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POW 8
There are five bales of hay, a, b, c, d, and e. Instead of weighing them individually they weighed them in combinations of two bales. The weights in numerical order in Kg are 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91. The question asks you to find the sum in the combinations of the bales of hay.
First I thought of every combination of bales. I ended up getting ten different combinations. The weights in Kg are listed in order of numbers not in order of weight in pairs of bales. So I knew I needed to figure out the heaviest and the lightest bales. So I switched the numbers 1-5 for the variables a-e. So a=1, b=2, c=3, d=4, and e=5. So now I am just going to mess with numbers until I find numbers that will work. The smallest combination of bales is 80 so I will divide that by 2 and I get 40 so I know the both numbers will be around 40. So I am just going to try moving the numbers around a little bit and see what I can get.
a + b
39 + 41 = 80
a + c
39 + 43 = 82
a + d
39 + 44 = 83
a + e
39 + 47 = 86
b + c
41 + 43 = 84
b + d
41 + 44 = 85
b + e
41 + 47 = 88
c + d
43 + 44 = 87
c + e
43 + 47 = 90
d + e
44 + 47 = 91
Well