• Mar122012

POSTED AT 03:31 PM

A few members of the class have already submitted their solutions to this problem.  Nice work!  Here are the hidden questions that were put together by members of the team.  As you answer these questions, you will get closer to the answer.

How many seconds does it take Ulrike to fill and tie one balloon?
How many seconds does it take Lars to fill and tie one balloon?
How many seconds are in 14 minutes?
How many balloons can Ulrike finish in 14 minutes (__________ seconds)?
How many balloons can Lars finish in 14 minutes (___________seconds)?
How many balloons do Ulrike and Lars together complete in 14 minutes?
How many balloons will be in each of 3 bags if there is an equal number of balloons in each bag?

Mar092012

POSTED AT 05:23 PM

Here are  Daniel G.'s and James M's solutions to the problem. Congratulations on having your answers chosen as examples!
Daniel's:

the jump rope was 30 feet long in the beginning of the day.

## Explanation:

10 divided by 2=5 5+10=15 15X2=30

the length of the jump rope started off being 30 feet long. First I paid no attention to the half of the rope that got thrown away. Next I realized the other half got cut into a third and the longer part was ten feet long. The third must be five feet long because one third is half of two thirds and five is half of ten. If one half is 15 feet long then the other half is 15 feet long so 15X2=30. Thats why the jump rope is 30 feet long.
James':

The jump rope was 30 feet when she began.

## Explanation:

First, the jump rope was cut in half.  Then, one half was cut 1/3 from the end.  The rest she had left to play with was 10 feet long, the longer section.  This means that the 10 feet piece of rope was equal to 2/3.  If 2/3 is 10 feet, then 1/3 is 5 ft.  This is because we know that each 1/3 of the rope is equal in length, and 2/3 of that half equals 10 feet, therefore each 1/3 is 5 ft.  This also means that 1/2 of the rope is equal to 15 ft. because 1/3 (5ft) + 2/3 (10ft) = 1 whole (15 ft).  If 15 ft is the length of half the rope then the entire rope is 2 x 15 ft or 30 ft.

Feb252012

POSTED AT 05:51 PM

Congratulations to Evan!  His solution has been, once again, highlighted as an example solution on the Math Forum web site.  Here is his solution:
Farmer Mead can build 9 rectangular pens that use all 36 meters. The amount of frogs that each pen can hold if each frog needs 1 square meter is equal to the area of the pen (area=length x width) The maximum number of frogs that can be in any pen is 81 frogs because the largest pen in 9mx9m - proving that a square is a rectangle, giving the most space.

To  create  at least 4 rectangular pens that all used 36 meters of fencing, first I picked lengths for each side that added up to 36 - remembering that rectangles had 2 sets of equal sides.
Side a and b were equal and side c and d were equal, so using that example,
a=17m and b=17m, c=1m and d=1m - adding all side = 36m. (17x2 + 1x2 = 36) .  So my first pen was 17mx1m.   I tried other combinations with the same strategy, and picked 2mx16m because 2x2+16x2=36.  Next I tried 3mx15m because 3x2+15x2=36.  I realized that this would work as long I kept decreasing the long side by 1 and increasing the short side by 1.  Therefore, pen #4, would be 4mx14m, the fifth pen option would be 5mx13m, pen six would be 6mx12m, pen 7 would be created by using 7mx11m, pen 8 would be 8mx10m and the last pen would be 9mx9m.

If each frog needs one square meter of area, then according the area formula of length x width, my pens can hold the following number of frogs:
pen 1 can hold 17 frogs (17m x 1m=17m2)
pen 2 can hold 32 frogs (16m x 2m=32m2)
pen 3 can hold 45 frogs (15m x 3m =45m2)
pen  4 can hold 56 frogs (14m x 4m=56m2)
pen 5 can hold 65 frogs (13m x 5m=65m2)
pen 6 can hold 72 frogs (12m x 6m=72m2)
pen 7 can hold 77 frogs (11m x 7m=77m2)
pen 8 can hold 80 frogs (10m x 8m=80m2)
pen 9 can hold 81 frogs (9mx9m=81m2)

Extra:  As you can see above the pen that holds the largest number of frogs is the pen that is 9m x 9m, pen #9.  Since the pen has to be a rectangle, which means that 2 of the sides must be equal and parallel and the other 2 sides must be equal and parallel, and a pen that is 9m x 9m is a square, then a square must also be a rectangle.  The rectangle that will be the largest is one that has equal sides all around. This pen would hold 81 frogs because 9m x 9m =81m2 = so this pen would hold 81 frogs.

Feb172012

POSTED AT 05:29 AM

We have some really good entries for Farmer Mead's Frogs.  Good entries include words like dimensions, length, width, frogs, rectangular, perimeter, area, strategy.

Be sure to include an explanation of how you know that each of your pens uses all 36 meters of fence and how you know the number of frogs each pen holds.  If you got stuck, explain how you got stuck and "unstuck".
Hint: a 4 x 9 pen does NOT use 36 meters of fence.  Can you see why?

Feb112012

POSTED AT 05:55 AM

Here are some hints for the POW this week from members of the class:
Make sure that each of your rectangular pens is enclosed by 36 meters of fence.
Remember that a square is a kind of rectangle.
The area inside the pens will change depending on the dimensions of the rectangles.
There are more than four possibilities.
Once you figure out what all the possible rectangular pens are, the extra is easier to figure out.

Feb092012

POSTED AT 05:12 PM

Our present Math Forum problem is about finding good dimensions for a frog pen.  Here are some I notice and I wonder comments from the class:
 I notice...that farmer Mead has to use 36 meters of fencing to make her pen.that the shape of the pen has to be a rectangle (and that includes possible squares).that one frog needs 1 square meter of living space.I wonder...what the possible lengths and widths of all the rectangles would behow many frogs farmer Mead hasif some rectangles with a perimeter of 36 meters would have more area than otherswhat operations I will need to use to solve this problem

Dec102011

POSTED AT 06:14 AM

Great news!  Two members of the team have had their solutions to the Anthony's Rolls problem highlighted as examples on the Math Forum site.  To see their solutions, and to see solutions from other students around the country, go to
Click on "Latest Solutions"  and read the introduction.

Dec082011

POSTED AT 05:13 AM

Your score on "reflection" is about your thinking.  One way to improve your reflection score is to explain step by step how you checked your answer.  Use sequence words like first, next, then, finally. Another way to improve your reflection score is to explain where you got stuck or "unstuck" on a problem.  Any time you talk about your own thinking, you are reflecting.

Dec052011

POSTED AT 02:44 PM

One strategy to use when solving the Anthony's rolls problem is to break the problem into pieces and solve the problem piece by piece.  We could, for instance, look at how many rolls Anthony has to make for the adults. There are 12 adults.  Anthony has to make three rolls for every two adults. 12 divided by 2 gives us the number of groups of two adults. There are six groups and each group gets three rolls.  So 6 x 3 = 18 rolls for the adults. We have solved part of the problem!

Dec032011

POSTED AT 09:01 AM