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Math Expert V
Joined: 02 Sep 2009
Posts: 57091
M03-30  [#permalink]

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3
24 00:00

Difficulty:   95% (hard)

Question Stats: 52% (02:51) correct 48% (02:38) wrong based on 189 sessions

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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

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Math Expert V
Joined: 02 Sep 2009
Posts: 57091
Re M03-30  [#permalink]

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4
5
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
_________________
Intern  Joined: 08 Jan 2015
Posts: 2
Re: M03-30  [#permalink]

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Hello,
I did not understand why the solution is only for the weight of gold?
Intern  Joined: 11 Sep 2013
Posts: 22
Location: India
Concentration: Marketing, International Business
GMAT 1: 660 Q48 V33 Re: M03-30  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 57091
Re: M03-30  [#permalink]

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sophianaf wrote:
Hello,
I did not understand why the solution is only for the weight of gold?

x in the solution above denotes the weight of the first bar, not gold in the first bar.
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 57091
Re: M03-30  [#permalink]

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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.
_________________
Intern  Joined: 23 Nov 2014
Posts: 12
Re: M03-30  [#permalink]

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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?
Intern  Status: Mr
Joined: 05 Jul 2015
Posts: 43
Location: India
Concentration: Entrepreneurship, General Management
GMAT 1: 720 Q48 V40 GMAT 2: 770 Q50 V46 WE: Business Development (Advertising and PR)
Re: M03-30  [#permalink]

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2
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.
Manager  Joined: 05 Jul 2015
Posts: 96
Concentration: Real Estate, International Business
GMAT 1: 600 Q33 V40 GPA: 3.3
Re: M03-30  [#permalink]

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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.
Intern  Joined: 26 Nov 2015
Posts: 9
GMAT 1: 710 Q51 V34 GPA: 3.3
Re: M03-30  [#permalink]

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1
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
Intern  Joined: 08 Jul 2015
Posts: 44
GPA: 3.8
WE: Project Management (Energy and Utilities)
Re: M03-30  [#permalink]

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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 _________________
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Intern  Joined: 26 May 2014
Posts: 39
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: M03-30  [#permalink]

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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
Intern  Joined: 20 Jul 2014
Posts: 2
Re M03-30  [#permalink]

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I think this is a high-quality question and I agree with explanation.
Manager  B
Joined: 31 Oct 2016
Posts: 106
Re: M03-30  [#permalink]

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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.
Manager  B
Joined: 26 Feb 2018
Posts: 76
Location: United Arab Emirates
GMAT 1: 710 Q47 V41 GMAT 2: 770 Q49 V47 Re: M03-30  [#permalink]

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You can solve by simultaneous equations

First look at the ration of the final bar - 5:11 - and see you have 16 'parts'. Divide 8kg by 16 and multiply to get the amounts of gold and silver - That means you have 2.5kg gold and 5.5kg silver

Let relative contribution from first bar be 'n' and the relative contribution from second be 'm'
First bar has 2 parts gold, second has 3 parts gold - 2n + 3m = 2.5
First bar has 3 parts silver, second has 7 parts silver - 3n + 7m = 5.5

From here you can multiply the first equation by 7 and the second by 3 and subtract to get n, or you can use slightly easier numbers and multiply the first by 3 and the second by 6. Subtracting gives you 5m = 3.5 and m = 0.7. Plug this back in to get n = 0.4.

The first bar has 5 (2+3) 'parts' and each part is 500g - 2.5kg total 0.4 *2.5 = 1kg

It's a bit long winded but the only other way is to start trying answer choices which is also rather time consuming
Intern  Joined: 03 Jul 2018
Posts: 1
Re: M03-30  [#permalink]

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hi bunuel,plz lemme know if i am wrong
first bar =2/5 (gold)
2nd bar =3/7
final ratio =5/16
now the difference between the second bar and final ratio would be the weighted first bar(allegation)
3/7-5/16=13/112
13/112 multiplied by 8 =7/8kg
is this approach wrong(maybe because we are not provided by the exact individual weights)
or this difference in answer is coz of fractions involved
urgent help needed will be really grateful if any1 can do the needful!!
Intern  S
Joined: 10 Jul 2016
Posts: 44
Re: M03-30  [#permalink]

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Bunuel wrote:
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

Just want to point out that I used to REALLY REALLY struggle with Mixture problems, and questions such as these would have me stumped!

But Bunuel and his questions/explanations have helped me solve this problem in 30 seconds!! and the difficulty level for this is 95%!!

Thanks a lot guys!
Manager  B
Joined: 30 Oct 2018
Posts: 67
Re: M03-30  [#permalink]

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I solved by weighted avg approach-
only for gold-
(3/10) - (5/16)
____________
(5/16) - (2/5)

=1/7
= (A)
_________________

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Plz Kudos! if my post helps....
Manager  G
Joined: 23 Jan 2018
Posts: 204
Location: India
WE: Information Technology (Computer Software)
Re: M03-30  [#permalink]

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chetan2u and Bunuel

Dear experts,

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

Regards,
Arup
Math Expert V
Joined: 02 Aug 2009
Posts: 7756
Re: M03-30  [#permalink]

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1
1
ArupRS wrote:
chetan2u and Bunuel

Dear experts,

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

Regards,
Arup

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
_________________ Re: M03-30   [#permalink] 11 May 2019, 23:46

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