November Problem Solving
Problem #1 This problem comes from a blog post of Dan Meyer and was contributed from
another teachers. The original activity can be found at Correlation Stations
 Measure or calculate the answer to each question.
 Write the answer in the box.
 PREDICT THE SHAPE OF THE GRAPH
 Plot your answers on the graph paper.
Station
Plot on X-axis
Plot on Y-axis
Predict the shape of
the graph.
Number of letters in
your last name
Station #1
Number of letters in
your first name
Station #2
Arm span in cm
Height in cm
Date of your
birthday.
EX: Oct 10 = 10
Number of days left
in the month after
your birthday.
EX: Oct 10 = 20
Station #4 You are
going to take a trip to
a town that is 240
miles away.
Your speed in miles
per hour
The time in hours
that it takes to get to
the town
Station #5 You have
120 feet of fence to
build a rectangular
garden.
The length of one
side of the garden
The area of the
garden.
Station #3
I purchased the flip chart graph paper can be purchased at Office Max.
What are some other questions that you could ask to generate graphs of interest to
students?
Answers:
Station #1 No correlation
Station #2 Positive
Station #3 Negative
Station #4 Inverse
Station #5 quadratic
Problem #2 This problem comes from Mathematics Teaching in
the Middle School, December 2011/January 2012, Solve it!
Student Thinking, “Squares and Pegs” p 266-267. This is a
regular feature where a problem that was given in the journal 1
year ago is shown with student responses. These are great
resource for “problem solving” problems.
Question #1 Use the 5 X 5 peg geoboard, like this. How many
different size squares can you create? Assume that you must use
pegs (or dots) as vertices.
Question #2: How many squares could you create?
The solution is found on the document from Mathematics Teaching in the Middle
School.
There are 8 different squares that can be created. It is helpful to label the squares by
the length of the sides.
1X1= 16
2X2 = 9
3X3 = 4
1X1 = 1
SQ root 2 = 9
SQ root 5 = 8
SQ root 8 = 1
SQ root 10 = 2
Total number of squares is 50.
Also see Chuck’s geogebra file Nov#2-SQS.ggb
Solutions http://mathforum.org/trscavo/geoboards/geobd3.html
Problem #3 This problem comes from the
December Mathematics Teaching in the
Middle School, Solve it! Shaded Rectangles p
271
Draw a rectangle:
 Divide the sides (left and right) of the
height into 3 congruent line segments.
 Divide the sides of the base (top and
bottom) into 4 congruent line
segments.
 Draw a line segment to each point in
the rectangle from the point just below
the top left corner.
 Shade every other section, starting at
the top left and moving clockwise.
What fraction of the rectangle is shaded?
Does it make a difference if the rectangle is
drawn as show on the left or right?
Answer is 7/12 Below is one of may ways to
solve the problem.
Also see Chuck’s geogebra solution “Nov#3Rect%Shaded.ggb”
Problem #4
A Classic Problem:
If a hen and a half lay an egg and a half in a day and a half, how many hens are needed to lay
a dozen eggs in one day?
Or
If 6 cats can catch and kill 6 rats in 6 minutes, how many cats will it take to catch and kill 100
rats in 50 minutes? Reference Mathematics Teaching in the Middle School, May 2012, p 538543, “Proportioning Cats and Rats.”
The key to working with the numbers is that only two quantities can be changed at a time.
Hens Eggs Days
1.5
1.5
1.5
original problem
3
3
1.5
double hens, double eggs, days stay same
3
6
3
double days and eggs hens stay same
9
6
1
1/3 of days, triple the hens days stay equal
18
12
1
double hens and eggs days stay same.
Moscow Problems:
4. “The Moscow Puzzles” Page 96, #218 The Idler and the Devil
An idler sighed, “Everyone says, ‘We don’t need idlers. You are always in the way. Go to the
devil.’ But will the devil tell me to get rich?”
No sooner did the idler say this than the devil himself stood in front of him.
“Well,” said the devil, “the work I have for you is light, and you will get rich. Do you see the
bridge? Just walk across and I will double the money you have now. In fact, each time you
cross I will double your money.”
“You don’t say!”
“But there is one small thing. Since I am so generous you must give me $24 after each
crossing.”
The idler agreed. He crossed the bridge, stopped to count his money …. A miracle! It had
doubled.
He threw $24 to the devil and crossed again. His money doubled, he paid another $24,
crossed a third time. Again his money doubled. But now he had only $24, and he had to
give it all to the devil. The devil laughed and vanished.
The moral: When anyone gives you advice you should think before you act.
How much money did the idler start with?
Additional Questions:
If the idler had more money to begin with, could the idler have become “rich?”
Is there a breakeven point in this problem?
Answer:
We will also solve this problem backward, but verbally. Before the third
crossing the idler had $12. Adding the $24 he gave the devil after the
second crossing, he had $36, twice the $18 he had before the second
crossing. Adding $24 again, he had $42, twice the $21 he started with.
Another solution:
x2x2-244x – 48  4x -72  8x – 144 = 24
Chuck also has a method to solve:
Nov#4-ProblemAnswer.xlsx
5. “The Moscow Puzzles” Page 119, #282 A Child’s Age
A child’s age increased by 3 years gives a number which has an integral square root.
Decreased by 3 years, the child’s age gives the square root.
How old is the child? Find a similar relationship with a difference other than 3.
Additional Questions:
Can you solve this problem algebraically? If so, what topic in an Algebra course would lend
itself to including this problem?
Answer:
Squares which can be considered 3 years older than a child’s age are 4, 9,
16. Of these, only 9 gives its square root when you subtract 3 and again
3. The child’s age is 6.
With differences other than 3, we might have:
2 + 1 = age 3; 3 + 1 = 2 x 2
4 + 6 = age 10; 10 + 6 = 4 x 4
5 + 10 = age 15; 15 + 10 = 5 x 5
Chuck also shows a solution in file
Nov#5ProblemAnswers.xlsx
6. “The Moscow Puzzles” Page 97, #219 A Smart Boy
Three brothers shared 24 apples, each getting a number equal to his age 3 years before.
The youngest proposed a swap:
“I will keep only half the apples I got, and divide the rest between you two equally. But
then the middle brother, keeping half his accumulated apples, must divide the rest equally
between the oldest brother and me, and then the oldest brother must do the same.”
They agreed. The result was that each ended with 8 apples.
How old were the brothers?
Answer:
Before the oldest gave half his apples to his brothers he had 16; the
middle and the youngest had 4 each. Before the middle brother divided
his apples he had 8; this means the oldest had 16 – (1/2)(4) = 14 and the
youngest had 2. Before the youngest divided his apples he had 4; the
middle brother had 8 – (1/2)(2) = 7, and the oldest 13.
The youngest brother is 7, the middle brother is 10, and the oldest is 16.