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

7 marks · Section B

Number systems · Numerical Analysis · asked 8× in 14 yrs · Read the full method →

Question

Evaluate $(11001.1011)_{10} + (33.24)_8 + (101101)_2$ and convert the final result in hexadecimal system.

Technique

Mixed-radix addition is always done by moving to one common base first. Here two of the three operands are in bases $8$ and $2$ and the required output is base $16$ — all powers of two — so binary is the natural meeting point: octal converts three-bits-per-digit, hexadecimal reads back four-bits-per-digit, and no repeated division is needed anywhere. Decimal is carried alongside purely as an arithmetic check. The one real decision the question forces is the base of the first operand, which the paper prints as $10$; that is settled explicitly in Step 0 before any arithmetic is done.

Solution

Step 0 — The base of the first operand: state the reading and justify it.

The paper prints the first operand as $(11001.1011)_{10}$, i.e. in base ten. Take that literally and the sum is

whose fractional part $0.4136=\dfrac{517}{1250}$ has $1250=2\cdot5^4$ in its lowest-terms denominator. A fraction $p/q$ (lowest terms) terminates in base $b$ iff every prime factor of $q$ divides $b$; here $5\nmid 16$, so the hexadecimal expansion of the answer never terminates. The printed instruction “convert the final result in hexadecimal system” then cannot be carried out exactly at all.

Read instead as $(11001.1011)_2$ and everything closes:

the fractional parts cancel to a whole number, and the answer is the clean two-digit $(62)_{16}$. That exact cancellation is engineered, not accidental; together with the fact that the string $11001.1011$ contains only the digits $0$ and $1$ and sits between two operands in bases $8$ and $2$, it makes the subscript $10$ a misprint for $2$.

Decision: solve $(11001.1011)_{2}+(33.24)_8+(101101)_2$ — that is the question the marking key will carry. The literal base-ten reading is then worked out in full in Step 5, because a candidate in the hall did have to choose, and a script that states the choice, solves it, and disposes of the alternative in three lines loses nothing.

Reading I (the intended question): first operand in base 2

Step 1 — Bring every operand to binary.

$(11001.1011)_2$ is already binary.

Octal $\to$ binary is three bits per octal digit, applied outward from the point:

$(101101)_2$ is already binary.

Decimal check of the conversions (kept for the arithmetic check in Step 3):

Step 2 — Add in binary.

First two operands. The fractional columns give $(0.1011)_2+(0.0101)_2=(1.0000)_2$ exactly — a carry of $1$ into the units column and nothing left over:

(units column: $1+1+1_{\text{carry}}=11_2$, and so on; the integer sum is $11001_2+11011_2+1_2=110101_2$).

Now add the third operand, an integer, so the fractional part stays zero:

Hence

Step 3 — Arithmetic check in decimal.

Step 4 — Convert to hexadecimal.

Binary $\to$ hexadecimal is four bits per hex digit, grouped outward from the binary point, padding with leading zeros:

Check: $(62)_{16}=6\cdot16+2=96+2=98 .\;\checkmark$

Reading II (literally as printed): first operand in base 10

Step 5 — The same computation with $(11001.1011)_{10}$.

Integer part $\to$ hexadecimal, by repeated division by $16$ (remainders read upwards):

Check: $2\cdot4096+11\cdot256+4\cdot16+1=8192+2816+64+1=11073 .\;\checkmark$

Fractional part $\to$ hexadecimal, by repeated multiplication by $16$ (integer parts read downwards), working with the exact fraction $0.4136=\dfrac{517}{1250}$:

and it does not stop: as shown in Step 0, $1250=2\cdot5^4$ carries the prime $5$, which does not divide $16$, so the expansion is non-terminating (being rational, it is eventually periodic). Hence

a hexadecimal fraction that can only be quoted to a stated number of places. $\qquad\blacksquare$

The contrast is itself the argument for Reading I: one reading gives a two-digit exact answer, the other gives an unending one to a question that asks for “the final result in hexadecimal system”.

Answer