This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Base Conversion”.

1. Which of the number is not allowed in Binary representation of a number:

a) 0

b) 1

c) 2

d) None of the mentioned

View Answer

Explanation: Binary numbers are formed with combination of 0 & 1 only.

2. Which of the number is not allowed in Octal representation of a number:

a) 0

b) 4

c) 8

d) None of the mentioned

View Answer

Explanation: Octal numbers are formed with combination of 0 to 7 only.

3. Hexadecimal number equivalent of decimal 10 is:

a) 10

b) A

c) F

d) None of the mentioned.

View Answer

Explanation: In hexadecimal representation A is represented as decimal 10.

4. Decimal equivalent of binary number 1010 is:

a) 11

b) A

c) 10

d) None of the mentioned.

View Answer

Explanation: 0X1 + 1X2 + 0X4 + 1X8 = 10 in decimal.

5. Decimal 13 in base 8 can be represented as :

a) 15

b) 12

c) 22

d) None of the mentioned

View Answer

Explanation: 1X8 + 5X1 = 12, 15 is the octal representaion of 13.

6. State whether the given statement is true or false

F in hexadecimal representaion is equivalent to 9 in decimal.

a) True

b) False

View Answer

Explanation: F in hexadecimal representaion is equivalent to 15 in decimal.

7. Octal number may contains digits from 1 to 8.

a) True

b) False

View Answer

Explanation: Octal number contain digits from 0 to 7, * is not allowed.

8. For some base r, the digits which are allowed in its representaion are?

a) Digits from 1 to r

b) Digits from 0 to r-1

c) Digits from 1 to r-1

d) None of the mentioned

View Answer

Explanation: A base r number may contain digits from 0 to r-1.

9. The binary number 100110 in octal is reprsented by _______________-

a) 45

b) 10012

c) 46

d) 58

View Answer

Explanation: Pairing 3 numbers from right hand side we get 110 as 6 and 100 as 4 in octal so number is 46.

10. A number greater than 32 would require minimum of how may bits in binary representation:

a) 5

b) 6

c) 4

d) 10

View Answer

Explanation: Since through 5 bits we can only represent numbers till 31 since 2

^{5}= 32 we need greater than 5 bits, so minimum would be 6.

**Sanfoundry Global Education & Learning Series – Discrete Mathematics.**

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