Problem on mixtures

[kprabhala.mba]

An alloy A has two materials in the ratio 1:2 and another alloy B has the same two materials in the ratio 2:3. How much content each of alloy A and alloy B should be mixed to produce another alloy that has the materials in the ratio 17:27?

[GMATGuruNY]

An alloy A has two materials in the ratio 1:2 and another alloy B has the same two materials in the ratio 2:3. How much content each of alloy A and alloy B should be mixed to produce another alloy that has the materials in the ratio 17:27?

The following method is called alligation. It's an easy way to handle weighted average problems.

One of the materials constitutes 1/3 of Alloy A and 2/5 of Alloy B. This material must constitute 17/44 of the resulting mixture.

To combine a 1/3 mixture with a 2/5 mixture to get a combined mixture that is 17/44:

The proportion needed of each starting fraction is the positive difference between the other 2 fractions.

First, let's put all the fractions over a common denominator:

A = 1/3 = 220/660
B = 2/5 = 264/660
A+B = 17/44 = 255/660

Proportion needed of A = (264/660) - (255/660) = 9/660.
Proportion needed of B = (255/660) - (220/660) = 35/660

A:B = (9/660):(35/660) = 9:35.

[kprabhala.mba]

An alloy A has two materials in the ratio 1:2 and another alloy B has the same two materials in the ratio 2:3. How much content each of alloy A and alloy B should be mixed to produce another alloy that has the materials in the ratio 17:27?

The following method is called alligation. It's an easy way to handle weighted average problems.

One of the materials constitutes 1/3 of Alloy A and 2/5 of Alloy B. This material must constitute 17/44 of the resulting mixture.

To combine a 1/3 mixture with a 2/5 mixture to get a combined mixture that is 17/44:

The proportion needed of each starting fraction is the positive difference between the other 2 fractions.

First, let's put all the fractions over a common denominator:

A = 1/3 = 220/660
B = 2/5 = 264/660
A+B = 17/44 = 255/660

Proportion needed of A = (264/660) - (255/660) = 9/660.
Proportion needed of B = (255/660) - (220/660) = 35/660

A:B = (9/660):(35/660) = 9:35.

Thank you very much for the solution.

I solved it after forming these two equations as I was not aware of the alligation technique.

x/3 + 2y/5 = 17/44 and
2x/3 + 3y/5 = 27/44.

I too, therefore, got x:y as 9:35 but the answer given is 3:7.

Someone named 'stormier' sent me this solution by email (Thank you!):

"Lets add alloy A and alloy B in the ratio 1:x to produce the final material.

material m = 1 (from alloy A) + 2x (from alloy B) = 1+2x
material n = 2+3x

(1+2x)/(2+3x) = 17/27

=> x = 7/3

required ratio = 1:x =1:7/3 = 3:7"

Both the answers seem correct (the second one especially as it matches the book answer), but which one is the wrong one and why?

[stormier]

An alloy A has two materials in the ratio 1:2 and another alloy B has the same two materials in the ratio 2:3. How much content each of alloy A and alloy B should be mixed to produce another alloy that has the materials in the ratio 17:27?

Hi kprabhala, I retracted my solution because it was incorrect. Mitch's and your's are right.


Let's work both cases backwords.

Let's say we mix the two alloys in the ratio 3:7

3 (kg) of A contains -> 1 kg m + 2 kg n

7 (kg) of B contains -> 14/5 kg m + 21/5 kg n

Mixture contains -> (1+14/5) kg m : (2+21/5) kg n = 19/5:31/5 = 19/31 Incorrect.


Let's look at the 9:35 case

9:35

9 (kg) of A contains -> 3 kg m + 6 kg n

35 (kg) of B contains-> 14 kg m + 21 kg n

mixture contains -> (3+14) kg m : (6+21) kg n = 17:27 Correct.


Here's the flaw in the original 3:7 solution

I say that let the ratio of materials A: B be 1:x

In final mixture,

material m = 1/3 (from A) + 2x/5 (from B) [ In the wrong solution I wrote this as 1+2x, which is incorrect!)
material n = 2/3 + 3x/5

(1/3+2x/5)/(2/3+3x/5) = 17/27
=> x = 35/9

so answer = 1 to x = 1 to 35/9 = 9:35

[kprabhala.mba]

An alloy A has two materials in the ratio 1:2 and another alloy B has the same two materials in the ratio 2:3. How much content each of alloy A and alloy B should be mixed to produce another alloy that has the materials in the ratio 17:27?

Hi kprabhala, I retracted my solution cause it was incorrect. Mitch's is right.


Let's work both cases backwords.

Let's say we mix the two alloys in the ratio 3:7

3 (kg) of A contains -> 1 kg m + 2 kg n

7 (kg) of B contains -> 14/5 kg m + 21/5 kg n

Mixture contains -> (1+14/5) kg m : (2+21/5) kg n = 19/5:31/5 = 19/31 Incorrect.


Let's look at the 9:35 case

9:35

9 (kg) of A contains -> 3 kg m + 6 kg n

35 (kg) of B contains-> 14 kg m + 21 kg n

mixture contains -> (3+14) kg m : (6+21) kg n = 17:27 Correct.

Thank you for clarifying the answers, stormier. The answer given in the text book is
apparently wrong!

[anshumishra]

Here is the algebraic (and stormier's) approach :

Lets say the two materials are m1 and m2, and we mix k1 part of A and k2 part of B to produce the materials in the ratio 17:27. We want to find the ratio k1/k2 = z.

=> Part of m1/Part of m2 = 17/27
=> (k1/3 + 2k2/5)/(2k1/3 + 3k2/5) = 17/27
=> (5k1+6k2)/(10k1+9k2) = 17/27
=> (5z + 6) / (10z +9) = 17/27 (dividing the numerator and denominator by k2 and replacing k1/k2 by z, which we need to find)
=> 135 z + 162 = 170z + 153
=> z = 9/35.

[kprabhala.mba]

The following method is called alligation. It's an easy way to handle weighted average problems.

The proportion needed of each starting fraction is the positive difference between the other 2 fractions.

Mitch, just curious how the alligation technique would work for more than 2 fractions.
e.g. 3 alloys with 3 materials in different proportions are mixed to form a compound alloy.
The equation method would definitely be tedious with multiple unknowns. I hope such questions
would not appear in GMAT as they are computationally intensive (and time-consuming)

[GMATGuruNY]

The following method is called alligation. It's an easy way to handle weighted average problems.

The proportion needed of each starting fraction is the positive difference between the other 2 fractions.

Mitch, just curious how the alligation technique would work for more than 2 fractions.
e.g. 3 alloys with 3 materials in different proportions are mixed to form a compound alloy.
The equation method would definitely be tedious with multiple unknowns. I hope such questions
would not appear in GMAT as they are computationally intensive (and time-consuming)

I wouldn't use alligation for a mixture that contains more than 2 elements. Most mixture problems can be solved easily by plugging in our own values or by plugging in the answer choices.

Here is a GMATPrep problem in which 3 different elements are combined. If you don't want to be exposed to the problem and its solution, please read no further.

Three grades of milk are 1%, 2%, 3% fat by volume. If x gallons of the 1% grade, y gallons of the 2% grade, and z gallons of the 3% grade are mixed to give x+y+z gallons of a 1.5 % grade, what is x in terms of y and z?

1) y+3z
2) (y+z)/4
3) 2y + 3z
4) 3y+z
5) 3y+4.5z

If we use 0 gallons of z, then equal amounts of x (1%) and y (2%) will yield a mixture that is 1.5% fat.

Plug in x=2, y=2, and z=0.
The question asks for the value of x=2. This is our target.
Now we plug y=2 and z=0 into all the answer choices to see which yields our target of 2.

Only answer choice A works:
y + 3z = 2 + 3*0 = 2.

The correct answer is A.

Here's an OG problem that can be solved easily by plugging in the answers:

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X?

(A) 10%
(B) 33.33 %
(C) 40%
(D) 50%
(E) 66.66 %

Answer choice C: Resulting mixture is 40% X
Let X=40, Y=60.
Ryegrass in X = .4*40 = 16.
Ryegrass in Y = .25*60 = 15.
Total ryegrass = 16+15 = 31.
Total ryegrass/Total mixture = 31/100 = 31%.
Since the resulting percentage of ryegrass must be just a bit smaller (30%), we need to use just a bit less of X (since it contains a smaller percentage of ryegrass than does Y).

The correct answer is B.

The problem above also could be solved by using alligation:

Proportion needed of X = 30-25 = 5.
Proportion needed of Y = 40-30 = 10.

Percentage of X in final mixture = 5/15 = 33.33%.

Hope this helps!