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There are 2 bars of copper-nickel alloy. One bar has 2 parts [#permalink]

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07 Jul 2011, 22:12

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There are 2 bars of copper-nickel alloy. One bar has 2 parts of copper to 5 parts of nickel. The other has 3 parts of copper to 5 parts of nickel. If both bars are melted together to get a 20 kg bar with the final copper to nickel ratio of 5:11. What was the weight of the first bar?

There are 2 bars of copper-nickel alloy. One bar has 2 parts of copper to 5 parts of nickel. The other has 3 parts of copper to 5 parts of nickel. If both bars are melted together to get a 20 kg bar with the final copper to nickel ratio of 5:11. What was the weight of the first bar?

(A) 1 kg (B) 4 kg (C) 6 kg (D) 14 kg (E) 16 kg

The best way to solve mixture problems are alligation.. Goofle it .. loads of data available on the nest. very easy concept.

Great diagram, but how do you work out the bottom, I tried 16 and 112 as common denominators and I am lost after that, could someone solve the fractions out. Thanks

Re: There are 2 bars of copper-nickel alloy. One bar has 2 parts [#permalink]

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18 Apr 2014, 21:17

Please let me know where I am going wrong in the below solution.

C:N

Bar 1 - 2:5 - X kg Bar 2 - 3:5 - Y kg

To make Bar 3 - 5:11

Whilst mixing bar1 and bar2 the C:N will change accordingly. Hence,

{2X+3Y}/{5X+5Y} = 5/11

3X=8Y

Also, we know that the Bar3 is 20 kg , so

X+Y=20

And X=160/11

Can you let me know where I am doing error.

Rgds, TGC!
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Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: There are 2 bars of copper-nickel alloy. One bar has 2 parts [#permalink]

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05 May 2014, 19:59

@TGC,

2X+3Y & 5X+5Y is the total quantity of C and N respectively in the 20 KG bar. So, you need to equate the above equations with their final values and NOT the ratio of the final mixture. 5/11 is the ratio of C to N in the 20KG Bar. So the C and N solution in the mixture = 5/16*20 and 11/16*20

You can now directly substitute this value in your earlier equation:

i.e, 2X+3Y=25/4 5X+5Y=55/4

Solve the above and you get the values of X and Y.

Please let me know where I am going wrong in the below solution.

C:N

Bar 1 - 2:5 - X kg Bar 2 - 3:5 - Y kg

To make Bar 3 - 5:11

Whilst mixing bar1 and bar2 the C:N will change accordingly. Hence,

{2X+3Y}/{5X+5Y} = 5/11

This equation is not correct. Bar 1 has 2/7 copper and 5/7 nickel. When you say Bar 1 is total X kg, the amount of copper in Bar 1 will be (2/7)*X. 2X in your fraction doesn't represent the amount of copper. When the total weight of the bar is X kg, how can the amount of copper in it be 2X kg? Does this make sense?

Re: There are 2 bars of copper-nickel alloy. One bar has 2 parts [#permalink]

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17 May 2016, 13:45

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: There are 2 bars of copper-nickel alloy. One bar has 2 parts [#permalink]

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07 Jul 2016, 00:45

VeritasPrepKarishma wrote:

TGC wrote:

Please let me know where I am going wrong in the below solution.

C:N

Bar 1 - 2:5 - X kg Bar 2 - 3:5 - Y kg

To make Bar 3 - 5:11

Whilst mixing bar1 and bar2 the C:N will change accordingly. Hence,

{2X+3Y}/{5X+5Y} = 5/11

This equation is not correct. Bar 1 has 2/7 copper and 5/7 nickel. When you say Bar 1 is total X kg, the amount of copper in Bar 1 will be (2/7)*X. 2X in your fraction doesn't represent the amount of copper. When the total weight of the bar is X kg, how can the amount of copper in it be 2X kg? Does this make sense?

Since total weight of mix is 20 kg, bar 1 must be (7/10)*20 = 14 kg

An easier approach would be to work on only one of Copper and Nickel. I have discussed that on the link given by subhashghosh above.

Quote:

Hi Karishma: You have mentioned the following in the link provided by subhashghosh:

I visit the GMAT Club forum regularly and discuss some ideas, some methodologies there. The weighted averages method I discussed in my previous two posts is one of my most highly appreciated inputs on the forum. People love how easily they can solve some of the most difficult questions by just drawing a scale or using a ratio. If you are not a Quant jock, I am sure you feel a chill run down your spine every time you see a mixtures problem. But guess what, they are really simple if you just use the same weighted average concepts we discussed in the previous two posts. Let’s look at a mixtures question in detail:

Can you please share the link of the previous two posts as well? Thanks again!

Please let me know where I am going wrong in the below solution.

C:N

Bar 1 - 2:5 - X kg Bar 2 - 3:5 - Y kg

To make Bar 3 - 5:11

Whilst mixing bar1 and bar2 the C:N will change accordingly. Hence,

{2X+3Y}/{5X+5Y} = 5/11

This equation is not correct. Bar 1 has 2/7 copper and 5/7 nickel. When you say Bar 1 is total X kg, the amount of copper in Bar 1 will be (2/7)*X. 2X in your fraction doesn't represent the amount of copper. When the total weight of the bar is X kg, how can the amount of copper in it be 2X kg? Does this make sense?

Since total weight of mix is 20 kg, bar 1 must be (7/10)*20 = 14 kg

An easier approach would be to work on only one of Copper and Nickel. I have discussed that on the link given by subhashghosh above.

Quote:

Hi Karishma: You have mentioned the following in the link provided by subhashghosh:

I visit the GMAT Club forum regularly and discuss some ideas, some methodologies there. The weighted averages method I discussed in my previous two posts is one of my most highly appreciated inputs on the forum. People love how easily they can solve some of the most difficult questions by just drawing a scale or using a ratio. If you are not a Quant jock, I am sure you feel a chill run down your spine every time you see a mixtures problem. But guess what, they are really simple if you just use the same weighted average concepts we discussed in the previous two posts. Let’s look at a mixtures question in detail:

Can you please share the link of the previous two posts as well? Thanks again!

Re: There are 2 bars of copper-nickel alloy. One bar has 2 parts [#permalink]

Show Tags

08 Sep 2017, 07:42

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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