UPSC Maths 2025 Paper 2 Q5c-i — Solution

5 marks · Section B

Question

Convert the number $(3479)_{10}$ into binary system and the number $(7AE \cdot 9F)_{16}$ into decimal system.

Technique

Use repeated division by 2 for the decimal-to-binary conversion, and positional (place-value) expansion for the hexadecimal-to-decimal conversion, with negative powers of 16 for the fractional hex digits.

Solution

Part 1: $(3479)_{10} \to$ binary

Repeated division by 2, reading remainders bottom-to-top:

division	quotient	remainder
$3479 \div 2$	$1739$	$1$
$1739 \div 2$	$869$	$1$
$869 \div 2$	$434$	$1$
$434 \div 2$	$217$	$0$
$217 \div 2$	$108$	$1$
$108 \div 2$	$54$	$0$
$54 \div 2$	$27$	$0$
$27 \div 2$	$13$	$1$
$13 \div 2$	$6$	$1$
$6 \div 2$	$3$	$0$
$3 \div 2$	$1$	$1$
$1 \div 2$	$0$	$1$

Reading remainders from last to first: $(3479)_{10} = (110110010111)_2.$

Check: $2048 + 1024 + 256 + 128 + 16 + 4 + 2 + 1 = 3479.$ ✓

Part 2: $(7AE.9F)_{16} \to$ decimal

Hex digits: $7, A=10, E=14$ (integer part); $9, F=15$ (fractional part).

Integer part: $7 \times 16^2 + 10 \times 16^1 + 14 \times 16^0 = 7(256) + 10(16) + 14 = 1792 + 160 + 14 = 1966.$

Fractional part: $9 \times 16^{-1} + 15 \times 16^{-2} = \frac{9}{16} + \frac{15}{256} = 0.5625 + 0.05859375 = 0.62109375.$

Total: $(7AE.9F)_{16} = 1966 + 0.62109375 = 1966.62109375.$

Answer

$\boxed{(3479)_{10} = (110110010111)_2,\qquad (7AE.9F)_{16} = (1966.62109375)_{10}}$