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) $30\sqrt2$

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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

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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

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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

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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

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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

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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

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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

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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

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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) $21- 12\sqrt3$

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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

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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) $4\sqrt{11}$

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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

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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

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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) $225\sqrt2$

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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

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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

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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

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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
$\frac{abc}{5} = -1$