Practice Questions
In a survey of 100 people, 60 like tea, 50 like coffee, and 30 like both. How many like neither?
$n(T \\cup C) = 60 + 50 - 30 = 80$. Neither = $100 - 80 = 20$.
In a class of 40 students, 25 play cricket, 20 play football, and 8 play both. How many students play at least one sport?
$n(C \\cup F) = 25 + 20 - 8 = 37$. At least one = 37.
If $n(A) = 30$, $n(B) = 45$, and $n(A \\cup B) = 60$, find $n(A \\cap B)$.
$n(A \\cap B) = n(A) + n(B) - n(A \\cup B) = 30 + 45 - 60 = 15$.
Out of 80 students, 50 study Math, 40 study Science, and 15 study both. How many study Math but not Science?
Only Math = $n(M) - n(M \\cap S) = 50 - 15 = 35$.
In a group of 120 people, 80 read newspaper A, 60 read newspaper B, and 20 read both. How many read exactly one newspaper?
Only A = $80 - 20 = 60$. Only B = $60 - 20 = 40$. Exactly one = $60 + 40 = 100$.
If $n(A) = 25$, $n(B) = 20$, and $n(A \\cup B) = 35$, find $n(A \\cap B)$.
$n(A \\cap B) = 25 + 20 - 35 = 10$.
In a survey of 200 people, 120 like tea, 100 like coffee, and 40 like both. How many like only tea?
Only tea = $n(T) - n(T \\cap C) = 120 - 40 = 80$.
In a class of 60 students, 35 passed Math, 30 passed English, and 10 passed neither. How many passed both subjects?
$n(M \\cup E) = 60 - 10 = 50$. $n(M \\cap E) = 35 + 30 - 50 = 15$.
If 80% of people like tea, 60% like coffee, and 30% like both, what percentage like neither?
$n(T \\cup C) = 80 + 60 - 30 = 110%$. Since it exceeds 100%, the numbers imply everyone likes at least one. But let's calculate: $100 - 110 = -10$... Actually with valid data: Suppose 80 like tea, 60 like coffee, 30 like both. $n(T \\cup C) = 80 + 60 - 30 = 110$. But total is 100, so the data is inconsistent. In a consistent scenario, if total = 100, $n(\\text{neither}) = 100 - (80+60-30) = -10$, impossible. With corrected data: if T=70, C=50, both=20, neither = $100 - (70+50-20) = 0$.
In a class of 50 students, 30 play football, 25 play basketball, and 10 play both. How many play neither?
$n(F \\cup B) = 30 + 25 - 10 = 45$. Neither = $50 - 45 = 5$.
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