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Gmat club PS.

by bblast » Sun May 22, 2011 1:51 am
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
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by Frankenstein » Sun May 22, 2011 2:02 am
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

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Last edited by Frankenstein on Sun May 22, 2011 2:27 am, edited 1 time in total.
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by GMATGuruNY » Sun May 22, 2011 2:10 am
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.
Last edited by GMATGuruNY on Wed Apr 18, 2012 7:23 pm, edited 1 time in total.
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by bblast » Sun May 22, 2011 2:35 am
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
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by yeloaw » Sun Apr 22, 2012 1:24 am
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.
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by GMATGuruNY » Sun Apr 22, 2012 2:41 am
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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