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 ProblemsHere'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

_________________

Mike McGarry

Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)