In a survey of $400$ students in a school, $100$ were listed as taking apple juice, $150$ as taking orange juice and $75$ were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.
Let $U$ denote the set of surveyed students and $A$ denote the set of students taking apple juice and $B$ denote the set of students taking orange juice. Then
$n(U) = 400,n(A) = 100,n(B) = 150$ and $n(A \cap B) = 75$
Now $n\left( {{A^\prime } \cap {B^\prime }} \right) = n{(A \cup B)^\prime }$
${ = n(U) - n(A \cup B)}$
${ = n(U) - n(A) - n(B) + n(A \cap B)}$
${ = 400 - 100 - 150 + 75 = 225\,}$
Hence $225$ students were taking neither apple juice nor orange juice.
Two newspaper $A$ and $B$ are published in a city. It is known that $25\%$ of the city populations reads $A$ and $20\%$ reads $B$ while $8\%$ reads both $A$ and $B$. Further, $30\%$ of those who read $A$ but not $B$ look into advertisements and $40\%$ of those who read $B$ but not $A$ also look into advertisements, while $50\%$ of those who read both $A$and $B$ look into advertisements. Then the percentage of the population who look into advertisement is
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the number of people who read exactly one newspaper.
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Chemical $C_{1}$ or chemical $C_{2}$
In a city $20$ percent of the population travels by car, $50$ percent travels by bus and $10$ percent travels by both car and bus. Then persons travelling by car or bus is......$\%$