Question

on a very hot day, david noticed that his swimming pool has lost a lot of water through evaporation. the water had dropped 1 cm during the day. at the start of the day, the pool was 1.25m deep. it is a rectangular pool, 4.5m wide and 8.5m long.
calculate the volume of water at the start of the day
calculate the volume of water at the end of the day
how much water had been lost
if the concentration of chlorine in the pool at the start of the day was 14 mg/L, what would the concentration be at the end of the day? give your answer to 2 decimal points

Answer 1

original volume = 1.25*4.5*8.5 m^3 = 47.8125 m^3
new volume = 1.24*4.5*8.5 = 47.43 m^3

loss = the difference between the 2 volumes above.

The rest of the question will depend on the physics of the situation, not
sure if the evaporation would be purely in water or if the chlorine would
also evaporate.
To make this a suitable math question, let's assume that only water
evaporates.
there are 1000 litres in 1 m^3
so originally you had 47,812.5 litres in the pool.
at 14mg/l, you had 14*47812.5 mg of chlorine
to get the new concentration, divide that by the new volume
I got 14.11 gm/L

to make a long story short, we could have just set up a simple ...
1.25 / 1.24 * 14 = 14.11