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Question

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?

1 kg
3 kg
5 kg
6 kg
7 kg

A

Frankenstein · Answer

Hi,
Let the weight of 1st bar be 5a kg, and the 2nd bar be 10b kg

-------- Gold-------Silver
bar1 --->2a ------- 3a
bar2 --->3b ------- 7b
------------------------
mixture -(2a+3b)----(3a+7b)

Given that weight of both bars melted is 8 kgs
i.e. 5a+10b = 8
Also Gold : Silver =5:11 => (2a+3b)/(3a+7b) = 5/11
Solving these 2 equations. we get 5a = 1kg
Hence, A

Cheers!

GMATGuruNY · Answer

bblast wrote: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?

1 kg
3 kg
5 kg
6 kg
7 kg

A

The first bar is 2/5 gold.
The second bar is 3/10 gold.
The 8kg bar is 5/16 gold.

We can use alligation to determine the ratio of the first bar to the second bar in the mixture:

The proportion needed of each bar is equal to the distance between the fraction attributed to the other bar and the fraction attributed to the mixture.

Proportion needed of the first bar = 5/16 - 3/10 = 25/80 - 24-80 = 1/80.
Proportion needed of the second bar = 2/5 - 5/16 = 32/80 - 25/80 = 7/80.
First bar:Second bar = (1/80) : (7/80) = 1:7.
Since 1+7=8, the first bar is 1/8 of the 8kg bar.
(1/8)*8 = 1.

The correct answer is A.

bblast · Answer

good work Frank

and thanks for the alligation approach mitch, Was looking for the same.

Its good idea transforming the ratios to a common denominator first ?

how to apply alligation to this :

https://www.beatthegmat.com/alligations- ... 83398.html

yeloaw · Answer

GMATGuruNY wrote:bblast wrote: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? 1 kg 3 kg 5 kg 6 kg 7 kg A The first bar is 2/5 gold. The second bar is 3/10 gold. The 8kg bar is 5/16 gold. We can use alligation to determine the ratio of the first bar to the second bar in the mixture: The proportion needed of each bar is equal to the distance between the fraction attributed to the other bar and the fraction attributed to the mixture. Proportion needed of the first bar = 5/16 - 3/10 = 25/80 - 24-80 = 1/80. Proportion needed of the second bar = 2/5 - 5/16 = 32/80 - 25/80 = 7/80. First bar:Second bar = (1/80) : (7/80) = 1:7. Since 1+7=8, the first bar is 1/8 of the 8kg bar. (1/8)*8 = 1. The correct answer is A. Does it matter whether I perform the alligation method on the gold or silver? I tried on got the same result, but I want to understand the concept. Will it ever matter which metal I do the alligation on? Or can I assume each metal is proportionally distributed from the first bar to the third bar to calculate the weight.

GMATGuruNY · Answer

yeloaw wrote:GMATGuruNY wrote:bblast wrote: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? 1 kg 3 kg 5 kg 6 kg 7 kg A The first bar is 2/5 gold. The second bar is 3/10 gold. The 8kg bar is 5/16 gold. We can use alligation to determine the ratio of the first bar to the second bar in the mixture: The proportion needed of each bar is equal to the distance between the fraction attributed to the other bar and the fraction attributed to the mixture. Proportion needed of the first bar = 5/16 - 3/10 = 25/80 - 24-80 = 1/80. Proportion needed of the second bar = 2/5 - 5/16 = 32/80 - 25/80 = 7/80. First bar:Second bar = (1/80) : (7/80) = 1:7. Since 1+7=8, the first bar is 1/8 of the 8kg bar. (1/8)*8 = 1. The correct answer is A. Does it matter whether I perform the alligation method on the gold or silver? I tried on got the same result, but I want to understand the concept. Will it ever matter which metal I do the alligation on? Or can I assume each metal is proportionally distributed from the first bar to the third bar to calculate the weight. If there are only two ingredients in the mixture -- in this case, gold and silver -- the alligation can be performed in terms of either ingredient.

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