M03-30

Question

There are two bars made of gold-silver alloy. The first bar contains 2 parts of gold and 3 parts of silver by weight, while the second contains 3 parts of gold and 7 parts of silver by weight. If both bars are melted and combined to form a single 8-kilogram bar with a gold-to-silver ratio of 5:11 by weight, what was the weight of the first bar?

A. 1 kilogram
B. 3 kilograms
C. 5 kilograms
D. 6 kilograms
E. 7 kilograms

A. 1 kilogram 
B. 3 kilograms
C. 5 kilograms
D. 6 kilograms
E. 7 kilograms

Bunuel · Accepted Answer

Official Solution:

There are two bars made of gold-silver alloy. The first bar contains 2 parts of gold and 3 parts of silver by weight, while the second contains 3 parts of gold and 7 parts of silver by weight. If both bars are melted and combined to form a single 8-kilogram bar with a gold-to-silver ratio of 5:11 by weight, what was the weight of the first bar?

A. 1 kilogram
B. 3 kilograms
C. 5 kilograms
D. 6 kilograms
E. 7 kilograms

A. 1 kilogram 
B. 3 kilograms
C. 5 kilograms
D. 6 kilograms
E. 7 kilograms

Let's assume that the weight of the first bar is  x  kilograms and the weight of the second bar is  8-x  kilograms. To determine the value of  x , we must construct an equation that we can solve for  x . One possible method is to equate the weights of gold in the first and second bars to the weight of gold in the final mixture.
 
The weight of gold in the first bar is  \frac{2}{2+3}*x  kilograms, and the weight of gold in the second bar is  \frac{3}{3+7}*(8-x)  kilograms. By adding these two quantities and setting the result equal to the weight of gold in the final bar, which is  \frac{5}{5+11} * 8 = 2.5  kilograms, we get the equation  \frac{2}{5}*x + \frac{3}{10}*(8-x)=2.5 .
 
To simplify the equation, we multiply both sides by 10, giving us  4x + 3*(8-x)=25 . Solving for  x , which represents the weight of the first bar, we find that  x=1  kilogram.

Answer: A

gmatophobia · Answer

https://gmatclub.com/forum/download/file.php?id=81883 GMAT-Club-Forum-thxhpbnk.png

ArupRS · Answer

chetan2u  and  Bunuel

Dear experts,

can you please help me to solve this question by using allegation.

Regards,
Arup

chetan2u · Answer

Yes, you can ..
Just take any of the metal, say Gold..
So in first G:S=2;3, so G is 2/(2+3)=2/5
In second G becomes 3/(3+7)=3/10
And overall it becomes 5/(5+11)=5/16..

So, the weight of the first bar is 8*(5/16-3/10)/(2/5-3/10)..
To is always easy to handle integers..
So take LCM of (5,10,16)= 80..
So 2/5 of 80 is 32, 3/10 of 80=24 and 5/16 of 80 is 25..

Thus first bar's weight=8*(25-24)/(32-24)=8*1/8=1

dcummins · Answer

Some may find this table easier to understand.

Essentially if we can determine the amount of weight of one component of the final mix, we can reverse engineer it to find the actual physical weight of the sub-components (Bar 1 and 2) that contributed.

Following my table, you will see that:
0.4x + 0.3(8-x) = 2.5kg

Solving for x gives 0.1x = 0.1 
x= 1

l1ll1ll1ll1ll1l · Answer

I found it easier to use teeter-totter method from MGMAT.
Think of this as a mixture/solution question.

1st bar is 2/5 part gold, which is 40%
2nd bar is 3/10 part gold, which is 30%.

Final product is 5/16 part gold.

Using a common denominator,

1st bar is 32/80
2nd bar is 24/80

Final mixture is 25/80

.....24..............25.......................................32 out of 80
(2nd bar)===(Final)==============(1st bar)
 ...............

The final mixture must be made of 7/8 of 2nd bar and 1/8 of 1st bar.
weight of 1st bar = 1/8 X 8kg = 1 kg

Therefore, 1kg is the answer.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

egunckel10 · Answer

Found an alternative way of getting to the answer.

Since we are told that both bars are melted into an 8-kg bar with a gold-to-silver ratio of 5:11, we can calculate the weight of gold and silver of the new bar:

5x+11x=16x

16x=8

x=0,5 ===) Therefore, the weight of gold in the bar is  0,5*5=2,5 kg

Since we are asked to calculate the weight of gold of the first bar, we know that the number CANNOT be less than 2,5kg.
Looking at the answer choices, a) is the only option that satisfies that condition.

sayan640 · Answer

gmatophobia  ,  chetan2u ,
Can you please share an alligation diagram for this question ? I want to solve it using alligation method.

sayan640 · Answer

Take the gold as = 2*x
silver as 3*x
Total weight of the 1st bar = 5*x

Total weight of the 2nd bar = 10*y with gold = 3*y and silver 7*y

1st equation => 
(2*x + 3*y ) / (3*x + 7*y) = 5/11

second equation =>
5*x + 10*y = 8

Solving , we get x= 2*1/10
and 5*x= weight of the 1st bar = 5*1/5 = 1 kg

A is the answer.

sayan640 · Answer

Thank you gmatophobia GMAT-Club-Forum-5hm3q18w.png

bumpbot · Answer

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M03-30

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