Numeral System

1) The symbol 25_b represents a two-digit number in base b. If the number 52_b is double the number 25_b, then b is: [Asked in “FMS”.]

A) 7 B) 8 C) 9 D) 11

 

2) Convert the decimal number 22 to binary number.

 

3) Convert the decimal number 110 to octal number.

 

4) Convert the decimal number 733 to hexadecimal number.

 

5) Convert the decimal number 0.153 to binary number.

 

6) Convert the decimal number \frac{5}{11} to binary number.

 

7) Convert the decimal number 0.153 to octal number.

 

8) Convert the decimal number 0.153 to hexadecimal number.

 

9) Convert the binary number 1111101000.10101 to octal number.

 

10) Convert the binary number 1111101000.10101 to hexadecimal number.

 

11) Convert the hexadecimal number 3E8.A8 to binary number.

 

12) Convert the octal number 1750.52 to binary number.

 

13) Convert the octal number 1750.52 to hexadecimal number.

 

14) Convert the hexadecimal number 3E8.A8 to octal number.

 

15) 111010_2+1110111_2= ?

 

16) 1010110001_2-11110111_2= ?

 

17) 110.11_2\times 11.01_2= ?

 

18) 156_8+34_8= ?

 

19) 106_8-37_8= ?

 

20) 1011011_2\div 111_2= ?

__________⊗__________

Detailed solutions and some general tricks:

 

1) 25 and 52 are two-digit numbers in base b. Therefore 2 and 5 are digits in the base b numeral system.

Also given that 25_b\times 2_b=52_b.

We have 2_b\times 2_b=4_n.

therefore 5_b\times 2_b=12_b

\Rightarrow 5_b\times 2_b=10_b+2_b.

Now 10_b=b_{10}, 5_b=5_{10} and 2_b=2_{10}.

therefore 5_{10}\times 2_{10}=b_{10}+2_{10}

\Rightarrow 10_{10}=(b+2)_{10}

\Rightarrow b=8_{10}.

 

2)

Convert the decimal number 22 to binary number

Hence 22_{10}=1\times 2^4+0\times 2^3+1\times 2^2+1\times 2^1+0\times 2^0=10110_2.

 

3)

Convert the decimal number 110 to octal number

Hence 110_{10}=1\times 8^2+5\times 8^1+6\times 8^0=156_8.

 

4)

Convert the decimal number 733 to hexadecimal number

Hence 733_{10}=2\times 16^2+D\times 16^1+D\times 16^0=2DD_{16}.

 

5)

Convert the decimal number 0.153 to binary number

Hence 0.153_{10}=0\times 2^{-1}+0\times 2^{-2}+1\times 2^{-3}+0\times 2^{-4}+0\times 2^{-5}+1\times 2^{-6}+1\times 2^{-7}+1\times 2^{-8}+\dots =0.00100111_2. (After we round and cut the number.)

 

6)

Convert the decimal number frac{5}{11} to binary number

Hence \frac{5}{11}=0.0111\dots _2.

 

7)

Convert the decimal number 0.153 to octal number

Hence 0.153_{10}=1\times 8^{-1}+1\times 8^{-2}+6\times 8^{-3}+2\times 8^{-4}+5\times 8^{-5}+4\times 8^{-6}+\dots =0.116254_8. (After we round and cut the number.)

 

8)

Convert the decimal number 0.153 to hexadecimal number

Hence 0.153_{10}=2\times 16^{-1}+7\times 16^{-2}+2\times 16^{-3}+11\times 16^{-4}+0\times 16^{-5}+2\times 16^{-6}+\dots =0.272B02_{16}. (After we round and cut the number.)

 

Image Source : Shutterstock

Image Source : Shutterstock

9)

1111101000.10101

1 111 101 000 . 101 010

1 7 5 0 . 5 2

Hence 1111101000.10101_2=1750.52_8.

 

10)

1111101000.10101

11 1110 1000 . 1010 1000

3 14 8 . 10 8

3 E 8 . A 8

Hence 1111101000.10101_2=3E8.A8_{16}.

 

11)

3E8.A8

3 E 8 . A 8

11 1110 1000 . 1010 1000

Hence 3E8.A8_{16}=1111101000.10101_2.

 

12)

1750.52

1 7 5 0 . 5 2

1 111 101 000 . 101 010

Hence 1750.52_8=1111101000.10101_2.

 

13)

1750.52

1 7 5 0 . 5 2

1 111 101 000 . 101 010

1111101000.10101

11 1110 1000 . 1010 1000

READ:   A closer look into a well-known quotient ring

3 14 8 . 10 8

3 E 8 . A 8

Hence 1750.52_8=3E8.A8_{16}.

 

14)

3E8.A8

3 E 8 . A 8

11 1110 1000 . 1010 1000

1111101000.10101

1 111 101 000 . 101 010

1 7 5 0 . 5 2

Hence 3E8.A8_{16}=1750.52_8.

 

15)

111010_2+1110111_2

 

16)

1010110001_2-11110111_2

 

17)

110.11_2 x 11.01_2

Hence required value = 10101.1111

 

18)

156_8+34_8

 

19)

106_8-37_8

 

20)

1011011_2 div 111_2

 

 

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Pankaj Jyoti Mahanta

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