Modern Math

Set Theory

Easy

In a group of 150 students, 52 like tea, 48 like juice and 62 like coffee. If each student in the group likes at least one among tea, juice and coffee, then the maximum number of students that like more than one drink is:

Entered answer:

✅ Correct Answer: 12

Solution figure for IPMAT Indore 2024 SA question 13 (Modern Math)

Let's denote the number of students who like tea as $n(T) = 52$ , juice as $n(J) = 48$ , and coffee as $n(C) = 62$ . The total number of students is $n(T \cup J \cup C) = 150$ .

Using the Principle of Inclusion-Exclusion for three sets:

$n(T \cup J \cup C) = n(T) + n(J) + n(C) - n(T \cap J) - n(T \cap C) - n(J \cap C) + n(T \cap J \cap C)$

$150 = 52 + 48 + 62 - n(T \cap J) - n(T \cap C) - n(J \cap C) + n(T \cap J \cap C)$

$150 = 162 - n(T \cap J) - n(T \cap C) - n(J \cap C) + n(T \cap J \cap C)$

Rearranging, we get:

$n(T \cap J) + n(T \cap C) + n(J \cap C) - n(T \cap J \cap C) = 162 - 150 = 12$

So we have: $d + e + f - k= 12$

Answer: 12 students

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