First Year Business Mathematics Chapter 6 Online MCQ Test for 1st Year Business Mathematics Chapter 6 (Binary Number Systems and its Operation)

This online test contains MCQs about following topics:

Introduction . Binary number system . Conversions of number systems . Arithmetic operations in binary number system

ICOM Part 1 Business Mathematics Ch. 6 Test
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MCQ's Test For Chapter 6 "Business Mathematics Icom Part 1 Chapter 6 Online Test"

Try The MCQ's Test For Chapter 6 "Business Mathematics Icom Part 1 Chapter 6 Online Test"

  • Total Questions15

  • Time Allowed20

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Business Mathematics Icom Part 1 Chapter 6 Online Test

00:00
Question # 1

(1001001)2 in decimal system is -----------------------

Question # 2

The decimal number "2" in binary number system is equivalent to:

Question # 3

(100011)2 x (1101)2 = ---------------------------

Question # 4

5 in binary system is:

Question # 5

In base 2 system digits are used

Question # 6

The decimal number 23 in simplest form is:

Question # 7

2143 in binary system is ------------------

Question # 8

(10110)2 in decimal number is:

Question # 9

(1101)2 + (1001)2 = --------------------------------

Question # 10

(1100000)2 - (111111)2= --------------------------- :

Question # 11

In decimal system (12)5 is equal to

Question # 12

29 in binary number system is

Question # 13

(145)10 = ( )2

Question # 14

Basically proportion is of:

Question # 15

Number of digits in decimal system

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11th Business Mathematics Chapter 6 Test

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ICom Part 1 Business Mathematics Chapter 6 Important MCQ's

Sr.# Question Answer
1 In decimal  (101)2 + (11)2 is equal to:
A. 2
B. 4
C. 8
D. None of these
2 The decimal number 23 in simplest form is:
A. 3(10)o + (10)1
B. 2(10)o + 3 (10)1
C. 2(10)o + 3 (10)2
D. 3(10) + 2 (10)2
3 In decimal system (12)5 is equal to
A. 17
B. 7
C. 15
D. 60
4 Number of digits in decimal system
A. 5
B. 8
C. 10
D. 9
5 (10110)2 in decimal number is:
A. 20
B. 22
C. 24
D. 26
6 In base 2 system digits are used
A. 0, 2
B. 0, 1
C. 2, 3
D. 0, 1, 2
7 (100011)2 x (1101)2 = ---------------------------
A. (111000111)2
B. (100011001)2
C. (100000001)
D. (110011001)2
8 (1100000)2 - (111111)2= --------------------------- :
A. (100001)2
B. (110001)2
C. (1000111)2
D. (111110)2
9 (145)10 = ( )2
A. 10010001
B. 10010111
C. 11100001
D. 10001001
10 29 in binary number system is
A. (110101)2
B. (10101011)2
C. (1011101)
D. (11101)2

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