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Essay text:
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)
Simplify that:
-a + 3b = 488
Adding the first equation:
-a+3b + (a+b) = 488 + 312
Simplify:
4b = 800
Dividing that one by 4:
b = 200 -- we now have our first hippo weight!
Using the very first equation above:
a+b = 312
Plug in the value for "b":
a+200 = 312
a = 112
Using the fourth equation:
b+c = 444
Plug in "b":
200+c = 444
c = 244
Using the fifth equation:
b+d = 466
Plug in "b":
200+d = 466
d = 266
So the individual weights are 112, 200, 244, 266...
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