Two alloys of gold

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talaangoshtari: Master | Next Rank: 500 Posts; Posts: 154; Joined: Wed May 21, 2014 4:29 amFollowed by: 1 member; Thanked: 8 times

by talaangoshtari » Tue May 12, 2015 9:54 am

Two alloys of gold, P and Q, prepared by mixing copper and gold in the ratio (2 : 7) and (11 : 7) respectively. If equal quantities of the alloys are melted to form a third alloy R, the ratio of copper and gold in R will be:

A.4:9
B.7:5
C.5:7
D.9:5
E.4:7

How can we solve this problem by using alligation approach?

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by GMATGuruNY » Tue May 12, 2015 11:19 am

talaangoshtari wrote:Two alloys of gold, P and Q, prepared by mixing copper and gold in the ratio (2 : 7) and (11 : 7) respectively. If equal quantities of the alloys are melted to form a third alloy R, the ratio of copper and gold in R will be:

A.4:9
B.7:5
C.5:7
D.9:5
E.4:7

Q: copper/gold = 11/7
Let Q = 11 grams copper + 7 grams gold, for a total of 18 grams.

P: copper/gold = 2/7
Since 2/7 = 4/14, let P = 4 grams copper + 14 grams gold, for a total of 18 grams.

P and Q combined:
(copper in P) + (copper in Q) = 4+11 = 15.
(gold in P) + (gold in Q) = 14+7 = 21.
Resulting ratio:
copper/gold = 15/21 = 5/7.

The correct answer is C.

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by Brent@GMATPrepNow » Tue May 12, 2015 1:30 pm

talaangoshtari wrote:Two alloys of gold, P and Q, prepared by mixing copper and gold in the ratio (2 : 7) and (11 : 7) respectively. If equal quantities of the alloys are melted to form a third alloy R, the ratio of copper and gold in R will be:

A.4:9
B.7:5
C.5:7
D.9:5
E.4:7

I prefer Mitch's approach, but here's another approach that works:

This can be viewed as a "weighted" averages where each alloy gets the same weight. By the way, we have a free video on weighted averages: https://www.gmatprepnow.com/module/gmat- ... ics?id=805

Alloy P: copper and gold in 2:7 ratio
This means that 2/9 of alloy P is copper
Or we can say that 4/18 of alloy P is copper

Alloy Q: copper and gold in 11:7 ratio
This means that 11/18 of alloy Q is copper

Since we're adding EQUAL QUANTITIES of the two alloys, the resulting alloy (R) will have a mixture that is the same as AVERAGE of the two alloys.

The AVERAGE of 4/18 and 11/18 = 7.5/18

So, in 18 pounds of alloy R, copper comprises 7.5 pounds, which means gold comprises the other 10.5 pounds
So, the ratio of copper and gold in alloy R = 7.5 : 10.5
This does not appear as one of the answer choices, so we'll need to examine EQUIVALENT ratios.
7.5 : 10.5 = 15 : 21 = [spoiler]5 : 7[/spoiler]

Answer: C

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Aman verma: Master | Next Rank: 500 Posts; Posts: 304; Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members; Location: International Space Station; Thanked: 11 times

by Aman verma » Wed May 13, 2015 1:15 am

talaangoshtari wrote:Two alloys of gold, P and Q, prepared by mixing copper and gold in the ratio (2 : 7) and (11 : 7) respectively. If equal quantities of the alloys are melted to form a third alloy R, the ratio of copper and gold in R will be:

A.4:9
B.7:5
C.5:7
D.9:5
E.4:7

How can we solve this problem by using alligation approach?

Hello talaangoshtari,

In this type of questions,it is helpful to take one element as common and set the equation accordingly, using alligation graphic as shown above. Here, we have taken copper as the common element and shown it as a part of total. You can set the equation using the alligation graphic as follows:-
x-2/9 = 11/18 - x
or,2x=15/18
or, x=5/12
Here, copper is shown as part of total mixture. So, subtract 5 from 12 to get the proportion of gold= 12-5=7. Therefore, the ratio of copper and gold in R will be 5:7. Option C

800. Arjun's-Bird-Eye

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