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# RRB NTPC Arithmetic Questions PDF

Download RRB NTPC Arithmetic Questions and Answers PDF. Top 30 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: In an exam, Amit is supposed to calculate the product of 34.1 * 4.67. He instead calculates the product of 0.0341 * 46.7 as 1.59. What is the actual answer in the exam?

a) 1.59

b) 15.90

c) 0.159

d) 159.0

Question 2: If $x^{2}-5x-24=0$, then find the value of x

a) +8,-3

b) +8,+3

c) -8,-3

d) -8,+3

Question 3: If A = 13, B = 24 and C = 19, which of the following is true? I. A+B-C = C
II. B+C-A = 30
III. A+B+C = 2B
IV. A+C-B = 8

a) II and IV only

b) I and III only

c) I, II and IV only

d) II, III and IV only

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

a) 8

b) 9

c) 10

d) 11 Question 5: The difference of two positive numbers is 10. 2 and 48 are the HCF and LCM of those two numbers, then find the smaller number ?

a) 12

b) 16

c) 6

d) 8

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

a) 2

b) 3

c) 4

d) 5

Question 8: 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 9: 29% of 6500 is what percentage of 2500?

a) 75.4%

b) 76.5%

c) 80.4%

d) 72.5%

Question 10: If the number 24a53b1 is divisible by 9, then what can be the value of a+b?

a) 3

b) 12

c) 21

d) Cannot be determined Question 11: 4.23232323… = ?

a) 422/99

b) 440/9

c) 419/99

d) 415/99

Question 12: $\frac{32}{3}+\frac{46}{9}-\frac{14}{3}$ = ?

a) 100/9

b) 10

c) 11

d) 103/9

Question 13: 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 14: What is the value of 21*2.9 + 72?

a) 122.9

b) 132.9

c) 133.8

d) 143.8

Instructions

What approximate value should come in place of question-mark (?) in the following questions ? (You are expected to calculate the exact value)

Question 15: $561204\times58 = ? \times 55555$

a) 606

b) 646

c) 586

d) 716

Question 16: $(9321+5406+1001)\div(498+929+660)$ = ?

a) 13.5

b) 4.5

c) 16.5

d) 7.5

Question 17: What is the HCF of the fractions: ½, 5/7, 8/11, ¾ ?

a) 2/121

b) 1/154

c) 1/121

d) 1/308

Question 18: Find the value of the following expression:126\div 9 + 4*3-17

a) 5

b) 8

c) 9

d) 10

Question 19: What will be the value of –
59% of 6190 + 37% of 5000 =

a) 5500

b) 5515

c) 5510

d) 5502 Question 20: Given that log2 = 0.3 approx, one billion would be approximately

a) $2^{9}$

b) $2^{10}$

c) $2^{20}$

d) $2^{30}$

Question 21: 75% of a number from which 10 has been deducted is 180. The number is

a) 200

b) 240

c) 330

d) 250

Question 22: 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 23: Ram was asked to calculate the sum of the first 40 numbers. He reported the sum as 799. He then realized that he missed adding a number during the summation. The number is

a) 19

b) 20

c) 21

d) 22

Question 24: $100 \times 10 – 100 + 2000 \div 100 = ?$

a) 29

b) 920

c) 980

d) 1000

Question 25: If ‘P’ means ‘+’, ‘Q’ means ‘-’, ‘R’ means ‘÷’ and ‘S’ means ‘x’, then which of the following equation is correct?

a) 14 R 7 S 6 P 4 Q 3 = 11

b) 3 S 6 P 2 Q 3 R 6 = 35/2

c) 11 R 12 S 48 P 10 Q 6 = 48

d) 9 S 8 P 6 R 4 S 8 = 80

Question 26: Find the least number among $\frac{5}{9},\sqrt\frac{9}{49},0.43$ and $(0.7)^{2}.$

a) $\frac{9}{5}$

b) $(0.7)^{2}$

c) $\sqrt\frac{9}{49}$

d) 0.43

Question 27: What is the value of $(\frac{\sqrt{2}}{3}-Cosec 60^\circ)?$

a) $\frac{(\sqrt{6}-6)}{3\sqrt{3}}$

b) $\frac{(2-2\sqrt{3})}{\sqrt{3}}$

c) $\frac{(1-\sqrt{6})}{\sqrt{2}}$

d) $\frac{(4-\sqrt{3})}{2\sqrt{3}}$

Question 28: Find the value of $\sqrt{(2x-5)^{2}}+2\sqrt{(x-1)^{2}}$, if 1<x<2.

a) 1

b) 2

c) 3

d) 4

Question 29: What is the value of $(cosec 60^\circ – \frac{1}{2})$

a) $\frac{(4-\sqrt{3})}{2\sqrt{3}}$

b) $\frac{(2\sqrt{3}-1)}{\sqrt{3}}$

c) $\frac{(3\sqrt{3}-1)}{3}$

d) $\frac{1}{\sqrt{3}}$

Question 30: Which of the following relation is CORRECT?
I. $(\sqrt{15}+\sqrt{7})<(2\sqrt{22})$
II. $(\sqrt{17}+\sqrt{5})<(\sqrt{20}+\sqrt{2})$

a) Only I

b) Only II

c) Neither I nor II

d) Both I and II 0.0341 * 46.7 = 1.59
So, 34.1 * 46.7 = 1.59 * 1000
34.1 * 4.67 = 1.59 * 100

$x^{2}-5x-24=0$

So, $x^{2}-8x + 3x -24=0$

ie x(x- 8) + 3(x-8) = 0

=> (x+3)(x-8)=0

or x = 8 or -3

A+B-C = 13+24-19 = 18 which is not equal to 19. Hence, statement I. is wrong.
B+C-A = 24+19-13 = 30. Hence, statement II. is true.
A+B+C = 13+24+19 = 56 which is not equal to 2*24=48. Hence, statement III. is wrong.
A+C-B = 13+19-24 = 8. Hence, statement IV. is true.

Therefore, statement II and IV are true and the correct 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

Let one number is X, then another number is 10+X

Product of two numbers = (LCM)*(HCF)

(X)(X+10) = (2)(48)

X^2+10X = 96

X^2+10X-96 = 0

(X+16)(X-6) = 0

X = -16 or 6

Since X is positive, X = 6 = smaller number

So the answer is option C.

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

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

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

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

∴5 is the remainder

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

29% of 6500 can be calculated as (30-1) % of 6500 i.e. 1950-65 = 1885
Hence, 1885 will be $\frac{1885}{2500} \times 100$ % of 2500
i.e. 75.4%

24a53b1 is divisible by 9 if (2+4+a+5+3+b+1) is divisible by 9.
(15+a+b) should be divisible by 9.
The maximum value of each of “a” and “b” should be 9.
a+b should not be more than 18.
15+a+b should not be more than 18+15 = 33.
15+a+b = 18 or 27
a+b = 18-15 or 27-15 = 3 or 12
a+b should be either 3 or 12.

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

$\frac{32}{3}+\frac{46}{9}-\frac{14}{3} = 6+ \frac{46}{9}$ = $\frac{100}{9}$

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.

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

$561204\times58 = ? \times 55555$

implies ? = $561204\times58 \div 55555$ = 586 9321+5406+1001= 15728
498+929+660= 2087
15728$\div$2087= 7.536
So option D is the right answer.

The HCF of fractions is equal to HCF of numerators/LCM of denominators.
HCF of numerators = HCF of 1, 5, 8 and 3 = 1
LCM of numerators = LCM of 2, 7, 11 and 4 = 7*11*4 = 308
So, the HCF of the fractions is 1/308

126\div 9 + 4*3-17 = 14+4*3-17 = 14+12-17 = 26-17 = 9

59% of 6190 can be calculated as (50+9)% of 6190 i.e. equal to 3652.1
And 37% of 5000 can be calculated as (30+7)% of 5000 i.e. equal to 1850
Hence, their summation will be = 1850+3652.1 = 5502.1

1 billion = $10^9$

log (1 billion) = 9

log 2 = 0.3

$log_2$1billion = 30

=> 1 billion = $2^{30}$

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

Mean = sum/3

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

= 2x + 10

Sum of the numbers from 1 to 40 = 40*41/2 = 20*41 = 820
So, difference = 820 – 799 = 21
The number that Ram forgot to add was 21.

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

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

Given : ‘P’ means ‘+’, ‘Q’ means ‘-’, ‘R’ means ‘÷’ and ‘S’ means ‘x’

(A) : 14 R 7 S 6 P 4 Q 3 = 11

L.H.S. = $14\div7\times6+4-3$

= $12+1=13\neq$ R.H.S.

(B) : 3 S 6 P 2 Q 3 R 6 = 35/2

L.H.S. = $3\times6+2-3\div6$

= $18+2-0.5=19.5=\frac{39}{2}\neq$ R.H.S.

(C) : 11 R 12 S 48 P 10 Q 6 = 48

L.H.S. = $11\div12\times48+10-6$

= $44+4=48=$ R.H.S.

=> Ans – (C)

Terms : $\frac{5}{9}\approx0.56$

$\sqrt{\frac{9}{49}}$ $=\frac{3}{7}\approx0.42$

$0.43$

$(0.7)^2=0.49$

Thus, the least number is = $\sqrt\frac{9}{49}$

=> Ans – (C)

Expression : $(\frac{\sqrt{2}}{3}-Cosec 60^\circ)$

= $\frac{\sqrt2}{3}-\frac{2}{\sqrt3}$

= $\frac{(\sqrt{6}-6)}{3\sqrt{3}}$

=> Ans – (A)

Expression : $(cosec 60^\circ – \frac{1}{2})$

= $\frac{2}{\sqrt3}-\frac{1}{2}$

= $\frac{(4-\sqrt{3})}{2\sqrt{3}}$

=> Ans – (A)

I : $(\sqrt{15}+\sqrt{7})<(2\sqrt{22})$

Squaring both sides, we get :

L.H.S. = $(\sqrt{15}+\sqrt{7})^2=15+7+2\sqrt{105}=(22+2\sqrt{105})\approx(22+2\times10)=42$

R.H.S. = $(2\sqrt{22})^2=88$

Thus, L.H.S. < R.H.S., which is correct.

II : $(\sqrt{17}+\sqrt{5})<(\sqrt{20}+\sqrt{2})$

Squaring both sides, we get :

L.H.S. = $(\sqrt{17}+\sqrt{5})^2=17+5+2\sqrt{85}=(22+2\sqrt{85})$

R.H.S. = $(\sqrt{20}+\sqrt{2})^2=20+2+2\sqrt{40}=(22+2\sqrt{40})$

$\because$ $\sqrt{85}>\sqrt{40}$, then L.H.S. > R.H.S.

Thus, only I is correct.

=> Ans – (A) We hope this Arithmetic questions for RRB NTPC Exam will be highly useful for your preparation.