UPSC 2026 Maths Optional Paper 2 Q6b-i — Step-by-Step Solution

8 marks · Section B

Representation of Integers, Signed Integers, and Reals (incl. Double Precision) · Numerical Analysis · asked 2× in 14 yrs · Read the full method →

Question

Find the decimal equivalent of the following floating point 10-bit numbers with 5 bits as fractional part:

1010111011 \quad\text{and}\quad 0100101010

Technique

“10 bits with 5 bits as fractional part” fixes the position of the binary point: 5 integer bits, then the point, then 5 fractional bits. That is a fixed-point word (a $Q5.5$ format), so the value is just the positional sum $\sum b_i 2^{i}$ with $i$ running from $+4$ down to $-5$ — equivalently, read all ten bits as one unsigned integer $N$ and divide by $2^{5}=32$ . The only judgement the question demands is whether the leading bit is a sign bit; nothing in the wording says it is, so read the words unsigned, and say so.

Solution

Step 1 — Fix the format explicitly before computing anything.

A 10-bit word $b_4b_3b_2b_1b_0\,\bullet\,b_{-1}b_{-2}b_{-3}b_{-4}b_{-5}$ with 5 fractional bits has value

\text{val}=\sum_{i=-5}^{4} b_i\,2^{\,i} = \underbrace{b_42^4+b_32^3+b_22^2+b_12^1+b_02^0}_{\text{integer part}} \;+\;\underbrace{b_{-1}2^{-1}+b_{-2}2^{-2}+b_{-3}2^{-3}+b_{-4}2^{-4}+b_{-5}2^{-5}}_{\text{fractional part}} . \tag{1}

Weights, written out once and reused:

2^4,2^3,2^2,2^1,2^0\;=\;16,\;8,\;4,\;2,\;1;\qquad 2^{-1},2^{-2},2^{-3},2^{-4},2^{-5}\;=\;0.5,\;0.25,\;0.125,\;0.0625,\;0.03125 .

Equivalently, and this is the fastest reliable route under exam pressure:

\text{val}=\frac{N}{2^{5}}=\frac{N}{32},\qquad N=\text{the whole 10-bit string read as a plain binary integer}, \tag{2}

because inserting a binary point five places from the right is exactly a division by $2^5$ . Formula (2) also gives a free arithmetic check on (1).

Convention actually used: the word is read as unsigned. The question describes only a bit-count and the position of the point — it names no sign bit, no bias and no exponent field — so an unsigned fixed-point reading is the one the wording supports. (The signed readings are tabulated in Step 4, for completeness.)

Step 2 — First number, $1010111011$ .

Split five and five:

1010111011\;\longrightarrow\;\underbrace{10101}_{\text{integer}}\;\bullet\;\underbrace{11011}_{\text{fraction}} .

Integer part:

(10101)_2=1\cdot16+0\cdot8+1\cdot4+0\cdot2+1\cdot1=16+4+1=21 .

Fractional part:

(0.11011)_2=1\cdot\tfrac12+1\cdot\tfrac14+0\cdot\tfrac18+1\cdot\tfrac1{16}+1\cdot\tfrac1{32} =\frac{16+8+0+2+1}{32}=\frac{27}{32}=0.84375 .

Hence

(1010111011)\;\text{in }Q5.5\;=\;21+0.84375\;=\;21.84375 .

Check by (2). As a plain integer,

(1010111011)_2=512+128+32+16+8+2+1=699,\qquad \frac{699}{32}=21.84375 . \;\checkmark

Step 3 — Second number, $0100101010$ .

0100101010\;\longrightarrow\;\underbrace{01001}_{\text{integer}}\;\bullet\;\underbrace{01010}_{\text{fraction}} .

Integer part:

(01001)_2=0\cdot16+1\cdot8+0\cdot4+0\cdot2+1\cdot1=8+1=9 .

Fractional part:

(0.01010)_2=0\cdot\tfrac12+1\cdot\tfrac14+0\cdot\tfrac18+1\cdot\tfrac1{16}+0\cdot\tfrac1{32} =\frac{0+8+0+2+0}{32}=\frac{10}{32}=\frac{5}{16}=0.3125 .

Hence

(0100101010)\;\text{in }Q5.5\;=\;9+0.3125\;=\;9.3125 .

Check by (2). As a plain integer,

(0100101010)_2=256+32+8+2=298,\qquad \frac{298}{32}=9.3125 . \;\checkmark

Both values are exact — every $Q5.5$ word is an exact multiple of $2^{-5}=0.03125$ , so a terminating decimal with at most five places is guaranteed, and any answer with a recurring tail is wrong by construction. That is a useful sanity rule to state.

Step 4 — The signed readings, stated as labelled alternatives (not as hedging).

The second word begins with $0$ , so it is positive under every one of the standard conventions and equals $9.3125$ in all of them. Only the first word, which begins with $1$ , is convention-sensitive:

Reading of $1010111011$	Value
Unsigned $Q5.5$ (the reading used above)	$\dfrac{699}{32}=\mathbf{21.84375}$
Sign–magnitude: $b_9$ = sign, remaining nine bits $= (010111011)$ with the same point position	$-\dfrac{699-512}{32}=-\dfrac{187}{32}=-5.84375$
Two’s complement with the same scaling $2^{-5}$	$\dfrac{699-1024}{32}=-\dfrac{325}{32}=-10.15625$

Which a marker expects: $21.84375$ . The question supplies exactly one piece of format information — where the binary point sits — and supplies it for both numbers alike; had a sign bit been intended, the effective integer field would be 4 bits, not 5, and the phrase “5 bits as fractional part” would no longer determine the layout. Moreover the second number is deliberately chosen with a leading $0$ , so that it reads the same either way — the paper is testing positional weighting, not two’s-complement arithmetic. The unsigned values are therefore given as the answer, with the alternatives recorded above so that the reading is a stated decision rather than an accident. $\qquad\blacksquare$

Answer

\boxed{\;(1010111011)_{Q5.5}=(10101.11011)_2=\frac{699}{32}=21.84375,\qquad (0100101010)_{Q5.5}=(01001.01010)_2=\frac{298}{32}=9.3125\;}