Answer:
The tank is 10cm high
Step-by-step explanation:
Given
[tex]L=60cm[/tex] -- length
[tex]W=60cm[/tex] -- width
[tex]x = \frac{1}{5}[/tex] --- water lever
[tex]Addition = 24L[/tex]
Required
The height of the tank
Let y represents the remaining fraction before water is added.
So:
[tex]y + x = 1[/tex]
Make y the subject
[tex]y = 1 - x[/tex]
[tex]y = 1 - \frac{1}{5}[/tex]
Solve
[tex]y = \frac{5 - 1}{5}[/tex]
[tex]y = \frac{4}{5}[/tex]
Represent the volume of the tank with v
So:
[tex]y * v = 24L[/tex]
Make v the subject
[tex]v = \frac{24L}{y}[/tex]
Substitute: [tex]y = \frac{4}{5}[/tex]
[tex]v = \frac{24L}{4/5}[/tex]
[tex]v = 30L[/tex]
Represent the height of the tank with h;
So, the volume of the tank is:
[tex]v = lwh[/tex]
Make h the subject
[tex]h = \frac{v}{lw}[/tex]
Substitute values for v, l and w
[tex]h = \frac{30L}{60cm * 50cm}[/tex]
Convert 30L to cm^3
[tex]h = \frac{30*1000cm^3}{60cm * 50cm}[/tex]
[tex]h = \frac{30000cm^3}{3000cm^2}[/tex]
[tex]h = 10cm[/tex]
