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?
Two alloys of gold
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- talaangoshtari
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Q: copper/gold = 11/7talaangoshtari 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
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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I prefer Mitch's approach, but here's another approach that works: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
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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Hello talaangoshtari,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?
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
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