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mixture problems with more than 1 variable

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mixture problems with more than 1 variable

by topspin360 » Sat Jul 07, 2012 7:45 am
is there a standard framework we can use to solve mixture problems with more than 1 variable such as follows:

A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is 2 mm thick and its diameter is 15 mm). If the weight of the coin is 30 grams and the volume of aluminum in the alloy equals that of silver, what will be the weight of a coin measuring 1 x 30 mm made of pure aluminum if silver is twice as heavy as aluminum?
36 grams
40 grams
42 grams
48 grams
50 grams

Thanks.
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by eagleeye » Sat Jul 07, 2012 8:31 am
I am not sure about whether there is a standard way of doing 2 or more variable mixture problems. I would just try to evaluate each on a case by case basis.

Here are two ways of solving the question. One is algebraic and second is reasoning-based.

Algebraic:
Volume of first coin = V1= pi*2*(15/2)^2
Volume of second coin = V2 = pi*1*(30/2)^2
V2/V1 = (pi*1*(30/2)^2) / (pi*2*(15/2)^2) = 30*30*1/(15*15*2) = 2
V2=2V1
In the first coin, volume of aluminum = volume of silver = V1/2
Weight of silver / Weight of Aluminum = 2:1

Hence for the first coin, Weight of Aluminum in the coin = 1/3*30 = 10 grams.

So V1/2 weighs = 10 grams.
Second coin is made entirely of aluminum.
Hence V2 = 2*V1 = 2*(2*V1/2) = 4*(V1/2) must weigh 4*10 = 40 grams.

Here's the non-standard reasoning approach that I used to solve the question without doing any pen paper calculation:

Volume of first one comes from 2*15. Second one is from 1*30. Volume is proportional to height and proportional to square of the diameter. Since 30 is twice 15, and height is halved, volume of second one must be 2^2*1/2 = twice that of first one.
Now first one has equal volume of silver and aluminum, and silver is twice as heavy as aluminum. Since they have equal volumes, silver is twice as heavy as aluminum, and total weight is 30, weight of aluminum must be 10 and that of silver must be 20.

So, aluminum's weight in half of first coin = 10 grams. If the whole of first coin were made of aluminum, it would weigh = 20 grams. Since the volume of second coin is double that of the first, it must weigh twice of 20 = 40 grams.
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by GMATGuruNY » Sat Jul 07, 2012 2:10 pm
topspin360 wrote:is there a standard framework we can use to solve mixture problems with more than 1 variable such as follows:

A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is 2 mm thick and its diameter is 15 mm). If the weight of the coin is 30 grams and the volume of aluminum in the alloy equals that of silver, what will be the weight of a coin measuring 1 x 30 mm made of pure aluminum if silver is twice as heavy as aluminum?
36 grams
40 grams
42 grams
48 grams
50 grams

Thanks.
This is more of a PROPORTION problem than a mixture problem.

Coin made of alloy:
Volume = �r²h = �(15²)(2) = 450�.
The total weight = 30 grams.
Since silver weighs twice as much as aluminum, silver = 20 grams and aluminum = 10 grams.

Since the volume of each metal is the same, if the silver -- with a weight of 20 grams -- is replaced with aluminum -- with a weight of 10 grams -- the total weight of the resulting coin will be 20 grams.
Thus, the weight of an aluminum coin with a volume of 450� = 20 grams.

Pure aluminum coin:
V = �r²h = �(30²)(1) = 900�.
Since the weight of an aluminum coin with a volume of 450� = 20 grams, the weight of an aluminum coin that is twice as big (with a volume of 900�) = 40 grams.

The correct answer is B.
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