Question

In a school $70\% $ students like oranges and $64\% $ like apples. If \[\] like both oranges and apples, then
a. $x \geqslant 34$
b. $x \leqslant 64$
c. $34 \leqslant x \leqslant 64$
d. $x \leqslant 70$

Answer 1

Hint: We will apply the formula of $n(A \cup B)$ here which is $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ where is $P(A \cup B)$ represents the specific value and $n(A \cap B)$ represents the common value among $n(A)$ and $n(B)$.

Complete step-by-step solution:
Given: A school is given where $70\% $ students like oranges and $64\% $ students like apples and $x\% $ students like both oranges and apples.
We have to find the value of \[\] according to the option given.
As we know $n\left( {A \cup B} \right) = n(A) + n(B) - n(A \cap B)$
Let us take $n(A) = $ oranges
And $n(B) = $ apples
$n(A \cup B) = $ the students who like at least one of the fruits.
$n(A \cap B) = $ no. of students who like both oranges and apples.
As given values, $n(A) = 70\% $
..
Now we will substitute the values in the formula.
$n(A \cap B) = 70\% + 64\% - x\% $
And we know $n(A \cup B) \leqslant 100\% $
That means $70\% + 64\% - x\% \leqslant 100\% $
$134\% - x\% \leqslant 100\% $
$ \Rightarrow x \geqslant 34$ --equation $\left( 1 \right)$
So the value of $x$ always equal or greater than $34$ but the common value cannot be greater than the smaller value of one of the specific values which means
$n(A \cap B) \leqslant n(A)\,{\text{or}}\,n(B)$
We will take the smaller value from the $n(A)$ and $n(B)$
So $x\% \leqslant 64\% $
$ \Rightarrow x \leqslant 64$ --equation $\left( 2 \right)$
So by equation $\left( 1 \right)$ and equation $\left( 2 \right)$ we get a result i.e. $34 \leqslant x \leqslant 64$
So option B is correct.

Note: In this type of question we always find the maximum possible value and also minimum possible value and the required value will always lie between the minimum possible value and maximum possible value.