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.
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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
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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
RRB Group D previous year papers
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
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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
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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
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Answers & Solutions:
1) Answer (D)
Mean = sum/3
= $ (2X^{2}-10 + 30-X^{2} -X^{2}+6X+10) $/3
= 2x + 10
2) Answer (A)
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
3) Answer (A)
61, 67, 71, 73, 79, 81, 87 and 97 are the prime numbers between 61 and 100.
=> There are 8 such numbers.
4) Answer (C)
$60^2$ = 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.
5) Answer (B)
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.
6) Answer (A)
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
7) Answer (C)
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
8) Answer (B)
By BODMAS, Division and Multiplication will be done before addition and subtraction.
Hence, 100 x 10 – 100 + 2000 / 100 = 1000 – 100 + 20 = 920
9) Answer (D)
$\frac{391}{667}$ = $\frac{17 \times 23}{29 \times 23}$ = $\frac{17}{29}$
10) Answer (A)
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.
11) Answer (D)
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
12) Answer (C)
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%
13) Answer (A)
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.
14) Answer (C)
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
15) Answer (B)
21*2.9 = 60.9
The expression boils down to 60.9 + 72 = 132.9
16) Answer (C)
$1001\div 91+27-12*4 = 11+27-48 = -10$
17) Answer (C)
$\sqrt{121}+\sqrt{1.44}+\sqrt{0.1024}+\sqrt{2.25} = 11+1.2+0.32+1.5 = 14.02$
18) 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.
19) 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.
20) Answer (D)
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 = $\frac{11^2(11+1)^2}{4} = \frac{17424}{4} = 4356$
So the answer is option D.
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