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M03-30

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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?

A. 1 kg
B. 3 kg
C. 5 kg
D. 6 kg
E. 7 kg
[Reveal] Spoiler: OA

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Official Solution:

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?

A. 1 kg
B. 3 kg
C. 5 kg
D. 6 kg
E. 7 kg


Find the total amount of one of the ingredients. We have the ratio of gold in each alloy as well as in the final alloy. Express the gold weight balance of the final alloy in an equation where \(x\) is the weight of the first bar and \(8-x\) is the weight of the second bar.
\(x+y = 8\)
\(\frac{2}{2+3}*x + \frac{3}{3+7}*(8-x) = \frac{5}{5+11} * 8\)

2.5 kg is the weight of gold in the final alloy resulted after the two bars were melted. Transform the second equation:
\(0.4x + 0.3(8-x) = 2.5\)
\(0.1x + 2.4 = 2.5\)
\(x = 1\)


Answer: A
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New post 11 Jun 2015, 04:04
Hello,
I did not understand why the solution is only for the weight of gold?

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New post 11 Jun 2015, 04:17
Hi,

Any other way to solve?

Does the following method right?
I bar: 2G+3S=5
II bar:3G+7S-10
Total=15 but weight is 8
To get 8= (15+1)/2
Similarly, to get the weight of first bar
(5+1)/2 but first bar is 1/3 of total so,
(5+1)/2.3=1

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New post 11 Jun 2015, 04:51
vgurukool wrote:
Hi,

Any other way to solve?

Does the following method right?
I bar: 2G+3S=5
II bar:3G+7S-10
Total=15 but weight is 8
To get 8= (15+1)/2
Similarly, to get the weight of first bar
(5+1)/2 but first bar is 1/3 of total so,
(5+1)/2.3=1


You can check several alternative approaches HERE.

Hope it helps.
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New post 07 Aug 2015, 18:44
how can u find the total weight of first bar when u consider only gold wt 2.5..... and where u use silver wt to get the total wt of first bar.... confused with ur soln bunuel?

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Bunuel please change the language of this question or delete it from the database. The only reason this question is 'hard' is because it is very ambiguous and it is very possible to get a different answer.

First, When you say 2 parts of Gold and 3 Parts of silver, are you talking about their Volume or Weight?
From the answers given here, you've obviously taken it as the weight. But common sense says that when you use 'parts' you usually mean volume.
Second, it does not just say that the ratio in the first bar was 2/3 and that the ratio in the second bar was 3/7. Instead, it says 2 parts of gold in the first bar and 3 parts of gold in the second bar. What if the 'parts' referred to an actual measurement. This would mean that in all there were 5 parts of Gold and 10 parts of Silver in the combined bar.

Going by these two ambiguities, one can solve for the densities of Gold and Silver and the answer for the question would be something around 2.5 kgs.

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New post 21 Feb 2016, 21:45
I disagree with the person above me and think this is a good question.

I solved it with Allegation.

First I focused on just the gold and I made all the fractions match up. Ended up with 1 part out of 8 so 1 kg.

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I agree with icefrog. This question can be perceived in multiple ways, thus leading to different answers. From the official answer, it is obvious you assume that the weight of gold part in the first bar is different from the weight of gold part in the second bar, however, we are not told something about this. Moreover, you assume that the weight of silver and gold in each bar are equal, otherwise we cannot write 2/5*x for the weight of gold in the first bar. But if we assume that all gold parts are the same in weight and all silver parts are the same, then we arrive at different answer:
total weight of gold = 5/2 kg (5 gold parts) ---> each gold part = 1/2 kg.
total weight of silver = 11/2 kg (10 gold parts) ---> each silver part = 11/20 kg.
weight of first bar = 2*1/2 + 3*11/20 = 2.65 kg

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New post 28 May 2016, 00:25
rmohammadi wrote:
I agree with icefrog. This question can be perceived in multiple ways, thus leading to different answers. From the official answer, it is obvious you assume that the weight of gold part in the first bar is different from the weight of gold part in the second bar, however, we are not told something about this. Moreover, you assume that the weight of silver and gold in each bar are equal, otherwise we cannot write 2/5*x for the weight of gold in the first bar. But if we assume that all gold parts are the same in weight and all silver parts are the same, then we arrive at different answer:
total weight of gold = 5/2 kg (5 gold parts) ---> each gold part = 1/2 kg.
total weight of silver = 11/2 kg (10 gold parts) ---> each silver part = 11/20 kg.
weight of first bar = 2*1/2 + 3*11/20 = 2.65 kg


Me too agree with icefrog, also rmohammadi - this is exactly the way I solved this question, end up no answer fit this solution - I thought the part refer to measurement 8-)
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New post 30 Aug 2016, 12:40
Weight of Bar 1 = X , Weight of Bar 2= y

Lets create a equation for weight of gold only

(2/5) X + (3/10) Y = (5/16) * 8

0.4X + 0.3Y=2.5
Multiply by 10
4X+3Y=25

Two possible values of X and Y
X=4 , Y=3
X=1,Y=7

If you take X=4, the total weight of Bar 1 comes out to be : 0.4 * 4 + 0.6 * 4 =4 Kg -- This is not an option so the Value of X and Y must be X=1.Y=7

With X=1 the weight comes out to be 1 KG

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New post 31 Aug 2016, 12:12
I think this is a high-quality question and I agree with explanation.

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New post 27 Oct 2017, 06:21
Agree with icefrog. Spent half of an hour to figure out where I'm wrong. But finally, I got 1 kg is the weight of 2 parts of gold. So, the weight of the first bar (2 gold + 3 silver) will be more than 1 kg, but obviously less than 5 kg. So, the answer is B.

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Re: M03-30   [#permalink] 27 Oct 2017, 06:21
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