# Number System Question and Answer Set 1

Hi students, welcome to **Amans Maths Blogs (AMB)**. On this post, you will get the Number System Question and Answer Set 1 is the questions with solution for SSC CGL CHSL CAT and other competative exams like NTSE NSEJS NMTC etc. It will help you to practice the questions on the topics of maths as Number System.

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**Number System Questions and Answer Set 1: Ques No 1**

The difference between two positive numbers is 12. If sum of their squares is 272, then the sum of the numbers is,

**Options:**

A. 20

B. 83

C. 25

D. 32

**Number System Questions and Answer Set 1: Ques No 2**

Find the number of ways in which the number 5105100 can be split into two factors which are relatively prime.

**Options:**

A. 63

B. 64

C. 31

D. 32

**Number System Questions and Answer Set 1: Ques No 3**

What least number must be added to 1056, so that the sum is completely divisible by 23?

**Options:**

A. 5

B. 3

C. 18

D. 21

**Answer: A**

**Number System Questions and Answer Set 1: Ques No 4**

Which of the following statements is INCORRECT?

**Options:**

A. If ‘a’ is a rational number and ‘b’ is irrational, then a + b is irrational.

B. The product of a non-zero rational number with an irrational number is always irrational.

C. Addition of any two irrational numbers can be rational.

D. Division of any two integers is an integer.

**Answer: D**

**Number System Question and Answer Set 1: Ques No 5**

N is a two-digit prime number. When the digits are interchanged, we get another prime number M. If N + M = 176, then find |N – M|

**Options:**

A. 18

B. 36

C. 54

D. 9

**Answer: B**

**Number System Question and Answer Set 1: Ques No 6**

Find the number of factors of 16800.

**Options:**

A. 48

B. 46

C. 96

D. 64

**Answer: A**

**Number System Question and Answer Set 1: Ques No 7**

Find the sum of divisors of 16800 (including 1 and 16800)

**Options:**

A. 56020

B. 56280

C. 62496

D. 70679

**Answer: C**

**Number System Question and Answer Set 1: Ques No 8**

Find the product of all divisors of 84.

**Options:**

A. 36^{9}7^{9}

B. 24^{4}14^{6}

C. 35^{6}28^{5}

D. 84^{4}

**Answer: A**

**Number System Question and Answer Set 1: Ques No 9**

The number of ways in which 7056 can be expressed as the product of two positive integers is.

**Options:**

A. 32

B. 64

C. 36

D. 29

**Answer: B**

**Number System Question and Answer Set 1: Ques No 10**

The number of ways in which 13! can be expressed as the product of two positive integers is

**Options:**

A. 342

B. 792

C. 685

D. 598

**Answer: B**

# Number System Question and Answer Set 1 : Solutions

**Answer 1: C**

Let two positive numbers a and b. Then, a – b = 12 and a^{2} + b^{2} = 272.

Now, (a – b)^{2} = 144 ⇒ a^{2} + b^{2} – 2ab = 144 ⇒ 272 – 2ab = 144 ⇒ 2ab = 128,

So, (a + b)^{2} = a^{2} + b^{2} + 2ab = 272 + 128 = 400 ⇒ (a + b) = 20

**Answer 2: B**

Since the number 5105100 = 2^{2} x 3^{1} x 5^{2} x 7^{1} x 11^{1} x 13^{1} x 17^{1}

This has n = 7 different prime factors (2, 3, 5, 7, 11, 13, 17)

Thus, the number 5105100 can be split into two factors which are relatively prime by number of ways = 2^{7-1} = 64

**Answer 3: A**

Since 1099 = (45 x 24) + 19, thus, to be divisible by 24, we need to add 24 – 19 = 5 to 1099.

**Answer 4: D**

Division of two integers may or may not be integer.

**Answer 5: A**

N = 10x + y, then M = 10y + x.

N + M = 11(x + y) ⇒ 176 = 11(x + y) ⇒ x + y = 16.

Thus, (x, y) = (9, 7) or (7, 9) as N = 10x + y = 97 or 79 both are prime numbers.

Therefore, |N – M| = |97 – 79| = 18

**Answer 6: C**

16800 = 2^{5} x 3^{1} x 5^{2} x 7^{1 }

Thus, Number of factors of 16800 = (5 + 1)(1 + 1)(2 + 2)(1 + 1) = 96

**Answer 7: C**

Since 16800 = 2^{5} x 3^{1} x 5^{2} x 7^{1}, then the sum of divisors is

[(2^{5+1} – 1)/(2-1)][(3^{1+1} – 1)/(3-1)][(5^{2+1} – 1)/(5-1)][(7^{1+1} – 1)/(7-1)]

= 63 x 4 x 31 x 8 = 62496

**Answer 8: A**

Since 252 = 2^{2} x 3^{2} x 7^{1}, then the product of all divisors of 252 is

252^{(2+1)(2+1)(1+1)/2} = 252^{9} = (36 x 7)^{9} = 36^{9} x 7^{9}

**Answer 9: B**

28224 = 2^{6} x 3^{2} x 7^{2} = A perfect square, thus the number of ways in which it can be resolved into two factors is (1/2){(6+1)(2+1)(2+1) + 1} = 64

**Answer 10: B**

Since 13! = 2^{10} x 3^{5} x 5^{2} x 7^{1} x 11^{1} x 13^{1}, then the number of ways 13! can be expressed as the product of two product integers is (1/2)(10+1)(5+1)(2+1)(1+1)(1+1)(1+1) = 792