UPSC 2018 Maths Optional Paper 2 Q6b — Step-by-Step Solution

15 marks · Section B

Number systems · Numerical Analysis · asked 7× in 13 yrs · Read the full method →

Question

Find the equivalent of numbers given in a specified number system to the system mentioned against them:

Positional expansion (binary→decimal, hex→decimal); nibble-grouping from the radix point (binary→hex); repeated division by 2 (decimal→binary).

Positional expansion (powers of $2$ , with the binary point):

\underbrace{1\cdot32+1\cdot16+1\cdot8+0\cdot4+1\cdot2+1\cdot1}_{\text{integer}}+\underbrace{1\cdot\tfrac12+0\cdot\tfrac14+1\cdot\tfrac18}_{\text{fraction}}.

=(32+16+8+2+1)+(0.5+0.125)=59+0.625.

\boxed{(111011.101)_2=(59.625)_{10}.}

Group bits into nibbles (4 bits) outward from the binary point, padding the ends with zeros.

Integer part $1000111110000$ → pad left to 16 bits: $0001\;0001\;1111\;0000$

0001=1,\quad 0001=1,\quad 1111=\text{F},\quad 0000=0\ \Rightarrow\ 11\text{F}0.

Fraction part $00101100$ (already 8 bits) → $0010\;1100$

0010=2,\quad 1100=\text{C}\ \Rightarrow\ 2\text{C}.

\boxed{(1000111110000.00101100)_2=(11\text{F}0.2\text{C})_{16}.}

With $\text{C}=12,\ \text{F}=15$ :

12\cdot16^3+4\cdot16^2+15\cdot16+2=12\cdot4096+4\cdot256+15\cdot16+2.

=49152+1024+240+2=50418.

\boxed{(C4F2)_{16}=(50418)_{10}.}

Repeated division by $2$ (remainders read bottom-up):

418\div2=209\ r\,0,\ 209\div2=104\ r\,1,\ 104\div2=52\ r\,0,\ 52\div2=26\ r\,0,

26\div2=13\ r\,0,\ 13\div2=6\ r\,1,\ 6\div2=3\ r\,0,\ 3\div2=1\ r\,1,\ 1\div2=0\ r\,1.

Reading remainders upward: $1\,1\,0\,1\,0\,0\,0\,1\,0$ .

\boxed{(418)_{10}=(110100010)_2.}