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.
Download Algebra Questions for RRB Group-D Set-3 PDF
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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
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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
RRB Group D previous year papers
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) $\frac{1}{\sqrt{3}}$
RRB Group-D Important Questions (download PDF)
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
General Science Notes for RRB Exams (PDF)
Answers & Solutions:
1) Answer (C)
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
2) Answer (B)
$x=2+\sqrt{3}$
$\frac{1}{x}=2-\sqrt{3}$
$(\sqrt{x} + \frac{1}{\sqrt{x}})^{2}$ = $x+\frac{1}{x}$ + 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.
3) Answer (D)
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
4) Answer (C)
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
5) Answer (A)
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
adding above equations we get,
$a^2 + b^2$ 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
6) Answer (B)
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)
7) Answer (C)
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 $\frac{x}{y}$ = $\frac{3}{2}$
8) Answer (C)
$x=\frac{1}{2+\sqrt{3}}$ , 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
9) Answer (A)
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
10) Answer (A)
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
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We hope this Algebra Questions set – 3 PDF for RRB Group-D Exam will be highly useful for your preparation.
