word problem

Note: 

I think this problem is a variation on this theme :

http://web2.0calc.com/questions/how-much-do-they-weigh

Here's my solution :

I'm assuming that we have 5 pumpkins and we are weighing them in couples and all the pumpkins are of different weights

Let  a > b > c > c > d > e       where these are the weights of the pumpkins

And  each is weighed 4 times.....

So we have that     ( sum of the weights)/4  = 1156/4  ⇒  a + b + c + d + e   = 289 

And here are the equations that we are sure of

The two heaviest  = a + b  = 121  (1)

Since a > b, then a + c > b + c....so a + c is the second heaviest weight

a + c  =  120  (2)

The two lightest =  d + e  =  110   (3)

Since c > d, then c + e >  d + e  ......so c + e is the second lightest weight

c + e  =  112  (4)

Subtract  (2) from (1)  =  b - c  = 1  ⇒  b  = c + 1

Subtract  (4) from (3)  =  c - d  = 2  ⇒  c  =  d + 2

So this implies that   b =  d + 3

Now.....add (1) and (2)

2a + b + c  =  241   ⇒   b + c    = 241 - 2a  ⇒  (d + 3) + (d + 2)  = 241 - 2a  ⇒

2d + 5  = 241 - 2a    →  2a + 2d  = 236  ⇒  a + d  = 118  ⇒  a = 118 - d

Similarly.....add (3) + (4)  and we have that

c + d  + 2e  =  222  ⇒  (d + 2) + d =  222 - 2e ⇒  2d = 220 - 2e ⇒ d = 110 - e ⇒ e = 110 - d

So

a + b + c + d + e  = 289    and substituitng, we have

(118 - d) + (d + 3) + (d + 2)  + d + (110 - d)  = 289

d + 233  =  289

d = 56 lbs

So

a = 118 - d  =  62 lbs

b = d + 3 =  59 lbs

c = d + 2  =  58 lbs

d = 56 lbs

e = 110 - d = 54 lbs

John weighed the pumpkins two at a time. After he is done weighing the pumpkins he observed the following weights:

110, 112, 113, 114, 115, 116, 117, 118, 120, and 121 pounds. 
How much did each pumpkin weigh? 
Explain Clearly how you got the answer?

see https://www.algebra.com/algebra/homework/Exponents/Exponents.faq.question.70198.html