#
Problem
of the Week

Problem
E and Solution

Pineapples
and Bananas

## Problem

In a recent survey, Grade \(12\)
students were asked if they like pineapples. They were then asked if
they like bananas. It was found that

\(30\%\) of the students do not
like pineapples,

\(36\) students do not like
bananas,

\(60\) students like both
fruits, and

\(48\) students like one fruit
but not the other.

How many students do not like pineapples and do not like bananas?

## Solution

Let \(x\) be the number of students
that like only pineapples.

Let \(y\) be the number of students
that like only bananas.

Let \(z\) be the number of students
that do not like pineapples and do not like bananas.

We’re given that \(60\) students like
both pineapples and bananas.

This information is summarized in the Venn diagram.

The number of students that do not like bananas is equal to \(x+z\). Therefore, \[x+z=36\tag{1}\] The number of students who
like one fruit but not the other is equal to \(x+y\). Therefore, \[x+y=48\tag{2}\] The total number of
students is equal to \(x+y+z+60\) and
\(30\%\) of this is \(0.3(x+y+z+60)\). This is also equal to the
number of students that do not like pineapples, \(y+z\). Therefore, \[0.3(x+y+z+60)= y+z\tag{3}\]

Subtracting equation \((1)\) from
equation \((2)\), we get \(y-z=12\). Therefore, \(y = z+ 12\).

Substituting equation \((1)\) into
equation \((3)\) we get \(0.3(36 + y +60)= y+z\), or \(0.3(96 + y)= y+z\).

Now substituting \(y=z+12\), we get
\[\begin{aligned}
0.3(96+ (z+12))&= (z+12)+z\\
28.8 + 0.3z + 3.6 &= 2z + 12\\
20.4 &= 1.7z\\
z&=12
\end{aligned}\] Therefore, there are \(12\) students who do not like pineapples
and do not like bananas.

Although we are not asked to do so, we could go on and solve for
\(x=24\) and \(y=24\).