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Time to Weigh the Hippos Martha, a caregiver at a wild animal park, is picking up four hippos. She must weigh each of them before she can take them. The only scale large enough to weigh them is a truck scale. The minimum weight that a truck scale can weigh is 300 kg. However, each individual hippo weighs less than 300 kg. She weighs the hippos in every possible pair so that she can figure out the individual weights afterwards. She gets these weights: 312 kg, 356 kg, 378 kg, 444 kg, and 466 kg, but the weight of the sixth pair breaks the scale. What are the weights of the individual hippos? What is the weight of the last pair of hippos who broke the scale?

In determining a solution for this problem we had to think of it as an algebraic equation. In our solution, we assigned variables to the weight of the hippos from the lightest “a” to to the heaviest “d”. Let the weight of one hippo be A(lightest), the weight of another hippo be B, and so forth for the weight of each hippo. Weight of hippos = A, B, C, D 2) She weighs all possible pairs.(a+b) ,(a+c)

a + d, b + c, b + d, and c + d.

So, based on the pairs of weights that are given in the problem:

We assume that:

a+b = 312 kg

a+c = 356 kg

a+d = 378 kg

b+c = 444 kg

b+d = 466 kg

c+d = ??? kg

By using logic a +b is the lightest combination, a+c is the

Adding the first five equations above:

a+b + a + c + a + d + b + c + b + d = 312 + 356 + 378 +444 + 466

Simplify that:

3a + 3b + 2c + 2d = 1956

Subtracting off the second and third equations twice from the equation directly above this:

3a + 3b + 2c + 2d - 2(a+c) - 2(a+d) = 1956 - 2(356) - 2(378)


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