UPSC 2023 Maths Optional Paper 2 Q5c-ii — Step-by-Step Solution

5 marks · Section B

Algebra of Binary Numbers · Numerical Analysis · asked 2× in 13 yrs · Read the full method →

Question

Evaluate, using the binary arithmetic, the following numbers in their given system: (ii) $(7AB.432)_{16}-(5CA.D61)_{16}$

Technique

4-bit-per-hex-digit conversion; binary subtraction with borrow propagation; regroup.

Solution

The procedure:

Convert each operand to binary (4 bits per hex digit).
Subtract in binary using the rule ” $a_i-b_i-bin_i$ ; borrow if negative”.
Re-group the binary result into hex.

Step 1 — Convert to binary (4 bits per hex digit):

Hex	Binary	Hex	Binary	Hex	Binary
7	0111	A	1010	B	1011
5	0101	C	1100	A	1010
4	0100	3	0011	2	0010
D	1101	6	0110	1	0001

(7AB.432)_{16}=0111\,1010\,1011\,.\,0100\,0011\,0010_2,

(5CA.D61)_{16}=0101\,1100\,1010\,.\,1101\,0110\,0001_2.

Step 2 — Binary subtraction.

Fractional part (positions $2^{-1}$ to $2^{-12}$ , MSB→LSB):

Position	$2^{-1}$	$2^{-2}$	$2^{-3}$	$2^{-4}$	$2^{-5}$	$2^{-6}$	$2^{-7}$	$2^{-8}$	$2^{-11}$	$2^{-12}$
minuend	0	1	0	0	0	0	1	1	1	0
subtrahend	1	1	0	1	0	1	1	0	0	1
difference	0	1	1	0	1	1	0	1	0	1

(Borrow propagates from $2^{-12}$ up. The final borrow out into the integer part is $1$ .)

Integer part (positions $2^0$ to $2^{11}$ , LSB→MSB), with borrow-in $=1$ from the fraction:

Position	$2^0$	$2^1$	$2^3$	$2^5$	$2^6$	$2^7$	$2^8$	$2^9$	$2^{10}$
minuend	1	1	1	1	0	1	1	1	1
subtrahend	0	1	1	0	1	1	1	0	1
difference	0	0	0	1	1	1	1	0	0

Integer bits from $2^{11}$ down: $0\,0\,0\,1\;\;1\,1\,1\,0\;\;0\,0\,0\,0=0001\,1110\,0000_2$ . Fractional bits from $2^{-1}$ down: $0\,1\,1\,0\;\;1\,1\,0\,1\;\;0\,0\,0\,1=0110\,1101\,0001_2$ .

Step 3 — Regroup in fours back to hex.

Integer: $0001=1,\;1110=E,\;0000=0\Rightarrow 1E0$ . Fractional: $0110=6,\;1101=D,\;0001=1\Rightarrow .6D1$ .

Answer

\boxed{\;(7AB.432)_{16}-(5CA.D61)_{16}=(1E0.6D1)_{16}.\;}