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Here are six solutions. Notice that among the six solutions,
there ar ethree different answers. Are they all correct?
Is it possible that this problem has more than one solution?
If not, where did the wrong ones go wrong?
The first three solutions (the most recent, June 2005)
come all the way from Hong Kong:
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Hi, I'm Ethan and I'm a sixth grader from Hong Kong International School.
I think the answer for Ballons is 4.
I think this because the data I collected from the notecards state that Anna has two more than Eric,
Hort has five more than David, Eric has less than 20, Hort has more than Anna,
and David has half of Anna's.
Immedietly I knew that David has the least.
So I started with David having 1.
However, the numbers don't fit for everyone. So I tried 2 for David, and it fit for everyone. |
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Hi.
We're students in grade 6 from Hong Kong International School,
and we think we have found the answer to Eric's Balloons.
We knew that Dave had half the amount of balloons has Anna.
If Hortensia had 5 more than Dave and more than Anna, we knew that Anna had to have less than 10 balloons.
We therefore figured that Anna had 8 balloons. That would mean Dave had 4 balloons.
If Hortensia had 5 more balloons than Dave, she would than have 9 balloons-more than Anna.
This would justify that Anna really did have 8 balloons,
which would mean that Eric had 6 baloons-2 fewer than Anna.
Thanks for taking time to read this. |
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My group figured out the answer to Balloons. Here it is.
Answer: Eric has 4 balloons.
I figured it out by changing all the clues to algebraic equations:
A/2=D
A-2=E
D+5=H
H>A
Sincerely,
Janet Kwan
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Great! Thanks!
Now, it's not clear how Janet's group went from the algebra to the answer. And how did they use the inequality "H > A"?
The next three solutions may be from the
same class in Clarkstown, Michigan, in Sashabaw (or Sashabae) Middle
School. Thanks to all the group members who submitted!
The first one, where perhaps they stopped
typing a little early...
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We figured the problem out by guess and check. We started
with the # 20 and found out that it didn't work then went to 10 and
found that it didn't work then finally we went to 6 and found out
that it did fit with all the clues.
Eric has 4 balloons
Tony Coleman
Shane Carlson
Bianca Gonzalez
Tyler Kenerson
Tim Stolzenfeld
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The second one, also guess and
check... |
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We found the answer by using the guess and check method. We
tried many answers until we came upon the write one. We knew that
Anna had to have an even amount of balloons. We also knew that Eric
couldn't have had more than 20 balloons. Here are our answers:
Eric-4
David-3
Anna-6
Hortensia-8
Sincerely,
Ryan Addis
Jordan Hefty
Katie Laskowska
Zack Polson
Sean Goebel |
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The third is different... |
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First we wrote equations for each card
and added them to the chart. Then we realized that Hortensia has to
have 5 more then David. To get Anna's amount then we double David's
amount then we subtracted two from Anna's. Hortensia has 10, Anna has
10, Eric has 8, and David has 5.
From: Taylor Rademacher,
Cathy Fick
Jasmin Olinger
Lisa Savino
Garrett Roberts
Mitchell Kuch |
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Okay, folks, what do you think? Here we have
more than one solution. Are they both right? Are there other
solutions? If somebody was wrong, what went wrong? Write to
with your comments! |
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