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?
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 \(y=24\).