RRB NTPC Expected Maths Questions Set-2 PDF

RRB NTPC Expected Maths Questions 2019 Set-2 PDF

Download RRB NTPC Expected Maths Questions and Answers 2019 Set-2 PDF. Top 15 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

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Question 1: If tan(θ/2) = 5/12, then find sinθ = ?

a) 5/13

b) 119/120

c) 120/169

d) 12/13

Question 2: $1/sinA – SinA/Cos^2A$ = ?

a) $2Cot2A.secA$

b) $2Cos2A.cosecA$

c) $2Cos2A.secA$

d) $2Cot2A.cosecA$

Question 3: $a^{2}-10a+16$=0 and $b^{2}-14b+48$=0, compare the values of a and b.

a) $ a>b $

b) $ a<b $

c) $a \leq b$

d) $a \geq b$

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Question 4: 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) 52

d) 43

Question 5: M is as much younger than N as he is older than O. If the sum of the ages of N and O is 48 years, then find the age of M.

a) 20 years

b) 12 years

c) 24 years

d) 16 years

Question 6: Two containers of equal volume contain milk and water in the ratio 3 : 2 and 5 : 3 respectively. If they are mixed together, find the new ratio.

a) 5 : 2

b) 25 : 4

c) 16 : 9

d) 49 : 31

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Question 7: $\sqrt{\sqrt{\sqrt{\sqrt{65536}}}}$

a) 2

b) 3

c) 4

d) 5

Question 8: $2\sqrt{48}-4\sqrt{12}+3\sqrt{3} = ?$

a) 0

b) $\sqrt{3}$

c) $3\sqrt{3}$

d) $2\sqrt{3}$

Question 9: 20% of 30% of 400% of 50% of 60 = ?

a) 8.4

b) 7.2

c) 6.8

d) 8.2

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Question 10: If circumference of a circle is increased by 10%, the area of the circle will increase by

a) 5%

b) 10%

c) 20%

d) 21%

Question 11: In an examination, a student gets 20% of total marks and fails by 30 marks. Another student gets 32% of total marks which is more than the minimum pass marks by 42 marks. The pass percentage is

a) 24%

b) 25%

c) 26%

d) 28%

Question 12: The ratio of present ages of Ratna and Shantanu is 5:4. After 19 years the ratio of their ages will be 10:9. What is Ratna’s present age?

a) 19

b) 15

c) 9

d) 72

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Question 13: If 42 persons consume 144 kg, of wheat in 15 days, then in how many days will 30 persons consume 48 kg of wheat ?

a) 8 days

b) 7 days

c) 12 days

d) 6 days

Question 14: Instead of dividing 391 cookies among 3 children A, B, C in the ratio $\frac{1}{5}:\frac{1}{4}:\frac{1}{8}$, it was divided in to the ratio 5: 4: 8. Who gains the most and how many ?

a) A, 21 cookies

b) B, 78 cookies

c) C, 99 cookies

d) C, 78 cookies

Question 15: In a circle of radius 30 cm, two parallel chords were drawn which are separated by a distance of 42 cm and length of one of the chord is 48 cm. What is the length of other chord ?

a) 36 cm

b) 48 cm

c) 32 cm

d) 40 cm

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Answers & Solutions:

1) Answer (C)

$Sinθ$ = $\frac{2tan(θ/2)}{1+tan^2(θ/2)}$ = $\frac{2(5/12)}{1+(5/12)^2}$ = $120/169$

So the answer is option C.

2) Answer (A)

$1/sinA – SinA/Cos^2A$
= $(cos^2A-sin^2A)/(sinA.cos^2A)$
= $2(Cos2A)/(sin2A.CosA)$
= $2Cot2A.secA$
So the answer is option A.

3) Answer (B)

$a^{2}-10a+25$=0
$a^{2}-5a-5a+25$=0
$a(a-5)-5(a-5)$=0
$(a-5)(a-5)$=0
$a$=5 and $a$=5
$b^{2}-14b+48$=0
$b^{2}-8b-6b+48$=0
$b(b-8)-6(b-8)$=0
$(b-8)(b-6)$=0
$b$=8 and $b$=6
Therefore $ a<b $
Hence, option B is the correct answer.

4) Answer (C)

let the number be yx
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 52
Hence, option C is the correct answer.

5) Answer (C)

Given that difference between the ages of M and N is equal to the difference between the ages of N and O.
N – M = M – O
⇒ 2M = N+O
Given, N+O = 48 years
Then, 2M = 48 ⇒ M = 24 years
Therefore, The age of M = 24 years.

6) Answer (D)

Let the total quantity of the mixture be x litres.
Given that both the containers have same volume.
Hence, Volume of container 1 = Volume of container 2 = x litres.
Quantity of milk in container 1 = $\dfrac{3}{5}x$ litres
Quantity of water in container 1 = $\dfrac{2}{4}x$ litres

Quantity of milk in container 2 = $\dfrac{5}{8}x$ litres
Quantity of water in container 2 = $\dfrac{3}{8}x$ litres

Quantity of milk in new container = $\dfrac{3}{5}x + \dfrac{5}{8}x = \dfrac{49x}{40}$ litres
Quantity of water in new container = $\dfrac{2}{5}x + \dfrac{3}{8}x = \dfrac{31x}{40}$ litres

Required ratio = $\dfrac{49x}{40} : \dfrac{31x}{40} = 49 : 31$

7) Answer (A)

$\sqrt{\sqrt{\sqrt{\sqrt{65536}}}}$ = $\sqrt{\sqrt{\sqrt{256}}}$ = $\sqrt{\sqrt{16}}$ = $\sqrt{4} = 2$

So the answer is option A.

8) Answer (C)

$2\sqrt{48}-4\sqrt{12}+3\sqrt{3}$ = $2(4\sqrt3)-4(2\sqrt3)+3\sqrt3 = 8\sqrt3-8\sqrt3+3\sqrt3 = 3\sqrt3$

So the answer is option C.

9) Answer (B)

$(0.2)(0.3)(4)(0.5)(60) = (0.2)(0.3)(120) = (0.06)(120) = 7.2$

So the answer is option B.

10) Answer (D)

Circumference of a circle is 2$ \pi$ r

So, if the circumference increases by 10%, the radius also increases by 10%

area of the circle = $ \pi r ^ {2}$

If radius becomes 1.1r, area increases by $1.1 ^ {2}$ = 1.21

11) Answer (B)

Let the total marks be x

and pass marks be z

So, x * 20% = z – 30

x * 32% = z + 42

Implies, x * 12% = 72 or x = 600

z = 150

The required ratio = 150/600 = 25%

12) Answer (A)

Let Ratna’s present age = $5x$ years and Shantanu’s present age = $4x$ years

According to ques, => $\frac{5x + 19}{4x + 19} = \frac{10}{9}$

=> $45x + 171 = 40x + 190$

=> $45x – 40x = 190 – 171$

=> $5x = 19$

=> $x = \frac{19}{5}$

$\therefore$ Ratna’s age = $5 \times \frac{19}{5} = 19$ years

=> Ans – (A)

13) Answer (B)

Using, $\frac{M_1D_1}{W_1}$ $=\frac{M_2D_2}{W_2}$, where $M$ is number of men, $D$ is number of days and $W$ is work done.

Let in $x$ days, 30 persons will consume 48 kg of wheat.

According to ques,

=> $\frac{42\times15}{144}=\frac{30\times x}{48}$

=> $\frac{7\times5}{8}=\frac{5x}{8}$

=> $x=7$

$\therefore$ 30 persons will consume 48 kg of wheat in 7 days.

=> Ans – (B)

14) Answer (C)

Case I = A : B : C = $\frac{1}{5}:\frac{1}{4}:\frac{1}{8}$

L.C.M. (5,4,8) = 40

= $\frac{40}{5}:\frac{40}{4}:\frac{40}{8}=8:10:5$

Sum of terms of ratio = $8+10+5=23$

Case II = A : B : C = $5:4:8$

Sum of term of ratio = $5+4+8=17$

Clearly, only C gains by = $(\frac{8}{17}-\frac{5}{23})\times391$

= $\frac{184-85}{391}\times391=99$

=> Ans – (C)

15) Answer (A)

let the two chords be AB and CD
Midpoints of AB and CD be X and Y
From the figure we have $OC^{2}$=$OY^{2}+YC^{2}$
$30^{2}$=$OY^{2}+24^{2}$
$OY^{2}$=54*6
OY=18 cm
Given XY=42
XO+OY=42
XO=42-18
XO=24
From the figure we have $OA^{2}$=$OX^{2}+XA^{2}$
900=576+$XA^{2}$
$XA^{2}$=324
XA=18 cm
AB=2 AX
AB=36 cm

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