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A 5 litre jug contains 4 litre of a saltwater solution that

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Senior Manager
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A 5 litre jug contains 4 litre of a saltwater solution that [#permalink] New post 09 Aug 2006, 02:54
A 5 litre jug contains 4 litre of a saltwater solution that is 15% salt. If 1.5 litre of the solution spills out of the jug and the jug is then filled to capacity with water, approximately what percent of the resulting solution in the jug is salt?
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 [#permalink] New post 09 Aug 2006, 08:43
Answer 30 %

3/5 salt , 2/5 water .. So 4-1.5 = 2.5 remains..

off this 1.5 l is salt and 1 is water. Additional 2.5 water is added . Therefore
1.5 / 5.0 = 30%
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 [#permalink] New post 09 Aug 2006, 08:54
7.5%

4L contains 15/100 salt i.e. 0.6l salt
2.5L will contain x

i.e. x = (2.5 * 0.6) / 4 = 1.5/4 = 3/8

Hence after adding 2.5L water, %age of salt = (3/8)/5 = 3/40 = 3/40 * 100 %= 7.5%
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 [#permalink] New post 09 Aug 2006, 09:15
Answer: 7.5%

Salt in 4 lts = 4x15/100 = 0.600 lts

Salt in 1.5 lts = 1.5x15/100 = 0.225

Salt remaining = 0.6-0.225=0.375

%tage = 0.375x100/5 = 7.5%
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 [#permalink] New post 09 Aug 2006, 09:18
After spill solution is still 15%
salt remaining = 2.5 * 15/100 = 3/8
After adding water total solution is 5 litres.

Percentage of salt = (3/8) * 1/5 * 100 = 7.5%
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 [#permalink] New post 09 Aug 2006, 11:37
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The answer is correct with all the math above, but, for any question, i think it helps to get down to what you are actually 'doing' in the situation at hand, independent of the general methods used to solve "this type of problem". this is probably where the biggest time savers are found.

notice here that the 2.5l left after the spill is HALF the volume of the 5l jug.

forget the 15% for a second.

if you have a glass half full of vodka (or empty, i guess that depends on how you feel about vodka), how much of it is vodka? trivially, "all of it." now, top off the glass with tonic. how much of it is vodka? "Half of it." You've put the same amount of vodka into twice the space- you've halved the concentration _no matter what it was initially_.

here, it was initially 15% salt. so it becomes (15/2 = 7.5)% salt.

if the question were more complicated, say there was some salt in the solution you added after the spill, then you might need to do some more math, but the same principle will still apply.

for example, what if the bottle, full with 2.5l of a 15% salt solution, was topped-off to 5l with a 20% salt solution?

very quickly, 17.5%.

why? the total concentration is the avearage of the two contitiuent concentrations because they contribute equally to the total volume.

(15+20/2)%=17.5%

this extends back to the original example where

(15/2 = 7.5)% == (15+0/2)%=7.5%

again, we are lucky here that the begin volume is half the end volume or else we would have to revert to the longer methods used above, but when a problem tosses you a fortunate scenario such as this, it usually pays to lever it in your solution.
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 [#permalink] New post 09 Aug 2006, 11:50
7.5% it is...

15% of 2.5lt is 0.375
=> it is 7.5% of 5lt (Since 0.375 remains constant...)
  [#permalink] 09 Aug 2006, 11:50
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