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# Algebra Questions For RRB JE PDF

Download RRB JE Algebra Questions and Answers PDF. Top 25 RRB JE Maths questions based on asked questions in previous exam papers very important for the Railway JE exam

Question 1: Sum of a natural number and 5 times its reciprocal is 14/3, then find that number ?

a) 6

b) 5

c) 4

d) 3

Question 2: Find $a \times b$ if $a+b = 36$ & $a^2+b^2 = 720 ?$

a) 288

b) 244

c) 284

d) 248

Question 3: Find the value of x if 5*(4x-3/2) + 2x = 4-3x/2

a) 23/47

b) 1/2

c) 2/3

d) 3/2

Question 4: Find $10^3+9^3+…….+3^3$ ?

a) 3015

b) 3020

c) 3025

d) 3016

Question 5: If $p, q$ are the sum and product of the roots of $3x^3-4x^2+5x-6$, then find $\frac{q}{p}$ ?

a) $3$

b) $\frac{1}{3}$

c) $\frac{2}{3}$

d) $\frac{3}{2}$

Question 6: If $x+\frac{1}{x} = 2$ then find $x^{32}+\frac{1}{x^{32}}$?

a) 2

b) 16

c) 4

d) 8

Question 7: 2[3x – 4/5] + 6x/7 = 8, then find x ?

a) 4/5

b) 5/6

c) 6/7

d) 7/5

Question 8: If $p, q$ are the sum and product of the roots of $4x^3-3x^2+2x-1$, then find $\frac{p^2}{q^2}$ ?

a) $4$

b) $9$

c) $\frac{1}{9}$

d) $\frac{1}{4}$

Question 9: If sum of a number and 4 times its fraction is 4, then find that number ?

a) 1

b) 2

c) 3

d) 4

Question 10: Find a-b if a+b = 23, ab = 132 ?

a) 1

b) -1

c) 2

d) Either (a) or (b)

Question 11: If $a+b+c = 0$, then $\frac{a^2+b^2+c^2}{ab+bc+ca} = ?$

a) 0

b) 1

c) -2

d) 2

Question 12: 4[$\frac{x}{3}$ -$\frac{3}{8}$ ] + $\frac{x}{6}$ = $\frac{x}{3}$ -$\frac{1}{2}$, then find x ?

a)  $\frac{6}{7}$

b)  $\frac{7}{6}$

c)  $\frac{4}{5}$

d)  $\frac{5}{4}$

Question 13: If $x^2+\frac{1}{x^2} = 11$ then find $x^4+\frac{1}{x^4}$?

a) 123

b) 119

c) 81

d) 108

Question 14: Find 7+8+9+10+……+50 = ?

a) 1275

b) 1254

c) 1257

d) 1244

Question 15: Find $a^3-b^3$, if $a-b = 1, ab = 12$ ?

a) 27

b) 37

c) 47

d) 57

Question 16: If p, q are the sum and product of the roots of $3x^3-4x^2+5x-6=0$, then find p*q ?

a) 3/8

b) 8/3

c) 2/3

d) 3/2

Question 17: Find the value of $a^2+b^2$, if $a-b = 3, ab = 108$ ?

a) 205

b) 215

c) 225

d) 235

Question 18: If $2 \times (\frac{3x}{4} – \frac{1}{2}) = \frac{x}{2} – 2$, then find the value $\frac{1}{x}$ ?

a) 1

b) -1

c) 1/2

d) -1/2

Question 19: If $x+\frac{1}{x} = 2$ then find $x^8+\frac{1}{x^8}$?

a) 8

b) 4

c) 2

d) 16

Question 20: Find the value of $a^3+b^3$, if $a+b = 13, ab = 40$ ?

a) 627

b) 637

c) 647

d) 657

Question 21: If √x, √y are the quadratic surds and if a+√x = √y , then ?

a) a = 1

b) x = 0

c) y = 0

d) x = y

Question 22: Find the value of $3^3+4^3+5^3+…..+10^3$ ?

a) 3015

b) 3020

c) 3025

d) 3016

Question 23: If p, q are the sum and product of the roots of $3x^3-4x^2+5x-6=0$, then find p/q ?

a) 3

b) 1/3

c) 2/3

d) 3/2

Question 24: If p , q are the roots of the equation $3x^2-4x-4 = 0$, then which of of the following is true ?

a) p, q are not equal and greater than 0.

b) p, q are imaginary numbers

c) p = q

d) p, q are real numbers and unequal

Question 25: Find the value of x if 9(8x/7 + 6/5) – 4x/3 = 2

a) 235/231

b) 231/235

c) -231/235

d) -235/231

Let the number be x

$x+5(1/x) = 14/3$

$x^2+5 = 14x/3$

$3x^2+15 = 14x$

$3x^2-14x+15 = 0$

$3x^2-9x-5x+15 = 0$

$(3x-5)(x-3) = 0$

$x = 5/3$ or $x = 3$

So the answer is option D.

$(a+b)^2 = a^2+b^2+2ab$

$(36)^2 = 720+2ab$

$1296 = 720+2ab$

$1296 – 720 = 2ab$

$576 = 2ab$

$288 = ab$

So the answer is option A.

5(4x-3/2) + 2x = 4-3x/2

20x – 15/2 + 2x = 4-3x/2

22x + 3x/2 = 4 + 15/2

47x/2 = 23/2

x = 23/47

So the answer is option A.

$10^3+9^3+…..+3^3=3^3+5^3+…..10^3=(1^3+2^3+….+10^3)-(1^3+2^3)=(\frac{10^2(11)^2}{4})-(1+8)$
=> $3025 – 9 = 3016$

So the answer is option D.

$p$ = sum of roots = $\frac{-b}{a} = \frac{4}{3}$

$q$ = product of roots = $\frac{-d}{a} = \frac{6}{3}$ = 2

$\frac{q}{p} = \frac{2}{(\frac{4}{3})} = \frac{3}{2}$

So the answer is option D.

If $x+\frac{1}{x} = 2$ then $x^{2^n}+\frac{1}{x^{2^n}} = 2$

put n = 5, $x^{32}+\frac{1}{x^{32}} = 2$

So the answer is option A.

2[3x – 4/5] + 6x/7 = 8

6x – 8/5 +6x/7 = 8

48x/7 = 8+8/5

48x/7 = 48/5

x = 7/5

So the answer is option D.

$p$ = Sum of roots = $\frac{3}{4}$ ===> $p^2 = 9/16$

$q$ = product of roots = $\frac{1}{4}$ ===> $q^2$ = $\frac{1}{6}$

$\frac{p^2}{q^2}$ = $\frac{9}{1}$

So the answer is option B.

$x+4/x = 4$

$x^2+4 = 4x$

$x^2-4x+4 = 0$

$(x-2)^2 = 0$

$x = 2$

So the answer is option B.

$(a-b)^2 = (a+b)^2 – 4ab$

$(a-b)^2 = (23)^2 – 4(132)$

$(a-b)^2 = 529 – 528$

$(a-b)^2 = 1$

$a-b = +1 or -1$

So the answer is option D.

Given $a+b+c = 0$, then

$(a+b+c)^2 = 0$

$a^2+b^2+c^2+2ab+2bc+2ca = 0$

$a^2+b^2+c^2 = -2(ab+bc+ca)$

$\frac{a^2+b^2+c^2}{ab+bc+ca} = -2$

So the answer is option C.

4[$\frac{x}{3}$ – $\frac{3}{8}$] + $\frac{x}{6}$ = $\frac{x}{3}$ – $\frac{1}{2}$

$\frac{4x}{3}$ – $\frac{3}{2}$ + $\frac{x}{6}$ = $\frac{x}{3}$ – $\frac{1}{2}$

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

$\frac{7x}{6}$ = 1

x = $\frac{6}{7}$

So the answer is option A.

$x^2+\frac{1}{x^2} = 11$

Squaring on both sides

$x^4+\frac{1}{x^4}+2 = 121$

$x^4+\frac{1}{x^4} = 121-2 = 119$

So the answer is option B.

WKT, sum of 1st n natural numbers = $\frac{n(n+1)}{2}$

$7+8+9+10+……+50 = (1+2+3+…..+50)-(1+2+3+….+6)$

This equals $\frac{(50)(51)}{2} – \frac{(6)(7)}{2} = 1275 – 21 = 1254$

So the answer is option B.

$a^3-b^3 = (a-b)^3+3ab(a-b) = 1^3+3(12)(1) = 1+36 = 37$

So the answer is option B.

p = sum of roots = -b/a = 4/3

q = product of roots = -d/a = 6/3 = 2

pq = (4/3)(2) = 8/3

So the answer is option B.

$a^2+b^2 = (a-b)^2+2ab = 3^2+2(108) = 9+216 = 225$

So the answer is option C.

2[3x/4 – 1/2] = x/2 – 2

3x/2 – 1 = x/2 – 2

2x/2 = -1

x = -1

So the answer is option B.

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

Squaring on both sides

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

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

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

Squaring on both sides

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

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

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

Squaring on both sides

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

$x^8+\frac{1}{x^8}+2 = 4$

$x^8+\frac{1}{x^8} = 2$

So the answer is option C.

$a^3+b^3 = (a+b)^3-3ab(a+b) = 13^3-3(40)(13) = 2197 – 1560 = 637$

So the answer is option B.

a+√x = √y+0

here √x, √y are the quadratic surds ( $\because$ quadratic surd is the square root of an irrational number, example $\sqrt2,\sqrt3$etc.)

so a = 0, √x = √y  => x = y

So the answer is option D.

$3^3+4^3+5^3+…..+10^3 = (1^3+2^3+…..+10^3)-(1^3+2^3) = (\frac{10^2(11)^2}{4}) – (1+8) = 3025 – 9 = 3016$

So the answer is option D.

Sum of roots = p = -b/a = 4/3

Product of roots = -d/a = 6/3  = 2

p/q = (4/3)/2 = 2/3

So the answer is option C.

Discriminant = $b^2-4ac = (-4)^2-4(3)(-4) = 16+48 = 64 > 0$

So roots are real and unequal.

The product of p and q is $-4/3$, hence one of them is greater than 0 and the other is less than 0.

So the answer is option D.

9[8x/7 + 6/5 ] – 4x/3 = 2

72x/7 + 54/5 -4x/3 = 2

72x/7 – 4x/3 = 2-54/5

216x/21 – 28x/21 = -44/5

188x/21 = -44/5

47x/21 = -11/5

x = -231/235

So the answer is option C.

We hope this Algebra Questions for RRB JE Exam will be highly useful for your preparation.