RRB JE Arithmetic Questions PDF

0
1540
RRB JE Arithmetic Questions PDF
RRB JE Arithmetic Questions PDF

RRB JE Arithmetic Questions PDF

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

Download RRB JE Arithmetic Questions PDF

10 RRB JE Mocks – Just Rs. 117

RRB JE Previous Papers [Download PDF]

Question 1: If ‘+’ means ‘-’, ‘-’ means ‘*’, ‘*’ means ‘/’ and ‘/’ means ‘+’, then what is the value of the following expression? 13 – 13 / 3 – 63 * 9 + 2 -5

a) 170

b) 180

c) 190

d) 200

Question 2: $\frac{3}{4}\times \frac{16}{27} \div \frac{2}{3} = ?$

a) 2/3

b) 3/2

c) 9/4

d) 4/9

FREE RRB JE YOUTUBE VIDEOS

RRB NTPC Free Mock Test

Question 3: $\sqrt{0.6+\sqrt{0.0123-0.0082-0.0025}}$ = ?

a) 0.6

b) 0.7

c) 0.8

d) 0.9

Question 4: $\frac{6}{8}\times \frac{16}{18} \times \frac{3}{4} \times X = \frac{4}{7}$, find $1/X$ ?

a) 4/7

b) 8/7

c) 7/8

d) 7/4

Question 5: Find $((4096)^{1/3})^{1/2}=?$

a) 2

b) 3

c) 4

d) 5

Question 6: Find LCM of 2/3 , 3/6, 4/12 ?

a) 4

b) 2

c) 3

d) 6

RRB NTPC Previous Papers Download PDF

Question 7: Find the LCM of 7/2, 9/4 and 5/3 ?

a) 252

b) 21

c) 126

d) 315

Question 8: Which of the following is divisible by 3, 7, 9 and 11 ?

a) 2645

b) 4158

c) 3791

d) 1188

Question 9: Find the angle between the hours hand and the minutes hand at 3:18 pm ?

a) 13°

b) 12°

c) 9°

d) 10°

Question 10: 4591.15 – 528.116 = x + 456.123. Find x ?

a) 3660.911

b) 3666.911

c) 3006.911

d) 3606.911

18000+ Questions – Free SSC Study Material

Question 11: Find the product of LCM and HCF of 48 and 72 ?

a) 432

b) 864

c) 1728

d) 3456

Question 12: $\frac{2}{3}\times \frac{27}{36}\times \frac{4}{7} (X) = 20$, find $X$ ?

a) 210

b) 140

c) 70

d) 35

Question 13: The difference of two positive numbers is 10. 2 and 48 are the HCF and LCM of those two numbers, then find the smaller number ?

a) 12

b) 16

c) 6

d) 8

Question 14: Sum of two numbers is 99 and difference is 63, then find the smaller number ?

a) 18

b) 36

c) 81

d) 54

RRB NTPC Free Mock Test

RRB JE Free Mock Test

Question 15: What is the value closest to $2.0101^2$?

a) 4.01

b) 4.02

c) 4.03

d) 4.04

Question 16: What is the approximate value of $\frac{(\sqrt{334} + \sqrt{223}) * (\sqrt{334} – \sqrt{223})}{(\sqrt{111} + \sqrt{11}) * (\sqrt{111} – \sqrt{11})}$?

a) 1.01

b) 1.10

c) 1.11

d) 1.21

Question 17: What is the value closest to $\sqrt{2028}$ – $\sqrt{1152}$?

a) 8

b) 9

c) 10

d) 11

Question 18: What is the greatest common divisor of 576 and 960?

a) 96

b) 240

c) 144

d) 192

Question 19: What is the value of $2^5$ + 16 * 4 + 4 /2 – 1?

a) 97

b) 67

c) 47

d) 27

Daily Free Online Tests for RRB Exams

Question 20: Find the total number of factors of $18^{25} \times 24^{32}$

a) 9360

b) 9600

c) 10126

d) 8480

Question 21: The mean of five consecutive odd numbers is 87. What is the sum of the smallest two numbers in the group?

a) 170

b) 172

c) 168

d) 164

Download RRB GK Material PDF

Question 22: Given that the number 1234P is divisible by 11 and P is a single natural number, what is the value of P?

a) 9

b) 5

c) 0

d) 2

Question 23: What is the square root of 156.25

a) 13.5

b) 15.25

c) 10.5

d) 12.5

Question 24: What is the value of $7^3 + 8^3 + 3 \times 7 \times 8 \times 15 – 15^3$?

a) 900

b) 169

c) 225

d) 0

Question 25: If $X$ is a real number, what is the minimum value of $X^2 – 6X + 20$ ?

a) 9

b) 11

c) 20

d) 4

General Science Notes for RRB Exams (PDF)

Answers & Solutions:

1) Answer (B)

Let us replace the symbols with the actual operators.

13 – 13 / 3 – 63 * 9 + 2 -5 = 13 * 13 + 3*63/9 – 2*5

Applying BODMAS rule, we get, 169 + 3*7 – 10 = 180
Therefore, option B is the right answer.

2) Answer (A)

$\frac{3}{4}\times \frac{16}{27} \div \frac{2}{3}$ = $\frac{3}{4}\times \frac{16}{27} \times \frac{3}{2}$ = $\frac{4}{9} \times \frac{3}{2} = \frac{2}{3}$

So the answer is option A.

3) Answer (C)

$\sqrt{0.6+\sqrt{0.0123-0.0082-0.0025}}$ = $\sqrt{0.6+\sqrt{0.0016}}$ = $\sqrt{0.6+0.04}$ = $\sqrt{0.64} = 0.8$

So the answer is option C.

4) Answer (B)

$\frac{6}{8}\times \frac{16}{18} \times \frac{3}{4} \times X = \frac{4}{7}$

$\frac{3}{4}\times \frac{8}{9} \times \frac{3}{4} \times X = \frac{4}{7}$

$\frac{1}{2} \times X = \frac{4}{7}$

$X = 8/7$

$1/X = 7/8$

So the answer is option B.

5) Answer (C)

$((4096)^{1/3})^{1/2} = 16^{1/2} = 4$

So the answer is option C.

6) Answer (B)

2/3 , 3/6, 4/12 = 2/3 , 1/2 , 1/3

LCM of fractions = (LCM of numerators)/(HCF of denominators) = (LCM of 2, 1, 1)/(HCF of 3,2,3) = 2/1 = 2

So the answer is option B.

7) Answer (D)

LCM of fractions = LCM of numerators/HCF of denominators

LCM of 7/2, 9/4 and 5/3 = (LCM of 7,9,5)/(HCF of 2,4,3) = 315/1 = 315

So the answer is option A.

8) Answer (B)

3*7*9*11 = 2079

2079*2 = 4158 is divisible by 3, 7, 9, 11

So the answer is option B.

9) Answer (C)

Hours hand rotates 30° in 1 hour($\because 360/12 = 30$°) and rotates 0.5° in 1 minute($\because 30/60 = 1/2 = 0.5$°)

Minute hand rotates 6° in 1 minute($\because 360/60 = 6$°)

At 3:18,

Angle made by hours hand = 3(30°)+18(0.5°) = 90°+ 9°= 99°

Angle made by minutes hand = 18(6°) = 108°

Hence the angle b/w these two hands = 108°-99° = 9°

So the answer is option C.

10) Answer (D)

4591.15 – 528.116 = x + 456.123

4063.034 = x + 456.123

4063.034 – 456.123 = x

3606.911 = x

So the answer is option D.

11) Answer (D)

Product of LCM & HCF of two numbers = product of those two numbers = 48*72 = 3456

So the answer is option D.

12) Answer (C)

$\frac{2}{3}\times \frac{27}{36}\times \frac{4}{7} (X) = 20$

$\frac{2}{7}(X) = 20$

$X = 10(7) = 70$

So the answer is option C.

13) Answer (C)

Let one number is X, then another number is 10+X

Product of two numbers = (LCM)*(HCF)

(X)(X+10) = (2)(48)

X^2+10X = 96

X^2+10X-96 = 0

(X+16)(X-6) = 0

X = -16 or 6

Since X is positive, X = 6 = smaller number

So the answer is option C.

14) Answer (A)

x+y = 99 —-(1)

x-y = 63 —–(2)

(1)-(2) ==> x+y-x+y = 99-63 ==> 2y = 36 ==> y = 18

So the answer is option A.

15) Answer (D)

Let’s calculate the value of 2.01 * 2.01 = 4.0401
Since 2.0101 is greater than 2.01, the value of its square will be higher as well. So, it will be greater than 4.04.

16) Answer (C)

The formula we use to simplify the expression is (a+b)(a-b) = $a^2- b^2$
So, $\frac{(\sqrt{334} + \sqrt{223}) * (\sqrt{334} – \sqrt{223})}{(\sqrt{111} + \sqrt{11}) * (\sqrt{111} – \sqrt{11})}$ = $\frac{334 – 223}{111-11}$ = 1.11

17) Answer (D)

The value of $\sqrt{2025}$ = 45
The value of $\sqrt{1152}$ = 34
So, the value closest to $\sqrt{2028}$ – $\sqrt{1152}$ = 11

18) Answer (D)

$576 = 24*24 = 3*2^3*3*2^3 = 3^2*2^6$
$960 = 24*40 = 3*2^3*5*2^3 = 3*5*2^6$
Therefore, their GCD equals $3*2^6 = 3*64 =192$

19) Answer (A)

We have to use BEDMAS rule in this situation.
So we resolve Brackets, Exponents, Division, Multiplication, Addition, Subtraction in that order.

=> 32 + 16 * 4 + 4/2 – 1
=> 32 + 16 * 4 + 2 – 1
=> 32 + 64 + 2 – 1 = 98 -1 = 97

20) Answer (C)

$18^{25} \times 24^{32} = 2^{25} \times 3^{50} \times 2^{96} \times 3^{32}$
This equals $2^{121} \times 3^{82}$
Hence, the number of factors equals $122 \times 83 = 10126$

21) Answer (C)

Let the five odd numbers be equal to a, a+2, a+4, a+6 and a+8.
Therefore, their sum equals 5a + 20 and their average equals a+4
Therefore, a+4 = 87 and a = 83.
Hence, the sum of the smallest two numbers in the group equals a + (a+2) = 83 + 85 = 168

22) Answer (D)

For a number to be divisible by 11, the sum of the odd digits should be equal to the sum of the even digits.
Therefore, 1+3+P = 2+4
Or P = 2

23) Answer (D)

$15625 = 5*3125 = 25*625 = 25^3 = 5^6$
Hence, $\sqrt{15625} = 5^3 = 125$
Therefore, $\sqrt{156.25} = 12.5$

24) Answer (D)

We know that $A^3 + B^3 + C^3 – 3ABC = (A+B+C) \times (A^2 + B^2 + C^2 – 3ABC)$
In this case, $A+B+C = 7+8-15 = 0$
Therefore, the value of $7^3 + 8^3 + 3 \times 7 \times 8 \times 15 – 15^3 = 0$

25) Answer (B)

$X^2 – 6X + 20 = X^2 – 6X + 9 + 11$
Therefore, this equals $(X-3)^2 + 11$
Note that $(X-3)^2 \geq 0$
Hence, the minimum value of $X^2 – 6X + 20$ is $11$

Free RRB Online Coaching

DOWNLOAD APP FOR RRB FREE MOCKS

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

LEAVE A REPLY

Please enter your comment!
Please enter your name here