# Number System Questions for RRB NTPC Set-4 PDF

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## Number System Questions for RRB NTPC Set-4 PDF

Download RRB NTPC Top-10 Number System Questions Set-4 PDF. Questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: Find the least number which should be subtracted from 9999 so that the difference is exactly divisible by 16?

a) 15

b) 12

c) 14

d) 13

Question 2: The LCM and HCF of two numbers are 4284 and 32, respectively. If one of the numbers is 672, then the second number is;

a) 102

b) 64

c) 204

d) 92

Question 3: Let n be the number of different 5-digit numbers divisible by 4, with the digits 4, 5, 6, 7, 8 and 9, no digit being repeated in the numbers. What is the value of n?

a) 8

b) 24

c) 168

d) 192

Question 4: The LCM of 56, 84 and 112 is:

a) 210

b) 336

c) 420

d) 168

Question 5: The HCF of 42, 63 and 105 is:

a) 7

b) 630

c) 63

d) 21

Question 6: Three bells ring at intervals of 15, 30 and 45 minutes respectively. At what time will they
ring together again, if they rang simultaneously at 8.00 AM ?

a) 8.30 AM

b) 9.30 AM

c) 9.00 AM

d) 8.45 AM

Question 7: Find the least number that is divisible by 12, 18, 21, and 30

a) 1020

b) 1260

c) 1620

d) 1060

Question 8: Find the LCM of $ab^{2}c^{2}$, $a^{2}bc$ and $a^{3}b^{3}c^{2}$.

a) $abc$

b) $a^{2}b^{2}c^{2}$

c) $a^{3}b^{3}c^{2}$

d) $a^{3}b^{3}c^{3}$

Question 9: Find the largest two-digit prime number.

a) 91

b) 97

c) 93

d) 89

Question 10: What is the difference between the place and face values of ‘5’ in the number 3675149?

a) 4995

b) 5000

c) 495

d) 4990

$\frac{9999}{16} = 624\frac{15}{16}$

$\frac{15}{16}\times 16 = 15$

This 15 is needed to be subtracted from 9999 to get a number which is exactly divisible by 16

Option A is correct.

HCF is 32

LCM is 4284

Let the other number be x

We know that HCF * LCM = produce of 2 numbers.

$32\times 4284 = 672 \times x$

$x =\frac{32\times4284}{672}$

$x = 204$

Option C is correct.

Find the prime factorization of 56,84,112
56 = 2 × 2 × 2 × 7 ………i)

84 = 2 × 2 × 3 × 7……….ii)

112= 2 × 2 × 2 ×2 ×7………iii)

Multiply each factor the greater number of times it occurs in steps i) , ii) or iii) above to find the lcm:

LCM = 2 × 2 × 2 × 2×3 × 7

=336

42 = $2 \times 3 \times 7$

63 = $3 \times 3 \times 7$

105 =  $3 \times 5 \times 7$

Taking the common factors, the HCF would be 21.

So , the answer would be option d)21.

Given that three bells rang at intervals of 15, 30 and 45 minutes respectively.
They will together ring again in 90 minutes (LCM of 15, 30, 45).
90 minutes = 01 : 30 hours
Then, After 8 : 00 AM, they will again ring together in 08 : 00 + 01 : 30 = 09 : 30 AM

$12=2\times 2\times 3$

$18=2\times 3\times 3$

$21 =3\times 7$

$30 =2\times 3\times 5$

Select the common terms

LCM = $2\times 2\times 3\times 3\times 5\times 7$

= 1260

$ab^{2}c^{2}$, $a^{2}bc$ and $a^{3}b^{3}c^{2}$.

LCM

$=(abc)\times(bc)\times(a)\times(ab)$

$=a^{3}b^{3}c^{2}$