# Thread: Problem Solving: Hefty Hippos Weight Watchers

1. ## Problem Solving: Hefty Hippos Weight Watchers

As an incentive to lose weight, 5 members of the Hefty Hippos Weight Watchers' Club decided that the hippo who lost the most weight by a certain date would win a prize. When the day came to determine the winner, the hippos went down to the nearby warehouse to use the heavy duty scales. But the scales started at 300kg, more than any of them weighted. What were they to do? Heloise, a hippo of some considerable mathematical prowess, came up with a solution. She said that they merely needed to weigh all possible pairs of hippos and then determine each hippo's weight from all these weighings. The weights of all possible pairs were as follows (in kg): 381, 383, 386, 388, 391, 391, 292, 295, 397, 402. How much did each hippo weigh? (You may assume that all weights are whole numbers.)

I would greatly appreciate any help on this problem. Thanks!

2. Pair of weight in ascending order are
292, 295, 381, 383, 386, 388, 391, 391, 397, 402

Let weight of 5 hipos in ascending order be a, b,c,d and e
i.e
a<b<c<d<e

Now a+b+c+d+e =one fourth of sum of all 10 weights given above (think why )

so a+b+c+d+e = 3706/4 = 926.5

But question says that all weight are in whole number.
So you have made a mistake in typing the pair of weights.

Anyways.
let me give you a brief approach to solve this problem.

From above we have a+b+c+d+e.

Now a+b = 292 (since a and b are two lowest weights, their sum will be the lowest paired weight

d+e = 402 (since d and e are largest, their sum is largest)

so we get a+b+d+e = 694

so c can be found out
c = 926.5 - 694 = 232.5

Now second highest sum 397 should be of pair e and c
so e + c = 397
so e = 397 - 232.5 = 164.5 which is less than c
so we can see that numbers are all messed up

But now you have an idea of approach to solve this question.
If you can re check your question and give me right numbers, I can provide complete solution.

3. Yeah, I accidentally typed my numbers in wrong.

The numbers are...

381, 383, 386, 388, 391, 392, 393, 395, 397, and 402.

Thanks for the help!

4. Sum of
381, 383, 386, 388, 391, 392, 393, 395, 397, and 402.

is 3908

So continuing the approach I used above
a+b+c+d+e = 3908/4 =977

d+e = 402
a+b = 381

so a+b+d+e = 381+402 = 783

so c = 977 - 783 = 194

Now e+c = 397 (second largest pair)

so e = 397 - 194 = 203

d = 402 - 203 = 199

Now a + c = 383 (second smallest pair)

so a = 383 - 194 = 189

b = 381 - 189 = 192

so five weights are
189, 192, 194, 199 and 203 Answer