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# There are 2 bars of gold-silver alloy; one piece has 2 parts

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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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09 Feb 2016, 03:58
icefrog wrote:
I think this question is very ambiguous!

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.

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.

The question is fine - it is not a Chemistry or Metallurgy question. It is a GMAT Quant question and in that context, the data given is very clear. Since there is no discussion of densities, the parts are by weight only.
Also, the question clearly says that the alloys are gold-silver. One bar has 2 parts of gold to 3 parts of silver. This gives you the ratio of gold to silver in bar 1 as 2:3. Similarly, the ratio of gold to silver in second bar is 3:7.
The term "parts" is commonly used to express ratios in GMAT world.
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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09 Feb 2016, 05:51
Its very obvious to anybody that Gold and Silver have different densities. I wouldn't expect the question to discuss about densities.
'Parts' is actually used for measurement and when you're describing just one scenario, it can be taken just as the ratio. Because that is the only information it gives. But when you're describing two scenarios with the same 'parts', it gives much more information. Actually the question is not ambiguous, its just wrong.
And I've solved quite a few official GMAT problems and written a GMAT already. I've never seen such ambiguity. This could never be an official problem.
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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27 Mar 2017, 10:57
1
trahul4 wrote:
There are 2 bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver and another has 3 parts of gold to 7 parts of silver. If both bars are melted into 8 kg bar with the final gold to silver ratio of 5:11. What was the weight of the first bar?

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

We are given that the first bar of gold-silver alloy has a ratio of gold to silver of 2x : 3x (thus it has a weight of 5x) and the second bar has a ratio of gold to silver of 3y : 7y (thus it has a weight of 10y). Both bars are melted into a new bar that has a final ratio of gold to silver of 5 : 11 and a weight of 8 kg. We can create the following two equations:

(2x + 3y)/(3x + 7y) = 5/11

11(2x + 3y) = 5(3x + 7y)

22x + 33y = 15x + 35y

7x = 2y

3.5x = y

We also know that 5x + 10y = 8. Since y = 3.5x, that gives us:

5x + 10(3.5x) = 8

40x = 8

x = 8/40 = 1/5

Since x = 1/5, the total weight of the first bar is 5(1/5) = 1 kg.

Answer: A
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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10 Jun 2017, 00:34
trahul4 wrote:
There are 2 bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver and another has 3 parts of gold to 7 parts of silver. If both bars are melted into 8 kg bar with the final gold to silver ratio of 5:11. What was the weight of the first bar?

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

1. (weight of first bar*proportion of gold) +(weight of second bar*proportion of gold) / (weight of first bar*proportion of silver) +(weight of second bar*proportion of silver)= 5/11
2. x*(2/5) + y*(3/10) / x*(3/5) + y*(7/10) =5/11
3. So x/y =1/7 and we have x+y=8
x=1 kg or weight of the first bar is 1 kg
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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27 Feb 2018, 09:58
1
Hi All,

This question is essentially just a wordy weighted average question.

We're given the composition of 2 types of alloy bar:

Bar A: 2 parts gold, 3 parts silver = 2/5 gold = 40% gold
Bar B: 3 parts gold, 7 parts silver = 3/10 gold = 30% gold

We're going to "mix" these two bar "types" and end up with a mixture that is 5/16 gold

We know that the total mixture will be 8kg. We're asked how many of the kg are Bar A.

A = # of kg of Bar A
B = # of kg of Bar B

Now we can set up the weighted average formula:

(.4A + .3B) / (A + B) = 5/16

And cross-multiply and simplify....

6.4A + 4.8B = 5A + 5B
1.4A = 0.2B

Multiply by 5 to get rid of the decimals...

7A = B

A/B = 1/7

This tells us that the ratio of A to B is 1:7. Since the total weight is 8kg, we must have 1 kg of Bar A and 7 kg of Bar B.

Final Answer:

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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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26 Mar 2019, 09:01
trahul4 wrote:
There are 2 bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver and another has 3 parts of gold to 7 parts of silver. If both bars are melted into 8 kg bar with the final gold to silver ratio of 5:11. What was the weight of the first bar?

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

Solving this algeabrically:

1st part: 2/3 (call it X weight)
2nd part: 2/7 (call it 8-x weight)

X & 8-X because if you call X 1st and Y 2nd, then X+Y=8, so reinstating Y=8-X. Now you got rid of a variable and you can express a 1 variable equation.

Now we can solve: 2/3*X+5/11*(8-X)=8*5/11

Solve for X you get 1
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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05 Mar 2020, 00:52
my answer is switching bars!...HALP!!

1st bar: x kg with 2/5 gold
2nd bar: y kg with 3/10 gold
melted: (x plus y) kg 5/16 gold

now......

2x/5 + 3y/10 = 5/16(x +y) --> LCM is 80 and simpllifying gives us

32x + 24y = 25x + 25y --> 32x - 25x = 25y - 24y --> 7x = 1y

why is the switch occuring where as it shd be 1x and 7y!
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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05 Mar 2020, 13:07
1
Mansoor50 wrote:
my answer is switching bars!...HALP!!

1st bar: x kg with 2/5 gold
2nd bar: y kg with 3/10 gold
melted: (x plus y) kg 5/16 gold

now......

2x/5 + 3y/10 = 5/16(x +y) --> LCM is 80 and simpllifying gives us

32x + 24y = 25x + 25y --> 32x - 25x = 25y - 24y --> 7x = 1y

why is the switch occuring where as it shd be 1x and 7y!

Hi Mansoor50,

You have correctly calculated the ratio of the two weights, but you still have think in terms of the weight of X vs the weight of Y.

You properly determined that 7(X) = 1(Y).... this means that "7 times the value of X" is equal to "1 times the value of Y"

For example, when X = 1.... Y = 7

The prompt tells us that the total weight of the mixed bar is 8km, which means we'll have 1kg of X and 7 kg of Y in the mixture.

GMAT assassins aren't born, they're made,
Rich
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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06 Mar 2020, 00:57
EMPOWERgmatRichC wrote:
Mansoor50 wrote:
my answer is switching bars!...HALP!!

1st bar: x kg with 2/5 gold
2nd bar: y kg with 3/10 gold
melted: (x plus y) kg 5/16 gold

now......

2x/5 + 3y/10 = 5/16(x +y) --> LCM is 80 and simpllifying gives us

32x + 24y = 25x + 25y --> 32x - 25x = 25y - 24y --> 7x = 1y

why is the switch occuring where as it shd be 1x and 7y!

Hi Mansoor50,

You have correctly calculated the ratio of the two weights, but you still have think in terms of the weight of X vs the weight of Y.

You properly determined that 7(X) = 1(Y).... this means that "7 times the value of X" is equal to "1 times the value of Y"

For example, when X = 1.... Y = 7

The prompt tells us that the total weight of the mixed bar is 8km, which means we'll have 1kg of X and 7 kg of Y in the mixture.

GMAT assassins aren't born, they're made,
Rich

thank you so much....... you corrected my thinking!
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts  [#permalink]

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16 Mar 2020, 04:28
1
VeritasKarishma wrote:
You can easily use the scale method here. The scale method is explained here:
http://gmatclub.com/forum/tough-ds-105651.html#p828579

Focus on any one of the two elements say Gold.
First bar has 2/5 gold. Second bar has 3/10 gold and mixture has 5/16 gold.
Make the fractions comparable for easy calculation i.e. give them the same denominator. LCM of 5, 10 and 16 is 80.
First bar has 32/80 gold. Second bar has 24/80 gold and mixture has 25/80 gold.
Attachment:
Ques3.jpg

So first bar weight:second bar weight should be in the ratio 1:7. Out of 8 total kgs, first bar must have been 1 kg.

Quote:
Hi Karishma,

For the scale method can we use gold to silver ratios rather than gold to total and work this problem out?
I understand in this case gold to total is an easier fraction to work with but trying to understand possibilities in scale method

No, you cannot use gold/silver ratio.
Gold/Total is the gold concentration (C1 and C2) in bar1 and bar2. You are finding the gold concentration in the two bars combined using volume as the weight of the weighted average.

Cavg = (C1 * Volume of bar1 + C2 * Volume of bar2) / (Total volume of the two bars)

Cavg = (Amount of gold in bar1 + Amount of gold in bar2) / (Total volume of the two bars)

So you can work with either gold concentration or silver concentration but not with gold/silver ratio. It doesn't give you any concentration.
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Re: There are 2 bars of gold-silver alloy; one piece has 2 parts   [#permalink] 16 Mar 2020, 04:28

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