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RRB Group-D Maths Questions PDF

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

Question 1: Find the LCM of 7/2, 9/4 and 5/3 ?

a) 252

b) 21

c) 126

d) 315

Question 2: 4591.15 – 528.116 = x + 456.123. Find x ?

a) 3660.911

b) 3666.911

c) 3006.911

d) 3606.911

Question 3: Sum of two numbers is 99 and difference is 63, then find the smaller number ?

a) 18

b) 36

c) 81

d) 54

Question 4: What is the value closest to $\sqrt{2028}$ – $\sqrt{1152}$?

a) 8

b) 9

c) 10

d) 11

Question 5: What is the value of $21^3 – 19^3$

a) 2602

b) 2202

c) 2204

d) 2402

Question 6: The simplest form of $\frac{391}{667}$ is

a) $\frac{23}{31}$

b) $\frac{19}{23}$

c) $\frac{15}{19}$

d) $\frac{17}{29}$

Question 7: What is the average of 56, 45, 47, 61, 49, 54 and 52

a) 52

b) 54

c) 49.12

d) 63

Question 8: If the number 254a8b is divisible by 10 as well as 9, find the value of a.

a) 8

b) 9

c) 6

d) 7

Question 9: Find the smallest positive integer that should be multiplied to 2352 to make it a perfect square.

a) 2

b) 3

c) 7

d) 11

Question 10: Find the greatest common divisor of the numbers 75, 155 and 195.

a) 5

b) 15

c) 25

d) None of the above

Question 11: The least number n, where n>4, which when divided by 9, 11 and 12 leaves a remainder 4 is

a) 396

b) 400

c) 796

d) None of the above

Question 12: 1818.12 + 1818.24 + 1919.19 = ?

a) 5545.45

b) 5333.44

c) 5555.55

d) 5444.44

Question 13: 75% of a number from which 10 has been deducted is 180. The number is

a) 200

b) 240

c) 330

d) 250

Question 14: If a speed of 24 kmph is converted into m/s, what would be the value obtained?

a) 6.67

b) 12.33

c) 13.33

d) 9.67

Question 15: 29% of 9000 + 35% of 5000 = ?

a) 4400

b) 4260

c) 4210

d) 4360

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.

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.

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.

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

We know that $a^3 – b^3 = (a-b) \times (a^2 + ab + b^2) = (a-b) \times ((a+b)^2 – ab)$
In this case, $21^3 – 19^3 = (21 – 19) \times (21^2 + 21 \times 19 + 19^2)$
Which equals $2 \times ((21+19)^2 – 21\times 19)$
This equals $2 \times (40^2 – 399) = 2 \times (1600-399) = 2 \times 1201 = 2402$

$\frac{391}{667}$ = $\frac{17 \times 23}{29 \times 23}$ = $\frac{17}{29}$

We see that the numbers are distributed around 50.
So we find the differences around 50
Estimated mean= 50
Sum = +6-5-3+11-1+4+2 =14
n=7

Actual mean = 50 + 14/7 = 52.

Since the number is divisible by 10, b = 0. Since the number is divisible by 9, the sum of the digits is also divisible by 9.
19+a+b should be divisible by 9. Since b = 0, only a = 8, satisfy this criteria.

Let’s factorize 2352 into its prime factors.
2352 = 16 * 3 * 49 = $2^4 * 3 * 7^2$.
To be a perfect square, the number should have prime factors with even indices. Hence, for 3 to have an even index, the number should be multiplied with 3.

The numbers are $3*5^2$, $5*31$ and $5*39$.
So, the HCF of the three numbers is 5.

$9 = 3^2$
$12 = 3 * 2^2$
So, the LCM of 9, 11 and 12 is $3^2 * 2^2 * 11$ = 9 * 4 * 11 = 396
So, the required number should be of the form 396k + 4
For k = 1, the number becomes 396 + 4 = 400

The summation of first two terms will be = 3636.36 and the whole summation will be 5555.55

Let the number be x. Hence, 75%*(x-10)=180. Hence, x=250.