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A dairy man pays $ 6.4 per litre of milk; he adds water and
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
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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.
Just to add my approach.
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
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.
Well, What can I say. I'm an addict, now! :drunk
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 :drunk
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.