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Mixture Alligation

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Mixture Alligation  [#permalink]

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New post 18 Jul 2017, 11:46
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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, 16:41
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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, 03:30
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


Moved to Quantitative forum.

P.S. Please read carefully and follow our RULES OF POSTING. Thank you.
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Re: Mixture Alligation  [#permalink]

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New post 19 Jul 2017, 06:42
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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, 08:15
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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New post 20 Jul 2017, 08:19
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, 05:04
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  [#permalink]

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New post 12 Jul 2019, 05:42
This Question can be done much easily , just ignore the kgs of a particular mixtures and imagine if both of them are mixed with these proportions then whatever i will take from resultant mixture either 18kg or x kg it will always be in that same proportion.

Now A is given in 1:7 therefore let’s assume it is 1+7 = 8 units (you can assume it as 8x as well )
B is 3:4 therefore (3+4 = 7 units ) now let’s make them equal solutions and then add in ratios of 1:2

Therefor let’s multiply the 1st ratio by 7 and 2nd one by 8 making them both equal to 56 units and then again multiplying 2nd by 2 as the need to be mixed in 1:2
This will give me resultant soln as (1x7 + 3x4x2): (7x7 + 4x8x2) => 55:113
Now if the question asked what is 1st quantity in the solution then it will be (55x18)/113.

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Re: Mixture Alligation  [#permalink]

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New post 16 Jul 2019, 05:34
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This is a simple mixture and alligation question, which is made complicated by providing the different quantities of the Alloys. Keep few basic rules in your mind and you will become pro in solving similar questions.

Rule 1: Remember that the quantity does not matter to get the mixture ratio. So, forget about the quantities at first place.

Rule 2: If the material to be mixed is also made up of different materials, convert the ratio into fraction by picking one of the components.

If you have understood this well, you are ready to go. :)

Step 1: Let's start by converting ratios into fractions. You can pick either copper or tin (It's all your wish). I am taking Tin in order to show that you will still get the same answer.

Tin in First Alloy: 7/8
Tin in Second Alloy: 4/7

Step 2: We do not know anything about the weightage of tin in the mixture. Let it be X.
Step 3: Final ratio is 1:2. Set an alligation.

7/8 4/7
\ /
\ /
\ /
X
/ \
/ \
/ \
1 2

Use formula for alligation:

7/8 - X = 2
X - 4/7 1
Solve to get, X = 113/168

=> Quantity of Tin in the mixture = 113
Therefore, Copper = 168 - 113 = 55

Thus, ratio of Cu to Tin in the mixture will be 55:113.

Give kudos if you liked the solution. :)


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Re: Mixture Alligation   [#permalink] 16 Jul 2019, 05:34
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