Question 1

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…

Accepted Answer

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

Question 2

Class size from percentages with a known overlap (both subjects): In a class, 72% took Biology and 44% took Mathematics. Each student took at least one of these subjects, and 40 students took both. Find the total number of students in the class.

Accepted Answer

Introduction / Context: Percentages of two overlapping groups sum to more than 100% due to the overlap. When the numeric overlap is given, we can solve for the total by equating inclusion-exclusion to the class size. Given Data / Assumptions: Let total = N Biology = 0.72N Mathematics = 0.44N Both = 40 At least one → union size = N Concept / Approach: Inclusion-exclusion for counts: |B ∪ M| = |B| + |M| − |B ∩ M|. Here |B ∪ M| = N, and |B ∩ M| = 40. Step-by-Step Solution: N = 0.72N + 0.44N − 40N − 1.16N = −40 → −0.16N = −40 → N = 250 Verification / Alternative check: Both percentage = 0.72 + 0.44 − 1 = 0.16 → 0.16N = 40 → N = 250 (same). Why Other Options Are Wrong: 200, 240, 320 do not satisfy 0.16N = 40. Common Pitfalls: Forgetting that the union equals the whole class when each student takes at least one subject. Final Answer: 250

Question 3

In a class of 50 students, 35 opted for mathematics and 37 opted for Biology. How many have opted for both Mathematics and Biology? How many have opted for only Mathematics? (Assume that each student has to opt for at least one of the subjects).

Accepted Answer

Given in the question, n(M ∪ B) = 50, n(M) = 35, n(B) = 37, n(M ∩ B) =? use the below formula n(M ∪ B) = n(M) + n(B) – n(M ∩ B) We get 50 = 35 + 37 – n(M ∩ B) ⇒ n(M ∩ B) = 35 + 37 – 50 = 72 – 50 = 22 ∴ 22 students have opted for both Mathematics and Biology. Again number of students who have opted for only Mathematics = n(M) – n(M ∩ B) = 35 – 22 = 13.

Sets and Functions Questions