Arithmetic Questions for RRB Group-D PDF

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

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Question 1: What is the value closest to $\sqrt{2028}$ – $\sqrt{1152}$?

a) 8

b) 9

c) 10

d) 11

Question 2: What is the square root of 156.25

a) 13.5

b) 15.25

c) 10.5

d) 12.5

Question 3: If $Y–\sqrt{Y}=30$, what is the value of $Y$?

a) 25

b) 36

c) Any of the above two options

d) Can’t be determined

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Question 4: What is the value of ${21}^3–{19}^3$

a) 2602

b) 2202

c) 2204

d) 2402

Question 5: $\sqrt{0.01+\sqrt{0.0967+0.0258}}$ = ?

a) 0.6

b) 0.5

c) 0.4

d) 0.3

Question 6: When polynomial $x^4-3x^2+2x+5$ is divided by (x-1), the remainder is

a) 2

b) 3

c) 4

d) 5

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Question 7: Evaluate :- $(-216\times 1728{)}^{\frac{1}{3}}$

a) -72

b) 27

c) 72

d) -27

Question 8: What is the average of 56, 45, 47, 61, 49, 54 and 52

a) 52

b) 54

c) 49.12

d) 63

Question 9: 45/13 * (19 + 7) + 5 = ?

a) 95

b) 105

c) 100

d) 87.5

Question 10: If the function $x^3-3x^2+2x-a$ is divisible by (x-2), then find the value of “a”.

a) 0

b) 1

c) 2

d) -1

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Question 11: If the number 2a3 is divisible by 11, then find the value of “a”?

a) 3

b) 4

c) 5

d) 6

Question 12: 4.23232323… = ?

a) 422/99

b) 440/9

c) 419/99

d) 415/99

Question 13: IF 5=10,6=18,7=35,8=56,10=?

a) 100

b) 130

c) 120

d) 110

Question 14: By interchanging which two signs the equation will be correct?
19 + 36 x 12 ÷ 4 – 26 = 5

a) + and –

b) x and ÷

c) ÷ and –

d) + and x

Question 15: The following equation is incorrect. Which two signs should be interchanged to correct the equation?

$15\times 12+40÷40–6=21$

a) + and x

b) + and ÷

c) – and +

d) ÷ and x

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Answers & Solutions:

1) Answer (D)

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

2) Answer (D)

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

3) Answer (B)

Let the value of $\sqrt{Y}=p$. Note that $p$ is positive.
Therefore, $p^2–p=30$
Or, $p^2–p–30=0$
Therefore, $p^2–6p+5p–30=0$
Therfore, $(p-6)(p+5)=0$
Hence, $p=6$ or $p=-5$
But as $p$ is positive, it implies that $p=6$ and $Y=p^2=36$

4) Answer (D)

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 $2\times ({40}^2–399)=2\times (1600-399)=2\times 1201=2402$

5) Answer (A)

$\sqrt{0.01+\sqrt{0.0967+0.0258}}$ = $\sqrt{0.01+\sqrt{0.1225}}$ = $\sqrt{0.01+0.35}$ = $\sqrt{0.36}$ = $0.6$

So the answer is option A.

6) Answer (D)

$f(x)=x^4-3x^3+2x+5$

to find the remainder put x=1 in $f(x)$

$f(1)$=1-3+2+5=5

∴5 is the remainder

7) Answer (A)

$(-216\times 1728{)}^{\frac{1}{3}}$

=$(-6^3\times {12}^3{)}^{\frac{1}{3}}$

=-72

8) 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.

9) Answer (A)

As per BODMAS rule, we would add the contents of the brackets first.
=> 45/13 *26 + 5
26 = 13 * 2
=> 45 * 2 + 5
=> 90 + 5
=> 95

10) Answer (A)

f(x) = $x^3-3x^2+2x-a$. f(2) = 0
$2^3-(3\ast 2\ast 2)+(2\ast 2)-a=0$
$8-12+4-a=0$
a = 0

11) Answer (C)

Since 2a3 is divisible by 11, therefore (2+3-a) should be divisible by 11.
5-a = 0 or divisible by 11
a = 5.

12) Answer (C)

x = 4.232323….
100x = 423.23232323….
100x-x = 423.2323…-4.2323… = 419
x = 419/99

13) Answer (B)

In this sequence each number is multiplied by a prime number 5*2=10,6*3=18,7*5=35,8*7=56,9*11=99,10*13=130.

14) Answer (B)

Expression : 19 + 36 x 12 ÷ 4 – 26 = 5

(A) : + and –

$\equiv 19-36\times 12÷4+26=5$

L.H.S. = $19-(36\times 3)+26=19-108+26=-63\ne$ R.H.S.

(B) : x and ÷

$\equiv 19+36÷12\times 4-26=5$

L.H.S. = $19+(3\times 4)-26=19+12-26=5=$ R.H.S.

=> Ans – (B)

15) Answer (A)

Expression : $15\times 12+40÷40–6=21$

(A) : + and x

$\equiv 15+12\times 40÷40–6=21$

L.H.S. = $15+(12\times \frac{40}{40})-6$

= $15+12-6=21=$ R.H.S.

=> Ans – (A)

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