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# RRB NTPC Maths Questions Set-2 PDF

Download RRB NTPC Maths Set-2 Questions and Answers PDF. Top 15 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1:Â Find the value of $tan 56^\circ – tan 11^\circ – tan 56^\circ.tan 11^\circ$

a)Â 0

b)Â 1

c)Â -1

d)Â 2

Question 2:Â Minimum value of $2\sin^{2}x+1\cos^{2}x$

a)Â 1.5

b)Â 2

c)Â 1

d)Â 0

Question 3:Â If (a+b) = 13 and ab = 36, then find (a-b).

a)Â 8

b)Â 5

c)Â 6

d)Â 9

Question 4:Â A number is increased by 20 and divided by 13 to get the result as 9 and if the number is decreased by 4 and divided by 27 then the remainder obtained ?

a)Â 11

b)Â 14

c)Â 12

d)Â 13

Question 5:Â The product of $(a+b+c) \times (a+2b-c)$ is ?

a)Â $a^2+2b^2-c^2+3ab+bc$

b)Â $a^2+2b^2-c^2+3ab+2bc$

c)Â $a^2+2b^2-c^2+2ab+bc$

d)Â $a^2+2b^2-c^2+3ab+3bc$

Question 6:Â 3[2x-4/3] + 4[2-3x/2] = 4, then find x ?

a)Â 3/2

b)Â 0

c)Â x can be any number

d)Â 1

Question 7:Â Find a-b if a+b = 23, ab = 132 ?

a)Â 1

b)Â -1

c)Â 2

d)Â Either (a) or (b)

Question 8:Â If p, q are the sum and product of the roots of $3x^3-4x^2+5x-6=0$, then find p*q ?

a)Â 3/8

b)Â 8/3

c)Â 2/3

d)Â 3/2

Question 9:Â Find the value of $3^3+4^3+5^3+…..+10^3$ ?

a)Â 3015

b)Â 3020

c)Â 3025

d)Â 3016

Question 10:Â Find the value of $1-\frac{1}{1-\frac{1}{1+\frac{1}{x}}}$ ?

a)Â $x$

b)Â $-x$

c)Â 1

d)Â -1

Question 11:Â Find x*y, if (x+y) = 14, (x-y) = 4 ?

a)Â 40

b)Â 56

c)Â 45

d)Â 50

Question 12:Â Consider the following set of numbers: A = {1, 2, 3, 2, 5, 2, 3, 4, 6, 2, 3, 5, 5, 6, 7, 8, 7, 8, 3, 5, 6, 1, 3, 8, 7}. What is the mode of the set?

a)Â 4

b)Â 5

c)Â 3

d)Â 2

Question 13:Â Find the absolute difference between the median and mode of the following data:
3, 6, 8, 6, 8, 7, 5, 4, 3, 8, 6, 7, 4, 9, 4, 2, 5, 1, 2, 8

a)Â 2

b)Â 3

c)Â 2.5

d)Â 2.75

Question 14:Â If 0.20 : 4 : : x : 120, then find the value of x.

a)Â 0.6

b)Â 60

c)Â 20

d)Â 6

Question 15:Â The run-rate of 12 overs of a cricket match is 8 rpo. What should be the run-rate for the next 8 overs to chase the target of 160 runs?

a)Â 8 rpo

b)Â 6 rpo

c)Â 7.5 rpo

d)Â 8.6 rpo

$56-11 = 45$
Applying â€˜tanâ€™ on both sides
$tan (56-11)^\circ = tan 45^\circ$ â†’ (1)
We know that $tan (A-B)^\circ = \dfrac{tan A^\circ – tan B^\circ}{1+tan A^\circ.tan B^\circ}$
Then, (1) becomes
$\dfrac{tan 56^\circ – tan 11^\circ}{1+tan 56^\circ.tan 11^\circ} = 1$ (Since, $tan 45^\circ = 1$)

$tan 56^\circ – tan 11^\circ = 1+tan 56^\circ.tan 11^\circ$
â‡’ $tan 56^\circ – tan 11^\circ – tan 56^\circ.tan 11^\circ = 1$

$(\sin^{2}x+\cos^{2}x)+\sin^{2}x$=1+$\sin^{2}x$
Minimum value of $\sin x$ is when $x$=0 i.e $\sin 0$=0
Therefore 1+0=1
Hence, option CÂ is the correct answer.

We know that $(a-b)^2 = (a+b)^2-4ab$
â‡’ $(a-b)^2 = 169-4*36 = 169-144 = 25$
â‡’ (a-b) = 5

let the number be x
(x+20)/13=9
x=117-20
x=97
And then (97-4)=93
93/27 we get 12 as the remainder.

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

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

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

So the answer is option A.

3[2x-4/3] + 4[2-3x/2] = 4

6x-4+8-6x = 4

0 = 0

here we have 6x and -6x, So for any value of x, 6x-6x = 0

So the answer is option C.

$(a-b)^2 = (a+b)^2 – 4ab$

$(a-b)^2 = (23)^2 – 4(132)$

$(a-b)^2 = 529 – 528$

$(a-b)^2 = 1$

$a-b = +1 or -1$

So the answer is option D.

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.

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

So the answer is option D.

$1-\frac{1}{1-\frac{1}{1+\frac{1}{x}}}$ =Â $1-\frac{1}{1-\frac{1}{\frac{x+1}{x}}}$ =Â $1-\frac{1}{1-\frac{x}{1+x}}$ =Â $1-\frac{1}{\frac{1+x-x}{1+x}}$ =Â $1-\frac{1}{\frac{1}{1+x}}$ = $1-1-x = -x$

So the answer is option B.

x+y = 14

x-y = 4

On solving these 2 equations

x = 9 & y = 5

Now, x*y = 9*5 = 45

So the answer is option C.

The frequency of repetition of the numbers is as follows:

1 – 2
2 – 4
3 – 5
4 – 1
5 – 4
6 – 3
7 – 3
8 – 3
So, the mode is 3.

8 occurs the most number of times => 8 is mode.
When the data is arranged in the ascending order, we get:
1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9.
10th and 11th digits are 5 and 6 respectively.
=> Median = (5+6)/2 = 5.5
Difference = |8 – 5.5| = 2.5

$4\times x = 0.20\times120$