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# RRB NTPC Algebra Questions PDF

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

Question 1: A two digit number when reverse decreases its value by 63 and the sum of the digits of the number is 11.Fin the value of the number.

a) 56

b) 47

c) 38

d) 29

Question 2: A two digit number when reverse decreases its value by 27 and the sum of the digits of the number is 7.Fin the value of the number.

a) 34

b) 16

c) 25

d) 43

Question 3: If $a+b=0$ then what is the value of $a^3+b^3$?

a) -1

b) 0

c) 1

d) Can’t be determined

Question 4: Find the value of $1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{7}{3}}}}$.

a) $\frac{37}{16}$

b) $\frac{25}{14}$

c) $\frac{26}{17}$

d) $\frac{36}{23}$

Question 5: A person can buy 54 gifts with the money he has and if the cost of the gift is decreased by 4 rupees then he can buy 60 gifts and he is left with 18 rupees. What is the total money he has ?

a) 1900

b) 2035

c) 1998

d) 1961

Question 6: A number is increased by 12 and divided by 19 to get the result as 9 and if the number is decreased by 40 and divided by 17 then the remainder obtained is ?

a) 3

b) 0

c) 1

d) 2

Question 7: If x+y+z = 20 and xy+yz+xz = 10, find the value of $x^3 + y^3 + z^3 -3xyz$.

a) 6800

b) 7525

c) 7400

d) 6400

Question 8: If $x+\dfrac{1}{x} = 3$, then find the value of $x^3+\dfrac{1}{x^3}$.

a) 16

b) 14

c) 18

d) 22

Question 9: Find the value of $1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{5}{4}}}}$.

a) $\frac{37}{11}$

b) $\frac{35}{22}$

c) $\frac{25}{44}$

d) $\frac{35}{11}$

Question 10: If 7/2(2x/3 – 1/2) + 11/2 = 2x/3, then what is the value of 1/x?

a) -9/4

b) -4/9

c) 4/9

d) 9/4

Question 11: If $a+\frac{1}{a} = 5$, then find the value of $a^2+\frac{1}{a^2}$.

a) 18

b) 23

c) 27

d) 14

Question 12: Find the value of $1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}}$.

a) $\dfrac{18}{11}$

b) $\dfrac{16}{11}$

c) $\dfrac{27}{11}$

d) $\dfrac{12}{11}$

Question 13: Find the value of $x$ (from the given options), if $5 = 40-12x+x^2$ ?

a) 3

b) 4

c) 5

d) 6

Question 14: Find $(a+b)(a^2-ab+b^2+3ab)$ = ?

a) $(a-b)^3$

b) $(b+a)^3$

c) $(a+b)^2$

d) $(b-a)^3$

Question 15: Find the value of $x$ (from the given options), if $5x^2-13x-6 = 0$ ?

a) -2

b) 3/5

c) -3

d) -2/5

Question 16: If $x+y+z = 9$, $xy+yz+zx = 18$, then find $x^2+y^2+z^2$

a) 81

b) 45

c) 56

d) 63

Question 17: The product of $(a+b+c) \times (a+2b-c)$ is ?

a) $a^2+2b^2-c^2+3ab+bc$

b) $a^2+2b^2-c^2+3ab+2bc$

c) $a^2+2b^2-c^2+2ab+bc$

d) $a^2+2b^2-c^2+3ab+3bc$

Question 18: If $b+1/b = 6$, then find $b^2+1/b^2$ ?

a) 3

b) 27

c) 83

d) 34

Question 19: FInd $x^4+1/x^4$, if $x+1/x = 2$ ?

a) 5

b) 3

c) 4

d) 2

Question 20: There are some sheep and hens. If there are a total of 36 legs and there are 14 animals and birds in total, how many sheep are there ?

a) 4

b) 10

c) 8

d) 12

Question 21: Sum of 4 times of a fraction and its reciprocal is 4, find that fraction ?

a) 2/3

b) 3/2

c) 3/4

d) 1/2

Question 22: Find $ab+bc+ca$, if $a^2+b^2+c^2 = 24$ & $a+b+c=4\sqrt3$ ?

a) 12

b) 13

c) 14

d) 15

Question 23: [4x-1/4]3 = x + 4/3, then find x ?

a) 123/25

b) 132/25

c) 25/132

d) 25/123

Question 24: Find which of $a$ & $b$ is smaller, $(a+b)^3 = 27$ & $(a-b)^2 = 1$ ?

a) a

b) b

c) a=b

d) cannot be determined

Question 25: 3[2x-4/3] + 4[2-3x/2] = 4, then find x ?

a) 3/2

b) 0

c) x can be any number

d) 1

let the number be xy
Given x+y=11
10y+x-(10x+y)=63
9(y-x)=63
y-x=7
2y=18
y=9
x=2
Required number is 29

let the number be xy
Given x+y=7
10y+x-(10x+y)=27
9(y-x)=27
y-x=3
2y=10
y=5
x=2
Required number is 25

Given, $a+b=0$
$a^3+b^3$ can be written as
$a^3+b^3=(a+b)(a^2-ab+b^2)$
⇒ $a^3+b^3=0*( a^2-ab+b^2)$
⇒ $a^3+b^3=0$

$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{7}{3}}}} = 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{\dfrac{10}{3}}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{3}{10}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{\dfrac{13}{10}}}$

$= 1+\dfrac{1}{1+\dfrac{10}{13}}$

$= 1+\dfrac{1}{\dfrac{23}{13}}$

$= 1+\dfrac{13}{23}$

$= \dfrac{36}{23}$

let the cost of gift be x
54x=((x-4)*60)+18
54x=60x-240+18
6x=222
x=37
If x=37 total money with him is 54*37=Rs 1998

let the number be x
(x+12)/19=9
x=171-12
x=159
And then (159-40)=119
119/17 we get 0 as the remainder.

We know that $x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2 + y^2 +z^2 -xy -yz -zx)$
Now, $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$
=> $x^2 + y^2 +z^2 -xy -yz -zx = (x+y+z)^2 – 3(xy+yz+zx)$
=> $x^3 + y^3 + z^3 -3xyz = (x+y+z)[ (x+y+z)^2 – 3(xy+yz+zx)] = 20 [400 – 30] = 20 \times 370 = 7400$

Given $x+\dfrac{1}{x} = 3$
Cubing on both sides

$(x+\dfrac{1}{x})^3 = 3^3$

$x^3+\dfrac{1}{x^3}+3\times x \times \dfrac{1}{x}(x+\dfrac{1}{x}) = 27$

⇒ $x^3+\dfrac{1}{x^3}+3\times3 = 27$

⇒ $x^3+\dfrac{1}{x^3}+9 = 27$

⇒ $x^3+\dfrac{1}{x^3} = 18$

$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{5}{4}}}} = 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{\dfrac{9}{4}}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{4}{9}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{\dfrac{13}{9}}}$

$= 1+\dfrac{1}{1+\dfrac{9}{13}}$

$= 1+\dfrac{1}{\dfrac{22}{13}}$

$= 1+\dfrac{13}{22}$

$= \dfrac{35}{22}$

7/2(2x/3 – 1/2) + 11/2 = 2x/3
7x/3 – 7/4 + 11/2 = 2x/3
7x/3 – 2x/3 = 7/4 – 11/2
5x/3 = -15/4
x = -9/4
1/x = -4/9
So the answer is option B.

Given, $a+\frac{1}{a} = 5$

Squaring on both sides
$(a+\frac{1}{a})^2 = 5^2$

$a^2+\frac{1}{a^2}+2\times a\times\frac{1}{a} = 25$

$a^2+\frac{1}{a^2}+2 = 25$

$a^2+\frac{1}{a^2} = 23$

$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{\dfrac{4}{3}}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{1+\frac{3}{4}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{\dfrac{7}{4}}}$

$= 1+\dfrac{1}{1+\dfrac{4}{7}}$

$= 1+\dfrac{1}{\dfrac{11}{7}}$

$= 1+\dfrac{7}{11} = \dfrac{18}{11}$

$5 = 40-12x+x^2$

$x^2-12x+35 = 0$

$x^2-7x-5x+35 = 0$

$x(x-7)-5(x-7) = 0$

$(x-7)(x-5) = 0$

$x = 7$ or $x = 5$

So the answer is option C.

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

So the answer is option B.

$5x^2-13x-6 = 0$

$5x^2-15x+2x-6 = 0$

$5x(x-3)+2(x-3) = 0$

$(x-3)(5x+2) = 0$

$x = 3 (or) x = -2/5$

So the answer is option D.

$(x+y+z)^2$ = $x^2+y^2+z^2+2(xy+yz+zx)$

$(9)^2$ = $x^2+y^2+z^2+2(18)$

$81$ = $x^2+y^2+z^2+2(18)$

$x^2+y^2+z^2 = 81-36 = 45$

So the answer is option B.

$(a+b+c)(a+2b-c)$

=$a^2+2ab-ac+ab+2b^2-bc+ac+2bc-c^2$

=$a^2+2b^2-c^2+3ab+bc$

So the answer is option A.

$b+1/b = 6$

Squaring on both sides

$(b+1/b)^2 = 36$

$b^2+1/b^2 +2 = 36$

$b^2+1/b^2 = 36-2 = 34$

So the answer is option D.

$x+1/x = 2$

squaring on both sides

$(x+1/x)^2 = 4$

$x^2+1/x^2+2 = 4$

$x^2+1/x^2 = 2$

squaring on both sides

$(x^2+1/x^2)^2 = 4$

$x^4+1/x^4+2 = 4$

$x^4+1/x^4 = 2$

So the answer is option D.

A sheep has 4 legs and hen has 2 legs

4x+2y = 36 ===> 2x+y = 18—-(1)

x+y = 14 —-(2)

(1)-(2) ==> 2x+y-x-y = 18-14 = 4

So the answer is option A.

Let the fraction be 1/x

$4(1/x)+x = 4$

$4+x^2 = 4x$

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

$(x-2)^2 = 0$

$x-2 = 0$

$x = 2$

Hence the fraction is 1/x = 1/2

So the answer is option D.

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

$(4\sqrt3)^2 = 24 + 2(ab+bc+ca)$

$48 = 24 + 2(ab+bc+ca)$

$24 = 2(ab+bc+ca)$

$(ab+bc+ca) = \frac{24}{2} = 12$

So the answer is option A.

[4x-1/4]3 = x + 4/3

12x – 3/4 = x + 4/3

11x =4/3 + 3/4

11x = 25/12

x = 25/132

So the answer is option C.

$(a+b)^3 = 27 \rightarrow a+b = 3$

$(a-b)^2 = 1 \rightarrow a-b = \pm 1$

If $a-b = -1$, & $a+b = 3$, then $a = 1$ & $b = 2$

If $a-b = +1$, & $a+b = 3$, then $a = 2$ & $b = 1$

So we can’t find exact values of $a$ & $b$

So the answer is option D.