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Director
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sells the mixture at $8 per litre, thereby making 37.5% profit. Find the proportion of water to milk received by the customers. 1) 1 : 15 2) 1 : 10 3) 1 : 20 4) 1 : 12 Senior Manager Joined: 02 Mar 2004 Posts: 327 Location: There Followers: 1 Kudos [?]: 0 [0], given: 0 ### Show Tags 23 Mar 2004, 01:24$0-------------------------------$64/11-----------------------------$6.4

(64/10)-(64/11) = 64/110 64/11 - 0 = 64/11

64/110 : 64/11 = 1:10
Director
Joined: 13 Nov 2003
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23 Mar 2004, 11:19
halle, I bow before you. Perfect.
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23 Mar 2004, 13:23
8*(m+w) = 1.375 * 6.4 * m
8w = m( 6.4 * (1.25 + 0.125) - 8 )
w/m = 0.8(1.25+0.125) - 1 = 0.1
so w:m = 1:10

I was trying not to solve quant problems for some time and concentrate on verbal. But kpadma dragged me into this. I think it is all about addiction. I wanr everyone about this addiction and suggest concentrating on certain known weak points.

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23 Mar 2004, 13:54
Price you would have to sell at in order to get 37.5% profit: 1.375*6.4$= 8.8$
In order for you to still keep a 37.5% profit while selling the diluted milk at 8$, then the .8$ difference must represent the water proportion. Then water to milk: .8$/ 8$ = 1 : 10
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Best Regards,

Paul

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23 Mar 2004, 14:37
anandnk wrote:

Well, What can I say. I'm an addict, now!

Here is another method!

Customer gets 1 liter for $8 Vender pays (1-x)*$6.4 for that adulterated 1 liter milk and makes
37.5% profit.

Where X = water in 1 liter adulterated milk

6.4(1-x) * 1.375 = 8

Solve for X, we get 1/11.
Pure milk in 1 liter adultrated milk is 10/11

W:M = 1/11 : 10/11 = 1:10

Well, time for another round
23 Mar 2004, 14:37
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