# Mathematics Questions For RRB Group-D PDF

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## Mathematics Questions For RRB Group-D PDF

Download Top 20 RRB Group-D Mathematics Questions and Answers PDF. RRB Group-D Mathematics questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

Question 1: Which of the following is the arithmetic mean of $2X^{2}-10, 30-X^{2} and -X^{2}+6X+10$ ?

a) 3X+15

b) 6X+10

c) 2X+30

d) 2X+10

Question 2: $(1-\frac{1}{2}) (1-\frac{1}{3}) (1-\frac{1}{4}) …..(1-\frac{1}{40}) = ?$

a) $\frac{1}{40}$

b) $\frac{1}{20}$

c) Countless

d) Zero

Question 3: How many prime numbers exist between 60 and 100?

a) 8

b) 9

c) 7

d) 6

Question 4: What is the least number that must be added to 3300 to make it a perfect square?

a) 4

b) 16

c) 64

d) 36

Question 5: Which of the following is divisible by 11?

a) 567899

b) 987657

c) 345656

d) 123456

Question 6: $\frac{13.33*13.33*13.33-6.33*6.33*6.33}{13.33*13.33+13.33*6.33+6.33*6.33}$ = ?

a) 7

b) 14

c) 20

d) 19.66

Question 7: 1000 $\div$ 20 x 10 + 50 x 10 – 100 = ?

a) 700

b) 800

c) 900

d) 1000

Question 8: $100 \times 10 – 100 + 2000 \div 100 = ?$

a) 29

b) 920

c) 980

d) 1000

Question 9: 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 10: What is the average of 56, 45, 47, 61, 49, 54 and 52

a) 52

b) 54

c) 49.12

d) 63

Question 11: What will be the value of –
45% of 550 + 39% of 600

a) 470.5

b) 480.5

c) 490.5

d) 481.5

Question 12: If 6 out of 150 people have lost the match then what percentage of the people have won the match?

a) 80%

b) 90%

c) 96%

d) 86%

Question 13: 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 14: Find the value of “a” if $x^4 – 5x^3 + 4x^2 + 3x – a$ is divisible by x-1?

a) 1

b) 2

c) 3

d) 4

Question 15: What is the value of 21*2.9 + 72?

a) 122.9

b) 132.9

c) 133.8

d) 143.8

Question 16: Find the value of the following expression:$1001\div 91+27-12*4$

a) 106

b) -25

c) -10

d) 104

Question 17: $\sqrt{121}+\sqrt{1.44}+\sqrt{0.1024}+\sqrt{2.25} = ?$

a) 13.53

b) 13.93

c) 14.02

d) 14.22

Question 18: 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 19: $\frac{6}{8}\times \frac{16}{18} \times \frac{3}{4} \times X = \frac{4}{7}$, find $1/X$ ?

a) 4/7

b) 8/7

c) 7/8

d) 7/4

Question 20: Find the sum of the cubes of the first 11 natural numbers ?

a) 4666

b) 4756

c) 4256

d) 4356

Mean = sum/3

= $(2X^{2}-10 + 30-X^{2} -X^{2}+6X+10)$/3

= 2x + 10

1 – 1/2 = 1/2

1 – 1/3 = 2/3

1 – 1/4 = 3/4

The denominator of the first term gets cancelled by the numerator of the second term and so on…

So, the final value = 1/40

61, 67, 71, 73, 79, 81, 87 and 97 are the prime numbers between 61 and 100.
=> There are 8 such numbers.

$60^2$ = 3600
$59^2$ = 3481
$58^2$ = 3364
$57^2$ = 3249
Hence, the least number that must be added to 3300 to make it a perfect square is 64.

Difference of sum of odd placed digits and sum of even placed digits must be divisible by 11.
Option A => (5+7+9) – (6+8+9) = -2
Option B => (9+7+5) – (8+6+7) = 0 => Divisible by 11.

This is in the form of $a^3-b^3$ = $(a-b)(a^2+ab+b^2)$
=> $\frac{13.33*13.33*13.33-6.33*6.33*6.33}{13.33*13.33+13.33*6.33+6.33*6.33}$ = 13.33 – 6.33 = 7

Use BODMAS rule here.
Division must be done first => 50 x 10 + 50 x 10 – 100
Multiplication must be done next => 500 + 500 – 100 = 900

By BODMAS, Division and Multiplication will be done before addition and subtraction.

Hence, 100 x 10 – 100 + 2000 / 100 = 1000 – 100 + 20 = 920

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

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.

45% of 550 can be calculated as (40+5)% of 550 i.e. 220+27.5 = 247.5
And 39% of 600 can be calculated as (40-1)% of 600 i.e. 240-6 = 234
Hence, their summation will be = 481.5

The number of people who lost the match is 6.
So, the number of people who won the match is 150-6 = 144
Hence, the percentage of people who won the match is 144/150 = 24/25 = 96%

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.

f(x) = $x^4 – 5x^3 + 4x^2 + 3x – a$
f(x) is divisible by (x-1) if f(1) = 0.
f(1) = 1-5+4+3-a = 0
3-a = 0
a = 3

21*2.9 = 60.9
The expression boils down to 60.9 + 72 = 132.9

$1001\div 91+27-12*4 = 11+27-48 = -10$

$\sqrt{121}+\sqrt{1.44}+\sqrt{0.1024}+\sqrt{2.25} = 11+1.2+0.32+1.5 = 14.02$

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.

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

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

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

$X = 8/7$

$1/X = 7/8$

So the answer is option B.

Sum of the cubes of 1st n natural numbers = $\frac{n^2(n+1)^2}{4}$
Sum of the cubes of 1st 11 natural numbers = $\frac{11^2(11+1)^2}{4} = \frac{17424}{4} = 4356$