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# Algebra Questions for RRB Group-D Set-3 PDF

Download Top-10 RRB Group-D Algebra Questions set – 3 PDF. RRB GROUP-D questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

Question 1: If a^3 – b^3 = 56 and a – b = 2, then the value of (a^2 + b^2 + ab) is :

a) ­10

b) ­12

c) 28

d) 18

Question 2: If x = 2 + √3 , then the value, $\sqrt{x} + \frac{1}{\sqrt{x}}$

a) √3

b) √6

c) 2 √6

d) 6

Question 3: If $2\sqrt{x} = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} + \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

a) 6

b) 30

c) √15

d) 16

Question 4: If a + 1/a = 1, then the value of a^2 + 1/a^2 is :

a) -2

b) 2

c) -1

d) 4

Question 5: If ax + by = 6, bx – ay = 2 and x^2 + y^2 = 4, then the value of (a^2 + b^2) would be:

a) 10

b) 2

c) 4

d) 5

Question 6: If $a =\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ then $\frac{a^2}{b}+\frac{b^2}{a}$

a) 900

b) 970

c) 1030

d) 930

Question 7: If (5x^2 – 3y^2): xy = 11:2, and x,y are positive, then the value of x/y is:

a) 7/2

b) 5/2

c) 3/2

d) 5/3

Question 8: If $x=\frac{1}{2+\sqrt{3}}$ and $y=\frac{1}{2-\sqrt{3}}$ then the value of $\frac{1}{x+1} + \frac{1}{y+1}$ is

a) $\frac{1}{2}$

b) $\sqrt{3}$

c) $1$

d) $\frac{1}{\sqrt{3}}$

Question 9: The value of $\sqrt{-\sqrt{3}+\sqrt{3+8\sqrt{7+4\sqrt{3}}}}$ is

a) 2

b) 3

c) 4

d) 1

Question 10: If $\sqrt{4x-9}+\sqrt{4x+9}=5+\sqrt{7}$, then the value of $x$ is

a) 4

b) 5

c) 7

d) 3

it is given that $a^3 – b^3$ = 56

we know $a^3 – b^3 = (a-b)(a^2 + b^2 + ab)$

a- b = 2 (Given)

hence

$(a^2 + b^2 + ab)$ = $\frac{56}{2}$ = 28

$x=2+\sqrt{3}$

$\frac{1}{x}=2-\sqrt{3}$

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

$(\sqrt{x} + \frac{1}{\sqrt{x}})^{2}$ = 4 + 2 = 6

$\sqrt{x} + \frac{1}{\sqrt{x}}$ = $\sqrt{6}$

so the answer is option B.

it is given that

$2\sqrt{x} = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} – \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

here , $\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$ = $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ x $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ = $\frac{(\sqrt5 + \sqrt3)^2}{2}$

similarly , $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ = $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ x $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ = $\frac{(\sqrt5 – \sqrt3)^2}{2}$

$\frac{(\sqrt5 + \sqrt3)^2}{2}$ + $\frac{(\sqrt5 – \sqrt3)^2}{2}$ = 2$\sqrt(x)$

8 = 2$\sqrt(x)$

x = 16

it is given that a + 1/a = 1

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

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

= $1^2$ – 2 = -1

it is given that

ax + by = 6……….(1)

bx – ay = 2……..(2)

and x^2 + y^2 = 4

now multiply 1 and 2nd equation by a and b respectively

we get

$a^2$x + $ab$y = 6a

$b^2$x – aby = 2b

$a^2 + b^2$ x = 6a +2b

x = $\frac{6a+2b}{a^2 + b^2}$

Similarly ,

we get y = $\frac{6a-2b}{a^2 + b^2}$

putting above values in x^2 + y^2 = 4

we get , a = 1 and b = 3

hence $1^2 + 3^2$ = 10

Give : $a =\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

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

=> $a=\frac{(\sqrt3-\sqrt2)^2}{3-2}$

=> $a=5-2\sqrt6$

Squaring both sides, we get : $a^2=49-20\sqrt6$

Similarly, $b=5+2\sqrt6$ and $b^2=49+20\sqrt6$

To find : $\frac{a^2}{b}+\frac{b^2}{a}$

= $\frac{a^3+b^3}{ab}=\frac{(a+b)(a^2+b^2-ab)}{ab}$

= $\frac{[(5-2\sqrt6)+(5+2\sqrt6)][(49-20\sqrt6)+(49+20\sqrt6)-(5-2\sqrt6)(5+2\sqrt6)]}{(5-2\sqrt6)(5+2\sqrt6)}$

= $\frac{10[49+49-(25-24)]}{25-24}$

= $10\times97=970$

=> Ans – (B)

it is given that (5x^2 – 3y^2): xy = 11:2

let divide the numerator and denominator of left hand side of the given equation by y^2

we will get ,

$\frac{5( \frac{x}{y})^2 – 3}{\frac{x}{y}}$ = $\frac{11}{2}$

let $\frac{x}{y}$ = k

So, $\frac{5k^2-3}{k}$ = $\frac{11}{2}$

Solving this we get , k = $\frac{3}{2}$

hence $\frac{x}{y}$ = $\frac{3}{2}$

$x=\frac{1}{2+\sqrt{3}}$ , on rationalizing x = 2 – $\surd3$

$y=\frac{1}{2-\sqrt{3}}$ , on rationalizing y = 2 + $\surd3$

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

using above values of x and y

$\frac{1}{x+1} + \frac{1}{y+1}$ = $\frac{1}{2 + \surd3 +1} + \frac{1}{2 – \surd3 +1}$

now on rationalizing ,

$\frac{1}{x+1} + \frac{1}{y+1}$ = $\frac{3 – \surd3 + 3 + \surd3}{6}$ = 1

we need to find value of $\sqrt{-\sqrt{3}+\sqrt{3+8\sqrt{7+4\sqrt{3}}}}$

$\sqrt{7 + \sqrt3}$ = $\sqrt{2^2 + (\sqrt3)^2 + 2 \times 2 \times \sqrt3}$ = $2 + \sqrt3$

$\sqrt{3 + 4^2 + 2 \times 4 \times \sqrt3}$ = $4 + \sqrt3$

$\sqrt{-\surd3 + 4 + \surd3}$ = 2\

hence $\sqrt{-\sqrt{3}+\sqrt{3+8\sqrt{7+4\sqrt{3}}}}$ = 2

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

We hope this Algebra Questions set – 3 PDF for RRB Group-D Exam will be highly useful for your preparation.