Question

In a locality of 100 families: 45 have radios, 75 have TVs, 25 have VCRs. 10 have all three. Every VCR owner has TV. 25 have radio only. How many families have only TV?

• A. 30
• B. 35
• C. 40
• D. 45

Hint:

Use Venn diagram logic and break down totals by exclusive groups and overlaps.

Answer 1

The Correct Option is C

Let: - \( R = \) number of families with radios = 45 - \( T = \) number with TVs = 75 - \( V = \) number with VCRs = 25 - All VCR owners also have TV → \( V \subseteq T \) - Families with all three: \( RTV = 10 \) - Radio only = 25 Let’s define: - \( R \cap T \cap V = 10 \) - Radio only = 25 → no TV or VCR - So other families with radio: \( 45 - 25 = 20 \) → have radio + (TV or VCR or both) Now: Total = 100 Let us calculate those with only TV (no radio, no VCR) TVs total = 75 Families with VCRs (25) — all counted in TV Also 10 have all three — already counted So subtract: - 10 (all three) - rest 15 with TV+VCR only Also, some families have Radio+TV (but not VCR) So let’s just sum: - Radio only: 25 - VCR + TV only: 15 - All three: 10 - Radio + TV (not VCR): say \( x \) Total so far: \[ 25 (R only) + 15 (T+V only) + 10 (R+T+V) + x + \text{TV only} = 100 \] We want to find TV only. TV total = 75 Out of that: - 10 (R+T+V) - 15 (T+V only) - \( x \) (R+T only) - Remaining = TV only = \( 75 - (10 + 15 + x) = 50 - x \) Now total: \[ 25 + 15 + 10 + x + (50 - x) = 100 100 \boxed{50 - x} = \boxed{40} \] \[ \boxed{40 \text{ families have TV only}} \]