Q.A survey shows that 63% of the people watch a News Channel whereas 76% watch another channel. If $x\%$ of the people watch both channel, then
(A) $x = 35$
(B) $x = 63$
(C) $39 \le x \le 63$
(D) $x = 39$

Concept: Set inclusion-exclusion and bounds on intersection.

Let $A$ be the set of people watching the first channel and $B$ the second. We know $|A| = 63\%$ and $|B| = 76\%$, and we need bounds on $|A \cap B| = x\%$.

By inclusion-exclusion, the percentage watching at least one channel is:

Since $|A \cup B| \le 100$ (cannot exceed the total population):

$139 - x \le 100 \implies x \ge 39$ …

When two groups overlap, the key insight comes from the inclusion-exclusion principle. If we know what fraction watches channel A, what fraction watches channel B, and what fraction watches both, we can find what fraction watches at least one channel:

The subtraction is necessary because people who watch both channels get counted twice when we add $P(A)$ and $P(B)$.

Now, the fraction watching at least one channel must satisfy two physical constraints:

• It cannot exceed 100% (you can't have more than everyone)
• It cannot be less than the larger of the two individual percentages (if 76% watch channel B, at least 76% watch at least one channel)

Let's translate this into inequalities with $x$ as the percentage watching both channels.

1. Set up the inclusion-exclusion formula Let $A$ be the set watching the first channel (63%) and $B$ the set watching the second (76%). Then:

2. Apply the upper bound constraint The percentage watching at least one channel cannot exceed 100%:

3. Apply the lower bound constraint …