Algebra Practice Questions For SSC CHSL PDF

0
215
algebra practice questions for ssc chsl pdf
algebra practice questions for ssc chsl pdf

Algebra Practice Questions For SSC CHSL PDF

SSC CHSL Algebra Practice Questions download PDF based on previous year question paper of SSC exams. 20 Very important Algebra Practice questions for SSC CHSL Exam.

Download Algebra Practice Questions For SSC CHSL PDF

Take a free mock test for SSC CHSL

Download SSC CHSL Previous Papers

Question 1: Find the value of $1+\Large\frac{1}{1+\frac{1}{1+\frac{1}{2}}}$

a) $\large\frac{6}{5}$

b) $\large\frac{8}{5}$

c) $\large\frac{8}{7}$

d) $\frac{7}{6}$

Question 2: Find the value of $1+\Large\frac{1}{1+\frac{1}{1+\frac{1}{6}}}$

a) $\large\frac{17}{5}$

b) $\large\frac{19}{6}$

c) $\large\frac{20}{13}$

d) $\frac{17}{13}$

Question 3: Find the value of $1+\Large\frac{1}{1+\frac{1}{1+\frac{3}{2}}}$

a) $\large\frac{13}{5}$

b) $\large\frac{17}{6}$

c) $\large\frac{17}{5}$

d) $\frac{12}{7}$

Question 4: Find the value of $\sqrt{3\sqrt{3\sqrt{3……}}}$

a) 9

b) 3

c) 27

d) 1.2

Question 5: Find the value of $\sqrt{4\sqrt{4\sqrt{4……}}}$

a) 4

b) 2

c) 16

d) 8

SSC CHSL PREVIOUS PAPERS

SSC CHSL Study Material (FREE Tests)

Question 6: Find the value of $\sqrt{7\sqrt{7\sqrt{7……}}}$

a) $\sqrt{7}$

b) 49

c) 7

d) 2.64

Question 7: Find the value of $\sqrt{6+\sqrt{6+\sqrt{6+……}}}$

a) 9

b) 3

c) 27

d) 16

Question 8: Find the value of $\sqrt{30+\sqrt{30+\sqrt{30+……}}}$

a) 12

b) 15

c) 6

d) 18

Question 9: Find the value of $\sqrt{42+\sqrt{42+\sqrt{42+……}}}$

a) 11

b) 7

c) 6

d) 10

Question 10: Find the value of $\sqrt{20-\sqrt{20-\sqrt{20-……}}}$

a) 8

b) 4

c) 6

d) 10

SSC CHSL FREE MOCK TEST

Question 11: Find the value of $\sqrt{56-\sqrt{56-\sqrt{56-……}}}$

a) 9

b) 8

c) 11

d) 14

Question 12: If $2X+\large\frac{2}{X}$ $= 6$, then find the value of $X^{5}+\large\frac{1}{X^{5}}$

a) 123

b) 121

c) 116

d) 107

Question 13: If $3X+\large\frac{3}{X}$ $= 6$, then find the value of $X^{6}+\large\frac{1}{X^{6}}$

a) 4

b) 3

c) 9

d) 2

Question 14: If a+b = 5 and a-b = 1, Then find the value of ab

a) 4

b) 6

c) 8

d) 12

Question 15: If $(a-b)^{2} = 16$ and $(a+b)^{2} = 36$, then find the value of $\frac{ab}{a+b}$

a) $\frac{5}{6}$

b) $\frac{8}{11}$

c) $\frac{6}{7}$

d) $\frac{7}{6}$

FREE SSC MATERIAL – 18000 FREE QUESTIONS

Question 16: If $x-\large\frac{1}{x}$ $= 3$, then $x^{3}-\large\frac{1}{x^{3}}$ $=$ ?

a) 24

b) 28

c) 36

d) 42

Question 17: Find the value of $1+\Large\frac{1}{1-\frac{1}{1+\Large\frac{1}{1-\frac{1}{7}}}}$

a) $\Large\frac{15}{7}$

b) $\Large\frac{19}{8}$

c) $\Large\frac{20}{7}$

d) $\Large\frac{17}{8}$

Question 18: If a = 48, b = 16, c = -64, then find the value of $\large\frac{a^{3}+b^{3}+c^{3}}{abc}$

a) 176

b) 64

c) 3

d) 12

Question 19: If a = 17, b = -4, c = -13, then find the value of $\large\frac{3a^{3}+3b^{3}+3c^{3}}{4abc}$

a) 3

b) $\frac{3}{4}$

c) 1

d) $\frac{9}{4}$

Question 20: If $(2^{x})(2^y) = 16$ and $(3^{x})(9^y) = 27$, then find (x,y)

a) (4,0)

b) (3,2)

c) (5,-1)

d) (6,-2)

DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE

SSC CHSL Important Q&A PDF

Answers & Solutions:

1) Answer (B)

$1+\Large\frac{1}{1+\frac{1}{1+\frac{1}{2}}}$ = $1+\Large\frac{1}{1+\frac{1}{\frac{3}{2}}}$

= $1+\Large\frac{1}{1+\frac{2}{3}}$

= $1+\Large\frac{1}{\frac{5}{3}}$

= $1+\Large\frac{3}{5}$

= $\large\frac{8}{5}$

2) Answer (C)

$1+\Large\frac{1}{1+\frac{1}{1+\frac{1}{6}}}$ = $1+\Large\frac{1}{1+\frac{1}{\frac{7}{6}}}$

= $1+\Large\frac{1}{1+\frac{6}{7}}$

= $1+\Large\frac{1}{\frac{13}{7}}$

= $1+\Large\frac{7}{13}$

= $\large\frac{20}{13}$

3) Answer (D)

$1+\Large\frac{1}{1+\frac{1}{1+\frac{3}{2}}}$ = $1+\Large\frac{1}{1+\frac{1}{\frac{5}{2}}}$

= $1+\Large\frac{1}{1+\frac{2}{5}}$

= $1+\Large\frac{1}{\frac{7}{5}}$

= $1+\Large\frac{5}{7}$

= $\large\frac{12}{7}$

4) Answer (B)

Let $\sqrt{3\sqrt{3\sqrt{3……}}}$ = X

Then, $\sqrt{3X} = X$

Squaring on both sides,
$3X = X^{2}$
⇒ X $= 3$

5) Answer (A)

Let $\sqrt{4\sqrt{4\sqrt{4……}}}$ = X

Then, $\sqrt{4X} = X$

Squaring on both sides,
$4X = X^{2}$
⇒ X $= 4$

6) Answer (C)

Let $\sqrt{7\sqrt{7\sqrt{7……}}}$ = X

Then, $\sqrt{7X} = X$

Squaring on both sides,
$7X = X^{2}$
⇒ X $= 7$

7) Answer (B)

Let $\sqrt{6+\sqrt{6+\sqrt{6+……}}}$ = X

Then, $\sqrt{6+X} = X$

Squaring on both sides,
$6+X = X^{2}$
⇒ $X^{2}-X-6 = 0$
⇒ $X^{2}-3X+2X-6 = 0$
⇒ $X(X-3)+2(X-3) = 0$
⇒ $(X-3)(X+2) = 0$
⇒ $X = 3$ or $X = -2$

X cannot be negative when all the terms are positive.
Hence, $X = 3$

8) Answer (C)

Let $\sqrt{30+\sqrt{30+\sqrt{30+……}}}$ = X

Then, $\sqrt{30+X} = X$

Squaring on both sides,
$30+X = X^{2}$
⇒ $X^{2}-X-30 = 0$
⇒ $X^{2}-6X+5X-30 = 0$
⇒ $X(X-6)+5(X-6) = 0$
⇒ $(X-6)(X+5) = 0$
⇒ $X = 6$ or $X = -5$

X cannot be negative when all the terms are positive.
Hence, $X = 6$

9) Answer (B)

Let $\sqrt{42+\sqrt{42+\sqrt{42+……}}}$ = X

Then, $\sqrt{42+X} = X$

Squaring on both sides,
$42+X = X^{2}$
⇒ $X^{2}-X-42 = 0$
⇒ $X^{2}-7X+6X-42 = 0$
⇒ $X(X-7)+6(X-7) = 0$
⇒ $(X-7)(X+6) = 0$
⇒ $X = 7$ or $X = -6$

X cannot be negative when all the terms are positive.
Hence, $X = 7$

10) Answer (B)

Let $\sqrt{20-\sqrt{20-\sqrt{20-……}}}$ = X

Then, $\sqrt{20-X} = X$

Squaring on both sides,
$20-X = X^{2}$
⇒ $X^{2}+X-20 = 0$
⇒ $X^{2}-4X+5X-20 = 0$
⇒ $X(X-4)+5(X-4) = 0$
⇒ $(X-4)(X+5) = 0$
⇒ $X = 4$ or $X = -5$

Hence, Option B is correct answer.

GK Questions And Answers PDF

11) Answer (B)

Let $\sqrt{56-\sqrt{56-\sqrt{56-……}}}$ = X

Then, $\sqrt{56-X} = X$

Squaring on both sides,
$56-X = X^{2}$
⇒ $X^{2}+X-56 = 0$
⇒ $X^{2}-8X+7X-56 = 0$
⇒ $X(X-8)+7(X-8) = 0$
⇒ $(X-8)(X+7) = 0$
⇒ $X = 8$ or $X = -7$
Hence, Option B is correct answer.

12) Answer (A)

Given $2X+\large\frac{2}{X}$ $= 6$

$\Rightarrow 2(X+\large\frac{1}{X})$ $= 6$

$\Rightarrow X+\large\frac{1}{X}$ $= 3$ –> (1)

Squaring (1) on both sides

$(x+\large\frac{1}{x})^{2}$ $= 9$

$\Rightarrow x^{2}+\large\frac{1}{x^{2}}$ $+2\times x\times\large\frac{1}{x}$ $= 9$

$\Rightarrow x^{2}+\large\frac{1}{x^{2}}$ $+2 = 9$

$\Rightarrow x^{2}+\large\frac{1}{x^{2}}$ $= 7$ –> (2)

Cubing (1) on both sides

$(x+\large\frac{1}{x})^{3}$ $= 27$

$\Rightarrow x^{3}+\large\frac{1}{x^{3}}$ $+3\times x\times\large\frac{1}{x}$ $\times(x+\large\frac{1}{x})$ $= 27$

$\Rightarrow x^{3}+\large\frac{1}{x^{3}}$ $+3\times3 = 27$

$\Rightarrow x^{3}+\large\frac{1}{x^{3}}$ $= 27-9 = 18$ –> (3)

Multiplying (2) and (3)

$x^{2}+\large\frac{1}{x^{2}}$ $\times x^{3}+\large\frac{1}{x^{3}}$ $= 18\times7$

$\Rightarrow x^{5}+\large\frac{1}{x^{5}}$ $+x^{2}\times\large\frac{1}{x^{3}}$ $+x^{3}\times\large\frac{1}{x^{2}}$ $= 126$

$\Rightarrow x^{5}+\large\frac{1}{x^{5}}$ $+x+\large\frac{1}{x}$ $= 126$

Substituting $x+\large\frac{1}{x}$ $= 3$ in above equation

$\Rightarrow x^{5}+\large\frac{1}{x^{5}}$ $+3 = 126$

$\Rightarrow x^{5}+\large\frac{1}{x^{5}}$ $= 123$

13) Answer (D)

Given $3X+\large\frac{3}{X}$ $= 6$

$\Rightarrow 3(X+\large\frac{1}{X})$ $= 6$

$\Rightarrow X+\large\frac{1}{X}$ $= 2$

Squaring on both sides
$(x+\large\frac{1}{x})^{2}$ $=4$

$\Rightarrow x^{2}+\large\frac{1}{x^{2}}$+$2\times x\times\large\frac{1}{x}$ $=4$

$\Rightarrow x^{2}+\large\frac{1}{x^{2}}$ $+2 = 4$

$\Rightarrow x^{2}+\large\frac{1}{x^{2}}$ $= 2$

Cubing on both sides

$(x^{2}+\large\frac{1}{x^{2}})^{3}$ $= 8$

$x^{6}+\large\frac{1}{6}$ $+3\times x^{2}\times\large\frac{1}{x^{2}}$ $\times(x^{2}+\large\frac{1}{x^{2}})$ $= 8$

$\Rightarrow x^{6}+\large\frac{1}{6}$ $+3\times2 = 8$

$\therefore x^{6}+\large\frac{1}{6}$ $= 8-6 = 2$

14) Answer (B)

Given, a+b = 5
a-b = 1

Then, 2a = 6 ==> a = 3

Substituting a = 3 in above equation
==> b = 2

Hence, ab = 3*2 = 6

15) Answer (A)

Given $(a-b)^{2} = 16$ and $(a+b)^{2} = 36$
$(a+b)^{2} = (a-b)^{2}+4ab$
$36 = 16+4ab$
=> $4ab = 20$
$ab = 5$

$(a+b)^{2} = 36$
=> $a+b = 6$

Hence, $\frac{ab}{a+b} = \frac{5}{6}$

16) Answer (C)

Given $x-\large\frac{1}{x}$ $= 3$

Cubing on both sides

$x^{3}-\large\frac{1}{x^{3}}$ $-3\timesx\times\large\frac{1}{x}$ $(x-\large\frac{1}{x})$ $= 27$

=> $x^{3}-\large\frac{1}{x^{3}}$ $-3\times3 = 27$

=> $x^{3}-\large\frac{1}{x^{3}}$ $-9 = 27$

=> $x^{3}-\large\frac{1}{x^{3}}$ $= 36$

17) Answer (C)

$1+\Large\frac{1}{1-\frac{1}{1+\Large\frac{1}{1-\frac{1}{7}}}}$ = $1+\Large\frac{1}{1-\frac{1}{1+\Large\frac{1}{\frac{6}{7}}}}$

= $1+\Large\frac{1}{1-\frac{1}{1+\Large\frac{7}{6}}}$

= $1+\Large\frac{1}{1-\frac{1}{\Large\frac{13}{6}}}$

= $1+\Large\frac{1}{1-\frac{6}{13}}$

= $1+\Large\frac{1}{\frac{7}{13}}$

= $1+\Large\frac{13}{7}$

= $\Large\frac{20}{7}$

18) Answer (C)

Given a = 48, b = 16, c = -64
Then, a+b+c = 48+16-64 = 0

We know that if a+b+c = 0, then $a^{3}+b^{3}+c^{3} = 3abc$

Hence, $\large\frac{a^{3}+b^{3}+c^{3}}{abc} = \frac{3abc}{abc}$ $= 3$

19) Answer (D)

Given a = 17, b = -4, c = -13
Then a+b+c = 0.
We know that if a+b+c = 0, then $a^{3}+b^{3}+c^{3} = 3abc$

Then, $\large\frac{3a^{3}+3b^{3}+3c^{3}}{4abc}$ $= \large\frac{3(a^{3}+b^{3}+c^{3})}{4abc} = \frac{3(3abc)}{4abc} = \frac{9}{4}$

20) Answer (C)

Given $(2^{x})(2^y) = 16$

=> $2^{x+y} = 2^{4}$

=> x+y = 4 — (1)

$(3^{x})(9^y) = 27$
=> $(3^{x})((3^2)^y) = 3^3$
=> $(3^x)(3^2y) =3^3$
=> $3^x+2y = 3^3$
=> $x+2y = 3$ — (2)
Solving (1) and (2)

=> y = -1
Substituting y = -1 in (1) –> x = 5

Therefore, (x,y) = (5,-1)

FREE SSC CHSL MOCK TEST SERIES

FREE SSC STUDY MATERIAL

LEAVE A REPLY

Please enter your comment!
Please enter your name here