There are 2 bars of copper-nickel alloy. One bar has 2 parts

Question

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

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

PareshGmat · Accepted Answer

Let the weight of first bar = x, the weight of the other = (20-x)

Copper composition in first bar = \frac{2}{2+5} * x

Copper composition in second bar  = \frac{3}{3+5} * (20-x)

Copper composition in resultant bar  = \frac{5}{11+5} * 20

Setting up the equation for total copper

\frac{2x}{7} + \frac{3(20-x)}{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

sudhir18n · Answer

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

subhashghosh · Answer

Karishma has posted an excellent analysis of this problem in her blog, please take a look :

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... -mixtures/

Policarpa · Answer

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

vinayak4u · Answer

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

vinayak4u · Answer

@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.

KarishmaB · Answer

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. Check out this video on Mixtures: https://www.youtube.com/watch?v=VdBl9Hw0HBg

pacifist85 · Answer

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).

yosita18 · Answer

KarishmaB · Answer

Here: https://anaprep.com/arithmetic-weighted-averages/ https://anaprep.com/arithmetic-mixtures/ https://www.youtube.com/watch?v=_GOAU7moZ2Q Further, this is the link to my blog: https://anaprep.com/blogs/ All my posts are here.

gracie · Answer

let x=weight of first bar
(2/7)(x)+(3/8)(20-x)=(5/16)(20)
x=14 kg

SVaidyaraman · Answer

first bar: c/n=2x/5x
second bar: c/n = 3y/5y
final bar: c/n = 5z/11z

weight of final bar is 20
=>5z+11z=20 => z= 5/4

copper in the final bar = 5z=25/4
simialrly nickel =11z= 55/4

we have 
2x+3y = 25/4 => 10x+15y = 125/4 (copper in first bar + in second bar)
5x+5y= 55/4 => 15x+15y= 165/4( nickel in first bar + in second bar)

solving the 2 equations we get,
=> x=2

weight of the first bar = 2x+5x=7x=14

pandajee · Answer

I actually went ahead by the total ratio of copper / nickel i.e. for Bar A- 2/5, Bar-B - 3/5, Final - 5/11.
It got the ratio of weights as 8/3. Steps are mentioned below:
Bar 1 ratio - 2/5, Bar 2 Ratio -3/5, Final ratio= 5/11
w1 / w2 = (3/5 - 5/11) / (5/11 - 2/5) = 8/3

w1 = 8/11 of 20 kgs = 160 / 11 kgs = 14.something but not clear cut 14 kgs
 Is it not right to take the total ratio of the bar as i have taken to solve the question ? Please comment

nisthagupta28 · Answer

Can someone please expand the alligations method? I'm unable to solve it

GinoRako · Answer

None of the above explainations makes any sense to me.

KarishmaB  your link to your explaination doest work anymore, atleast not from the Netherlands. I cannot open it.

Hopefully someone will be able to explain this question as simple as possible with no tedious and unnecessary calculations.

KarishmaB · Answer

GinoRako Though I am able to access previous links too but if you are not, you can access the relevant content on my own website. Check them here: https://anaprep.com/arithmetic-weighted-averages/ https://anaprep.com/arithmetic-mixtures/ https://www.youtube.com/watch?v=_GOAU7moZ2Q

tickledpink001 · Answer

how did you get 7/3?

luisdicampo · Answer

Deconstructing the Question

The first bar has copper to nickel ratio   2:5  , so the copper fraction is   \frac{2}{7}  .

The second bar has copper to nickel ratio   3:5  , so the copper fraction is   \frac{3}{8}  .

The final 20 kg bar has copper to nickel ratio   5:11  , so the copper fraction is   \frac{5}{16}  .

We use weighted average to find the ratio of the two original bars.

Step-by-step

Copper fraction in the first bar is

\frac{2}{2+5}=\frac{2}{7}

Copper fraction in the second bar is

\frac{3}{3+5}=\frac{3}{8}

Copper fraction in the final mixture is

\frac{5}{5+11}=\frac{5}{16}

Now compare distances from the final mixture.

\frac{3}{8}-\frac{5}{16}=\frac{6}{16}-\frac{5}{16}=\frac{1}{16}

\frac{5}{16}-\frac{2}{7}=\frac{35}{112}-\frac{32}{112}=\frac{3}{112}

So the ratio of first bar to second bar is

\frac{1}{16}:\frac{3}{112}

Since   \frac{1}{16}=\frac{7}{112}  , the ratio becomes

7:3

Total weight is   20   kg, so   10   parts correspond to   20   kg.

Each part is   2   kg.

Therefore, the first bar weighs

7 \cdot 2 = 14   kg

Answer: D

There are 2 bars of copper-nickel alloy. One bar has 2 parts

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There are 2 bars of copper-nickel alloy. One bar has 2 parts

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards