Home » Algebra Questions in Hindi PDF for all Competitive Exams

Algebra Questions in Hindi PDF for all Competitive Exams

Algebra Questions in Hindi PDF for all Competitive Exams
Share this post with your friends

Algebra Questions in Hindi PDF for all Competitive Exams

 Hello Students,

 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.

Sscresult.in is an online free notes pdf website, where we are sharing all types of pdf Sumitra Tricky Math Book PDF and New Syllabus, Competitive exams for UPSC, SSC, BANK, RAILWAY, LDC, NTPC and many other exams. Our Tricky mathematics PDF 2022 is very simple and easy to understand. We also cover basic subjects like Mathematics, Geography, History, General Science, Politics, etc. We also share study material including previous year question papers, current affairs, important sources, etc. for upcoming government exams.

This Tricky Mathematics in Hindi  is being provided to you free of charge, which you can download by clicking on the download button given below, and you can also DOWNLOAD some more new PDFs related to this Vedic Maths Book PDF in Hindi by going to the related notes. You can learn about all the new updates on SSCRESULT.IN by clicking on the Allow button on the screen.

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

More Related PDF Download

SSC Exams Latest News and Updates Quantitative Aptitude (Maths)
Current Affairs Reasoning and General Intelligence
English Grammar and Comprehension General Knowledge (GK)
SSC CGL Exam CPO SI Exam
SSC CHSL (10+2) Exam Stenographer Group C and D Exam

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(x– 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(x– 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

Join our Social Media Platforms

Dear friends you can help us by joining and shearing our social media platforms with your friends. Here we are sharing the links of Telegram channelFacebook page and YouTube Channel.

Tags :- algebra questions and answers hindi pdf, rrb ntpc algebra questions pdf in hindi, algebra notes for competitive exams pdf, algebra short tricks in hindi pdf, algebra questions for competitive exams, algebra questions and answers pdf, cds algebra questions pdf, average math question in hindi pdf

Leave a Comment

Your email address will not be published.

error: Content is protected !!
Scroll to Top