does anyone know the answer to this?

Question

The cooling system of a car has a capacity of 25 liters. If the system is filled with a mixture that is 60% antifreeze, how much of this mixture should be drained and replaced by pure antifreeze so that the system is filled with 90% antifreeze?

Answer 1

25 liters of 60% antifreeze means you have...

60%*25=15 L of antifreeze and 25-15=10 L of water

you want 25 L of 90% antifreeze which means you want 90%*25=22.5 L of antifreeze and 2.5 L of water

you need to remove 10-2.5=7.5 L of water

the mixture is 40% water

let x=amount of mixture to be removed, 40% of which is water

40% of x=7.5

0.4x=7.5

4x=75

x=75/4

x=18.75 L of mixture must be removed and replaced with 18.75 L of pure antifreeze

check:

60% of 18.75=0.6*18.75=11.25 L of antifreeze is removed

you originally had 15 L of antifreeze and now you have 15-11.25=3.75 L of antifreeze left

you need to have 22.5 L of antifreeze, so 22.5-3.75=18.75 L of pure antifreeze must be added

Answer 2

Let x = number of liters that is drained and replaced

 

We start with 25 liters of solution that is 60% antifreeze.  We then drain x liters and replace it with a x liters that is 100% antifreeze.  Our end product consists of 25 liters of solution that is 90% antifreeze.

 

The antifreeze content of a given solution is found by multiplying the % antifreeze by the number of liters of solution.

 

Antifreeze content of original solution = 0.60(25)

Antifreeze content of drained solution = 0.60x

Antifreeze content of replaced solution = 1.00x

Antifreeze content of end product = 0.90(25)

 

So, 0.60(25) - 0.60x + 1.00x = 0.90(25)

 

             15 + 0.40x = 22.5

 

                    0.40x = 7.5

 

                         x = 18.75 liters

 

Replace 18.75 liters of the original 60% antifreeze solution with the same amount of pure antifreeze to obtain 25liters of 90% antifreeze solution.