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UPSC 2026 Maths Optional Paper 2 Q5c — Step-by-Step Solution

10 marks · Section B

Logic gates and truth tables · Numerical Analysis · asked 3× in 14 yrs · Read the full method →

Question

Consider the logic circuit $L$: inputs $A$, $B$, $C$ feed three AND gates whose outputs feed a single OR gate producing $Y$. The first AND gate receives $A$, $B$ and $C$ directly; the second receives $A$, $\bar{B}$ (via an inverter on the $B$ line) and $C$; the third receives $\bar{A}$ (via an inverter on the $A$ line) and $B$.

(i) Express $Y$ as a Boolean expression.

(ii) Write 8-bit special sequences for $A$, $B$ and $C$.

(iii) Find the truth table of $L$ using 8-bit special sequences.

(In the printed paper this is the only figure: a gate diagram. The wiring above is the transcription of that diagram — three AND gates feeding one OR gate, with AND-3 taking exactly two inputs, $\bar A$ and $B$.)

The circuit $L$ as printed in the paper (UPSC CSM 2026 Paper II, Q5(c)). The prose above is the accessible-text transcription of this diagram; the two must agree exactly — in particular AND-3 has two inputs, not three.

Technique

Read the sum-of-products straight off the circuit — one product term per AND gate, OR-ed together — then simplify before tabulating: the first two terms differ only in $B$ and $\bar B$, so the complementation law collapses them at once. For part (iii) the “special sequences” technique replaces row-by-row evaluation with eight parallel evaluations done as bitwise operations on 8-bit words; the whole truth table then falls out as a single output word, and the answer must show the bitwise columns, because that is the technique being examined.

Solution

Step 0 — Read the circuit.

(i) The Boolean expression.

An AND gate outputs the product of its inputs and an OR gate the sum of its inputs, so

Simplify. The first two terms share the factor $AC$:

using the complement law $B+\bar B=1$ and the identity $X\cdot1=X$. Hence

This is minimal. The minterms of $Y$ are $\Sigma m(2,3,5,7)$ (established in (iii) below); on a three-variable K-map the prime implicants are $\bar AB$ (cells $2,3$), $AC$ (cells $5,7$) and $BC$ (cells $3,7$). Cell $2$ is covered only by $\bar AB$ and cell $5$ only by $AC$, so both are essential, and together they already cover all four cells. Therefore $\bar AB+AC$ is the unique minimal sum of products, and the consensus term $BC$ is redundant:

(Redundant logically, but the term one would add back to make the hardware free of a static-1 hazard when $A$ switches with $B=C=1$ — one line, and it shows the circuit is understood, not just the algebra.)

Interpretation. $Y=AC+\bar AB$ is precisely a 2-to-1 multiplexer with $A$ as the select line: $Y=C$ when $A=1$ and $Y=B$ when $A=0$. Gate count drops from $3$ AND $+\,1$ OR $+\,2$ NOT (with a 3-input OR-free 3-input AND) to $2$ AND $+\,1$ OR $+\,1$ NOT.

(ii) The 8-bit special sequences.

Convention (must be stated, since the sequences depend on it). For $n=3$ variables there are $2^{3}=8$ input combinations. List them as the rows of the truth table in ascending binary order with $A$ as the most significant bit:

Reading each variable down its column gives its 8-bit special (standard/distinctive) sequence. For the $k$-th of $n$ variables the sequence is alternating blocks of $2^{\,n-k}$ zeros and $2^{\,n-k}$ ones; with $n=3$:

Bit position $j$ ($j=0,\dots,7$, read left to right) of every sequence refers to the same input combination, namely the binary representation of $j$ — that is the whole point of the device.

(iii) The truth table by bitwise operations on the sequences.

Complements are bitwise NOT:

Now form the three AND-gate outputs by bitwise AND and the OR-gate output by bitwise OR. Working the unsimplified expression (1) — which is what the circuit actually builds — and, in parallel, the simplified expression (2), as a self-check:

In compact sequence form:

and by the simplified route

The two routes give the identical word, which confirms the simplification (2) — a free check the technique hands you.

Reading the truth table off the output sequence. Bit $j$ of $Y$ is the output for the row whose $(A,B,C)$ is the binary form of $j$:

So $Y=\Sigma m(2,3,5,7)$, as used in the K-map argument of part (i). Sanity spot-checks against $Y=AC+\bar AB$: row $2$ $(0,1,0)$ gives $\bar AB=1\Rightarrow Y=1$ ✓; row $6$ $(1,1,0)$ gives $AC=0,\ \bar AB=0\Rightarrow Y=0$ ✓; row $5$ $(1,0,1)$ gives $AC=1\Rightarrow Y=1$ ✓. And the multiplexer reading is visible in the table: for $A=0$ (rows $0$–$3$) $Y$ copies $B$; for $A=1$ (rows $4$–$7$) $Y$ copies $C$. $\qquad\blacksquare$

Answer

with truth table $Y=0,0,1,1,0,1,0,1$ for $(A,B,C)=000,001,010,011,100,101,110,111$ respectively.