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Algebra Questions in Hindi PDF for all Competitive Exams

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Algebra Questions in Hindi PDF for all Competitive Exams

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We all know that Math Tricks Notes PDF in Hindi comes in every exams and math subject is important for all types of exams so today, we are sharing Math Trick in Hindi Notes (2022*) for you. We all know that every exam has 20-25 maths questions and also we know every exam paper for limited time so improve your perpetration with Maths Tricks Book PDF in Hindi by Sumitra.

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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)

Show 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: 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)

Show 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: 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)

Show Answer

Ans : (D) 20
$({x -\frac 2x)^2} = 4$
$x^2 +\frac {4}{x^2} - 2\times2 = 16$
$x^2 +\frac {4}{x^2} = 16 + 4 = 20$

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Maths Questions Answers

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)

Show 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: 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)

Show 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: 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)

Show 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: 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)

Show 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: 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)

Show 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: 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)

Show 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: 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)

Show Answer

Ans : (C) $\pm 4$
$(x^2 + \frac{1}{49x^2}) = 15\frac57$
$(x^2 + \frac{1}{7x})^2 + 2\times x\times \frac {1}{7x} =\frac{110}{7} + \frac27$
$(x + \frac{1}{7x})^2 = \frac{112}{7} = 16$
$x + \frac{1}{7x} =\pm 4$
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)

Show 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: 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)

Show 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: 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)

Show 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: 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)

Show 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: 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)

Show 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: 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)

Show 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: 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)

Show Answer

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)

Show 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: 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)

Show 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: 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)

Show 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
$(x -\frac1x)^2 = (\sqrt 3 - 4)^2$
$x^2 +\frac{1}{x^2} - 2 = 3 + 16 - 8\sqrt3$
$x^2 +\frac{1}{x^2} = 21 - 8\sqrt3$
Algebra Questions with solutions in Hindi for SSC Exams

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