Modern Math

Set Theory

Medium

In a group of $250$ students, the percentage of girls was at least $44\%$ and at most $60\%$ . The rest of the students were boys. Each student opted for either swimming or running or both. If $50\%$ of the boys and $80\%$ of the girls opted for swimming while $70\%$ of the boys and $60\%$ of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are

$75$ and $96$ , respectively

$75$ and $90$ , respectively

$72$ and $88$ , respectively

$72$ and $80$ , respectively

✅ Correct Option: 4

We have 250 students total. The percentage of girls varies between 44% and 60%, with boys making up the remainder. Every student does at least one activity (swimming or running), and some do both.

Minimum girls = 44% of 250 = $0.44 \times 250 = 110$ girls

Maximum girls = 60% of 250 = $0.60 \times 250 = 150$ girls

When girls are minimum (110), boys are maximum = $250 - 110 = 140$ boys

When girls are maximum (150), boys are minimum = $250 - 150 = 100$ boys

Inclusion-exclusion principle:

$|S \cup R| = |S| + |R| - |S \cap R|$

$|S \cap R|$ represents students who do both.

Rearranging the equation (as we need to find students who do both).

$|S \cap R| = |S| + |R| - |S \cup R|$

When we add students swimming + students running, we're counting students who do both activities twice. Since every student does at least one activity, the total count equals all 250 students plus the extra count of "both" students. So we subtract 250 to get just the "both" students.

Lastly: $|S \cup R| =$ total $= 250$ (as given in question)

Case 1: Minimum girls case (110 girls, 140 boys):

Girls' participation:

Swimming: $80\%$ of $110$ = $0.8 \times 110 = 88$ girls

Running: $60\%$ of $110$ = $0.6 \times 110 = 66$ girls

Boys' participation:

Swimming: $50\%$ of 140 = $0.5 \times 140 = 70$ boys

Running: $70\%$ of 140 = $0.7 \times 140 = 98$ boys

Total participation:

Total swimming = $88 + 70 = 158$ students

Total running = $66 + 98 = 164$ students

Students doing BOTH = $158 + 164 - 250 = 72$ students

Case 2: maximum girls case (150 girls, 100 boys):

Girls' participation:

Swimming: $80\%$ of 150 = $0.8 \times 150 = 120$ girls

Running: $60\%$ of 150 = $0.6 \times 150 = 90$ girls

Boys' participation:

Swimming: $50\%$ of 100 = $0.5 \times 100 = 50$ boys

Running: $70\%$ of 100 = $0.7 \times 100 = 70$ boys

Total participation:

Total swimming = $120 + 50 = 170$ students

Total running = $90 + 70 = 160$ students

Students doing BOTH = $170 + 160 - 250 = 80$ students

Therefore, the minimum is 72 and the maximum is 80 students who opted for both swimming and running.

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