There are 55 bags in front of you. Each bag is of a different color -- either red, blue and yellow -- and there is a number of stones inside each bag. The average number of stones per bag is 30 7/11 (i.e. thirty and seven-elevenths).

If you consider only the red and blue bags, they have an average of 27 stones per bag. If you consider only the blue and yellow bags, you have an average of 29 2/3 (i.e. twenty-nine and two-thirds) stones per bag. If you consider only the red and yellow bags, you have an average of 35 stones per bag.

If you add 21 stones to each red bag, the average number of stones for red and blue bags is 7 stones higher, and the average number of stones for red and yellow bags becomes 6 stones higher.

Given only this information, how many bags of each color are there? What is the average number of stones in the red bags alone? The blue bags alone? The yellow bags alone?

The above problem was paraphrased from a graduate-level refresher course. Let's get to work.

Define our variables as follows:

x = number of red bags

y = number of blue bags

z = number of yellow bags

a = total number of stones in all red bags

b = total number of stones in all blue bags

c = total number of stones in all yellow bags

Solving a set of equations in six variables tends to be a long and tedious process, so what we should really do is assign the variables now and try to cut them down through simplification.

We know that the total number of bags is 55:

x + y + z = 55

And we know that the average number of stones among all bags is 30 7/11:

(Total number of stones in all bags) / (Total number of bags) = 30 7/11 = 337/11

>> (a + b + c) / (x + y + z) = 337/11

>> (a + b + c) / 55 = 337/11

>> (a + b + c) = 1685

Okay, now the average number of stones among all red and blue bags is 27:

(Total number of stones in red and blue bags) / (Total red and blue bags) = 27

>> (a + b) / (x + y) = 27

>> (a + b) = 27 * (x + y)

Similarly, the average number of stones among all blue and yellow bags is 29 2/3:

(Total number of stones in blue and yellow bags) / (Total blue and yellow bags) = 29 2/3

>> (b + c) / (y + z) = 29 2/3 = 89/3

>> (b + c) = (89/3) * (y + z)

And the average number of stones among all red and yellow bags is 35:

(Total number of stones in red and yellow bags) / (Total red and yellow bags) = 35

>> (a + c) / (x + z) = 35

>> (a + c) = 35 * (x + z)

Adding all resultant equations gets us:

(a + b) + (b + c) + (a + c) = [27 * (x + y)] + [(89/3) * (y + z)] + [35 * (x + z)]

>> 2 * (a + b + c) = 27x + 27y + (89y/3) + (89z/3) + 35x + 35z

>> 2 * (1685) = (27 + 35)x + [27 + (89/3)]y + [(89/3) + 35]z

>> 3370 = 62x + (170/3)y + (194/3)z

Now if we added 21 stones to each red bag, then:

Total number of additional stones in red bags = 21x, where x is the number of red bags.

Total number of stones in red bags now = a + 21x

So now the average number of stones among all red and blue bags is 27 + 7 = 34:

(New total number of stones in red and blue bags) / (Total red and blue bags) = 34

>> (a + 21x + b) / (x + y) = 34

>> (a + 21x + b) = 34 * (x + y)

>> (a + b) = [34 * (x + y)] - 21x

And the new average number of stones among all red and yellow bags is 35 + 6 = 41:

(New total number of stones in red and yellow bags) / (Total red and yellow bags) = 41

>> (a + 21x + c) / (x + z) = 41

>> (a + 21x + c) = 41 * (x + z)

>> (a + c) = [41 * (x + z)] - 21x

The average number of stones among all blue and yellow bags remains the same as above, which results in the equation:

(b + c) = (89/3) * (y + z)

Adding all resultant equations gets us:

(a + b) + (b + c) + (a + c) = [34 * (x + y)] - 21x + [(89/3) * (y + z)] + [41 * (x + z)] - 21x

>> 2 * (a + b + c) = 34x + 34y - 21x + (89y/3) + (89z/3) + 41x + 41z - 21x

>> 2 * (1685) = (34 + 41 - 21 - 21)x + [34 + (89/3)]y + [(89/3) + 41]z

>> 3370 = 33x + (191/3)y + (212/3)z

This gives us the following three equations in three variables:

I: 55 = x + y + z

II: 3370 = 62x + (170/3)y + (194/3)z

III: 3370 = 33x + (191/3)y + (212/3)z

Eliminating x from equations I and II:

I: 55 = x + y + z

>> 3410 = 62x + 62y + 62z

>> 3410 = 62x + (186/3)y + (186/3)z

Subtracting II: 3370 = 62x + (170/3)y + (194/3)z:

>> 40 = (16/3)y - (8/3)z

>> 120 = 16y - 8z

Eliminating x from equations I and III:

I: 55 = x + y + z

>> 1815 = 33x + 33y + 33z

>> 1815 = 33x + (99/3)y + (99/3)z

Subtracting from III: 3370 = 33x + (191/3)y + (212/3)z:

>> 1555 = (92/3)y + (113/3)z

>> 4665 = 92y + 113z

This results in the following two equations in two variables:

IV: 120 = 16y - 8z

V: 4665 = 92y + 113z

Eliminating z:

IV: 120 = 16y - 8z

>> 13560 = 1808y - 904z

V: 4665 = 92y + 113z

>> 37320 = 736y + 904z

Adding IV and V together:

>> 50880 = 2544y

>> y = 20

Now that we know that y = 20, we can solve for z:

IV: 120 = 16y - 8z

>> 120 = 320 - 8z

>> 8z = 200

>> z = 25

And we can solve for x:

I: 55 = x + y + z

>> 55 = x + 20 + 25

>> x = 10

Now we need to find a, b, and c. Based on previous equations:

(a + b) = 27 * (x + y)

>> (a + b) = 27 * (10 + 20)

>> (a + b) = 810

(b + c) = (89/3) * (y + z)

>> (b + c) = (89/3) * (20 + 25)

>> (b + c) = (89/3) * (45)

>> (b + c) = 1335

(a + c) = 35 * (x + z)

>> (a + c) = 35 * (10 + 25)

>> (a + c) = 1225

Thus three variables in three equations once more:

I: 810 = (a + b)

II: 1335 = (b + c)

III: 1225 = (a + c)

We can eliminate c by subtracting III from II:

110 = b - a

Then eliminate a by adding this with equation I:

920 = 2b

>> b = 460

Now that we know that b = 460, we can solve for a:

I: 810 = (a + b)

>> 810 = (a + 460)

>> a = 350

And we can solve for c:

II: 1335 = (b + c)

>> 1335 = (460 + c)

>> c = 875

Therefore, there are 10 red bags, 20 blue bags, and 25 yellow bags.

The average among red bags is (a/x) = (350/10) = 35 stones.

The average among blue bags is (b/y) = (460/20) = 23 stones.

The average among yellow bags in (c/z) = (875/25) = 35 stones.

...

...

I still remember how to do all this, it seems. :)

## 4 comments:

Of course you do :D Well and good.

You helped someone with homework?

Yup. I'm surprised that people still take up this stuff in graduate studies.

Then again, maybe it's a normal part of the curriculum, and I've just been out of school for too long...

It's probably part of the curriculum. What's her course?

Applied Math, I think. Odd stuff, really. Somewhat complex.

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