Vemuri wrote:scoobydooby wrote:let there be 100 litres to start with. volume of acid: 50 ltrs.
since equal amount removed and replaced, total volume remains intact.
let x be the fraction removed and replaced.
(50-50x+30x)/100=40/100
=>50-20x=40
=>x=1/2
hence, C
Hi Scoobydooby, can you explain how you arrived at the equation. I think I am having the usual trouble understanding what the question is asking...in this case I think I am confused with the removed and replaced concept.[/quot
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Vemuri another way to see this is solve by intuition if the algebra is confusing. Imagine you have a container of the solution. Since the problem says Salt is 50%. Make the tolal Volume 10 and the salt valume 5. So your original salt fraction is
5/10 ( Pure Salt/ Total Volume)( don't simplify so that it feels more natural).
Now if you remove any liquid from this container, you know you will have to remove it in the appropriate ratio. If you take 1 Litre for example, YOu are not removing 1 litre of pure salt but 1 litre of salt and somthing else. You need the pure salt part for your numerator. This is 1/2 liter of salt since that is the original ratio for salt. So if you remove x liter from your 10 liter you are removing x/2 liter of pure salt. Your new fraction after removal looks like:
(5-x/2)/10-x.
The problem doesn't tell you what this fraction is. It tells you to add x back. But again when you add x back you are adding in the APPROPRIATE RATIOS. This time they give you the ratio as 3x/10 ( This is pure salt) . The other 7x/10 goes to the other constituent/s which we don't care about. Your fraction after adding x back looks like
(5-x/2+3x/10)/10-x+x. The problem gives this as 2/5 or 40%.
50-5x +3x =40
x=5. Don't panic. This is not your fraction but how much you took out.
The appropriate fraction is 5/10= 1/2.
Note as an aside that when you did your first removal you only removed 5/2 litres of salt.
