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# SSC CGL Algebra Questions

Download the most important practice questions on algebra for SSC CGL exam 2020. Most important SSC CGL practice questions based on asked questions in previous exam papers for SSC CGL. These questions with solutions are also helpful for all competitive exams.

Question 1: If $x^{2}+\frac{1}{5}x+a^{2}$ is a perfect square then a is

a) $\frac{1}{100}$

b) $\frac{1}{5}$

c) $\frac{1}{10}$

d) 10

Question 2: Given that $1^{2}+2^{2}+3^{2}+…..+10^{2}=385$ the value of $2^{2}+4^{2}+6^{2}+….20^{2}=$

a) 770

b) 1540

c) 1155

d) (385)

Question 3: If ab + bc+ ca = 0, then the value of $\frac{1}{a^2-bc}+\frac{1}{b^2-ac}+\frac{1}{c^2-ab}$

a) 2

b) -1

c) 0

d) 1

Question 4: If the equation $2x^{2}-7x+12=0$ has two roots $\alpha$ and $\beta$ then the value of $\frac{\alpha}{\beta}$+$\frac{\beta}{\alpha}$ is

a) $\frac{7}{2}$

b) $\frac{1}{24}$

c) $\frac{7}{24}$

d) $\frac{97}{24}$

Question 5: Find the value of x for which the expression $2 – 3x- 4x^{2}$ has the greatest value.

a) $-\frac{41}{16}$

b) $\frac{3}{8}$

c) $-\frac{3}{8}$

d) $\frac{41}{16}$

Question 6: The expression $x^4- 2x^2 + k$ will be a perfect square if the value of k is

a) $1$

b) $0$

c) $\frac{1}{4}$

d) $\frac{1}{2}$

Question 7: If (x-1) and (x+3) are the factors of $x^{2}+k_{1}x + k_{2}$ then

a) $k_{1}=-2 ,k_{2}=-3$

b) $k_{1}=2 ,k_{2}=-3$

c) $k_{1}=2 ,k_{2}=3$

d) $k_{1}=-2 ,k_{2}=3$

Question 8: If $\frac{5x}{2x^2+5x+1}=\frac{1}{3}$ then the value of $(x+\frac{1}{2x})$ is

a) 15

b) 10

c) 20

d) 5

Question 9: If $x^{2}+\frac{1}{x^2}=66$ then the value of $\frac{x^2-1+2x}{x}=$ ?

a) ± 8

b) 10, – 6

c) 6, -10

d) ± 4

Question 10: If $a^{2} + a+ 1 = 0$, then the value of $a^{9}$ is

a) 2

b) 3

c) 1

d) 0

Question 11: Find the value of $\sqrt{(x^2+y^2+2)(x+y-32)}\div \sqrt{xy^3 2^2}$ when $x=+1,y=-3$

a) 1

b) 0

c) $\frac{34}{3}$

d) 2

Question 12: If a = 2 + √3 ,then the value of $(a^{2}+\frac{1}{a^{2}})$ is

a) 12

b) 14

c) 16

d) 10

e) Non of These

Question 13: If x + y = 15, then $(x-10)^{3}+(y-5)^{3}$ is

a) 25

b) 125

c) 625

d) 0

Question 14: If a + b = 6, a – b = 2, then the value of 2*(a^2 + b^2 ) is :

a) 20

b) 30

c) 40

d) 10

Question 15: The value of x when 5% of √2x is 0.01 will be :

a) 0.03

b) 0.02

c) 0.01

d) 0.05

Question 16: If $\sqrt{\frac{x-a}{x-b}}+\frac{a}{x}=\sqrt{\frac{x-b}{x-a}}+\frac{b}{x}$ and $b \neq a$, then the value of $x$ is

a) $\frac{b}{a+b}$

b) $\frac{ab}{a+b}$

c) 1

d) $\frac{a}{a+b}$

Question 17: If $x = \frac{2\sqrt{24}}{\sqrt{3}+\sqrt{2}}$, then the value of $\frac{x+\sqrt{8}}{x-\sqrt{8}}+\frac{x+\sqrt{12}}{x-\sqrt{12}}$ is

a) 1

b) 2

c) 3

d) -2

Question 18: If $a=\frac{2+\sqrt{3}}{2-\sqrt{3}}$ and $b=\frac{2-\sqrt{3}}{2+\sqrt{3}}$, then the value of $a^2+b^2+a \times b$ is

a) 185

b) 195

c) 200

d) 175

Question 19: If $x = \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}$, then the value of $\frac{x+\sqrt{2}}{x-\sqrt{2}} + \frac{x+\sqrt{3}}{x-\sqrt{3}}$ is

a) $\sqrt{2}$

b) $\sqrt{3}$

c) $\sqrt{6}$

d) $2$

Question 20: If $\sqrt{4x-9}+\sqrt{4x+9}=5+\sqrt{7}$, then the value of $x$ is

a) 4

b) 5

c) 7

d) 3

Expression : $x^{2}+\frac{1}{5}x+a^{2}$

= $[(x^2) + (2 * \frac{1}{10} * x) + (\frac{1}{10})^2] + a^2 – (\frac{1}{10})^2$

= $(x + \frac{1}{10})^2 + a^2 – (\frac{1}{10})^2$

Now, for the above expression to be perfect square,

=> $a^2 – (\frac{1}{10})^2 = 0$

=> $a^2 = (\frac{1}{10})^2$

=> $a = \frac{1}{10}$

Given that $1^{2}+2^{2}+3^{2}+…..+10^{2}=385$

Now,

$2^{2}+4^{2}+6^{2}+….20^{2}=(1^{2}+2^{2}+3^{2}+…..+10^{2}) \times 2^2 = 4 \times 385 = 1540$

Since, $ab + bc + ca = 0$

If we take $a = b = 1$, => $c = \frac{-1}{2}$

To find : $\frac{1}{a^2-bc}+\frac{1}{b^2-ac}+\frac{1}{c^2-ab}$

Substituting values of $a,b,c$ we get :

= $\frac{1}{1 – (\frac{-1}{2})} + \frac{1}{1 – (\frac{-1}{2})} + \frac{1}{(\frac{1}{4}) – 1}$

= $\frac{1}{\frac{3}{2}} + \frac{1}{\frac{3}{2}} + \frac{1}{\frac{-3}{4}}$

= $\frac{2}{3} + \frac{2}{3} – \frac{4}{3}$

= $\frac{4}{3} – \frac{4}{3} = 0$

Equation : $2x^{2}-7x+12=0$

=> Sum of roots = $\alpha + \beta = \frac{7}{2}$

=> Product of roots = $\alpha \beta = \frac{12}{2} = 6$

To find : $\frac{\alpha}{\beta}$+$\frac{\beta}{\alpha}$

= $\frac{\alpha^2 + \beta^2}{\alpha \beta}$

= $\frac{(\alpha + \beta)^2 – 2 \alpha\beta}{\alpha \beta}$

= $\frac{\frac{49}{4} – 12}{6}$

= $\frac{\frac{1}{4}}{6} = \frac{1}{24}$

NOTE :- To find the min/max value of an expression, we need to differentiate it, and put the first derivative equal to ‘0’ to find value of $x$

Then, we need to differentiate it again and put value of $x$, if second derivative is less than zero, then $x$ is maxima otherwise $x$ is minima.

Expression : $y$ = $2 – 3x- 4x^{2}$

=> $\frac{dy}{dx} = -3 – 8x$

Now, putting it equal to 0, we get :

=> $y’ = -3 -8x = 0$

=> $x = \frac{-3}{8}$

Differentiating it again :

=> $\frac{d^2y}{dx^2} = -8$

Since, it is less than ‘0’ ,=> $x = \frac{-3}{8}$ is maximum value.

It is clear that we have to write the expression in the form $(a-b)^2$.
The second term in the given expression in $-2x^2$. This expression must correspond to the form 2ab. Hence, $a = x^2$ and $b=1$.
Hence, k must be equal to 1 for the expression to be a perfect square.
Option A is the right answer.

METHOD1:

If (x-1) and (x+3) are the factors of $x^{2}+k_{1}x + k_{2}$

then its zeros are 1, -3

sum of zeros = $-k_{1}=-2\rightarrow k_{1}=2$

product of zeros = $k_{2}=-3\rightarrow k_{2}=-3$

so the answer is option B.

METHOD2:

Multiply (x-1) and (x+3)

= $x^{2}-x+3x-3$

= $x^{2}+2x-3$

now, compare  $x^{2}+2x-3$ with $x^{2}+k_{1}x + k_{2}$

$k_{1}=2$ and $k_{2}=-3$

so the answer is option B.

Expression : $\frac{5x}{2x^2+5x+1}=\frac{1}{3}$

=> $2x^2 + 5x + 1 = 15x$

=> $2x^2 + 1 = 10x$

To find : $(x+\frac{1}{2x})$

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

= $\frac{10x}{2x}$

= 5

Expression : $x^{2}+\frac{1}{x^2}=66$

=> $(x – \frac{1}{x})^2 + 2 = 66$

=> $(x – \frac{1}{x})^2 = 64$

=> $x – \frac{1}{x} = \pm 8$

To find : $\frac{x^2-1+2x}{x}$

= $x – \frac{1}{x} + 2$

= $(8 + 2)$ and $(-8 + 2)$

= 10 and -6

Expression : $a^{2} + a+ 1 = 0$

Multiplying both sides by $(a-1)$ , we get :

=> $(a-1)(a^2 + a + 1) = 0$

=> $a^3 – 1 = 0$

=> $a^3 = 1$

Cubing both sides

=> $a^9 = 1$

we need to find value of $\sqrt{(x^2+y^2+2)(x+y-32)}\div \sqrt{xy^3 2^2}$ when $x=+1,y=-3$

putting value of x , y

$\sqrt{(1^2 + (-3)^2 + 2)\times(1 – 3 – 32)}\div \sqrt{1 \times (-3)^3 \times 4}$

= 34/3

a = 2 + √3

$(a^{2}+\frac{1}{a^{2}})$ = $(a + \frac{1}{a})^2$ – 2

here ,

$a = 2+\sqrt{3}$

$\dfrac{1}{a} = \dfrac{1}{2+\sqrt{3}} \times\dfrac{2-\sqrt{3}}{2-\sqrt{3}} = 2-\sqrt{3}$

$a+\dfrac{1}{a} = 2+\sqrt{3} + 2-\sqrt{3}$

$a + \dfrac{1}{a} = 4$

So, $(a + \frac{1}{a})^2$ – 2 = 14

given that x + y = 15

So reaaranging the above equation

(x -10 ) + (y-5) =0

Now , x-10 =-( y -5 )

On cubing both sides

$(x-10)^3$ =- $(y-5)^3$

And hence

$(x-10)^3$ + $(y-5)^3$ =0

here we are given with

a + b = 6…….(1)

a – b = 2…(2)

2a = 8

a = 4

so b = 2

we need to calculate $2 \times (a^2 + b^2 ) = 2 \times (4^2 + 2^2) = 40$

It is given that 5% of √2x is 0.01

$\frac{5}{100} \times \surd(2x)$ = 0.01

x = 0.02

Given : $x = \frac{2\sqrt{24}}{\sqrt{3}+\sqrt{2}}$

=> $x = \frac{2\sqrt{24}}{\sqrt{3}+\sqrt{2}}\times(\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2})$

=> $x=\frac{2\sqrt{24}(\sqrt3-\sqrt2)}{3-2}$

=> $x=2\sqrt{72}-2\sqrt{48}$

=> $x=6\sqrt8-4\sqrt{12}$ —————(i)

To find : $\frac{x+\sqrt{8}}{x-\sqrt{8}}+\frac{x+\sqrt{12}}{x-\sqrt{12}}$

= $\frac{6\sqrt8-4\sqrt{12}+\sqrt{8}}{6\sqrt8-4\sqrt{12}-\sqrt{8}} + \frac{6\sqrt8-4\sqrt{12}+\sqrt{12}}{6\sqrt8-4\sqrt{12}-\sqrt{12}}$     [Using (i)]

= $\frac{7\sqrt{8}-4\sqrt{12}}{5\sqrt{8}-4\sqrt{12}} + \frac{6\sqrt8-3\sqrt{12}}{6\sqrt8-5\sqrt{12}}$

= $\frac{(336-35\sqrt{96}-24\sqrt{96}+240)+(240-15\sqrt{96}-24\sqrt{96}+144)}{240-25\sqrt{96}-24\sqrt{96}+240}$

= $\frac{960-98\sqrt{96}}{480-49\sqrt{96}}$

= $\frac{2(480-49\sqrt6)}{480-49\sqrt6}=2$

=> Ans – (B)

$a=\frac{2+\sqrt{3}}{2-\sqrt{3}}$ on rationalising we will get a = $(2 + \surd3)^2$

$b=\frac{2-\sqrt{3}}{2+\sqrt{3}}$ on rationalizing we will get b = $(2 – \surd3)^2$

now putting values of a and b in , $a^2+b^2+a \times b$

$a^2+b^2+a \times b$ = 195

Given : $x = \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}$

=> $x = \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times(\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2})$

=> $x=\frac{2\sqrt6(\sqrt3-\sqrt2)}{3-2}$

=> $x=2\sqrt{18}-2\sqrt{12}$

=> $x=6\sqrt2-4\sqrt3$ —————(i)

To find : $\frac{x+\sqrt{2}}{x-\sqrt{2}} + \frac{x+\sqrt{3}}{x-\sqrt{3}}$

= $\frac{6\sqrt2-4\sqrt3+\sqrt{2}}{6\sqrt2-4\sqrt3-\sqrt{2}} + \frac{6\sqrt2-4\sqrt3+\sqrt{3}}{6\sqrt2-4\sqrt3-\sqrt{3}}$     [Using (i)]

= $\frac{7\sqrt{2}-4\sqrt3}{5\sqrt{2}-4\sqrt3} + \frac{6\sqrt2-3\sqrt{3}}{6\sqrt2-5\sqrt{3}}$

= $\frac{(84-35\sqrt6-24\sqrt6+60)+(60-15\sqrt6-24\sqrt6+36)}{60-25\sqrt6-24\sqrt6+60}$

= $\frac{240-98\sqrt6}{120-49\sqrt6}$

= $\frac{2(120-49\sqrt6)}{120-49\sqrt6}=2$

=> Ans – (D)

given that

$\sqrt{4x-9}+\sqrt{4x+9}=5+\sqrt{7}$

$\sqrt{4x-9} – 5 = \sqrt{7} – \sqrt{4x+9}$

on squaring both sides

4x – 9 + 25 + 10$\sqrt{4x+9}$ = 7 + 4x + 9 – 2$\sqrt{28x+63}$

10$\sqrt{4x-9}$  = 2$\sqrt{28x+63}$

on squaring both sides again

400x – 900 = 112x + 252

288x = 1152

x = 4