# 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

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

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

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

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) $\sqrt2$

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) $30\sqrt2$

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

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) $22\sqrt2$

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

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

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

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

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

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

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

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

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

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

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

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) $21- 12\sqrt3$

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

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) $4\sqrt{11}$

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

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

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) $225\sqrt2$

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

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

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

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

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

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

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

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

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

Thanks for Study the SSC CHSL Algebra Questions with solutions

Algebra Questions with solutions in Hindi for SSC CHSL Exams