SSC CHSL ALGEBRA QUESTIONS WITH SOLUTIONS

SSC CHSL Algebra Questions with solutions. 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

Answer

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

Answer

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

Answer

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

Answer

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) $\sqrt2$

Answer

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) $30\sqrt2$

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

Answer

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) $22\sqrt2$

Answer

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

Answer

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

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

Answer

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

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

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

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

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

Answer

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

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

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

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) $21- 12\sqrt3$

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

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) $4\sqrt{11}$

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

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

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) $225\sqrt2$

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

Answer

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

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

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

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

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