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milk and water

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by coolhabhi » Sun Dec 01, 2013 1:44 pm
A dishonest milkman mixes water and milk and claims to sell the milk at cost price, thereby making a profit of 25%. How much water is there in one litre that he delivers to the customers?

A)200ml
B)250ml
C)180ml
D)166.66ml

I am struck between A and B
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by yash.522 » Sun Dec 01, 2013 4:37 pm
My answer "B".
Assume cost price = 100 for 1 litre milk.
Mix 750ml milk & 250ml water to make 1 litre...sell for 100 => 25% profit
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by Brent@GMATPrepNow » Sun Dec 01, 2013 4:55 pm
coolhabhi wrote:A dishonest milkman mixes water and milk and claims to sell the milk at cost price, thereby making a profit of 25%. How much water is there in one litre that he delivers to the customers?

A)200ml
B)250ml
C)180ml
D)166.66ml

I am struck between A and B
Let's test the answers.

To begin, let's say that the milkman pays $1 per liter of milk.
In other words, the milkman pays $0.10 for every 100 ml of milk.

Since the milkman CLAIMS to sell the milk at cost, he must charge $1 for a liter of milk.


Answer choice A.
If there's 200 ml of water in the mixture, then there must be 800ml of milk
The 800ml of milk would have cost the milkman $0.80
So, if the milkman charges $1 for the liter of diluted milk, his profit is $0.20
This represents a 25% profit

Answer: A

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Brent
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by Matt@VeritasPrep » Sun Dec 01, 2013 9:22 pm
I agree with Brent on this one - the trick is to remember that percentages do different things going up and going down! If you take 1000 ml and subtract 25%, you're left with 750 ml ... but you aren't going down from 1000 ml, you're going UP from the amount of milk that's actually in the mixture, so you need to add 25% to your (unknown) base. So the equation should be

x ml + 25% of x ml = 1000 ml

or

1.25x = 1000

or

x = 800

So there are 800 ml of milk in the mix (and hence 200 ml of water).
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by coolhabhi » Sun Dec 01, 2013 10:55 pm
Brent

if the milkman charges $1 for the liter of diluted milk, his profit is $0.20
This represents a 25% profit

Here $0.2 of $1 or $0.2 of $0.75??
I am still unable to get it.. Could you please elaborate.

I did it this way..
Suppose 1000ml of milk costs him 1000
Now (Taking option B) he is selling 750 ml of milk at the same 1000 (Which should have costed him 750)

So his profit is 250/1000 = 25%.
Hence option B
What is wrong in my approach?
Brent@GMATPrepNow wrote:
coolhabhi wrote:A dishonest milkman mixes water and milk and claims to sell the milk at cost price, thereby making a profit of 25%. How much water is there in one litre that he delivers to the customers?

A)200ml
B)250ml
C)180ml
D)166.66ml

I am struck between A and B
Let's test the answers.

To begin, let's say that the milkman pays $1 per liter of milk.
In other words, the milkman pays $0.10 for every 100 ml of milk.

Since the milkman CLAIMS to sell the milk at cost, he must charge $1 for a liter of milk.


Answer choice A.
If there's 200 ml of water in the mixture, then there must be 800ml of milk
The 800ml of milk would have cost the milkman $0.80
So, if the milkman charges $1 for the liter of diluted milk, his profit is $0.20
This represents a 25% profit

Answer: A

Cheers,
Brent
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by coolhabhi » Sun Dec 01, 2013 10:59 pm
Matt I am sorry but your words are tricky

I really did not understand why we need to add 25% to the unknown base. Please Explain.

Matt@VeritasPrep wrote:I agree with Brent on this one - the trick is to remember that percentages do different things going up and going down! If you take 1000 ml and subtract 25%, you're left with 750 ml ... but you aren't going down from 1000 ml, you're going UP from the amount of milk that's actually in the mixture, so you need to add 25% to your (unknown) base. So the equation should be

x ml + 25% of x ml = 1000 ml

or

1.25x = 1000

or

x = 800

So there are 800 ml of milk in the mix (and hence 200 ml of water).
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by Matt@VeritasPrep » Sun Dec 01, 2013 11:45 pm
coolhabhi wrote:Matt I am sorry but your words are tricky

I really did not understand why we need to add 25% to the unknown base. Please Explain.
Sure! Let's try it this way.

If I buy something for $100 and sell it for a profit of 25%, I sell it for $125.

If I buy something for $125 and sell it for a loss of 25%, I sell it for $93.75.

As you can see from those two examples, going up 25% to $125 is different from going down 25% from $125: if I go DOWN 25% from $125, I DON'T arrive back at $100!

In this problem, the milkman is making 25% MORE than his cost, so we have to go up 25% from his cost (which we don't know, so we have to call it x) to arrive at the result (1000). We can't start at 1000 and deduct 25% of 1000, since 1000 is our result, NOT our base.

Does that make more sense?
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by Mathsbuddy » Mon Dec 02, 2013 6:46 am
Another way to look at the problem, is this:
As we aren't given any prices, we can choose whatever price that is most convenient for us.

So let's imagine that milk is sold at $1.25 per litre. This figure comfortably accommodates the 25% hike in price.

Then the proportion of milk (to milk and water) is 1/1.25 = 0.8
Hence 0.2 of a litre is water

Answer = A)200ml
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by Brent@GMATPrepNow » Mon Dec 02, 2013 8:28 am
coolhabhi wrote:Brent

if the milkman charges $1 for the liter of diluted milk, his profit is $0.20
This represents a 25% profit

Here $0.2 of $1 or $0.2 of $0.75??
I am still unable to get it.. Could you please elaborate.

I did it this way..
Suppose 1000ml of milk costs him 1000
Now (Taking option B) he is selling 750 ml of milk at the same 1000 (Which should have costed him 750)

So his profit is 250/1000 = 25%.
Hence option B
What is wrong in my approach?
% profit = 100(selling price - original price)/(original price)
So, in my solution, % profit = 100($1 - $0.80)/($0.80)
= 25%

Let's examine your solution.
750 ml of milk cost the milkman $0.75 (so, original price = $0.75)
If he sells that milk (plus 250ml of water) for $1.00, then we get...
% profit = 100(selling price - original price)/(original price)
% profit = 100($1.00 - $0.75)/($0.75)
= 100(0.25)/(0.75)
= 33 1/3%
We want a 25% profit.

Cheers,
Brent
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by Mathsbuddy » Mon Dec 02, 2013 8:44 am
yash.522 wrote:My answer "B".
Assume cost price = 100 for 1 litre milk.
Mix 750ml milk & 250ml water to make 1 litre...sell for 100 => 25% profit
If it costs $100 per litre, he pays $75 for 750ml and adds 250ml of free water to make a diluted litre.

25% of $75 = $18.75
$75 + $18.75 = $93.75 which is not the original $100.


Therefore it can't be 250ml.

Does that help?
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by Mo2men » Fri Jan 18, 2019 12:01 am
Matt@VeritasPrep wrote:I agree with Brent on this one - the trick is to remember that percentages do different things going up and going down! If you take 1000 ml and subtract 25%, you're left with 750 ml ... but you aren't going down from 1000 ml, you're going UP from the amount of milk that's actually in the mixture, so you need to add 25% to your (unknown) base. So the equation should be

x ml + 25% of x ml = 1000 ml

or

1.25x = 1000

or

x = 800

So there are 800 ml of milk in the mix (and hence 200 ml of water).

Hi Matt,

I'm really interested in your solution. But I want to know its logic.

You multiplied the 25% by the quantity while it is about profit and money. Why? I do not understand the relation.

Thanks in advance
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