5
(C) 3
(D) 4

Show Answer
Ans : (A) 6
3x + y = 12
xy = 9
put x = 3, y = 3 Both Equation
So its satify
3x – y = 3 x 3 – 3
= 9 – 3 = 6

Q.13: If a2 + b2 + c2 = 576 and (ab + bc + ca) = 50 , then what is the value of (a + b + c), if (a+b+c) < 0 ?
(A) -24
(B) \pm 24
(C) \pm 26
(D) -26

Show Answer
Ans : (D) -26
(a + b + c)2 = a2 + b2 +c2 + 2 (ab + bc + ca)
576 + 2 x 50
(a + b + c)2 = 676
a + b + c = \pm 26
a + b + c < 0
so a+b+c = -26

Q.14: If x +\frac{1}{3x} = 5 , then the value of 27x^3 +\frac{1}{x^3} will be:
(A) 3024
(B) 3420
(C) 3042
(D) 3240

Show Answer
Ans : (D) 3240
x +\frac{1}{3x} = 5
3x +\frac{1}{x} = 15 [ Multiply by 3]
(3x +\frac{1}{x})^3 = 15^3
27x^3 +\frac{1}{x^3} + 135 = 3375
27x^3 +\frac{1}{x^3} = 3375 - 135
=3240

Q.15:If 3x + 5y = 14 and xy = 6 , then what is the value of 9x2 + 25y2 ?
(A) 182
(B) 16
(C) 14
(D) 20

Show Answer
Ans : (B) 16
(3x + 5y)2 = (14)2
9x2 + 25y2 + 30xy = 196
9x2 + 25y2 + 30 x 6 = 196
9x2 + 25y2 = 196 – 180
9x2 + 25y2 = 16

Q.16: If a – b = 7 and a2 + b2 = 169 where a,b >0 , then the value of 3 a+b is:
(A) 41
(B) 46
(C) 38
(D) 44

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Ans : (A) 41
a – b = 7
(a – b)2 = 72
a2 + b2 – 2ab = 49
169 – 2ab = 49
2ab = 169 – 49
2ab = 120
ab = 60 [12 x 5]
a = 12, b = 5
3a + b
= 3 x 12 + 5
= 41

Q.17: If a + 5b = 25 and ab = 20 , then one of the values of (a-5 b) is:
(A) 16
(B) 15
(C) 13
(D) 14

Show Answer
Ans : (B) 15
a + 5b = 25, ab = 20
Put a = 20
b = 1
so a – 5b
= 20 – 5 x 1 = 15

Q.18: If \sqrt {x} + \frac{1}{x} =2\sqrt{3} , then what will be the value of x^4 +\frac{1}{x^4} ?
(A) 10406
(B) 10402
(C) 9602
(D) 9606

Show Answer
Ans : (C) 9602
\sqrt {x} +\frac{1}{x} = 2\sqrt3
(\sqrt {x} +\frac{1}{x})^2 = (2\sqrt3)^2
x +\frac{1}{x} + 2 = 12
x +\frac{1}{x} = 10
x^2 + \frac{1}{x^2} = 100 - 2 = 98
x^4 + \frac{1}{x^4} = 98^2 - 2
= 9604 – 2
x^4 +\frac{1}{x^4} = 9602

Q.19: If (7x – 10y) = 8 and xy = 5 , then what is the value of 49x2 + 100y2 ?
(A) 632
(B) 623
(C) 746
(D) 764

Show Answer
Ans : (D) 764
7x – 10y = 8
(7x – 10y)2 = (8)2
49x2 + 100y2 – 140xy = 64
49x2 + 100y2 = 64 + 140xy
= 64 + 140xy
= 64 + 140 x 5
= 764

Q.20: If x^2 +(4 - \sqrt {3})x - 1 = 0 , then what is the value of x^2 +\frac{1}{x^2} ?
(A) 21- 8\sqrt3
(B) 17- 8\sqrt3
(C) 9- 8\sqrt3
(D) 21- 12\sqrt3

Show Answer
Ans : (A) 21- 8\sqrt3
x^2 +(4 -\sqrt3)x - 1 = 0
\div x Both side
x +(4 -\sqrt{3}) - \frac1x = 0
x -\frac1x = \sqrt3 - 4
square Both side
(x -\frac1x)^2 = (\sqrt 3 - 4)^2
x^2 +\frac{1}{x^2}  - 2 = 3 + 16 - 8\sqrt3
x^2 +\frac{1}{x^2} = 21 - 8\sqrt3

SSC CHSL Algebra Questions with solutions for Competitive Exams

Q.21: If x^2 +\frac{1}{x^2} = 83, x > 0 , then the value of x^3 +\frac{1}{x^3} is:
(A) 675
(B) 756
(C) 746
(D) 576

Show Answer
Ans : (B) 756
x^2 +\frac{1}{x^2} = 83
x^2 +\frac{1}{x^2} - 2 = 83 - 2
(x -\frac{1}{x})^2 = 81
x -\frac{1}{x} = 9
(x -\frac{1}{x})^3 = 9^3
x^3 -\frac{1}{x^3} -3x \frac1x (x -\frac1x) = 729
x^3 -\frac{1}{x^3} = 729 + 27
= 756

Q.22: If x +\frac1x = \sqrt{13} , then one of the values of x^3 -\frac{1}{x^3} is:
(A) 36
(B) 32
(C) 4\sqrt{13}
(D) 4\sqrt{11}

Show Answer
Ans : (A) 36
x +\frac1x = \sqrt{13}
x^2 +\frac{1}{x^2} + 2 = 13
x^2 +\frac{1}{x^2} = 11
x^2 +\frac{1}{x^2} - 2 = 11 - 2
x^2 +\frac{1}{x^2} - 2 = 9
(x -\frac{1}{x})^2 = 9
x -\frac1x = 3
x^3 +\frac{1}{x^3} - 3\times3 = 27
x^3 -\frac{1}{x^3} = 27 + 9
= 36

Q.23: The coefficient of x3 y in (x – 2y) x (5x + y)3 is:
(A) -150
(B) 75
(C) -175
(D) 250

Show Answer
Ans : (C) -175
(x – 2y) x (5x + y)3
(x – 2y) x [ 125x3 + y3 + 15xy (5x + y)
(x – 2y) x [ 125x3 + y3 + 75x2y + 15xy2 ]
coefficient of x3y
= 75x3 – 250x3y
= -175x3y
so coefficient of x3y
= -175

Q.24: If 9x2 – 6x + 1 = 0 , then the value of 27x3 + (27x3)-1 will be:
(A) 1
(B) 4
(C) 2
(D) 8

Show Answer
Ans : (C) 2
9x2 – 6x + 1 = 0
27x3 + (27x3)-1
9x2 – 6x + 1 = 0
(3x – 1)2 = 0
3x – 1 = 0
3x = 1
x =\frac13
put x =\frac13
27x^3 +\frac{1}{27^3}
27\times\frac{1}{27} +\frac{{1}\times{27}}{{27}\times{1}}
= 1 + 1 = 2

Q.25: What is the coefficient of y2 in the expansion of (\sqrt2y^2 - 5\sqrt3)^3 ?
(A) 30\sqrt3
(B) -225\sqrt2
(C) -30\sqrt3
(D) 225\sqrt2

Show Answer
Ans : (D) 225\sqrt2
(\sqrt2y^2 - 5\sqrt3)^3
(a – b)3 = a3 – b3 – 3ab(a – b)
(\sqrt2y^2)^3 - (5\sqrt3)^3 - 3\sqrt2y^2\times5\sqrt3 (\sqrt2y^2 - 5\sqrt3)
केवल y2 का गुणांक लेने पर
= -3\sqrt2y^2 \times 5\sqrt3\times - 5\sqrt3
= +225\sqrt2y^2
y2 का गुणांक = 225\sqrt2

Q.26: If a + b + c = 2 and ab + bc + ca = -1 , then the value of a3 + b3 + c3 – 3abc is:
(A) 5
(B) 10
(C) 2
(D) 14

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Ans : (D) 14
a + b + c = 2
(a + b + c)2 = ( 2 )2
a2+b2+c2 + 2(ab + bc + ca) = 4
a2+b2+c2 + 2 x(-1) = 4
a2+b2+c2 = 6
a3+b3+c3 – 3abc = (a+b+c) [(a2+b2+c2) – (ab+bc+ca)]
= 2[6 – (-1)]
= 2 x 7 = 14

Q.27: If x +\frac{1}{15x} =3 , then the value of 9x^3 +\frac{1}{375x^3} will be:
(A) 237.6
(B) 273.6
(C) 367.2
(D) 376.2

Show Answer
Ans : (A) 237.6
x +\frac{1}{15x} =3
(x +\frac{1}{15x})^3 =(3)^3
x^3 +\frac{1}{3375x^3} +3x\times\frac{1}{15x} \times3 = 27
x^3 +\frac{1}{3375x^3} +\frac35 = 27
x^3 +\frac{1}{3375x^3} = 27 -\frac35 =\frac{132}{5}
9 से गुणा Both side
9x^3 +\frac{1}{375x^3} =\frac{132}{5}\times 9
9x^3 +\frac{1}{375x^3} =\frac{1188}{2} = 237.6

Q.28: If a+b+c = 11 and ab+bc+ca = 15 , then what is the value of a3+b3+c3 – 3abc ?
(A) 386
(B) 836
(C) 368
(D) 638

Show Answer
Ans : (B) 836
a+b+c = 11
(a+b+c)2 = 112
a2+b2+c2 + 2 (ab+bc+ca) = 121
a2+b2+c2 + 2 x 15 = 121
a2+b2+c2 = 121 – 30
a2+b2+c2 = 91
a3+b3+c3-3abc =(a+b+c) [(a2+b2+c2 )-(ab+bc+ca)]
= 11[91 – 15]
= 11 x 76
a3+b3+c3-3abc = 836

Q.29: If a+b+c = 5, a2+b2+c2 = 27, and a3+b3+c3 = 125 , then the value of \frac{abc}{5} is:
(A) -5
(B) -1
(C) 5
(D) 1

Show Answer
Ans : (B) -1
a+b+c = 5
(a+b+c)2 = (5)2
a2+b2+c2 +2(ab+bc+ca) = 25
27 + 2(ab+bc+ca) = 25
ab+bc+ca = -1
a3+b3+c3 -3abc = (a+b+c) [(a2+b2+c2 -(ab+bc+ca)
125 – 3abc = 5 [27 – (-1)
= 5 x 28
125 – 3abc = 140
3abc = -15
abc = -5
\frac{abc}{5} = -1

Q.30: If x4 + x-4 = 47, x>0 , then the value of (2x – 3)2 is:
(A) 9
(B) 5
(C) 3
(D) 7

Show Answer
Ans : (B) 5
x^4 + \frac{1}{x^4} = 47
x^4 + \frac{1}{x^4} + 2 = 47+ 2
(x^2 + \frac{1}{x^2})^2 = 49
x^2 + \frac{1}{x^2} = 7
x^2 + \frac{1}{x^2} + 2 = 7+2
(x + \frac{1}{x})^2 = 9
x +\frac1x = 3
x2 + 1 = 3x
4x2 – 12x = -4
(2x – 3)2
= 4x2 + 9 – 12x
4x2 – 12x + 9
-4 + 9 = 5

SSC CHSL Algebra Questions with solutions for Competitive Exams

Q.31: If a-\frac{24}{a} = 5 , where a > 0, then the value of a^2 +\frac{64}{a^2} is:
(A) 45
(B) 60
(C) 65
(D) 56

Show Answer
Ans : (C) 65
a-\frac{24}{a} = 5
Put x = 8 its satisfy this equation
So a^2-\frac{64}{a^2}
64 +\frac{64}{64}
64 + 1 = 65

Q.32: If x = 555, y = 556 and z = 557 , then find the value of x3+y3+z3 – 3xyz.
(A) 5002
(B) 5008
(C) 5006
(D) 5004

Show Answer
Ans : (D) 5004
x3+y3+z3 – 3xyz = \frac12 (x+y+z)[(x – y)2 +(y – z)2 +(z – x)2 ]
\frac12\times 1668\times[1+1+4]
= 3 x 1668
= 5004

Q.33: If 3x-2y+3=0 , then what will be the value of 27x3+54xy+30-8y3 ?
(A) 57
(B) -57
(C) -27
(D) 3

Show Answer
Ans : (D) 3
3x-2y+3=0
3x-2y=-3
(3x-2y)3 = (-3)3
27x3 – 8y3+54xy =-27
27x3 – 8y3 =-27 – 54xy
27x3 +54xy+30 -8y3
-27-54xy+54xy+30
= 30-27 = 3

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