# Number System Questions for RRB NTPC Set-3 PDF

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## Number System Questions for RRB NTPC Set-3 PDF

Download RRB NTPC Number System Questions and Answers set-3 PDF. Top 15 RRB NTPC Number System questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1:Â Find the greatest number that divides 82, 147 and 199 so as to leave the same remainder in each case.

a)Â 7

b)Â 11

c)Â 13

d)Â 17

Question 2:Â If on subtracting 32 from a number, the remainder will be 2/3rd of the number, what will be 25% of the number?

a)Â 24

b)Â 48

c)Â 72

d)Â 96

Question 3:Â If one-sixth of $\frac{3}{8}$ of $\frac{2}{5}$ of a number is equal to 96 then what will be 10% of the number?

a)Â 284

b)Â 384

c)Â 484

d)Â 684

Question 4:Â Sum of three numbers is 120 and the ratio of first and second number is equal to 4:5. Ratio of second number to third number is 15:13. What is the third number?

a)Â 36

b)Â 45

c)Â 39

d)Â 38

Question 5:Â If the sum of the digits of a number is 8 and difference of the digits is 2 then what will be the product of the digits of the number?

a)Â 12

b)Â 15

c)Â 20

d)Â 24

Question 6:Â The difference between a two digit number and the number obtained by interchanging its digit is 54. What is the difference between the digits of the number?

a)Â 5

b)Â 4

c)Â 7

d)Â 6

Question 7:Â Find the smallest positive number which on division by 2,7,3 leaves remainder 1,6,2 respectively in each case.

a)Â 42

b)Â 41

c)Â 83

d)Â 82

Question 8:Â Find the 3rd smallest positive number which on division by 2,7,5 leaves remainder 1 in each case?

a)Â 71

b)Â 1

c)Â 211

d)Â 141

Question 9:Â If the number on division by 3, 5, 6 and 8 leaves the remainder 1, 3, 4, 6 respectively, then find the 2nd smallest positive number of such a series?

a)Â 358

b)Â 235

c)Â 238

d)Â 240

Question 10:Â A number on division by 5,7,10,12 gives remainder 3 in each case. Find the 5th smallest positive number of such a series.

a)Â 1683

b)Â 2103

c)Â 1263

d)Â 2523

Question 11:Â If the number 9a53b25 is divisible by 99, then find the value of a and b respectively.

a)Â 2,1

b)Â 3,0

c)Â 1,2

d)Â 0,1

Question 12:Â Find the second smallest positive number which when divided by 5,7,9 leaves remainder 3 in each case?

a)Â 3

b)Â 315

c)Â 317

d)Â 318

Question 13:Â What is the HCF of the numbers 8, 20, 24 and 36?

a)Â 4

b)Â 8

c)Â 12

d)Â 16

Question 14:Â Which of the following is not a prime number?

a)Â 167

b)Â 313

c)Â 509

d)Â 1001

Question 15:Â Find the digit in the ten’s place of $19^2+21^2$.

a)Â 0

b)Â 2

c)Â 4

d)Â 6

147 – 82 = 65
199 – 147 = 52
The HCF of these differences is the number that divides all the three numbers leaving the same remainder.
65 = 13 * 5
52 = 13 * 4
=> HCF = 13
Hence, 13 is the answer.

Letâ€™s say number is x
Hence, x-32 = 2x/3
Or x = 96
So 25% of the number will be 24

Letâ€™s say number is x then $x \times\frac{1}{6} \times \frac{3}{8} \times \frac{2}{5} = 96$
Hence, after solving the above equation we will get x = 3840
So 10% of the number will be 384

The ratio of the first two numbers is 4:5 which equals 12:15

Hence, the ratio of all three numbers will be = 12:15:13
And letâ€™s say numbers are 12x, 15x and 13x
So the summation of the numbers = 40x = 120
Or x= 3
Hence, numbers are 36, 45 and 39

Letâ€™s say number is 10x+y (Where digits of the number will be x and y)
x-y = 2
x+y = 8
Hence, x = 5 and y = 3
Product of the digits will be = 15

Letâ€™s say number is 10x+y. Hence, after interchanging their digits, new number will be 10y+x. So difference will be 9(x-y) = 54
Difference between the digits will be x-y = 6

This number can be expresssed in the form = LCM(2,7,3)k-1 = 42k-1. The smallest number is 42*1-1 = 41

The number is of the form LCM(2,7,5)k+1 = 70k+1. The numbers are 1,71,141 and so on. The 3rd smallest number isÂ 141.

In this case the difference between the divisor and the remainder is constant in each case = 3-1 = 2, 5-3 = 2, 6-4 = 2 and 8-6 = 2.
The number would be of the form LCM (3, 5, 6, 8) k – 2 = 120 k – 2
The second smallest positive number would be 120*2 – 2 = 238.

As mentioned in the concept, the number would be of the form LCM (5, 7, 10, 12) k + 3 = 420 k + 3.
The smallest term is 3. The second smallest term is (420*1)+3 = 423.
The 5th smallest term is (420*4) + 3 = 1680+3 = 1683.

Since the number is divisible by 99, the number is divisible by 9 and 11 as 99 = 9*11
9+a+5+3+b+2+5 should be divisible by 9
24+a+b should be divisible by 9.
24+a+b = 27 or 36
a + b = 3 or 12
Similarly (9+5+b+5)-(a+3+2) = 14+b-a should be divisible by 11
14+b-a = 11 or 22
a – b = 3 or -8
When a + b = 12 and a – b = 3, then a = 7.5 which is not possible
When a + b = 12 and a – b = -8, then a = 10 which is not possible
When a + b = 3Â and a – b = -8, then a = -2.5 which is not possible
When a + b = 3 and a – b = 3, then a = 3 and b = 0 which is the solution.

The number is of the form LCM(5,7,9)k+3 = 315k+3 = 318.

$8 = 2^3$
$20 = 2^2 * 5$
$24 = 2^3 * 3$
$36 = 2^2 * 3^2$

So, the HCF of the four numbers is $2^2$ = 4

$19^{2}+21^{2}$