# Algebra Questions for RRB Group – D Set – 2 PDF

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## Algebra Questions for RRB Group – D Set – 2 PDF

Download Top-15 RRB Group-D Algebra Questions set-2 PDF. RRB GROUP-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

Question 1: If $\frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} = 1$ the the value of $\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}$

a) 1

b) 3

c) 4

d) 0

Question 2: If $a^x = (x+y+z)^y$ , $a^y =(x+y+z)^z$ and $a^z = (x + y + z)^x$ , then the value of x + y + z (given a ≠ 0) is

a) 0

b) $a^3$

c) 1

d) a

Question 3: If $x^2 – 3x + 1= 0$ and x > 1, then the value of $(x – \frac{1}{x})$

a) √5 only

b) 1

c) ­√5 only

d) ±√5

Question 4: If $x = \frac{\sqrt{5}-2}{\sqrt{5}+2}$, then $x^4 + x^{-4}$ is

a) a surd

b) a rational number but not an integer

c) an integer

d) an irrational number but not a surd

Question 5: The value of $(\sqrt{5}+\sqrt{3})(\frac{3\sqrt{3}}{\sqrt{5}+\sqrt{2}} – \frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}})$ is

a) 0

b) $2-\sqrt{2}$

c) $2\sqrt{2}$

d) $3$

Question 6: If x and y are positive real numbers and xy = 8, then the minimum value of 2x + y is

a) 9

b) 17

c) 10

d) 8

Question 7: If x = 3 + 2√2 , then find the value of $x^2 + \frac{1}{x^2}$

a) 36

b) 30

c) 32

d) 34

Question 8: if $\frac{a}{b} + \frac{b}{a} – 1$ = 0, then the value of $a^3 + b^3$ is

a) 3

b) 0

c) 1

d) -1

Question 9: If (a – b) = 3, (b – c) = 5 and (c – a) = 1, then the value of $\frac{a^3 + b^3 + c^3 – 3abc}{a + b + c}$ is

a) 17.5

b) 20.5

c) 10.5

d) 15.5

Question 10: $3 – \frac{3+\sqrt{5}}{4} – \frac{1}{3 + \sqrt{5}}$

a) 0

b) 3/2

c) √5/2

d) √5

Question 11: If $x + \frac{1}{x} = 12$, the value of $x^2 + \frac{1}{x^2}$ is

a) 142

b) 126

c) 113

d) 129

Question 12: If $x^4 + \frac{1}{x^4} = 23$, then the value of $(x-\frac{1}{x})^2$ will be

a) 7

b) – 7

c) – 3

d) 3

Question 13: If $x = \sqrt{\frac{\sqrt{5} + 1}{\sqrt{5} – 1}}$, then the value of $5x^2 – 5x -1$ is

a) 0

b) 3

c) 4

d) 5

Question 14: The value of $\frac{3\sqrt{2}}{\sqrt{3} + \sqrt{6}} – \frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}} + \frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$ is

a) 4

b) 0

c) √2

d) 3√6

Question 15: If a + 1/a+2 = 0, then the value of $(a+2)^{2}+\frac{1}{(a+2)^{3}}$ is

a) 2

b) 6

c) 4

d) 3

Expression : $\frac{a}{1-a} + \frac{b}{1-b} + \frac{c}{1-c} = 1$

Let’s put each term equal to each other

=> $3\frac{a}{1 – a} = 1$

=> $3a = 1 – a$

=> $a = \frac{1}{4} = b = c$

To find : $\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}$

= $\frac{1}{1 – \frac{1}{4}} + \frac{1}{1 – \frac{1}{4}} + \frac{1}{1 – \frac{1}{4}}$

= $3 \times \frac{4}{3} = 4$

Expressions : $a^x = (x+y+z)^y$

$a^y =(x+y+z)^z$

$a^z = (x + y + z)^x$

Multiplying above equations, we get :

=> $a^x \times a^y \times a^z = (x + y + z)^x \times (x + y + z)^y \times (x + y + z)^z$

=> $(a)^{x + y + z} = (x + y + z)^{x + y + z}$

Since the power on both sides is same, thus :

=> $x + y + z = a$

Expression : $x^2 – 3x + 1= 0$

=> $x^2 + 1 = 3x$

Dividing by $(x)$ on both sides

=> $x + \frac{1}{x} = 3$

$\because (x – \frac{1}{x})^2 = (x + \frac{1}{x})^2 – 4$

=> $x – \frac{1}{x} = \sqrt{9 – 4}$

=> $(x – \frac{1}{x}) = \pm\sqrt{5}$

Given : $x = \frac{\sqrt{5}-2}{\sqrt{5}+2}$

=> $x = \frac{\sqrt{5}-2}{\sqrt{5}+2}\times(\frac{\sqrt5-2}{\sqrt5-2})$

=> $x=\frac{(\sqrt5-2)^2}{5-4}$

=> $x=9-4\sqrt5$ ————(i)

Similarly, $\frac{1}{x}=9+4\sqrt5$ ————-(ii)

To find : $x^4 + x^{-4}=x^4+\frac{1}{x^4}$

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

= $[(x+\frac{1}{x})^2-2]^2-2$

Substituting values from equations (i) and (ii), we get :

= $[(9-4\sqrt5+9+4\sqrt5)^2-2]^2-2$

= $[324-2]^2-2$

= $103684-2=103682$, which is an integer.

=> Ans – (C)

$(\sqrt{5}+\sqrt{3})(\frac{3\sqrt{3}}{\sqrt{5}+\sqrt{2}} – \frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}})$

=$(\sqrt{5}+\sqrt{3})(\frac{3\sqrt{3}}{\sqrt{5}+\sqrt{2}} * \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}- {\frac{\sqrt{5}}{\sqrt{3}+\sqrt{2} }x \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}})$

= $(\sqrt{5}+\sqrt{3})(\sqrt{15} – \sqrt{6} – \sqrt{15} + \sqrt{10})$ = $(\sqrt{5}+\sqrt{3})(\sqrt{10}-\sqrt{6})$ = $2\sqrt{2}$

As $x$ & $y$ are positive real numbers, we can use AM $\geq$ GM

Let two numbers be $2x$ and $y$

=> $\frac{2x + y}{2} \geq \sqrt{2xy}$

=> $2x + y \geq 2\sqrt{16}$

=> $2x + y \geq 8$

=> Minimum value of $(2x + y)$ is 8

Expression : $x = 3 + 2\sqrt{2}$

=> $\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}}$

=> $\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 – 2\sqrt{2}}{3 – 2\sqrt{2}}$

=> $\frac{1}{x} = 3 – 2\sqrt{2}$

$\therefore$ $x + \frac{1}{x} = 3 + 2\sqrt{2} + 3 – 2\sqrt{2} = 6$

Squaring both sides, we get :

=> $(x + \frac{1}{x})^2 = 6^2$

=> $x^2 + \frac{1}{x^2} + 2 = 36$

$\therefore x^2 + \frac{1}{x^2} = 34$

Expression : $\frac{a}{b} + \frac{b}{a} – 1 = 0$

=> $\frac{a^2 + b^2}{ab} = 1$

=> $a^2 + b^2 = ab$ ———-Eqn(1)

To find : $a^3 + b^3$

= $(a + b) (a^2 + b^2 – ab)$

Using eqn(1), we get :

= $(a + b) (ab – ab)$

= 0

it is given that (a – b) = 3, (b – c) = 5 and (c – a) = 1

we need to find the value of $\frac{a^3 + b^3 + c^3 – 3abc}{a + b + c}$

as we know that ${a^3 + b^3 + c^3}$ = $(a+b+c)({a^2 + b^2 + c^2 – ab – bc – ac})$ ……..(5)

and ${(a-b)^2 = a^2 + b^2 – 2ab}$…….(1)

${(b-c)^2 = b^2 + c^2 – 2cb}$………..(2)

${(c-a)^2 = a^2 + c^2 – 2ac}$……….(3)

adding 1 , 2 and 3

17.5 =  $({a^2 + b^2 + c^2 – ab – bc – ac}$ …..(4)

Now using 4 and 5 statement

$\frac{a^3 + b^3 + c^3 – 3abc}{a + b + c}$  = 17.5

Expression : $3 – \frac{3+\sqrt{5}}{4} – \frac{1}{3 + \sqrt{5}}$

= $3 – \frac{3+\sqrt{5}}{4} – [\frac{1}{3 + \sqrt{5}}\times(\frac{3-\sqrt5}{3-\sqrt5})]$

= $\frac{12-3-\sqrt5}{4}-(\frac{3-\sqrt5}{9-5})$

= $\frac{9-\sqrt5}{4}+ (\frac{-3+\sqrt5}{4})$

= $\frac{9-3-\sqrt5+\sqrt5}{4}=\frac{6}{4}=\frac{3}{2}$

=> Ans – (B)

it is given that $x + \frac{1}{x}$ = 12

we need to find value of $x^2 + \frac{1}{x^2}$

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

= $12^2 – 2$ = 142

it is given that $x^4 + \frac{1}{x^4} = 23$

$x^2 + \frac{1}{x^2}$ = $\surd(25)$ = 5

we need to calculate $(x-\frac{1}{x})^2$ = $x^2 + \frac{1}{x^2} – 2$

= 5 – 2 = 3

Given : $x = \sqrt{\frac{\sqrt{5} + 1}{\sqrt{5} – 1}}$

=> $x = \sqrt{\frac{\sqrt{5} + 1}{\sqrt{5} – 1}\times (\frac{\sqrt5+1}{\sqrt5+1})}$

=> $x=\sqrt{\frac{(\sqrt5+1)^2}{5-1}}$

=> $x=\frac{\sqrt5+1}{2}$ ————–(i)

Squaring both sides, we get : $x^2=\frac{6+2\sqrt5}{4}$ ————–(ii)

To find : $5x^2 – 5x -1$

= $5(x^2-x)-1$

Substituting values from equations (i) and (ii), we get :

= $5[(\frac{6+2\sqrt5}{4})-(\frac{\sqrt5+1}{2})]-1$

= $5\times(\frac{6+2\sqrt5-2\sqrt5-2}{4})-1$

= $5\times1-1=4$

=> Ans – (C)

Expression : $\frac{3\sqrt{2}}{\sqrt{3} + \sqrt{6}} – \frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}} + \frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$

= $\frac{3\sqrt2(\sqrt6+\sqrt2)(\sqrt3+\sqrt2)-4\sqrt3(\sqrt3+\sqrt6)(\sqrt3+\sqrt2)+\sqrt6(\sqrt3+\sqrt6)(\sqrt6+\sqrt2)}{(\sqrt3+\sqrt6)(\sqrt6+\sqrt2)(\sqrt3+\sqrt2)}$

= $\frac{1}{(\sqrt3+\sqrt6)(\sqrt6+\sqrt2)(\sqrt3+\sqrt2)}\times$ $[3\sqrt2(\sqrt{18}+\sqrt{12}+\sqrt6+\sqrt4)-4\sqrt3(\sqrt9+\sqrt6+\sqrt{18}+\sqrt{12}) + \sqrt6(\sqrt{18}+ \sqrt6+\sqrt{36}+\sqrt{12})]$

= $\frac{1}{(\sqrt3+\sqrt6)(\sqrt6+\sqrt2)(\sqrt3+\sqrt2)}\times$ $[3\sqrt2(3\sqrt2+2\sqrt3+\sqrt6+2)-4\sqrt3(3+\sqrt6+3\sqrt2+2\sqrt3)+ \sqrt6(3\sqrt2+\sqrt6+6+2\sqrt3)]$

= $\frac{1}{(\sqrt3+\sqrt6)(\sqrt6+\sqrt2)(\sqrt3+\sqrt2)}\times$ $[(18+6\sqrt6+6\sqrt3+6\sqrt2)+(-12\sqrt3-12\sqrt2-12\sqrt6-24)+ (6\sqrt3+6+6\sqrt6+6\sqrt2)]$

= $\frac{1}{(\sqrt3+\sqrt6)(\sqrt6+\sqrt2)(\sqrt3+\sqrt2)}\times$ $[(18+6-24)+(6\sqrt2-12\sqrt2+6\sqrt2)+(6\sqrt3-12\sqrt3+6\sqrt3)+(6\sqrt6-12\sqrt6+6\sqrt6)]$

= $\frac{1}{(\sqrt3+\sqrt6)(\sqrt6+\sqrt2)(\sqrt3+\sqrt2)}\times0=0$

=> Ans – (B)

it is given that $a + \frac{1}{a}$ = -2
$(a+2)^{2}+\frac{1}{(a+2)^{3}}$ = $-1^2 + \frac{1}{-1^2}$ = 2