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Weighing Kittens - posted March 4, 2002

My neighbor, George, owns seven kittens. The three oldest kittens are Liberty, Charlie and Alexander. The sum of their weights equals the sum of the weights of the four younger kittens (Sassy, Moonlight, Bubba and Randy).

Alexander weighs 2 oz. more than Charlie, who in turn weighs 6 oz. more than Liberty.

Bubba weighs 12 oz. more than Randy, who is 3 oz. more than Moonlight and 13 oz. more than Sassy.

Charlie and Bubba are almost the same weight, but Charlie is the heavier of the two.

If the kittens' weights are all whole ounces, what is the least that each kitten could weigh and still meet all of the given requirements?

The idea for this problem was suggested by George Miller.

Hint 1:
If you represent each kitten's weight by its first initial and make as many equations as possible from the information given in the problem, you could try to write L + C + A all in terms of C and write S + M + B + R all in terms of B.
Hint 2:
Once you have a statement in terms of C and B, there are a couple of different ways that you can figure out what C or B might be. Since you know that C is a little bigger than B (since Charlie and Bubba are almost the same weight, but Charlie is heavier), you can figure out some reasonable possibilities to try. Or you could actually let C equal B plus a little bit, say D (for the difference). Now C = B + D.
Hint 3:
If you decide to stick with C and B, look carefully at the statement you ended up with. When you know that you're going to have to make guesses and check them, knowing that the answer has to be divisible by a certain number can dramatically reduce the range of possibilities.
Hint 4:
If you said that C = B + D, you should end up with a statement in terms of B and D. You know that D, the difference between them, has to be a small whole number.
Ans 1:
When we put everything in terms of C and B, we got
  (C - 6) + C + (C + 2) = (B - 25) + (B - 15) + B + (B - 12)
Ans 2:
We eventually got 3C - 4 = 4B - 52. This tells us that C is divisible by 4 and greater than 48 and that B is divisible by 3.
Ans 3:
We found that Charlie's weight is 52 ounces.
Solution:
There are two solutions shown below. Compare them to your own and tell us what you think - is yours like either one, or a bit different? Did anything about these solutions surprise you? Did you learn anything? Do you have any questions?


Solution 1:

Liberty weighs 46 ounces. Charlie weighs 52 ounces. Alexander weighs 54 ounces.

Sassy weighs 26 ounces. Moonlight weighs 36 ounces. Bubba weighs 51 ounces. Randy weighs 39 ounces.

Represent the weight of each kitten by the first initial of their name. We know that

  1) l + c + a = s + m + b + r

  2) a = c + 2
  3) c = l + 6

  4) b = r + 12
  5) r = m + 3
  6) r = s + 13

  7) c almost equals b, but c > b
Since we know that c and b are almost equal, let's take equation 1 and put everything on the left in terms of c and everything on the right in terms of b and see what happens.

From 3, we know that l = c - 6, and we already know from 1 that a = c + 2.

From 4, we know that r = b - 12, and we can stick that into 5 and 6 to get m = b - 15 and s = b - 25. So we have

  (c - 6) + c + (c + 2) = (b - 25) + (b - 15) + b + (b + 12)
                 3c - 4 = 4b - 52
                3c
            8)  -- + 12 = b
                4
I needed to find values for c and b that are really close that make that a true statement. I know that c has to be divisible by 4 so that we'll get a whole number answer.
    c -> b
   32 -> 36     b needs to be smaller than c
   60 -> 57     too much smaller (I think)
   56 -> 54     not bad, but maybe I can get the weights even closer
   52 -> 51
That looks just right to me! I worked out the rest of the weights from there using the equations that I had set up.

Liberty weighs 46 ounces. Charlie weighs 52 ounces. Alexander weighs 54 ounces.

Sassy weighs 26 ounces. Moonlight weighs 36 ounces. Bubba weighs 51 ounces. Randy weighs 39 ounces.

To check it, we add together the older kittens and the younger kittens and see if those sums of their weights are equal.

               ?
  46 + 52 + 54 = 26 + 36 + 51 + 39
           152 = 152


Solution 2:

Liberty = 46 ounces, Charlie = 52 ounces, Alexander = 54 ounces, Sassy = 26 ounces, Moonlight = 36 ounces, Bubba = 51 ounces, Randy = 39 ounces.

I know that Charlie and Bubba weigh almost the same amount, but that Charlie is heavier. I will let b be Bubba's weight and b + d be Charlie's weight. d is the difference between them.

l + (b + d) + a = s + m + b + r.                                    
[1]

Alexander is 2 ounces more than Charlie, so that's a = (b + d) + 2. 
[2]
Charlie is 6 ounces more than Liberty, so (b + d) = l + 6.          
[3]
Bubba is 12 ounces more than Randy, so b = r + 12.                  
[4]
Randy is 3 ounces more than Moonlight, so r = m + 3.                
[5]
Randy is also 13 ounces more than Sassy, so r = s + 13.             
[6]
I want to make the left side of 1 all in terms of (b + d), and the right side all in terms of b. I can rewrite [3] as l = (b + d) - 6. The left side is now
  (b + d) - 6 + (b + d) + (b + d) + 2
I can also rewrite [4] as r = b - 12. Then [5] becomes b - 12 = m + 3, or m = b - 15. [6] becomes b - 12 = s + 13, or s = b - 25. The left side of [1] now becomes
  (b - 25) + (b - 15) + b + (b - 12)
[1] is now
  b + d - 6 + b + d + b + d + 2 = b - 25 + b - 15 + b + b - 12
                    3b + 3d - 4 = 4b - 52
                   3b + 3d + 48 = 4b
                        3d + 48 = b
Since we know that d is pretty small, I will assume that d is 1 ounce (since the weights are all in whole ounces, the smallest difference between two weights could be 1 ounce). This means that b is 51 ounces, and c is b + 1, which is 52 ounces. We can figure out the rest of the kittens' weights from there.

Liberty = 46 ounces
Charlie = 52 ounces
Alexander = 54 ounces
Sassy = 26 ounces
Moonlight = 36 ounces
Bubba = 51 ounces
Randy = 39 ounces

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