Cruz emptied 20% of a 5-liter container of antifreeze solution with a 60% concentration of antifreeze. How many liters of pure water must he add so that the new solution will have a 50% concentration of antifreeze?

 

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1 Answer

Since, 20% of 5 litre = 1 litre  or 1000 ml of chemical was emptied. so 4 litres or 4000 ml is left.

And 60% of 4000 ml of antifreeze = 2400 ml

So percent of water in the chemical = 4000 - 2400 = 1600 ml

Now,  Amount of antifreeze in final mixture is 50%

So, Amount of water in final mixture is also 50%

Now, let the X amount of pure water must be added.

So, on taking ratio of antifreeze to water we get

2400/(1600+X) = 50/50

=> 2400/(1600+X) = 1

=> 2400 = 1600 +X

=> X = 800 ml or 4/5 litres.

 

Another method,

Since,

Now 2400 is 60% of 4000 ml solution.

And antifreeze is 50% of Y ml solution, where Y is new total of antifreeze and water, after water being added to the solution

So,

Y * 50% = 2400

=> Y = 2400 * 100 / 50 = 4800

Therefore, the amount of water added to the solution = final solution - initial solution = 4800 - 4000 = 800ml or 4/5 litres

 

Hope this is what you want
by Level 4 User (6.1k points)
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