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Mixture/ratio problem .

by guerrero » Thu Feb 28, 2013 8:08 am
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

Kindly help me with a step by step approach to the solution. Experts - Can you also help me understand the concept to tackle such questions . I am frequently getting wrong answers in ratio/ mixture problems .. Appreciate your help !
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by Anurag@Gurome » Thu Feb 28, 2013 8:17 am
guerrero wrote: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?
Amount of copper in the 20 kg bar of alloy = 5*20/(5 + 11) = 100/16 = 25/4 kg

Let us assume that the weight of the first bar was x kg. Hence, weight of the 2nd bar was (20 - x) kg.
Amount of copper in the x kg bar of alloy = 2*x/(2 + 5) = 2x/7 kg
Amount of copper in the (20 - x) kg bar of alloy = 3*(20 - x)/(3 + 5) = 3*(20 - x)/8 kg

So, 2x/7 + 3*(20 - x)/8 = 25/4
--> [16x + 21*(20 - x)]/56 = 25/4
--> [16x + 21*20 - 21x] = 25*14
--> 5x = 21*20 - 25*14
--> x = 21*4 - 5*14 = 84 - 70 = 14

The correct answer is D.
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by Brent@GMATPrepNow » Thu Feb 28, 2013 8:23 am
guerrero wrote: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
Let x = weight of the first bar (our goal is to find the value of x)
20-x = weight of the second bar

Let's just keep track of the weight of copper in each bar.
First bar: 2 parts of copper to 5 parts of nickel
So, for every 7 parts, 2 parts are copper and 5 parts are nickel
In other words this bar is 2/7 copper
So, the weight of the copper in this bar is (2/7)x

Second bar: 3 parts of copper to 5 parts of nickel
So, for every 8 parts, 3 parts are copper and 5 parts are nickel
In other words this bar is 3/8 copper
So, the weight of the copper in this bar is (3/8)(20-x)

Final bar (after melting: 5 parts of copper to 11 parts of nickel
So, for every 16 parts, 5 parts are copper and 11 parts are nickel
In other words this final bar is 5/16 copper
If the weight of the entire bar is 20 kg, the weight of the copper in this bar is (5/16)(20)

Now, we'll write an equation using the fact that:
(weight of copper in 1st bar) + (weight of copper in 2nd bar) + (weight of copper in final bar)
So, (2/7)x + (3/8)(20-x) = (5/16)(20)
Solve for x, to get . . .

x = 14
Answer = D

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by GMATGuruNY » Thu Feb 28, 2013 12:34 pm
guerrero wrote: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 following approach is called ALLIGATION -- a very good way to handle MIXTURE PROBLEMS.
Alligation can be performed only with fractions or percentages.
Thus, the ratios here must be converted.

Let F = the first bar and S = the second bar.

Step 1: Convert the ratios to FRACTIONS.
F:
Since copper:nickel = 2:5, and 2+5=7, copper/total = 2/7.
S:
Since copper:nickel = 3:5, and 3+5=8, copper/total = 3/8.
Mixture:
Since copper:nickel = 5:11, and 5+11=16, copper/total= 5/16.

Step 2: Put the fractions over a COMMON DENOMINATOR.
F = 2/7 = 32/112.
S = 3/8 = 42/112.
Mixture = 5/16 = 35/112.

Step 3: Plot the 3 numerators on a number line, with the two starting numerators (32 and 42) on the ends and the goal numerator (35) in the middle.
F 32-------------35----------42 S

Step 4: Calculate the distances between the numerators.
F 32-------3-----35----7---- 42 S

Step 5: Determine the ratio in the mixture.
The ratio of F to S in the mixture is the RECIPROCAL of the distances in red.
F : S = 7:3.

The weight of the resulting bar is 20kg.
Since F : S = 7:3 = 14:6, F=14 and S=6, for a total weight of 20kg.

The correct answer is D.

For two other problems that I solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... tml#484583
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by Jim@StratusPrep » Thu Feb 28, 2013 1:28 pm
1st bar is 2/7 or about 29% copper the second is 3/8 or 37.5% copper.

The combination is 5/16 or 31.25%. There should be a ration of about 6:2 (37.5-31.25):(31.25-29). This means that 6/8 or 75% was from bar 1 - this is 15kg. Since we rounded down ratio (6.25:2.25) there should be less than 15kg.

14 is your answer.
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by Jim@StratusPrep » Thu Feb 28, 2013 1:31 pm
1st bar is 2/7 or about 29% copper the second is 3/8 or 37.5% copper.

The combination is 5/16 or 31.25%. There should be a ration of about 6:2 (37.5-31.25):(31.25-29). This means that 6/8 or 75% was from bar 1 - this is 15kg. Since we rounded down ratio (6.25:2.25) there should be less than 15kg.

14 is your answer.
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by hemant_rajput » Fri Mar 01, 2013 9:02 am
OK, so experts already posted various methods over here. Most of these methods are based on average and weight, I'll provide approach using Allegation.

1. copper to nickel in first alloy is 2/5 => copper to alloy ration = 2/(2 + 5)
2. copper to nickel in second alloy is 3/5=> copper to alloy ration = 3/(3 + 5)
3. copper to nickel in fused alloy is 5/11=> copper to alloy ration = 5/(5 + 11)

allegation formula:
cheaper/ dearer = (% of dearer - % of mixture)/(% of mixture - % of cheaper)
or
Q1/Q2 = (P2 - Pm)/(Pm - P1)
so,

Q1 = quantity of first alloy
Q2 = quantity of second alloy

Q1/Q2 = [( 3/8) - (5/16)]/[(5/16) - (2/7)]

= 7/3

so Q1 = 20* 7/(7 + 3) = 14
Bingo!!!

Answer is D.
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by freyesinsb » Fri Mar 01, 2013 12:25 pm
Step 5: Determine the ratio in the mixture.
The ratio of F to S in the mixture is the RECIPROCAL of the distances in red.


How do you determine that the mixture is the reciprocal? Is it always the reciprocal?
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by GMATGuruNY » Fri Mar 01, 2013 12:50 pm
freyesinsb wrote:Step 5: Determine the ratio in the mixture.
The ratio of F to S in the mixture is the RECIPROCAL of the distances in red.


How do you determine that the mixture is the reciprocal? Is it always the reciprocal?
Yes: the required ratio is always the RECIPROCAL of the distances.
Let's explore why:

F 32---3---35-----------7----------- 42 S

Here, because the mixture's value (35) is CLOSER to F's value than to S's value, F must occupy a GREATER PROPORTION of the mixture than does S.
As a result, the required ratio is the RECIPROCAL of the distances:
F : S = 7:3.

Another example:

F 32-1-33-------------9------------- 42 S

Here, the mixture's value (33) is VERY CLOSE to F's value (32).
Thus, F must occupy ALMOST ALL of the mixture, while S occupies very little.
As noted above, the required ratio is the reciprocal of the distances:
F : S = 9:1.
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by gmatbeater1989 » Tue Oct 20, 2015 3:16 pm
GMATGuruNY wrote:
guerrero wrote: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 following approach is called ALLIGATION -- a very good way to handle MIXTURE PROBLEMS.
Alligation can be performed only with fractions or percentages.
Thus, the ratios here must be converted.

Let F = the first bar and S = the second bar.

Step 1: Convert the ratios to FRACTIONS.
F:
Since copper:nickel = 2:5, and 2+5=7, copper/total = 2/7.
S:
Since copper:nickel = 3:5, and 3+5=8, copper/total = 3/8.
Mixture:
Since copper:nickel = 5:11, and 5+11=16, copper/total= 5/16.

Step 2: Put the fractions over a COMMON DENOMINATOR.
F = 2/7 = 32/112.
S = 3/8 = 42/112.
Mixture = 5/16 = 35/112.

Step 3: Plot the 3 numerators on a number line, with the two starting numerators (32 and 42) on the ends and the goal numerator (35) in the middle.
F 32-------------35----------42 S

Step 4: Calculate the distances between the numerators.
F 32-------3-----35----7---- 42 S

Step 5: Determine the ratio in the mixture.
The ratio of F to S in the mixture is the RECIPROCAL of the distances in red.
F : S = 7:3.

The weight of the resulting bar is 20kg.
Since F : S = 7:3 = 14:6, F=14 and S=6, for a total weight of 20kg.

The correct answer is D.

For two other problems that I solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... tml#484583
Sorcery!!!
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