Volume in problem situations
Often volume questions at this level are set in problem situations.
Example
Five litres of white paint are to be mixed with two and a half litres of green paint.
The mixture is to be poured into cylindrical jars which each have a base radius of \(2.4cm\) and a height of \(12.1cm\)
There is to be a space of \(3cm\) at the top of each jar above the level of the paint.
Question
How many jars can be fully filled in this way?
Each jar is a cylinder. Volume of cylinder\(=\pi\,r^{2}h \)
The height of the paint in each cylinder \(= 12.1- 3 = 9.1cm\)
Volume of paint in each cylinder \(=\pi\times2.4\times2.4\times9.1=164.59cm^{3} \)
A jar of volume \(164.59cm^{3}\) holds \(164.59 ml\).
Total volume of paint \(= 5 + 2.5 = 7.5\,l = 7\,500ml\)
Number of jars to use all of the paint \(= 7\,500\div 164.59 = 45.57\) jars (to \(2\,d.p\))
Number of full jars \(= 45\)
Question
How much paint will be left over?
There is \(0.57\) of a jar of paint left over.
One jar holds \(164.59ml\)
Amount of paint left over \(= 164.59 \times 0.57 = 93.81ml\)(to \(2\,d.p\))
(This could also be done by multiplying \(45\) and \(164.59ml\) then subtracting the result from \(7\,500ml\) to give \(93.45ml\). The slight difference of \(0.36ml\) is due to rounding but both methods are allowed as long as the method that you used is shown in your answer.)