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Director
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A mixture contain 17% spirit and rest water. 10 liters are [#permalink]
 01 Mar 2005, 21:48
A mixture contain 17% spirit and rest water. 10 liters are drawn and the vessel is filled up with water. The proportion of spirit now is 15 1/9%. Whats the vessel capacity?
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Intern
Joined: 17 Feb 2005
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Original volume of spirit = 17V/100
After removing 10 L, it becomes 17(V-10)/100
so 17(V-10)/100=136V/(9*100)
Solve for V=90L
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Director
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Let x be the volume of the vessel
(17x/100 -1.7)/x = 136x/9*100
solving x = 90
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VP
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Well another way to think about the problem is this...
If after 10L of water is drawn out of the mixture, the proportion of spirit is now 15 1/9%, then 17% - 15 1/9% = 1 8/9% = (17/9)%
This further implies that every 10L of the mixture contained (17/9)% of spirit.
17 divide by (17/9) = 9, 9 * 10L = 90L
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Manager
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Assuming that the vessel was initially full.
And it's volume is x
17/100(x-10)=136/900(x)
x=90 l
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