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.

Download RRB NTPC Maths Questions Set-2 PDF

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

a) 0

b) 1

c) -1

d) 2

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

a) 1.5

b) 2

c) 1

d) 0

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

a) 8

b) 5

c) 6

d) 9

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Question 4: 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) 11

b) 14

c) 12

d) 13

Question 5: 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) $a^2+2b^2-c^2+3ab+3bc$

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Question 6: 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

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

a) 1

b) -1

c) 2

d) Either (a) or (b)

Question 8: 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

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

a) 3015

b) 3020

c) 3025

d) 3016

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

a) $x$

b) $-x$

c) 1

d) -1

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

a) 40

b) 56

c) 45

d) 50

Question 12: 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) 4

b) 5

c) 3

d) 2

Question 13: 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) 2

b) 3

c) 2.5

d) 2.75

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

a) 0.6

b) 60

c) 20

d) 6

Question 15: 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) 8 rpo

b) 6 rpo

c) 7.5 rpo

d) 8.6 rpo

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Answers & Solutions:

1) Answer (B)

$56-11 = 45$
Applying ‘tan’ on both sides
$tan (56-11)^\circ = tan 45^\circ$ → (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, $tan 45^\circ = 1$)

$tan 56^\circ – tan 11^\circ = 1+tan 56^\circ.tan 11^\circ$
⇒ $tan 56^\circ – tan 11^\circ – tan 56^\circ.tan 11^\circ = 1$

2) Answer (C)

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

3) Answer (B)

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

4) Answer (C)

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.

5) Answer (A)

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

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

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

So the answer is option A.

6) Answer (C)

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.

7) Answer (D)

$(a-b)^2 = (a+b)^2 – 4ab$

$(a-b)^2 = (23)^2 – 4(132)$

$(a-b)^2 = 529 – 528$

$(a-b)^2 = 1$

$a-b = +1 or -1$

So the answer is option D.

8) Answer (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.

9) Answer (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.

10) Answer (B)

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

So the answer is option B.

11) Answer (C)

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.

12) Answer (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.

13) Answer (C)

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

14) Answer (D)

$4\times x = 0.20\times120$
⇒ $x = \dfrac{0.20\times120}{4} = 6

15) Answer (A)

Total score made in first 12 overs = 12*8 = 96 runs
Target = 160 runs
Remaining score to be made = 160-96 = 64 runs
Required run-rate = 64/8 = 8 rpo.

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We hope this Maths Set-2  Questions for RRB NTPC Exam will be highly useful for your Preparation