In town of 500 people, 285 read Hindu and 212 read Indian Express, 127 read…

Key insight: the problem is ambiguous about whether the pairwise counts (20, 29, 35) count people who may also read the third paper or count only those who read exactly those two. The result depends on that interpretation.

Interpretation 1 — pairwise counts are intersections that include people who read all three (standard mathematical reading):

• Total people who read at least one paper = 500 - 50 = 450.

• Sum of single-paper totals = 285 + 212 + 127 = 624. Sum of pairwise intersections = 20 + 29 + 35 = 84.

• By inclusion–exclusion: 450 = 624 - 84 + x, where x is the number who read all three.

• So 450 = 540 + x ⇒ x = -90, which is impossible. Therefore the given numbers are inconsistent under this interpretation and no valid answer can be determined.

Interpretation 2 — pairwise counts are numbers of people who read exactly those two newspapers (exclude those who read all three):

• Let x be the number who read all three. Then

• Only Hindu = 285 - 20 - 29 - x = 236 - x

• Only Times of India = 127 - 20 - 35 - x = 72 - x

• Only Indian Express = 212 - 29 - 35 - x = 148 - x

• Sum of all groups: (236 - x) + (72 - x) + (148 - x) + 20 + 29 + 35 + x + 50 = 500

• Combine constants: 590 - 2x = 500 ⇒ x = 45.

• Then only-one totals: 236 - 45 = 191; 72 - 45 = 27; 148 - 45 = 103. Sum = 191 + 27 + 103 = 321.

Conclusion: If the problem means the pairwise numbers are counts of people who read exactly those two newspapers (excluding the third), then the number of people who read exactly one paper is 321. If the problem uses the standard intersection meaning (pairwise counts include those who read all three), the supplied numbers are inconsistent and no valid answer exists. The question should be clarified or corrected.