Puzzle:
Suppose five bales of hay are weighed two at a time in all possible ways. The weights in pounds are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121.
How much does each bale weigh?
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Solution:
They weigh 54, 56, 58, 59, 62 pounds.
Let's assume that the weight of five bales are B1, B2, B3, B4 and B5 pounds respectively. Also, B1 <= B2 <= B3 <= B4 <= B5
It is given that five bales of hay are weighed two at a time in all possible ways. It means that each of the bale is weighted four times.
Thus,
4*(B1 + B2 + B3 + B4 + B5) = (110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121)
4*(B1 + B2 + B3 + B4 + B5) = 1156
(B1 + B2 + B3 + B4 + B5) = 289 pounds
Now, B1 and B2 must add to 110 as they are the lightest one.
B1 + B2 = 110
Similarly, B4 and B5 must add to 121 as they are the heaviest one.
B4 + B5 = 121
From above three equation, we get B3 = 58 pounds
Also, it is obvious that B1 and B3 will add to 112 - the next possible higher value. Similarly, B3 and B5 will add to 120 - the next possible lower value.
B1 + B3 = 112
B3 + B5 = 120
Substituting B3 = 58, we get B1 = 54 and B5 = 62
From 2 & 3 equations, we get B2 = 56 and B4 = 59
Hence, the weight of five bales are 54, 56, 58, 59 and 62 pounds.
They weigh 54, 56, 58, 59, 62 pounds.
Let's assume that the weight of five bales are B1, B2, B3, B4 and B5 pounds respectively. Also, B1 <= B2 <= B3 <= B4 <= B5
It is given that five bales of hay are weighed two at a time in all possible ways. It means that each of the bale is weighted four times.
Thus,
4*(B1 + B2 + B3 + B4 + B5) = (110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121)
4*(B1 + B2 + B3 + B4 + B5) = 1156
(B1 + B2 + B3 + B4 + B5) = 289 pounds
Now, B1 and B2 must add to 110 as they are the lightest one.
B1 + B2 = 110
Similarly, B4 and B5 must add to 121 as they are the heaviest one.
B4 + B5 = 121
From above three equation, we get B3 = 58 pounds
Also, it is obvious that B1 and B3 will add to 112 - the next possible higher value. Similarly, B3 and B5 will add to 120 - the next possible lower value.
B1 + B3 = 112
B3 + B5 = 120
Substituting B3 = 58, we get B1 = 54 and B5 = 62
From 2 & 3 equations, we get B2 = 56 and B4 = 59
Hence, the weight of five bales are 54, 56, 58, 59 and 62 pounds.
18 comments:
Answers r
54, 56, 58, 59, 62
Sum of all totals provided = 1156 which is 4 times sum of all the numbers.
hence A+B+C+D+E = 289
sum of two lowest == 110
sum of two highest == 121
Hence middle number C = 58
WIth the given sum of numbers, we can find the other numbers.
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