qlx wrote:
A dairyman pays Rs. 6.40 per liter of milk. He adds water and sells the mixture at Rs. 8per liter, thereby making 37.5% profit. The proportion of water to milk received by the customers is
(a) 1:10
(b) 1:12
(c) 1:15
(d) 1:20
(e) 1:25
Solution:We can let x = the amount of water in 1 liter of the mixture and kx = the amount of milk in 1 liter of the mixture (notice that the ratio of water to milk is x : kx = 1 : k and we need to determine the value of k).
Since x + kx = 1, we have:
x(1 + k) = 1
x = 1/(1 + k)
We see that kx = k/(1 + k) liter of milk is in 1 liter of the mixture. Since the profit margin is 37.5% or 3/8 and assuming water incurs no cost, we can create the equation:
8 - k/(1 + k) * 6.4 = k/(1 + k) * 6.4 * 3/8
8 = 6.4k/(1 + k) * 3/8 + 6.4k(1 + k)
8 = 6.4k/(1 + k) * (3/8 + 1)
80 = 64k/(1 + k) * 11/8
640/11 = 64k/(1 + k)
10/11 = k/(1 + k)
We see that k must be 10.
Alternate Solution:If the dairyman sold 1 liter of (pure) milk at a 37.5% profit, he would sell it for:
6.40 + (6.40)(0.375) = $8.80
But he is adding in some amount of water such that he can still generate 37.5% profit by selling one liter of a water/milk mixture for $8.00.
Since 8.00/8.80 = 80/88 = 10/11, we see that he is selling 10/11 liter of milk and 1/11 liter of water in each liter of milk/water mixture. Thus, the ratio of milk to water is 10 : 1, and the ratio of water to milk is 1 : 10.
Answer: A
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