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?

For Solution SCROLL DOWN...

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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.

## 20 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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