Question

1. (9) in a school with 300 people there is a finite math and chemistry class. in the school there are 110 students that are taking at least one of these two classes. there are 50 students that are taking the chemistry class. there are 200 students who are not taking finite math. determine the following.

(a) the number of students who are taking both classes.
(b) the number of students who are only taking chemistry.
(c) the number of students who are only taking finite math.

Explanation:

Step1: Find number of students taking finite - math

The total number of students is $N = 300$, and the number of students not taking finite - math is $200$. So the number of students taking finite - math, denoted as $n(F)=300 - 200=100$.

Step2: Use the inclusion - exclusion principle

Let $n(F)$ be the number of students taking finite - math, $n(C)$ be the number of students taking chemistry, and $n(F\cup C)$ be the number of students taking at least one of the two classes. The inclusion - exclusion principle states that $n(F\cup C)=n(F)+n(C)-n(F\cap C)$. We know that $n(F\cup C) = 110$, $n(C)=50$, and $n(F) = 100$. Substitute these values into the formula: $110=100 + 50 - n(F\cap C)$.

Step3: Solve for $n(F\cap C)$

Rearrange the equation from Step 2: $n(F\cap C)=100 + 50-110=40$.

Step4: Find number of students only taking chemistry

The number of students only taking chemistry is $n(C)-n(F\cap C)=50 - 40 = 10$.

Step5: Find number of students only taking finite - math

The number of students only taking finite - math is $n(F)-n(F\cap C)=100 - 40=60$.

Answer:

(a) 40
(b) 10
(c) 60