Only TV owners – radios, TVs, VCRs with given constraints: Out of 100 families, 50 have radios, 75 have TVs, 25 have VCRs. Exactly 10 families have all three devices, and every VCR owner also has a TV. If some families have radio only, how many have only a TV (no radio, no VCR)?

Introduction / Context:We apportion a three-set Venn diagram under constraints: V ⊆ T, 10 in all three, and some radio-only families exist. The goal is the number with only TV.

Given Data / Assumptions:

• Total families = 100
• |R| = 50, |T| = 75, |V| = 25
• |R ∩ T ∩ V| = 10
• V ⊆ T (every VCR owner has a TV)
• There exist radio-only families (so radio-only > 0)
• We assume no family lacks all three (n = 0) to make the count determinate under common exam convention

Concept / Approach:Let a = only T, b = only R, d = (R ∩ T) only, e = (T ∩ V) only, g = all three. Then V = e + g = 25 → e = 15. Also R = b + d + g = 50 → b + d = 40. T = a + d + e + g = 75 → a + d + 15 + 10 = 75 → a + d = 50.

Step-by-Step Solution:From above: (a + d) − (b + d) = 50 − 40 → a − b = 10Assuming no family with none: totals give n = 0 → a = 35 (and b = 25, d = 15, e = 15, g = 10)

Verification / Alternative check:Check T: 35 + 15 + 15 + 10 = 75; R: 25 + 15 + 10 = 50; V: 15 + 10 = 25; totals = 100.

Why Other Options Are Wrong:30, 40, 45 break one or more set totals under the same assumptions.

Common Pitfalls:Not using V ⊆ T, or neglecting the all-three count.

Final Answer:35