Problem solving : profit and loss

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Problem solving : profit and loss

by vaibhav101 » Sun Jun 04, 2017 2:40 am
A dairyman pays $6.40 per litre of milk. He adds water and sells the mixture at $8 per litre, thereby making 37.5% profit. The proportion of water to milk received by he customers is
A 1:10
B 1:12
C 1:15
D 1:20
E none of these[/u][/i]

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by Jay@ManhattanReview » Sun Jun 04, 2017 3:11 am
vaibhav101 wrote:A dairyman pays $6.40 per litre of milk. He adds water and sells the mixture at $8 per litre, thereby making 37.5% profit. The proportion of water to milk received by he customers is
A 1:10
B 1:12
C 1:15
D 1:20
E none of these[/u][/i]
Say the proportion of water to milk received by he customers is W : M :: x : 1

Cost of water and milk = 8 / 137.5% = (8/137.5)*100 = 8000 / 1375 = 64/11

Again cost of water and milk = (6.4*1 + 0*x) / (1+x); Water is free of cost

Thus, 6.4 / (1+x) = 64/11

=> x = 1/10

Thus, the proportion of water to milk received by he customers is W : M :: (1/10) : 1 => W : M :: 1 : 10.

The correct answer: A

Hope this helps!

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by GMATGuruNY » Sun Jun 04, 2017 7:20 am
On the GMAT, answer choice E would be an actual ratio instead of "none of these."
For the reason, I've replaced option E with an actual ratio:
vaibhav101 wrote:A dairyman pays $6.40 per litre of milk. He adds water and sells the mixture at $8 per litre, thereby making 37.5% profit. The proportion of water to milk received by he customers is
A 1:10
B 1:12
C 1:15
D 1:20
E 1:25
An alternate approach is to PLUG IN THE ANSWERS, which represent the ratio of water to milk.
When the correct answer choice is plugged in, the profit must be 37.5%.

B: 1 liter water per 12 liters milk, for a total of 13 liters
At a price of $6.40 per liter, the total cost price for 12 liters of milk = $76.80.
At a selling price of $8 per liter, the total selling price for 13 liters of mixture = 8*13 = 104.
Profit = (selling price) - (cost price) = 104 - 76.80 = $27.20.
Profit/cost = (27.20)/(76.80) = 2720/7680 = 272/768 = 136/384 = 68/192 = 34/96 = a little more than 34%.
The profit is TOO LOW.
To increase the profit, LESS MILK must be mixed with 1 liter of water.

The correct answer is A.
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by [email protected] » Sun Jun 04, 2017 10:16 am
Hi vaibhav101,

This question can be approached in a couple of different ways, depending on how you 'see' the math involved.

We're told that milk is purchased for $6.40/liter, mixed with some water, then the mixture is resold at $8/liter. We're told that this process leads to a 37.5% profit.

37.5% = 3/8, so we're looking for total revenue that is $6.40 + (3/8)($6.40)....

1/8 of $6.40 = $0.80, so....
3/8 of $6.40 = $2.40

Thus, that one liter of milk (and whatever water is mixed in) has to generate $6.40 + $2.40 = $8.80 of revenue.

Comparing $8 and $8.80, you'll notice that the second number is exactly 10% greater than the first number. Thus, the amount of water that we need to add must be 10% of the volume of milk. A ratio of water:milk of 1:10 is needed.

Final Answer: A

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by Matt@VeritasPrep » Mon Jun 05, 2017 11:15 pm
If he's paying $6.40 per liter and making a 37.5% profit, he's selling each liter of milk for $8.80.

If he's selling a liter of milk and water for $8.00, then the mixture he's selling is $8/$8.80 milk, or 10/11 milk, so the ratio of milk to water is 10 : 1.

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vaibhav101 wrote:
Sun Jun 04, 2017 2:40 am
A dairyman pays $6.40 per litre of milk. He adds water and sells the mixture at $8 per litre, thereby making 37.5% profit. The proportion of water to milk received by he customers is
A 1:10
B 1:12
C 1:15
D 1:20
E none of these[/u][/i]
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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