Question

3. A car's radiator can hold 12 tons of an inte solution (mix of water and purecoolant) at a 25% concentration,How much of the existing solution must be drained from the radiator and replacedwith pure coolant to bring the concentration back to the manutadturer'snecommendation of 50% water and 50% pure coolant?You anly have the ability to measure in US liquid igallons, se report your answer ingallons. Assume all values are exact.Enter your answer to the thousandthis place Do not indude units in your answer.

Answer 1

First of all we need to make a draw to understand the problem:

It means a 25% of concentration in all capacity of the car's radiator.

Now we need to make a equation that represents the conditions of the problem.

Let x = amount of 25% solution drained and amount of pure antifreeze to be added:

0.25*(12-x) +x: represents the quantity of tons to be drained of 25% (25%=0.25) and the quantity of coolant in tons to be added (the last x added)

The last equation will be equal to 12 tons of mix to 50% (50% = 0.5), it means:

0.50*(12)

Finally, we solve the system of equations:

`\begin{gathered} 0.25(12-x)+x=0.50(12) \\ 0.25(12)-0.25x+x=6 \\ 3+0.75x=6 \\ 0.75x=6-3 \\ 0.75x=3 \\ x=(3)/(0.75)=4 \\ x=4 \end{gathered}`

Then, the answer will be that you need drain 4 tons of mix at 25% and add 4 tons of pure coolant.

Now, if you need convert tons to gallons or liters.. you need to know that:

1 ton = 31.75471 gallon

4 tons = 4*(31.75471) = 127.01884, we can approximate to 127 gallons

Then you answer in gallons will be: you need drain 127 gallons of mix at 25% and add 127 gallons of pure coolant.