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 $latex (x +\frac 1x)^3 = 27$ , then what is the value of $latex (x^2 +\frac {1}{x^2})$ ? Given the x is real.
(A) 9
(B) 25
(C) 7
(D) 11
Answer
Ans : (C) 7
$latex (x +\frac 1x)^3 = 27$
$latex x +\frac 1x = (27)^{1/3} = 3$
$latex x^2 +\frac {1}{x^2} + 2 = 9$
$latex x^2 +\frac {1}{x^2} = 9 – 2 = 7$
Q.3: If $latex x -\frac 2x = 4$ , then what will be the value of $latex x^2 + \frac {4}{x^2}$ ?
(A) 18
(B) 8
(C) 12
(D) 20
Answer
Ans : (D) 20
$latex ({x -\frac 2x)^2} = 4$
$latex x^2 +\frac {4}{x^2} – 2\times2 = 16$
$latex x^2 +\frac {4}{x^2} = 16 + 4 = 20$
Q.4: If $latex \sqrt{x} + \frac {1} {\sqrt x} = \sqrt 6$ , then the value of $latex x^6 + \frac {1}{x^6}$ will be:
(A) 2712
(B) 2502
(C) 2270
(D) 2702
Answer
Ans : (D) 2702
$latex \sqrt{x} + \frac {1} {\sqrt x} = \sqrt6$
$latex (\sqrt{x} + \frac {1} {\sqrt x})^2 = (\sqrt6)^2$
$latex x +\frac 1x +2 = 6$
$latex x +\frac 1x = 4$
$latex (x +\frac 1 x)^2 = (4)^2$
$latex x^2 +\frac {1}{x^2} = 16 – 2 = 14$
$latex (x^2 +\frac {1}{x^2})^3 = (14)^3$
$latex x^6 +\frac {1}{x^6} + 3x^2\times\frac {1}{x^2}\times (x^2 + (\frac {1}{x^2}) = 2744$
$latex x^6 +\frac {1}{x^6} = 2744 – 42 = 2702$
Q.5: If x2 + 1 – 2x = 0, x > 0 , then x2(x2 – 2) =_______.
(A) 0
(B) -1
(C) 1
(D) $latex \sqrt2$
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 $latex x^2 -3\sqrt2x + 1 =0$ , then what is the value of $latex x^3 + (\frac {1}{x^3})$ ?
(A) $latex 15\sqrt6$
(B) $latex 30\sqrt6$
(C) $latex 45\sqrt2$
(D) $latex 30\sqrt2$
Answer
Ans : (C) $latex 45\sqrt2$
$latex x^2 – 3\sqrt2x + 1 = 0$
$latex x – 3\sqrt2 +\frac1x = 0$
$latex x +\frac 1x =3\sqrt2$
$latex (x +\frac 1x)^3 = (3\sqrt2)^3$
$latex x^3 +\frac {1}{x^3} + 3x\times\frac1x\times (x +\frac1x) = 54\sqrt2$
$latex x^3 +\frac{1}{x^3} = 54\sqrt2 – 9\sqrt2$
$latex = 45\sqrt2$
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 $latex x -\frac 1x = 2\sqrt2$ , then what will be the value of $latex x^3 -\frac {1}{x^3}$ ?
(A) $latex 12\sqrt2$
(B) $latex 10\sqrt2$
(C) $latex 20\sqrt2$
(D) $latex 22\sqrt2$
Answer
Ans : (D) $latex 22\sqrt2$
$latex x -\frac 1x =2\sqrt2$
$latex (x -\frac 1x)^3 = (2\sqrt2)^3$
$latex x^3 -\frac {1}{x^3} – 3\times x\times\frac 1x (x -\frac1x) = 16\sqrt2$
$latex x^3 -\frac {1}{x^3} – 3\times 2\sqrt2 = 16\sqrt2$
$latex x^3 -\frac {1}{x^3} = 16\sqrt2 + 6\sqrt2$
$latex = 22\sqrt2$
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 =$latex \frac {6498}{19\times6}$
xy = 57
Q.10: If $latex (x^2 + \frac{1}{49x^2}) = 15\frac 57$ , then what is the value of $latex (x +\frac{1}{7x})$ ?
(A) 4
(B) $latex \pm 7$
(C) $latex \pm 4$
(D) 7
Answer
Ans : (C) $latex \pm 4$
$latex (x^2 + \frac{1}{49x^2}) = 15\frac57$
$latex (x^2 + \frac{1}{7x})^2 + 2\times x\times \frac {1}{7x} =\frac{110}{7} + \frac27$
$latex (x + \frac{1}{7x})^2 = \frac{112}{7} = 16$
$latex x + \frac{1}{7x} =\pm 4$
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) $latex \pm 24$
(C) $latex \pm 26$
(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 = $latex \pm 26$
a + b + c < 0
so a+b+c = -26
Q.14: If $latex x +\frac{1}{3x} = 5$ , then the value of $latex 27x^3 +\frac{1}{x^3}$ will be:
(A) 3024
(B) 3420
(C) 3042
(D) 3240
Answer
Ans : (D) 3240
$latex x +\frac{1}{3x} = 5$
$latex 3x +\frac{1}{x} = 15$ [ Multiply by 3]
$latex (3x +\frac{1}{x})^3 = 15^3$
$latex 27x^3 +\frac{1}{x^3} + 135 = 3375$
$latex 27x^3 +\frac{1}{x^3} = 3375 – 135$
=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 $latex \sqrt {x} + \frac{1}{x} =2\sqrt{3}$ , then what will be the value of $latex x^4 +\frac{1}{x^4}$ ?
(A) 10406
(B) 10402
(C) 9602
(D) 9606
Answer
Ans : (C) 9602
$latex \sqrt {x} +\frac{1}{x} = 2\sqrt3$
$latex (\sqrt {x} +\frac{1}{x})^2 = (2\sqrt3)^2$
$latex x +\frac{1}{x} + 2 = 12$
$latex x +\frac{1}{x} = 10$
$latex x^2 + \frac{1}{x^2} = 100 – 2 = 98$
$latex x^4 + \frac{1}{x^4} = 98^2 – 2$
= 9604 – 2
$latex x^4 +\frac{1}{x^4} = 9602$
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 $latex x^2 +(4 – \sqrt {3})x – 1 = 0$ , then what is the value of $latex x^2 +\frac{1}{x^2}$ ?
(A) $latex 21- 8\sqrt3$
(B) $latex 17- 8\sqrt3$
(C) $latex 9- 8\sqrt3$
(D) $latex 21- 12\sqrt3$
Answer
Ans : (A) $latex 21- 8\sqrt3$
$latex x^2 +(4 -\sqrt3)x – 1 = 0$
$latex \div x$ Both side
$latex x +(4 -\sqrt{3}) – \frac1x = 0$
$latex x -\frac1x = \sqrt3 – 4$
square Both side
$latex (x -\frac1x)^2 = (\sqrt 3 – 4)^2$
$latex x^2 +\frac{1}{x^2} – 2 = 3 + 16 – 8\sqrt3$
$latex x^2 +\frac{1}{x^2} = 21 – 8\sqrt3$
SSC CHSL Algebra Questions with solutions for Competitive Exams
Q.21: If $latex x^2 +\frac{1}{x^2} = 83$, x > 0 , then the value of $latex x^3 +\frac{1}{x^3}$ is:
(A) 675
(B) 756
(C) 746
(D) 576
Answer
Ans : (B) 756
$latex x^2 +\frac{1}{x^2} = 83$
$latex x^2 +\frac{1}{x^2} – 2 = 83 – 2$
$latex (x -\frac{1}{x})^2 = 81$
$latex x -\frac{1}{x} = 9$
$latex (x -\frac{1}{x})^3 = 9^3$
$latex x^3 -\frac{1}{x^3} -3x \frac1x (x -\frac1x) = 729$
$latex x^3 -\frac{1}{x^3} = 729 + 27$
= 756
Q.22: If $latex x +\frac1x = \sqrt{13}$ , then one of the values of $latex x^3 -\frac{1}{x^3}$ is:
(A) 36
(B) 32
(C) $latex 4\sqrt{13}$
(D) $latex 4\sqrt{11}$
Answer
Ans : (A) 36
$latex x +\frac1x = \sqrt{13}$
$latex x^2 +\frac{1}{x^2} + 2 = 13$
$latex x^2 +\frac{1}{x^2} = 11$
$latex x^2 +\frac{1}{x^2} – 2 = 11 – 2$
$latex x^2 +\frac{1}{x^2} – 2 = 9$
$latex (x -\frac{1}{x})^2 = 9$
$latex x -\frac1x = 3$
$latex x^3 +\frac{1}{x^3} – 3\times3 = 27$
$latex x^3 -\frac{1}{x^3} = 27 + 9$
= 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
$latex x =\frac13$
put $latex x =\frac13$
$latex 27x^3 +\frac{1}{27^3}$
$latex 27\times\frac{1}{27} +\frac{{1}\times{27}}{{27}\times{1}}$
= 1 + 1 = 2
Q.25: What is the coefficient of y2 in the expansion of $latex (\sqrt2y^2 – 5\sqrt3)^3$ ?
(A) $latex 30\sqrt3$
(B) $latex -225\sqrt2$
(C) $latex -30\sqrt3$
(D) $latex 225\sqrt2$
Answer
Ans : (D) $latex 225\sqrt2$
$latex (\sqrt2y^2 – 5\sqrt3)^3$
(a – b)3 = a3 – b3 – 3ab(a – b)
$latex (\sqrt2y^2)^3 – (5\sqrt3)^3 – 3\sqrt2y^2\times5\sqrt3 (\sqrt2y^2 – 5\sqrt3)$
केवल y2 का गुणांक लेने पर
$latex = -3\sqrt2y^2 \times 5\sqrt3\times – 5\sqrt3$
$latex = +225\sqrt2y^2$
y2 का गुणांक $latex = 225\sqrt2$
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 $latex x +\frac{1}{15x} =3$ , then the value of $latex 9x^3 +\frac{1}{375x^3}$ will be:
(A) 237.6
(B) 273.6
(C) 367.2
(D) 376.2
Answer
Ans : (A) 237.6
$latex x +\frac{1}{15x} =3$
$latex (x +\frac{1}{15x})^3 =(3)^3$
$latex x^3 +\frac{1}{3375x^3} +3x\times\frac{1}{15x} \times3 = 27$
$latex x^3 +\frac{1}{3375x^3} +\frac35 = 27$
$latex x^3 +\frac{1}{3375x^3} = 27 -\frac35 =\frac{132}{5}$
9 से गुणा Both side
$latex 9x^3 +\frac{1}{375x^3} =\frac{132}{5}\times 9$
$latex 9x^3 +\frac{1}{375x^3} =\frac{1188}{2} = 237.6$
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 $latex \frac{abc}{5}$ 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
$latex \frac{abc}{5} = -1$
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
$latex x^4 + \frac{1}{x^4} = 47$
$latex x^4 + \frac{1}{x^4} + 2 = 47+ 2$
$latex (x^2 + \frac{1}{x^2})^2 = 49$
$latex x^2 + \frac{1}{x^2} = 7$
$latex x^2 + \frac{1}{x^2} + 2 = 7+2$
$latex (x + \frac{1}{x})^2 = 9$
$latex x +\frac1x = 3$
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 $latex a-\frac{24}{a} = 5$ , where a > 0, then the value of $latex a^2 +\frac{64}{a^2}$ is:
(A) 45
(B) 60
(C) 65
(D) 56
Answer
Ans : (C) 65
$latex a-\frac{24}{a} = 5$
Put x = 8 its satisfy this equation
So $latex a^2-\frac{64}{a^2}$
$latex 64 +\frac{64}{64}$
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
Answer
Ans : (D) 5004
x3+y3+z3 – 3xyz = $latex \frac12$ (x+y+z)[(x – y)2 +(y – z)2 +(z – x)2 ]
$latex \frac12\times 1668\times[1+1+4]$
= 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
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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