Sets, Relations & Functions
Practical Problems on Union and Intersection
Grade 11
Question:
<p>In a certain town, 25% of the families own a phone and 15% own a car; 65% families own neither a phone nor a car and 2,000 families own both a car and a phone. Consider the following three statements:<br><strong>(a)</strong> 5% families own both a car and a phone.<br><strong>(b)</strong> 35% families own either a car or a phone.<br><strong>(c)</strong> 40,000 families live in the town.<br>Then,</p>
<p>Only (a) and (b) are correct.</p>
<p>Only (a) and (c) are correct.</p>
<p>Only (b) and (c) are correct.</p>
<p>All (a), (b) and (c) are correct.</p>
Step-by-Step Solution
Key Concept: Use the principle of set union: n(P∪C) = n(P) + n(C) - n(P∩C), combined with the complement condition that 65% own neither, to find the total population and verify each statement.
<p><strong>Step 1: Set up the given information</strong></p><p>Let P = families owning a phone, C = families owning a car, Total = N families.</p><p>Given: n(P) = 0.25N, n(C) = 0.15N, n(P∩C) = 2000, and n(P∪C)' = 0.65N</p><p><strong>Step 2: Find families owning either phone or car</strong></p><p>Families owning neither = 0.65N, so families owning either = 1 - 0.65 = 0.35N</p><p>Therefore: n(P∪C) = 0.35N</p><p><strong>Step 3: Use the union formula</strong></p><p>n(P∪C) = n(P) + n(C) - n(P∩C)</p><p>0.35N = 0.25N + 0.15N - n(P∩C)</p><p>0.35N = 0.40N - n(P∩C)</p><p>n(P∩C) = 0.05N</p><p><strong>Step 4: Find total families</strong></p><p>Given n(P∩C) = 2000 and we found n(P∩C) = 0.05N:</p><p>0.05N = 2000 → N = 40,000</p><p><strong>Step 5: Verify each statement</strong></p><p>(a) Families with both = 0.05 × 40,000 = 2000 = 5% ✓ TRUE</p><p>(b) Families with either = 0.35 × 40,000 = 14,000 = 35% ✓ TRUE</p><p>(c) Total families = 40,000 ✓ TRUE</p><p>∴ Answer: All three statements (a), (b), and (c) are correct → <strong>D</strong></p>
Correct Answer: D