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Could you please explain how to solve this question?

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Answer

by project800 » Tue Sep 05, 2006 2:16 am
I'm sure there are easier ways to do this but this is how I went about it:
15% of 4l is salt
Therefore, 15% of 4= .6l of the 4l solution is salt
Now 1.5 l spills
Therefore, 15% of 1.5=.225l of salt also spills out
Therefore, amount of salt remaining = .375l
Total amount of water in the jug is now 5l but amount of salt remains the same at .375l
Percentage of remaining salt = .375/5= 7.5%
Option B is the closest to the answer
any comments?
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Answer

by project800 » Tue Sep 05, 2006 2:20 am
I'm sure there are easier ways to do this but this is how I went about it:
15% of 4l is salt
Therefore, 15% of 4= .6l of the 4l solution is salt
Now 1.5 l spills
Therefore, 15% of 1.5=.225l of salt also spills out
Therefore, amount of salt remaining = .375l
Total amount of water in the jug is now 5l but amount of salt remains the same at .375l
Percentage of remaining salt = .375/5= 7.5%
Option B is the closest to the answer
any comments?
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by Bloomsbury » Tue Sep 05, 2006 3:38 am
Umm, The answer is A. But, I don't know whether it is the correct key.

Actually, I try the same method like your way.

Anyway, Thanks for your help.
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by project800 » Tue Sep 05, 2006 4:36 am
Another way to look at it:
The removal of 1.5l does not decrease the percentage of salt
Therefore, After 1.5l spills, the % of Salt stays at 15%
There remains 2.5l in the jug.
Since the capacity of the jug is 5l it would require the addition of another 2.5l of water to fill the jug completely.
Since no addition salt is added and exactly the same amount of water(2.5l) is added as there is solution in the jug(2.5l) the % of salt will he halved to 7.5%
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by diebeatsthegmat » Fri May 07, 2010 10:05 am
Bloomsbury wrote:Umm, The answer is A. But, I don't know whether it is the correct key.

Actually, I try the same method like your way.

Anyway, Thanks for your help.
i did the same and got the answer B but how come its A? why dont you update its explaination?
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