Arithmetic Questions for Railway Exams PDF

0
356

Arithmetic Questions for Railway Exams PDF

Download Top-15 Railway Exams Arithmetic Questions PDF. Railway Exams Arithmetic questions based on asked questions in previous exam papers very important for the Railway  exams.

Question 1: 36% of 4800 x 0.2% of 1320 =?

a) 4535.52

b) 4551.36

c) 4561.92

d) 4572.48

Question 2: Evaluate :- $\frac{16 \times2^{n+1 }-4\times2^{n}}{16\times 2^{n+2}-2\times2^{n+2}}$

a) $\frac{1}{2}$

b) $0$

c) $4$

d) $\frac{1}{3}$

Question 3: $\sqrt{6400}$ = ?

a) 40

b) 80

c) 60

d) None of these

Question 4: 3251 + 587 + 369 – ? = 3007

a) 1250

b) 1300

c) 1375

d) 1200

Question 5: The square root of 19881 is:

a) 149

b) 129

c) 141

d) 131

Question 6: Find the square root of 15625 ?

a) 125

b) 145

c) 150

d) 135

Question 7: Which of the following is NOT a perfect square?

a) 1250

b) 16641

c) 2025

d) 9801

Question 8: Select the option that can replace the question mark (?) in the following equation.
$2 + 5\div [5+8\div( 1+\frac{1}{3}) – 1]$ = ?

a) $\frac{1}{2}$

b) $\frac{5}{2}$

c) $\frac{3}{2}$

d) 2

Question 9: $4 + \frac{1}{6} \times$ [ { $– 12 \times ( 24 – 13 – 3)$ } $\div ( 20 – 4)$ ] = ?

a) 4

b) 6

c) 5

d) 3

Question 10: If $\sqrt{(x-1) +(y+2)} = 7$ , then find the values of x and y.

a) 12, 16

b) 8, 5

c) 15, 12

d) 50, 47

Question 11: If $x = \left(\sqrt{2} + 1\right)^{\frac{-1}{3}}$, then find the value of $x^3 – \frac{1}{x^3}$.

a) $\sqrt3$

b) -2

c) 0

d) 2

Question 12: If the midvalue of a class is 20 and the width of the class is 5, find the lower and upper limits of the class.

a) 17 – 23

b) 17 – 22

c) 17.5 – 22.5

d) 18 – 22

Question 13: Find the value of $5y \sqrt{y^3 – y^2}$ when $y = 5$

a) 500

b) 250

c) $50\sqrt{2}$

d) 50

Question 14: simplify $\sqrt{25+10\sqrt{6}}+ \sqrt{25-10\sqrt{6}}$?

a) $2\sqrt{15}$

b) $2\sqrt{5}$

c) $\sqrt{50}$

d) $\sqrt{55}$

Question 15: $45-[38-(80\div4-(8-12\div3)\div4)]=?$

a) 25

b) 27

c) 26

d) 28

36% of 4800 = 1728

.2% of 1320 = 2.64

So, 36% of 4800 * .2% of 1320 = 1728 *2.64 = 4561.92

$\frac{16 \times2^{n+1 }-4\times2^{n}}{16\times 2^{n+2}-2\times2^{n+2}}$

=$\frac{2^{n}(32-4)}{2^{n+2}(16-2)}$

=$\frac{28}{4(14)}$=$\frac{1}{2}$

$\sqrt{6400} = \sqrt {80 \times 80}$ = 80

Let’s assume ? = Y.

3251 + 587 + 369 – Y = 3007

4207 – Y = 3007

4207 – 3007 = Y

Y = 1200

$\sqrt{\ 19881}=141$

Hence option ‘C’ is correct.

For finding square root, first obtain the multiple of a number.

15625 = $5\times5\times5\times5\times5\times5$

Now pairing the number and got $5\times5\times5$. Which is equal to 125.

1250 (It is not a perfect square of any number.)

16641 (It is a perfect square of 129.)

2025 (It is a perfect square of 45.)

9801 (It is a perfect square of 99.)

= $2 + 5\div [5+8\div( 1+\frac{1}{3}) – 1]$

= $2 + 5\div [5+8\div( \frac{4}{3}) – 1]$

= $2 + 5\div [5+8 \times \frac{3}{4} – 1]$

= $2 + 5\div 10$

= $\frac{5}{2}$

Hence, option B is the correct answer.

$4 + \frac{1}{6} \times$ [ { $– 12 \times ( 24 – 13 – 3)$ } $\div ( 20 – 4)$ ]
=$4 + \frac{1}{6} \times$ [ { $– 12 \times 8$ } $\div 16$ ]
=$4 + \frac{1}{6} \times -6$
=4-1
=3

Substituting y=5

Equation becomes

$5 \times 5 \sqrt{5^3-5^2}$

$= 25 \sqrt{125-25}$

$= 25 \sqrt{100}$

$= 25 \times 10$

=250

$45-[38-(80\div4-(8-12\div3)\div4)] = 45 – (38 – (20 – \dfrac{8-4}{4}))$
$= 45 – (38 -(20-\dfrac{4}{4})) = 45-(38-(20-1))$
$= 45-(38-19) = 45-19 = 26$