Set Theory / Venn Diagrams Questions and Answers

Answer: A

The number who play at least two sports is 25.

Every one of the 100 members plays at least one sport, so each member is in exactly one of two groups: those who play exactly one sport and those who play two or more.

\[ \begin{aligned} \text{exactly one} &= 20 + 30 + 25 = 75 \\ \text{at least two} &= 100 - 75 = 25 \end{aligned} \]

The pairwise figures are not needed: everyone who does not play exactly one sport plays at least two. Option 22 is the number who play exactly two sports, \(25 - 3\), which leaves out the 3 who play all three.

Answer: B

9 people read both French and English.

Of the 18 people, 6 read neither language, so 12 read at least one. By the inclusion-exclusion rule:

\[ \begin{aligned} n(F \cup E) &= n(F) + n(E) - n(F \cap E) \\ 12 &= 10 + 11 - n(F \cap E) \\ n(F \cap E) &= 9 \end{aligned} \]

Check: 1 reads French only, 2 read English only, 9 read both and 6 neither, making 18. 3 is the sum of the 'only' groups, and 6 and 4 do not satisfy the totals.

Answer: C

The number that is not possible is 28.

At least 12 of set M's 42 students are outside N, so at most 42 - 12 = 30 students are in both sets. Set N has 60 students, so the number in N but not in M is

\[ \begin{aligned} 60 - (\text{common}) &\ge 60 - 30 \\ &= 30 \end{aligned} \]

Any value from 30 to 60 is possible, depending on how many students the two sets share. 32, 42 and 45 all lie in that range, but 28 is below the minimum of 30, so it cannot occur.

Answer: C

36 students study all three languages.

Let x study all three and y exactly two. Those studying more than one number 1.5x, so y + x = 1.5x and y = 0.5x. Let a study exactly one language.

\[ \begin{aligned} a + y + x &= 180 \\ a + 2y + 3x &= 80 + 90 + 100 = 270 \\ y + 2x &= 90 \\ 0.5x + 2x &= 90 \Rightarrow x = 36 \end{aligned} \]

Check: 18 study exactly two, 126 exactly one, and 126 + 2(18) + 3(36) = 270.

18 is the number studying exactly two, a likely confusion; 24 and 42 fail the 270 total.

Answer: B

The number is 19.

The rectangle is the wide shape at the lower left and stands for males. The square is the larger near-square on the right and stands for rural workers. The triangle is for armed forces and the circle for doctors.

Females are the region outside the rectangle, urban areas are outside the square, so the number needed lies inside both the triangle and the circle but outside both four-sided shapes. That is the region marked 19, which sits in the triangle and the circle to the left of the square and above the rectangle.

24 is also in the triangle and circle but inside the square (rural). 21 is inside the rectangle and the square, and 36 is in the circle and the square but outside the triangle.

Answer: C

X ∪ Y contains 5 elements.

The union contains every element that is in X or Y or both, each counted once:

\[ X \cup Y = \{2, 3, 4, 5, 6\} \]

That is 5 elements. 4 is the size of either set on its own, 7 counts some shared elements twice, and 3 is the size of the intersection, {3, 4, 5}, not the union.

The formula gives the same result: \(n(X \cup Y) = n(X) + n(Y) - n(X \cap Y) = 4 + 4 - 3 = 5\).

Answer: A

A - B has 3 elements.

A - B is the set of elements that are in A but not in B. Removing from A = {a, e, i, o, u} the elements it shares with B, namely e and o, leaves {a, i, u}, which has 3 elements.

2 is the number of common elements, and 4 or 5 forget to remove one or both of the shared letters.

For comparison, B - A = {g, h, k} also has 3 elements, while A ∩ B = {e, o} has 2 and A ∪ B has 8.

Answer: B

45 students passed in at least one subject.

Adding the two pass lists counts the 10 students who passed both subjects twice, so subtract them once:

\[ n(M \cup E) = 30 + 25 - 10 = 45 \]

So 45 passed at least one subject and the other 5 of the 50 failed both.

55 is the plain sum 30 + 25, which double-counts the 10. 50 is the class size, which would mean no one failed both, and 40 subtracts the overlap twice.