its nedded to have 40% antifreez solution in the radiator

Question

Of a car.  The radiator now has 50 liters of 20% solution.  How many liters should be drained and replaced with 100% antifreez to get the desired strength?

Answer 1

you have 50 L of a 20% solution of antifreeze so 80% is water

you want a 40% solution

40% of 50=20

you need 20 L of antifreeze to have a 40% solution

you already have 10 L, so you need 10 more L of pure antifreeze

every 1 L of solution contains 0.2 L of antifreeze and 0.8 L of water (0.2+0.8=1)

every 1 L that you drain you drain 0.2 L of antifreeze

therefore when you replace that 1 L of solution with pure antifreeze you are only putting in 0.8 L of antifreeze because

0.2 L is being drained (1-0.2=0.8)

again you need 10 more L of pure antifreeze

again draining and replacing 1 L only gives you 0.8 L of antifreeze

how many drainings and replacements do you need to get a total of 10 L of antifreeze ??

how many times does 0.8 go into 10 ??

10/0.8=100/8=12.5

you need to drain 12.5 L of antifreeze SOLUTION and replace it with 12.5 L of pure antifreeze because you lost 12.5*0.2=2.5 L in the draining so you have only 10-2.5=7.5 L of antifreeze

when you add the 12.5 L of pure antifreeze you will now have 7.5+12.5=20 L of antifreeze which is a 40% solution

20/50=40/100=40%