Masalah Non-Rutin
Masalah 1
OPEN - Jumping Rope
My jumping rope was cut in half
half was thrown away.
The other half was cut again
one third along the way.
The longer part (ten feet long)
is what I use to play.
How long was my jumping rope
when I began today?
Comments and Sample Solutions
Here are a few of the best solutions we received this week.
They were chosen because they are well written and/or show an interesting
approach to the problem. Be sure to check back later for more mentor comments,
the full highlights, and the list of correct submitters.
The jump rope started out as 30 ft. long.
1. First, I will start out by making a prediction.I predict that
the jump rope is about 30 ft. long because I took the longer part of
the rope when it was cut into thirds and multiplied that 10 ft. by
the 3 thirds.
2. I will start out by making a drawing of the rope being cut with
a fraction and how long it is using feet.
3. __________________ 1/1
_________ 1/2 -1/2
____________ 2/3 10 ft.
______ 1/3 5 ft.
I started with one whole rope. I subtracted 1/2 like it says in
the problem. Then, I divided the 1/2 left over of the rope into
thirds. Since the problem states that they cut 1/3 of the left over
rope, the longer part would be 2/3 and it would add up to 10 feet
long. I divided the 10 feet by the 2 to find out how many feet were
in each third. 10 divided by 2=5 feet per third. 3/3 would then be 15
feet because 5 feet per third multiplied by 3 thirds equals 15 feet. I
multiplied 15 feet by 2 because of the 1/2 of the jump rope that was
thrown away. 15x2=30 feet. Therefore, the original jump rope would be
30 feet long.
The original rope was 30 feet long.
I drew a rope. Then I put a line in the middle of the rope and
colored the side of the rope that was thrown away.Then I divided the
rope into thirds. Then I cut off 1/3 of the rope. The longer part is
ten feet long. When I cut the rope into thirds, there were three even
sections that were each five feet long. There is fifteen feet on each
side of the original rope because three times five equals fifteen.
The rope is 30 feet long since one half of the rope has three
segments of rope that equal five feet each. So, if you add each half
of the rope together it equals 30 feet.
Masalah 2
LET
ME COUNT THE WAYS
Tina has 3 shirts-red, blue, and
yellow. She also has 2 skirts-black and white. Each day, she
wears a different shirt with a
different skirt.
For how many days can she do this before she must repeat and
Draw pictures of Tina and color if
you wish.
There is also a chart called a TREE
that can help Tina to know which outfit is next to wear. It looks like this:
Shirts
Skirts
|-----Black
|---Red-------|
|
|-----White
|
|
|
|-----Black
|--Blue-------|
|
|-----White
|
|
|
|-----Black
|--Yellow-----|
|-----White
So, you can see all the
possibilities and even count them.
Uh oh, Tina accidentally ripped her yellow shirt and cannot wear
it anymore. How many outfits does
she have now? Draw a tree.
Masalah 3
Counting Chicken Wings
At Annie's Home-Cooked Chicken Wings Restaurant, chicken
wings are served by the bucket. The Biggest Bucket O' Wings is
really big! Let's figure out how many wings are in it.
If they're removed two at a time, one wing will be left. If
they're removed three at a time, two wings will remain. If
they're removed four, five, or six at a time, then three,
four, and five wings, respectively, will remain. If they're
taken out seven at a time, no wings will be left over.
What's the smallest possible number of wings that could be
in the bucket? How do you know?
Comments
Counting Chicken Wings was our first problem written with hints,
ways to check your answer, examples of possible solutions, and a page
with more ideas. We knew that this new format would take some getting
used to! When we wrote back to you this time, we sometimes asked you
to look at the hints that we had already written instead of giving you
personal hints about how to improve your explanation. Thank you to
those of you who really looked through everything and a special thank
you to those students who gave us feedback by writing in the comment
areas.
One thing agreed on by students solving the Counting Chicken Wings
problem was that it certainly is a big bucket o' wings! There were a
variety of ways that the facts given in the problem could be used.
Many students started with the fact that the number of wings must be a
multiple of 7 because when you took out the wings 7 at a time there
would be nothing left in the bucket.
Sayuri Khandavilli of Woodrow Wilson Middle School and the team of
Kayla G., Christi S, and Kanushri W. of Quest Academy started with
that fact, wrote down a list of the multiples of 7, and then used each
of the other facts to eliminate numbers from the list. Sathya Sankaran
of Whitney Elementary School started by listing the multiples of 7 but
then showed in a chart why each one didn't work until reaching 119.
Isa Marin of National Cathedral School-Girls lets you hear her
thinking as she works through the problem. She starts with a short
list of multiples of 7 and tells you why she makes it longer as she
tries to solve the problem. In comparison to these rather long
explanations, there is the very efficient method written by Ayushman
Ghosh of Bourne Middle School. He thought of using the number facts
that narrow down the possibilities the quickest.
Make sure to take a look at the solutions written by Andrei Lazanu
of School No. 205, Natalie Krzyzanowski of Washington Junior High
School, and Abhay Dang of Manav Sthali School. They are all examples
of how using a little algebraic notation can also help to shorten the
problem.
As we continue with problems this year, you'll notice many will be
in this "new" format while a few will be in our previous
format. Either way, we hope you enjoy the challenge.
Solutions: 1
The smallest number of wings that could be in the bucket is 119.
The most important clue was that if 7 were taken at a time then none would be left. So, the number of
wings should be a multiple of 7.
I listed the multiples of 7 till 140
07 14 21 28 35 42 49 56 63 70 77 84
91 98 105 112 119 126 133 ................
From the problem:
When 2 wings are taken at a time 1 was left (this gives the clue that the number must be odd)
So this eliminates all the even numbers
07 21 35 49 63 77
91 105 119 133 ................
When 3 wings are taken at a time 2 were left (this gives the clue that it should not be the multiple of 3)
So this eliminates all the multiples of 3 :
07 35 49 77
91 119 133 ...
When 4 wings are taken at a time 3 were left :
So this eliminates the numbers which do not have a remainder of 3:
07 35 119
When 5 wings are taken at a time 4 were left :
So this give the clue that the number should not be the multiple of 5 and has a remainder of 4 .
So this eliminates the numbers which do not have a remainder of 4 and multiples of 5 :
119
When 6 wings are taken at a time 5 were left :
119 when divided by 6 leaves a remainder of 3
ANSWER : The smallest number of wings that could be in the bucket is 119.
Solutions: 2
I started by making a list of multiples of seven, since it says if
you take it out by sevens you get nothing left over. I checked each
number to see if it was the number I was looking for. After a wrong
number I will give the first number that is not compatible.
O = Right X = Wrong.
1. 7 X 3
2. 14 X 2
3. 21 X 3
4. 28 X 2
5. 35 X 4
6. 42 X 2
7. 49 X 3
8. 56 X 2
9. 63 X 3
10. 70 X 2
11. 77 X 4
12. 84 X 2
13. 91 X 3
14. 98 X 2
15. 105 X 5
16. 112 X 2
17. 119 O
So now I know that there are 119 chicken wings (at the least)
in the biggest bucket of wings.