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There are 2 bars of coppernickel 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 coppernickel 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
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Last edited by Bunuel on 29 Oct 2013, 00:03, edited 1 time in total.
Renamed the topic and edited the question.



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Re: Mixture problem [#permalink]
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07 Jul 2011, 22:53
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Karishma has posted an excellent analysis of this problem in her blog, please take a look : http://www.veritasprep.com/blog/2011/04 ... mixtures/
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Re: Mixture problem [#permalink]
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08 Jul 2011, 00:53
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harshavmrg wrote: There are 2 bars of coppernickel 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. Use the diagram.. calculate the ratio its 7:3 thus weight of first bar = 7/10* 20 = 14 hence D
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Re: Mixture problem [#permalink]
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28 Oct 2013, 18:21
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



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Re: There are 2 bars of coppernickel alloy. One bar has 2 parts [#permalink]
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03 Dec 2013, 08:11
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In the 20 KG bar,
Copper constitutes 5/16*20 = 25/4 of the mixture Nickel constitutes 11/16*20 = 55/4 of the mixture
Now, Let C/N ratio in Bar 1 of the solution be 2a : 5a Let C/N ratio in Bar 2 of the solution be 3b : 5b
Therefore, 2a+3b = 25/4 and 5a+5b = 55/4
Solving for the above you get,
b= 3/4 and a=2
Now weight of the first bar = 2a+5a = 7*2 = 14



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Re: There are 2 bars of coppernickel 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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Re: There are 2 bars of coppernickel 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.



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Re: There are 2 bars of coppernickel alloy. One bar has 2 parts [#permalink]
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05 May 2014, 20:59
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? So your equation will be \(\frac{(2/7)*X + (3/8)*Y}{(5/7)*X + (5/8)*Y} = \frac{5}{11}\) You get X/Y = 7/3 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.
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Re: There are 2 bars of coppernickel alloy. One bar has 2 parts [#permalink]
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24 Feb 2015, 05:04
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For this problem I used the unknown multiplier, twice, as I usually do... Hope it is logical to do so! Let me explain.
\(\frac{2x}{5}\) \(+\)\(\frac{3y}{5}\) \(=\)\(\frac{5(x+y)}{11}\)
\(\frac{2x+3y}{5}\) \(=\)\(\frac{5(x+y)}{11}\)
\(11 (2x+3y) = 5 (5x+5y)\)
\(22x + 33y = 25x + 25y\)
\(8y = 3x\)
\(\frac{y}{x}\) \(=\)\(\frac{3}{8}\)
Then we use the unknown multiplier again to find the amount:
\(8x + 3x = 20\)
\(11x = 20\)
\(x =\) \(\frac{20}{11}\)
So, for A: \(8*\)\(\frac{20}{11}\) \(=\)\(\frac{160}{11}\) \(= 14\) (about).



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Re: There are 2 bars of coppernickel alloy. One bar has 2 parts [#permalink]
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26 Mar 2015, 03:06
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Let the weight of first bar = x, the weight of the other = (20x) Copper composition in first bar\(= \frac{2}{2+5} * x\) Copper composition in second bar \(= \frac{3}{3+5} * (20x)\) Copper composition in resultant bar \(= \frac{5}{11+5} * 20\) Setting up the equation for total copper \(\frac{2x}{7} + \frac{3(20x)}{8} = \frac{100}{16}\) \(\frac{3x}{8}  \frac{2x}{7} = \frac{60}{8}  \frac{100}{16}\) \(x = \frac{20}{16}* \frac{56}{5} = 14\) Answer = D
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Re: There are 2 bars of coppernickel 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? So your equation will be \(\frac{(2/7)*X + (3/8)*Y}{(5/7)*X + (5/8)*Y} = \frac{5}{11}\) You get X/Y = 7/3 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!
Regards, Yosita



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Re: There are 2 bars of coppernickel alloy. One bar has 2 parts [#permalink]
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07 Jul 2016, 01:23
yosita18 wrote: 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? So your equation will be \(\frac{(2/7)*X + (3/8)*Y}{(5/7)*X + (5/8)*Y} = \frac{5}{11}\) You get X/Y = 7/3 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!
Regards, Yosita
Here: http://www.veritasprep.com/blog/2011/03 ... averages/http://www.veritasprep.com/blog/2011/04 ... gebrutes/Further, this is the link to my blog: http://www.veritasprep.com/blog/categor ... erwisdom/All my posts are here.
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There are 2 bars of coppernickel alloy. One bar has 2 parts [#permalink]
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07 Jul 2016, 10:51
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let x=weight of first bar (2/7)(x)+(3/8)(20x)=(5/16)(20) x=14 kg



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