
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. Then, which of the following is/are true?
1. $150$ students were taking at least one juice.
2. $225$ students were taking neither apple juice nor orange juice.
A. Only (1) is true
B. Only (2) is true
C. Both (1) and (2) are true
D. None of these
Answer
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Hint: A set is a well-defined group of items. In contrast to integers, we can define and investigate the properties of sets. An operation in set theory is an exercise in combining different sets so that a new set with distinct features is obtained. In this question, for first part apply the concept of union and intersection of sets and for second complement of sets.
Formula Used:
Sets formula –
$n\left( {X \cup Y} \right) = n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right)$
Complement of the set –
$n{\left( A \right)^C} = n\left( U \right) - n\left( A \right)$
Complete step by step solution:
Let, $U$ be the set of total number of students in a survey
$X$ be the set of students taking apple juice and $Y$ taking orange juice
$ \Rightarrow n\left( U \right) = 400,n\left( X \right) = 100,n\left( Y \right) = 150,n\left( {X \cap Y} \right) = 75$
Using, $n\left( {X \cup Y} \right) = n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right)$
$n\left( {X \cup Y} \right) = 100 + 150 - 75$
$n\left( {X \cup Y} \right) = 175$
Which implies that $175$ students were taking at least one juice.
Now, $n\left( {{X^C} \cap {Y^C}} \right) = n{\left( {X \cup Y} \right)^C}$
$n\left( {{X^C} \cap {Y^C}} \right) = n\left( U \right) - n\left( {X \cup Y} \right)$
$n\left( {{X^C} \cap {Y^C}} \right) = 400 - 175$
$n\left( {{X^C} \cap {Y^C}} \right) = 225$
$ \Rightarrow 225$ students were taking neither apple nor orange juice.
Option ‘C’ is correct
Note: The key concept involved in solving this problem is the good knowledge of Sets. Students must remember that the union of two sets $A$ and $B$ is the set of all elements that are either in $A$ or in $B$, i.e. $A \cup B$, whereas the intersection of two sets $A$ and $B$ is the set of all common elements. The complement of a set is the set that contains all of the universal set's elements that are not present in the provided set.
Formula Used:
Sets formula –
$n\left( {X \cup Y} \right) = n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right)$
Complement of the set –
$n{\left( A \right)^C} = n\left( U \right) - n\left( A \right)$
Complete step by step solution:
Let, $U$ be the set of total number of students in a survey
$X$ be the set of students taking apple juice and $Y$ taking orange juice
$ \Rightarrow n\left( U \right) = 400,n\left( X \right) = 100,n\left( Y \right) = 150,n\left( {X \cap Y} \right) = 75$
Using, $n\left( {X \cup Y} \right) = n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right)$
$n\left( {X \cup Y} \right) = 100 + 150 - 75$
$n\left( {X \cup Y} \right) = 175$
Which implies that $175$ students were taking at least one juice.
Now, $n\left( {{X^C} \cap {Y^C}} \right) = n{\left( {X \cup Y} \right)^C}$
$n\left( {{X^C} \cap {Y^C}} \right) = n\left( U \right) - n\left( {X \cup Y} \right)$
$n\left( {{X^C} \cap {Y^C}} \right) = 400 - 175$
$n\left( {{X^C} \cap {Y^C}} \right) = 225$
$ \Rightarrow 225$ students were taking neither apple nor orange juice.
Option ‘C’ is correct
Note: The key concept involved in solving this problem is the good knowledge of Sets. Students must remember that the union of two sets $A$ and $B$ is the set of all elements that are either in $A$ or in $B$, i.e. $A \cup B$, whereas the intersection of two sets $A$ and $B$ is the set of all common elements. The complement of a set is the set that contains all of the universal set's elements that are not present in the provided set.
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