Algebra Questions in Hindi for SSC Exams

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Algebra Questions in Hindi, selected from recently conducted SSC CGL, CHSL Exams. Important and repeated बीजगणित Maths questions for upcoming Competitive Exams.

Algebra Questions with solutions in Hindi for SSC Exams

Q.1: यदि a + b = p, ab = q है , तो (a⁴ + b⁴ ) का मान ज्ञात कीजिए I
(A) p⁴ – 4p²q + q²
(B) p⁴ – 4p²q² + 2q²
(C) p⁴ – 2p²q² + q²
(D) p⁴ – 4p²q + 2q²
(SSC CHSL Aug 2021)

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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: यदि $(x +\frac 1x)^3 = 27$ है , तो $(x^2 +\frac {1}{x^2})$ का मान क्या होगा ? दिया गया है कि जहां x वास्तविक है I
(A) 9
(B) 25
(C) 7
(D) 11
(SSC CHSL Aug 2021)

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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: यदि $x -\frac 2x = 4$ है , तो $x^2 + \frac {4}{x^2}$ का मान ज्ञात करें I
(A) 18
(B) 8
(C) 12
(D) 20
(SSC CHSL Aug 2021)

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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: यदि $\sqrt{x} + \frac {1} {\sqrt x} = \sqrt 6$ है , तो $x^6 + \frac {1}{x^6}$ का मान ज्ञात करें I
(A) 2712
(B) 2502
(C) 2270
(D) 2702
(SSC CHSL Aug 2021)

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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: यदि x² + 1 – 2x = 0, x > 0 है, तो x²(x²– 2) का मान ज्ञात करें I
(A) 0
(B) -1
(C) 1
(D) $\sqrt2$
(SSC CHSL Aug 2021)

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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: यदि $x^2 -3\sqrt2x + 1 =0$ है , तो $x^3 + (\frac {1}{x^3})$ का मान ज्ञात करें I
(A) $15\sqrt6$
(B) $30\sqrt6$
(C) $45\sqrt2$
(D) $30\sqrt2$
(SSC CHSL Aug 2021)

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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: यदि x – y = 4 और xy = 3 है , तो x³ – y³ का मान ज्ञात करें I
(A) 88
(B) 100
(C) 64
(D) 28
(SSC CHSL Aug 2021)

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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: यदि $x -\frac 1x = 2\sqrt2$ है , तो $x^3 -\frac {1}{x^3}$ का मान ज्ञात करें I
(A) $12\sqrt2$
(B) $10\sqrt2$
(C) $20\sqrt2$
(D) $22\sqrt2$
(SSC CHSL Aug 2021)

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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: यदि x + 2y = 19 और x³ + 8y³ = 361 है , तो xy का मान क्या होगा ?
(A) 58
(B) 56
(C) 55
(D) 57
(SSC CHSL Aug 2021)

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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: यदि $(x^2 + \frac{1}{49x^2}) = 15\frac 57$ है , तो $(x +\frac{1}{7x})$ का मान क्या होगा ?
(A) 4
(B) $\pm 7$
(C) $\pm 4$
(D) 7
(SSC CHSL Aug 2021)

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Ans : (C) $\pm 4$
$(x^2 + \frac{1}{49x^2}) = 15\frac57$

$(x^2 + \frac{1}{7x})^2 + 2\times x\times \frac {1}{7x} =\frac{110}{7} + \frac27$

$(x + \frac{1}{7x})^2 = \frac{112}{7} = 16$

$x + \frac{1}{7x} =\pm 4$

Algebra Questions with solutions in Hindi for SSC Exams

Q.11: यदि x + y = 27 और x² + y² = 425 है , तो (x – y)² का मान ज्ञात करें
(A) 121
(B) 225
(C) 169
(D) 144
(SSC CHSL Aug 2021)

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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: यदि 3x + y = 12 और xy = 9 है, तो (3x – y) का मान क्या होगा ?
(A) 6
(B) 5
(C) 3
(D) 4
(SSC CHSL Aug 2021)

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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: यदि a² + b² + c² = 576 और (ab + bc + ca) = 50 है, तो (a + b + c) का मान क्या होगा, यदि a+b+c < 0 है ?
(A) -24
(B) $\pm 24$
(C) $\pm 26$
(D) -26
(SSC CHSL Aug 2021)

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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: यदि $x +\frac{1}{3x} = 5$ है , तो $27x^3 +\frac{1}{x^3}$ का मान ज्ञात करें I
(A) 3024
(B) 3420
(C) 3042
(D) 3240
(SSC CHSL Aug 2021)

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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: यदि 3x + 5y = 14 और xy = 6 है, तो 9x² + 25y²का मान कितना होगा ?
(A) 182
(B) 16
(C) 14
(D) 20
(SSC CHSL Aug 2021)

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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: यदि a – b = 7 और a² + b² = 169 है, जहाँ a,b >0 है , तो 3a + b का मान ज्ञात करें
(A) 41
(B) 46
(C) 38
(D) 44
(SSC CHSL Aug 2021)

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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: यदि a + 5b = 25 और ab = 20 है, तो (a – 5b) का एक मान _________ होगा
(A) 16
(B) 15
(C) 13
(D) 14
(SSC CHSL Aug 2021)

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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: यदि $\sqrt {x} + \frac{1}{x} =2\sqrt{3}$ है, तो $x^4 +\frac{1}{x^4}$ का मान ज्ञात करें
(A) 10406
(B) 10402
(C) 9602
(D) 9606
(SSC CHSL Aug 2021)

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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: यदि (7x – 10y) = 8 और xy = 5 है, तो 49x² + 100y² का मान क्या होगा ?
(A) 632
(B) 623
(C) 746
(D) 764
(SSC CHSL Aug 2021)

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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: यदि $x^2 +(4 - \sqrt {3})x - 1 = 0$ है, तो $x^2 +\frac{1}{x^2}$ का मान ज्ञात करें
(A) $21- 8\sqrt3$
(B) $17- 8\sqrt3$
(C) $9- 8\sqrt3$
(D) $21- 12\sqrt3$
(SSC CHSL Aug 2021)

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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
$(x -\frac1x)^2 = (\sqrt 3 - 4)^2$

$x^2 +\frac{1}{x^2} - 2 = 3 + 16 - 8\sqrt3$

$x^2 +\frac{1}{x^2} = 21 - 8\sqrt3$

Algebra Questions with solutions in Hindi for SSC Exams

Q.21: यदि $x^2 +\frac{1}{x^2} = 83$ , x > 0 है, तो $x^3 +\frac{1}{x^3}$ का मान ज्ञात करें
(A) 675
(B) 756
(C) 746
(D) 576
(SSC CHSL Aug 2021)

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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: यदि $x +\frac1x = \sqrt{13}$ है, तो $x^3 -\frac{1}{x^3}$ का एक मान ज्ञात करें I
(A) 36
(B) 32
(C) $4\sqrt{13}$
(D) $4\sqrt{11}$
(SSC CHSL Aug 2021)

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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: (x – 2y) x (5x + y)³ में x³y का गुणांक (coefficient) ज्ञात करें
(A) -150
(B) 75
(C) -175
(D) 250
(SSC CHSL Aug 2021)

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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: यदि 9x² – 6x + 1 = 0 है, तो 27x³ + (27x³)^-1 का मान ज्ञात करें
(A) 1
(B) 4
(C) 2
(D) 8
(SSC CHSL Aug 2021)

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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: व्यंजक $(\sqrt2y^2 - 5\sqrt3)^3$ के विस्तार
(A) $30\sqrt3$
(B) $-225\sqrt2$
(C) $-30\sqrt3$
(D) $225\sqrt2$
(SSC CHSL Aug 2021)

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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: यदि a + b + c = 2 और ab + bc + ca = -1 है तो a³ + b³ + c³ – 3abc का मान ज्ञात कीजिए
(A) 5
(B) 10
(C) 2
(D) 14
(SSC CHSL Aug 2021)

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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: यदि $x +\frac{1}{15x} =3$ है तो $9x^3 +\frac{1}{375x^3}$ का मान ज्ञात करें
(A) 237.6
(B) 273.6
(C) 367.2
(D) 376.2
(SSC CHSL Aug 2021)

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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: यदि a+b+c = 11 और ab+bc+ca = 15 है, तो a³+b³+c³ – 3abc का मान ज्ञात करें
(A) 386
(B) 836
(C) 368
(D) 638
(SSC CHSL Aug 2021)

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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: यदि a+b+c = 5, a²+b²+c² = 27, और a³+b³+c³ = 125 है, तो $\frac{abc}{5}$ का मान ज्ञात करें
(A) -5
(B) -1
(C) 5
(D) 1
(SSC CHSL Aug 2021)

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

Q.30: यदि x⁴ + x^-4 = 47, x>0 है, तो (2x – 3)² का मान ज्ञात करें
(A) 9
(B) 5
(C) 3
(D) 7
(SSC CHSL Aug 2021)

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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
Algebra Questions in Hindi for SSC Exams

Q.31: यदि $a-\frac{24}{a} = 5$ है, जहाँ a > 0, है, तो $a^2 +\frac{64}{a^2}$ का मान ज्ञात करें
(A) 45
(B) 60
(C) 65
(D) 56
(SSC CHSL Aug 2021)

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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: यदि x = 555, y = 556 और z = 557 है, तो x³+y³+z³ – 3xyz का मान क्या होगा ?
(A) 5002
(B) 5008
(C) 5006
(D) 5004
(SSC CHSL Aug 2021)

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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: यदि 3x-2y+3=0 है, तो 27x³+54xy+30-8y³ का मान ज्ञात करें
(A) 57
(B) -57
(C) -27
(D) 3
(SSC CHSL Aug 2021)

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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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