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by nafiul9090 » Wed Apr 18, 2012 6:00 pm
There are two bars of gold-silver alloy. The first bar has 2 parts of gold and 3 parts of silver, and the other has 3 parts of gold and 7 parts of silver. If both bars are melted into an 8 kg bar with the final ratio of 5:11 (gold to silver), what was the weight of the first bar?

1kg
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is there any short cut method??
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by GMATGuruNY » Wed Apr 18, 2012 7:24 pm
I posted a solution here:

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by aneesh.kg » Wed Apr 18, 2012 9:34 pm
let the weight of the first and the second alloy to be X and Y.
Then, X + Y = 8 ---- (1)
Total weight of gold = (5/16)th of the total weight
Thus, (2/5)*X + (3/10)*Y = 5/16*8 ---- (2)
Substituting the value of Y from equation (1) in equation (2)
2/5*X + 3/10*(8 - X) = 5/2
0.4X + 2.4 - 0.3X = 2.5
0.1X = 0.1
X = 1

(A) 1 kg is the answer.
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by IshanGhose » Thu Apr 19, 2012 5:37 am
The final equation :- gold:silver :: 5:11 on an 8 kg bar tells us that, totally there is 2.5 kg of gold (5/16 of 8 kg) and 5.5 kg of silver (11/16 of 8 kg).

Based on this information, lets check each option:

a) 1 kg: If the 1st bar is 1 kg, then there should be .4kg gold (2/5 of 1 kg) and .6 kg silver (3/5 of 1 kg)......i)
Based on this value, the 2nd bar should weigh 7 kgs, for which there should be (3/10 of 7 kg ) = 2.1 kg gold and (7/10 of 7 kg) = 4.9 kg silver...ii)
i) and ii) add upto to 2.5 kg gold and 5.5 kg silver, thus a) is the correct answer.
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