Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Education » math-teach

Topic: "bales of hay" problem
Replies: 5   Last Post: Nov 10, 1997 4:54 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Ted Alper

Posts: 118
Registered: 12/6/04
Re: more "bales of hay" urgh!
Posted: Nov 10, 1997 6:15 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

well, here's the way I would do it, but i'm sure
there are many variants:



> suppose five bales of hay are weighed two at a time in all
> possible ways. The weights in lbs are 110, 112, 113,114,115,
> 116,117,118,120 and 121. How much does each bale weigh?


as usual, call the five individual weights a,b,c,d, and e
where a is less than or equal to b which is
less than or equal to c which is
less than or equal to d which is
less than or equal to e

some things I know for sure -- the lightest two-bale
combination is a+b and the heaviest is d+e. In fact,
I even know that the SECOND lightest two-bale combination must be a+c
[do you see why?] and the SECOND heaviest must be c+e.

[unfortunately, I can't tell the third-lightest, or third-heaviest,
without knowing the specific weights -- though I do know
that the third-lightest must be EITHER a+d or b+c, etc.]

Still, I already know quite a lot:
a+b=110
a+c=112
c+e=120
d+e=121

from the first two I can see that c=b+2;
from the last two I can see that d=c+1 [and since c=b+2, I can say d=b+3]
from the MIDDLE two I can see that e=a+8

putting this all together, I have
a is the smallest weight, b comes next,
then b+2 (i.e. "c"), b+3 ("d") and finally a+8 ("e").

Since I also know that a+b=110, this tells me
a+d=a+b+3=113 and b+e=a+b+8=118.

I still don't know quite which pairs of bales yield
the totals 114,115,116,117, but I do know that they
must come from the four remaining pairs:

a+e=2a+8= one of {114,115,116,117}
b+c=2b+2= one of {114,115,116,117}
b+d=2b+3= one of {114,115,116,117}
c+d=2b+5= one of {114,115,116,117}

but note that c+d must be 3 greater than b+c and, of
the four numbers as yet unassigned, the only one that's
3 greater than another is 117.
this quickly shows that b+c must be 114 and c+d must be 117,
i.e. 2b+2=114, so b=56. a=110-56, i.e. 54, and everything checks.
the weights are 54,56,58,59,62



Ted Alper





Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.