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.
There are $200$ individuals with a skin disorder, $120$ had been exposed to the chemical $C _{1}, 50$ to chemical $C _{2},$ and $30$ to both the chemicals $C _{1}$ and $C _{2} .$ Find the number of individuals exposed to
Chemical $C_{1}$ or chemical $C_{2}$
In a committee, $50$ people speak French, $20$ speak Spanish and $10$ speak both Spanish and French. How many speak at least one of these two languages?
In a classroom, one-fifth of the boys leave the class and the ratio of the remaining boys to girls is $2: 3$. If further $44$ girls leave the class, then class the ratio of boys to girls is $5: 2$. How many more boys should leave the class so that the number of boys equals that of girls?
In a town of $10,000$ families it was found that $40\%$ family buy newspaper $A, 20\%$ buy newspaper $B$ and $10\%$ families buy newspaper $C, 5\%$ families buy $A$ and $B, 3\%$ buy $B$ and $C$ and $4\%$ buy $A$ and $C$. If $2\%$ families buy all the three newspapers, then number of families which buy $A$ only is
In a survey of $60$ people, it was found that $25$ people read newspaper $H , 26$ read newspaper $T, 26$ read newspaper $I, 9$ read both $H$ and $I, 11$ read both $H$ and $T,$ $8$ read both $T$ and $1,3$ read all three newspapers. Find:
the number of people who read exactly one newspaper.