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Mixture Alligation

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Mixture Alligation  [#permalink]

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New post 18 Jul 2017, 10:46
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Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C?

Please explain how to solve it by alligation method
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Re: Mixture Alligation  [#permalink]

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New post 18 Jul 2017, 15:41
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Sahil15 wrote:
Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C?

Please explain how to solve it by alligation method

Dear Sahil15,

I'm happy to respond. :-)

My friend, you are relatively new to GMAT Club. Welcome. A few comments.

This subforum is not really the appropriate forum for this particular forum. The "Ask GMAT Experts" subforum is for general questions--strategies, retake advice, rule changes on the GMAT itself, etc. What you have asked is an individual math question. An individual math question should be posted in the Quant subforum.

I will also say, it's a particularly bad approach to post a problem and say that you want to see it solved in a way that's familiar to you. Instead of clinging to the familiar, you should welcome as many different ways to think about a problem as possible. The whole point of the GMAT is to be intellectually flexible. Here's a blog you may find germane.
GMAT Solution and Mixing Problems

Here's how I would think about the problem. In what follows, I will use the elemental abbreviations (Copper = Cu, Tin = Sn)
Alloy A of 24 kilograms contains Copper and Tin in 1:7.
Alloy B of 28 kilograms contains Copper and Tin in 3:4.

Hold on to those two facts for a moment.
Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created.
Well, that's 6 kg of A and 12 kg of B. Let's look at those individual.

First, look at 6k of A. In Alloy A, Cu:Sn = 1:7, so copper is 1/8 of the sample. We know 1/8 of 6 kg would be 6/8 = 3/4 kg. That's the amount of copper, and tin would be the rest, 5.25 kg = 21/4 kg.

Now, look at 12 kg of B. In Alloy B, Cu:Sn = 3:4, so copper is 3/7 and tin is 4/7. These are the really ugly ratios in this problem.
amount of copper = 12(3/7) = 36/7 kg
amount of tin = 12(4/7) = 48/7 kg
I will leave those as improper fractions for the moment.

In Alloy C, the total amounts we have, in kg, are
amount of copper = 3/4 + 36/7 = 21/28 + 144/28 = 165/28
amount of tin = 21/4 + 48/7 = 147/28 + 192/28 = 339/28

Thus, in Alloy C, Cu:Sn = 165/339 = 55/113

That's a particularly ugly problem.

Does all this make sense?
Mike
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Re: Mixture Alligation  [#permalink]

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New post 19 Jul 2017, 02:30
Sahil15 wrote:
Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C?

Please explain how to solve it by alligation method


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Re: Mixture Alligation  [#permalink]

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New post 19 Jul 2017, 05:42
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Sahil15 wrote:
Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C?

Please explain how to solve it by alligation method


I can show you how to solve this using alligation (and the below might not make much sense to anyone reading this who doesn't know that method), but this isn't like a real GMAT weighted average problem. Alligation is normally really easy on a real GMAT problem, but here it's really messy because the fractions become a bit horrible to work with.

The actual amounts (24 kg, 28 kg, 18 kg) are completely irrelevant, so we can ignore those numbers. The first mixture is 1/8 copper, the second is 3/7 copper. When we mix them in a 1:2 ratio, we'll be finding the fraction that is 2/3 of the way along between 1/8 and 3/7, so we want the value of X on the number line below:

---1/8--------------X-------3/7---

Here the distance between 1/8 and X is twice as big as the distance between X and 3/7. In other words, this equation is true:

X - 1/8 = 2 * (3/7 - X)

and if you solve that, you'll find X. Or you can get the same denominator for both fractions, and then multiply by 3 on the top and bottom (so that the distance between them is easier to divide in a 2 to 1 ratio) to get this number line, which is identical to the one above but easier to work with:

---21/168-------------X-----72/168----

The total distance between 21 and 72 is 51, so 2/3 of that distance is 34, and X = 21/168 + 34/168 = 55/168.

X is the proportion of the final mixture that is copper, but we want the ratio of copper to tin, so the answer is 55 to 168-55, or 55 to 113.
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Re: Mixture Alligation  [#permalink]

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New post 20 Jul 2017, 07:15
IanStewart wrote:
Sahil15 wrote:
Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C?

Please explain how to solve it by alligation method


I can show you how to solve this using alligation (and the below might not make much sense to anyone reading this who doesn't know that method), but this isn't like a real GMAT weighted average problem. Alligation is normally really easy on a real GMAT problem, but here it's really messy because the fractions become a bit horrible to work with.

The actual amounts (24 kg, 28 kg, 18 kg) are completely irrelevant, so we can ignore those numbers. The first mixture is 1/8 copper, the second is 3/7 copper. When we mix them in a 1:2 ratio, we'll be finding the fraction that is 2/3 of the way along between 1/8 and 3/7, so we want the value of X on the number line below:

---1/8--------------X-------3/7---

Here the distance between 1/8 and X is twice as big as the distance between X and 3/7. In other words, this equation is true:

X - 1/8 = 2 * (3/7 - X)

and if you solve that, you'll find X. Or you can get the same denominator for both fractions, and then multiply by 3 on the top and bottom (so that the distance between them is easier to divide in a 2 to 1 ratio) to get this number line, which is identical to the one above but easier to work with:

---21/168-------------X-----72/168----

The total distance between 21 and 72 is 51, so 2/3 of that distance is 34, and X = 21/168 + 34/168 = 55/168.

X is the proportion of the final mixture that is copper, but we want the ratio of copper to tin, so the answer is 55 to 168-55, or 55 to 113.



Thank you so much for the answer :)
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Re: Mixture Alligation  [#permalink]

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New post 20 Jul 2017, 07:19
mikemcgarry wrote:
Sahil15 wrote:
Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C?

Please explain how to solve it by alligation method

Dear

I'm happy to respond. :-)

My friend, you are relatively new to GMAT Club. Welcome. A few comments.

This subforum is not really the appropriate forum for this particular forum. The "Ask GMAT Experts" subforum is for general questions--strategies, retake advice, rule changes on the GMAT itself, etc. What you have asked is an individual math question. An individual math question should be posted in the Quant subforum.

I will also say, it's a particularly bad approach to post a problem and say that you want to see it solved in a way that's familiar to you. Instead of clinging to the familiar, you should welcome as many different ways to think about a problem as possible. The whole point of the GMAT is to be intellectually flexible. Here's a blog you may find germane.


Here's how I would think about the problem. In what follows, I will use the elemental abbreviations (Copper = Cu, Tin = Sn)
Alloy A of 24 kilograms contains Copper and Tin in 1:7.
Alloy B of 28 kilograms contains Copper and Tin in 3:4.

Hold on to those two facts for a moment.
Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created.
Well, that's 6 kg of A and 12 kg of B. Let's look at those individual.

First, look at 6k of A. In Alloy A, Cu:Sn = 1:7, so copper is 1/8 of the sample. We know 1/8 of 6 kg would be 6/8 = 3/4 kg. That's the amount of copper, and tin would be the rest, 5.25 kg = 21/4 kg.

Now, look at 12 kg of B. In Alloy B, Cu:Sn = 3:4, so copper is 3/7 and tin is 4/7. These are the really ugly ratios in this problem.
amount of copper = 12(3/7) = 36/7 kg
amount of tin = 12(4/7) = 48/7 kg
I will leave those as improper fractions for the moment.

In Alloy C, the total amounts we have, in kg, are
amount of copper = 3/4 + 36/7 = 21/28 + 144/28 = 165/28
amount of tin = 21/4 + 48/7 = 147/28 + 192/28 = 339/28

Thus, in Alloy C, Cu:Sn = 165/339 = 55/113

That's a particularly ugly problem.

Does all this make sense?
Mike



Dear Mike,
Thank you for the reply.
Apologies for posting the question incorrectly.
I was able to get the answer in this way, but I was having problem with the alligation method hence asked for that.
I know both ways now. Thanks again :)
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Re: Mixture Alligation  [#permalink]

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New post 07 Oct 2018, 04:04
Sahil15 wrote:
Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C?

Please explain how to solve it by alligation method


Alloy C : 18 Kg contains \(\frac{1}{3}*18=6\) Kg Alloy A and \(\frac{2}{3}*18=12\) Kg Alloy B

Alloy A : 24 Kg contains \(\frac{1}{8}*24=3\) Kg Copper and \(\frac{7}{8}*24=21\) Kg Tin

6 Kg contains \(\frac{1}{8}*6=\frac{3}{4}\) Kg Copper and \(\frac{7}{8}*6=\frac{21}{4}\) Kg Tin

Alloy B : 28 Kg contains \(\frac{3}{7}*28=12\) Kg Copper and \(\frac{4}{7}*28=16\) Kg Tin

12 Kg contains \(\frac{3}{7}*12=\frac{36}{7}\) Kg Copper and \(\frac{4}{7}*12=\frac{48}{7}\) Kg Tin

Alloy C contains \(\frac{3}{4}+\frac{36}{7}\) Kg Copper and \(\frac{21}{4}+\frac{48}{7}\) Kg Tin.

Alloy C contains \(\frac{165}{28}\) Kg Copper and \(\frac{339}{28}\) Kg Tin.

Ratio of Copper and Tin in Alloy C \(=\frac{\frac{165}{28}}{\frac{339}{28}}=\frac{165}{339}=\frac{55}{113}\)
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Re: Mixture Alligation &nbs [#permalink] 07 Oct 2018, 04:04
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