Set Theory / Venn Diagrams Questions and Answers

Answer: C

The group has 24 students.

Those who like at least one subject number, by inclusion and exclusion,

\[ n(M \cup E) = 10 + 12 - 4 = 18 \]

Adding the 6 who like neither gives \(18 + 6 = 24\).

18 forgets the 6 students who like neither subject. 20 adds 10 and 12 but forgets both corrections, and 32 adds every number in the question, counting the 4 students who like both twice and treating them as a separate group.

Answer: D

The number who failed is 4,20,000.

Work in thousands. Let \(x\) be the number answering all three correctly, which equals the number answering none. By inclusion-exclusion, the number answering at least one is

\[ \begin{aligned} &330 + 250 + 260 - 100 - 90 - 80 + x \\ &= 570 + x \end{aligned} \]

so the number answering none is \(630 - 570 - x = 60 - x\). Setting this equal to \(x\) gives \(x = 30\). Exactly two correct: \((100 + 90 + 80) - 3x = 180\). Those who pass have two or three correct: \(180 + 30 = 210\). So \(630 - 210 = 420\) thousand fail.

30,000 is the number answering all (or none), and 2,70,000 is the sum of the pairwise counts, which double-counts those answering all three.

Answer: B

The number with both diabetes and high BP is 15.

Everyone with high cholesterol also has high BP, and nobody has both high cholesterol and diabetes, so the 10 cholesterol patients sit inside the high-BP group and outside the diabetes group. The 45 high-BP patients then split into three parts:

\[ \begin{aligned} 45 &= 10 + 20 + x \\ x &= 15 \end{aligned} \]

where 10 have cholesterol and BP, 20 have only BP, and \(x\) have BP and diabetes. The remaining 75 − 45 = 30 patients have diabetes alone, which is consistent with the total of 75.

0 and 10 ignore part of the high-BP count, and 20 confuses this group with the 'only high BP' patients.

Answer: A

35 faculty members are connected only through Facebook.

Of the 150, 30 use neither, so 120 use at least one. By inclusion-exclusion, those on both are

\[ \begin{aligned} |F \cap W| &= 55 + 85 - 120 = 20 \\ \text{only } F &= 55 - 20 = 35 \end{aligned} \]

65 is the number connected only through WhatsApp (85 - 20). 45 and 90 do not fit the counts: with 45 on Facebook only, the 55 Facebook users would include just 10 on both, and the union would be 130, not 120.

Answer: A

The number who like their core branches is 1,800.

Of the 10,000 students, \(10000 - 1500 = 8500\) like at least one kind of branch. Let \(C\) like their core branches and \(O\) like other branches, with \(C = O/4\) and 500 liking both. By inclusion and exclusion,

\[ \begin{aligned} C + O - 500 &= 8500 \\ C + 4C &= 9000 \\ C &= 1800 \end{aligned} \]

So 1,800 like their core branches and 7,200 like other branches. 1,600 comes from subtracting the 500 who like both instead of adding it back (\(5C = 8000\)), and 3,500 and 1,500 are numbers from the question rather than results.

Answer: C

The answer is 35 per cent.

By inclusion and exclusion, the share drinking at least one of the two is

\[ \begin{aligned} 35 + 40 - 10 &= 65\% \\ 100 - 65 &= 35\% \end{aligned} \]

so 35% drink neither. 25% comes from adding the overlap instead of subtracting it (35 + 40 + 10 = 85), and 15 or 40 do not follow from any consistent count. The 10% who drink both are inside both the 35% and the 40%, which is why they are subtracted once.

Answer: C

60 students like romantic movies.

Of the 100 students, 30 like neither kind, so 70 like at least one. Let C be the number who like comedy; then \(2C\) like romance and 20 like both. By inclusion and exclusion:

\[ \begin{aligned} 2C + C - 20 &= 70 \\ 3C &= 90 \Rightarrow C = 30 \end{aligned} \]

So \(2C = 60\) students like romantic movies. 30 is the number who like comedy, 40 is the number who like romance only (60 - 20), and 20 is the number who like both.

Answer: D

The number interested in Humanities is 45.

By inclusion and exclusion, the students interested in at least one of the three sciences are

\[ \begin{aligned} &150 + 200 + 175 - 50 - 60 - 40 + 30 \\ &= 525 - 150 + 30 = 405 \end{aligned} \]

The remaining \(450 - 405 = 45\) are interested in Humanities. The triple overlap must be added back because subtracting the three pairwise overlaps removes the 30 students interested in all three subjects once too often. 30 and 40 are figures taken from the list, and 10 does not come from any consistent count.

Answer: D

40 students took none of the three subjects.

Use the inclusion-exclusion rule for three sets to count students taking at least one subject.

\[ \begin{aligned} n &= 56 + 62 + 84 \\ &\quad - 18 - 20 - 12 + 8 \\ &= 202 - 50 + 8 = 160 \end{aligned} \]

So 160 students took at least one subject, and \(200 - 160 = 40\) took none.

Forgetting to add back the 8 who took all three gives 152 and 48 with none, which is not offered. Simply adding the three subject totals (202) exceeds the 200 students, which shows why the overlaps must be subtracted.

Answer: D

The answer is None of these.

Of the 200 students, 10 passed nothing, so 190 passed at least one subject. Those 190 are split into exactly one, exactly two and all three subjects:

\[ \begin{aligned} 50 + 75 + x &= 190 \\ x &= 65 \end{aligned} \]

That gives 65 students passing all three, which is not among 50, 60 and 40.

The subject totals do not rescue any option either: counting passes, \(150 + 110 + 125 = 385\) should equal \(50 + 2 \times 75 + 3x\), which gives \(x = 61\tfrac{2}{3}\), not a whole number. Either way none of the listed numbers works, so 'None of these' is the answer.