Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Blue – Whole Number Problem Solving
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
1. Mrs. Crowe bought 4 shirts at $29 each, 6 pairs of pants at $17 per pair and a dress
for $54 before a sale. During the sale, each shirt was sold at $19, each pair of pants
at $12 and each dress at $49. How much money could she have saved if she had
bought all those items during the sale?
2. Mrs. Lin bought 20 m of cloth at $4 per meter. She tried to sew as many dresses as
possible from this piece of cloth, using 3 m of cloth for each dress. If she sold each
dress at $69, how much profit did she make altogether?
3. James bought 5 books and 6 magazines. Alvin bought 6 books and 5 magazines. Each
book cost $14 and each magazine cost half as much as each book. How much money
did both boys spend in all?
4. A transport company charges $1 for every pot safely delivered but pays a penalty of
$28 for every pot broken or lost. If it delivered 500 pots for a customer, lost 3 pots
and broke 5 pots, how much money did it collect altogether?
5. A tailor takes twice as much time to make 3 dresses, as he would need to make 5
shirts. If he takes 90 hours altogether to make 9 dresses and 15 shirts, how long
does he take to make each shirt?
6. Janet bought a box of cookies for her friends. If she gives each of them 4 cookies,
she will have 4 cookies left. If she gives each of them 5 cookies, she will need 5 more
cookies. How many cookies are there in the box?
7. A tour to Japan is twice as long as a tour to Hong Kong but half as long as a tour to
Europe. If the three tours last 8 weeks altogether, how long is the tour to Japan?
8. David took one and a half times as long as Billy to complete a project. If both of
them took 20 days altogether, how long did David take to complete the project?
9. Mary has twice as many beads as Lily. Lily has three times as many beads as Tracy. If
the three girls have 60 beads altogether, how many more beads does Mary have than
Tracy?
10. Mark has three times as many marbles as Bob. Bob has twice as many marbles as Paul.
If the three boys have 180 marbles altogether, how many more marbles does Mark
have than Paul?
11. Gary has five times as many stickers as Linda. Linda has 15 more stickers than May.
If the three students have 230 stickers altogether, how many more stickers does
Gary have than May?
12. Golden Gate Bridge
The roadway of the Golden Gate Bridge is suspended from two main cables. These
cables pass over the tops of the two main towers and are secured at each end in
giant anchorages.
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Each cable is 2,332 meters long. It is made up of 27,572 parallel wires compressed
together inside a casing that is nearly a meter in diameter.
Find the total length of the wire used in the two cables to the nearest kilometer.
Explain how you solved the problem.
Extra: I've read that's more than enough wire to wind around the Equator three
times! Is that true?
Tell how you figured that out.
13. School Spirit
Meryem, Emily, and Tristan gave the School Spirit order form to their dad and asked
him if he would give them money so each could buy a Cherry Street School T-shirt.
In addition, Meryem wanted a Fightin' Maraschino ball cap and Emily and Tristan
each wanted a school sticker.
"I'll give Meryem money to place the order. Tell me how much I should give her so
there won't be too much change for candy," Dad said, winking.
All three quickly looked at the numbers and did some figuring in their heads. Tristan
said, "Give her $59." Emily said, "That's not enough, give her $65." Meryem said, "It
will be less than $65, but that's pretty close."
How did Tristan and Emily get their estimates?
Extra: Without actually adding up all of the prices, find another way to estimate the
total to find out approximately how much change Meryem will have for buying candy.
Which estimation would you use in this case? Why?
14. The Place Value Game
Each Friday Ms. Descartes' class plays the Place Value Game. Each student has 3
cards. Each card has a random 5-digit number on it. Ms. Descartes pulls a number
from 1-9 from a hat and the students find the value of her number on all of their
cards. The student with the highest total score wins.
Here's an example:
If a card contains the number 47229 and Ms. Descartes picks a 7, the card is worth
7000 points. If she picks a 2, it's worth 220 points.
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Sylvia has the following cards:
If Ms. Descartes pulls a 3, how many points will Sylvia have?
What digit does she hope Ms. Descartes will pull next week so that she will have the
highest possible score?
Extra:
Sylvia's friend Carina has these cards:
What's one digit that Ms. Descartes could pull next week that would give Carina a
higher score than Sylvia? (Remember that both Sylvia and Carina have to compare
scores using the number Ms. Descartes draws.)
15. Heidi's Hometown
The population of Heidi's hometown at the end of 2006 was 18,079. In the next year
it grew by more than 1000, but less than 1500. Heidi noticed that the new population
had the exact same digits as it had at the end of 2006.
What was the population at the end of 2007? Explain how you solved the problem.
Extra: During 2008 the town's population grew by an amount that was a multiple of 9.
The population was still less 20,000, and still used the same five digits. What could
the population have been at the end of 2008? Can you find all possible answers?
Explain how you know.
16. Just the Facts
Mr. Nelson's students play a game called Just the Facts to help them learn
multiplication facts. Each player in turn rolls a pair of 10-sided dice and announces
the product of the two numbers. Another player checks the product with a
calculator. If the answer given is correct, the player scores 5 points.
Mr. Nelson wants to discourage his students from guessing, so the penalty for an
incorrect answer is losing 7 points!
Zach and Jorge played 24 rounds. At the end Zach's score was 0.
Out of his 24 turns, how many multiplication facts did Zach get correct?
Be sure to explain how you solved the problem.
Extra: The next day the boys played again. Zach got the same number of facts
correct, but this time his final score was 28. How many turns did he take? Explain
how you know.
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
17. Eating Grapes
On Monday Angela ate some grapes. On Tuesday she was hungrier and ate six more
grapes than she ate on Monday. Each day that week she ate six more grapes than the
day before. After she had eaten her grapes on Friday she had eaten 100 grapes in
all.
How many grapes did she eat on Monday?
Extra: If she continues this pattern, on each day eating six more grapes than the
day before, on which day would she eat her 300th grape?
18. Baseball Cards
Kevin, Dustin, and Mike collect baseball cards.



Together Dustin and Kevin have 81 cards.
If Dustin and Mike combined their cards, they would total 96.
The sum of Kevin's and Mike's cards is 93.
How many cards does each boy have?
Be sure to explain how you got your answer and show how you know it works.
Extra: The boys want to store their cards in albums. The pages hold 9 cards each.
How many pages will each boy need to hold his current collection?
19. March Madness!
College basketball playoffs are happening in the United States, making basketball
pretty popular right now! One afternoon after school, four friends named Abby, Bill,
Carla, and Damien decide to play a game of basketball. There are two players on each
team. Your job is to figure out which team won the game given the following facts.
1. Abby outscored her partner Bill by six points.
2. The number of points scored by Damien can be found by reversing the digits
of his favorite number, thirty-one.
3. Bill scored half as many points as Carla.
4. Carla scored three more than three times the points scored by her partner,
Damien.
Which team won and by how many points?
20. Trapezoid Teatime
Lipton Elementary School holds an annual tea to honor the
parent volunteers who work in the school. The trapezoid
tables they use can seat one person on each of the three
short sides and two people on the long side. In other words,
one table standing alone seats five people.
The tables are arranged in one long row in the cafeteria. When they connect two
tables together, here's how the seating looks:
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
1. How many guests can sit at 5 tables connected in a row?
2. How many guests can sit at 20 tables connected in a row?
Explain how you found your answers. Describe any observations or patterns that
helped you.
Extra 1: Use words or numbers and symbols to write a rule for calculating the
number of volunteers that can sit at any given number of tables.
Extra 2: How many tables would it take, arranged in a straight line, to seat 85
volunteers?
21. The Teddy Bears' Banquet
Ursinus Hotel is one of the world's few hotels just for bears.
The tables in its banquet room are regular hexagons with room
for one seat along each side. In other words, one table standing
alone seats six bears.
To make more room for dancing at the Teddy Bears' wedding
banquet, the staff arranges the tables in a long row along one side of the room.
When they connect two tables together, here's how the seating looks:
1. How many guests can sit at 10 tables?
2. How many guests can sit at 25 tables?
3. How many guests can sit at 100 tables?
Explain how you found your answers and how you know you are right. Describe any
patterns that helped you.
Extra 1: Use either words or numbers and symbols to write a rule for calculating the
number of bears that can sit at any given number of tables.
Extra 2: How many tables would it take, arranged in one straight row, to seat 120
bears?
22. Siblings
Alex has the same number of brothers as he has sisters. His sister Megan has twice
as many brothers as she has sisters. How many children in the family are boys and
how many are girls?
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Explain how you found the answer. Show how you can tell that you are right.
Extra: At a family reunion Alex and his siblings represented 1/4 of all the children
there. How many children attended the reunion?
Explain how you found the answer.
23. Sharing Birthdays
Alana and her mother share the same birthday. On Alana's 8th birthday her mother
turned 32. Alana noticed that her mother's age was exactly 4 times her own age.
Alana wondered if it were ever possible for her mother's age to be 3 times her own
age.
Is that possible? If not, how do you know? If so, how old would Alana and her mother
be at that time?
Be sure to explain how you solved the problem and how you know you are correct.
Extra: Alana then wondered if it would ever be possible for her mother's age to be
twice her own age. If so, she wants to know how old she and her mother would be at
that time.
24. Pirate Treasure
Two pirates stand on the shore of an island with the 4 treasure chests they have
just dug up. They have a small dinghy to carry them and their treasure back to their
ship anchored off shore. The dinghy can hold 1 pirate or 2 pirates or 1 pirate and 1
chest.
1. How many one-way trips will it take to get both pirates and all the treasure
back to the ship?
2. How many one-way trips would it take to get both pirates and 8 treasure
chests back to the ship?
Explain how you solved the problem.
Extra: Make a rule you could use to figure out how many trips it would take to carry
2 pirates and any number of chests back to the ship. Use your rule to determine out
how many trips it would take to carry 2 pirates and 93 chests. Explain your thinking.
25. Crossing the River with Ogres
Six ogres and two gnomes set out on an expedition. They came to a river they wanted
to cross.
They found a small boat tied to a tree. It could carry two gnomes or one ogre. The
boat must always be rowed by someone.
1. How many times must the boat cross the river to get everyone to the other
side?
2. How many crossings would it take to get 9 ogres and 2 gnomes across the
river?
Explain how you found your answers.
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Extra: Describe in words how you would find the number of crossings required by two
gnomes and any number of ogres. Show how you would use your rule for 50 ogres and
2 gnomes.
26. Building Bouquets
Jana owns and operates an independent flower shop. To help sell more flowers (and
cut down on waste), she offers a daily special of bouquets or corsages created from
"leftovers" from larger orders.
With today's batch of leftovers, Jana decides to try 4 flowers in each bouquet, but
she has three extra flowers left. To make pricing easier, she wants the specials to
be the same size, so she tries making the bouquets with 7 flowers each... but then
there are two flowers left. If she makes bundles of 8 flowers, she has three flowers
left.
1. What is the smallest possible number of flowers that Jana has to work with?
2. With the number of flowers determined above, how many bouquets or corsages
and of what size can Jana make?
Extra: Do you think that Jana would ever have enough leftovers that it wouldn't
matter if she chose to make bouquets of 7, 8, or 9 flowers? (Don't forget - Jana
doesn't want to waste a single flower.)
27. Bouncing Babies
I attended a friend's baby shower, and we started discussing baby statistics. One of
us had heard a report that stated that for every 100 babies born, there are 6 more
boys than girls.
If we were to pick one child randomly from a representative group, what would be
the probability that the child would be a girl?
Bonus: Would the probability that my friend will have a girl be the same? Why or
why not?
28. Birthday Trip
Jade's family is planning to travel to her Aunt Mazie's house to
celebrate Aunt Mazie's 102nd birthday. On the first day of
the trip they'll drive halfway there and then stop to set up
camp for the night. On the second day of the trip they'll drive
two-thirds of the remaining distance before stopping for some
sightseeing and camping. On the last day they'll have 145 miles
left to drive.
Question: How far is the trip to Aunt Mazie's house?
Extra: If the family averages 50 miles per hour while driving on the return trip, do
you think the family will make the trip in one day? Why or why not?
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
29. Timber!!!
High above the forest on a steep rocky slope, a large boulder is
struck by lightning. Dislodged with a thunderous explosion, the
rock tumbles wildly down the slope. Just before entering the
forest, it is catapulted into the air.
The rock hits a 222-foot-tall redwood tree, and the top is
snapped off part way up the trunk. The following day, when the
forest rangers look over the area for damage, they note that
the severed top is 1/5 the length of the remaining "stump."
How long is each piece?
30. The Math of the Irish
Sean's mother invited relatives to a St Patrick's Day dinner party. Sean used his new
calculator to find that the mean (average) age of the 13 people sitting around the
table was 27 years.
Uncle Paddy left the party early. Sean found that the mean age of the group then
was 25 years.
How old was Uncle Paddy?
Explain how you solved the problem.
Extra: Sean was the youngest at the party. Once Uncle Paddy left, the ages of the
12 remaining people were consecutive even numbers. How old was Sean?
31. Shop 'Til You Drop
Jason earned $600 over the summer by babysitting. It took him 10 shopping trips to
purchase the things he needed for the new school year – school supplies, new clothes
and soccer gear. On each trip he spent $10 more than on the previous trip. After the
10th trip, he had spent all his summer earnings.
How much did Jason spend on the first shopping trip?
Explain how you found your answer.
Extra: How many trips did it take him to spend 2/3 of his total earnings?
Explain your thinking.
32. You Think Your Teacher is Tough!
Mr. Garcia doesn't like students to rush through their work or take wild guesses. He
gave his class a test of 20 math problems. For each correct answer, a student earned
5 points. For every incorrect answer, Mr. Garcia subtracted 2 points.
Tyler answered every problem and his score for the test was 58. How many problems
did he get correct?
Extra: Is it possible to answer every problem and score 0 (zero) on this test? Explain
how you know. Be sure to show your math.
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
33. Greta's Garden
Greta has a vegetable garden. She sells her extra produce at the local Farmer's
Market. One Saturday she sold $200 worth of vegetables — peppers, squash,
tomatoes and corn.
 Greta received the same amount of money for the peppers as she did for the
squash.
 The tomatoes brought in twice as much as the peppers and squash together.
 The money she made from corn was $8 more than she made from the other
three kinds of vegetables combined.
How much did Greta receive for each kind of vegetable?
Be sure to explain how you solved the problem and how you know you are correct.
Extra: What percent of the total sales did each kind of vegetable represent?
34. Common Cents
Monique and Misty wanted to buy their mother a box of candy for Valentine's Day.
They emptied their piggy banks and discovered that Monique had twice as much in
her bank as Misty had in hers. Together they had a total of $10.50.
How much did each girl contribute to the total?
Be sure to explain how you found the answer and show how you know you are right.
Extra: The girls pooled their money and bought a box containing 36 pieces of candy.
Mom shared with Dad, but she ate 3 times the number of pieces that Dad ate.
How many pieces did Mom and Dad each eat?
Explain how you found the answer and show how you know you are right.
35. Video Trading
The local 4-H club has ten members. At the most recent fair, each member
videotaped a different part of the fair. There were many animals, vegetables, baked
goods, and more that were exhibited throughout the fair.
When the club meets, each member trades their video with another member. At the
next meeting, the videos are returned to their original owners to trade again.
How many individual trades must be made for every member to have seen every
video?
36. Mystery Numbers
A few archeologists were digging for a lost tomb when they found the entrance.
Inside there was an old scroll with numbers on it. Through the years in the cold,
damp tomb, some of the numbers had disappeared (they have been "replaced" by
letters in the diagram below to help you in your explanation).
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Can you help them find the mystery numbers?
7
5
12
3
(A)
8
4
(B)
(C)
5
4
9
1
8
(D)
(E)
2
4
2
1
3
2
7
(F)
1
(G)
2
10
4
(H)
2
(I)
15
3
14
8
37. Buy the Numbers
Tabitha's job is putting numbers on the apartments in a new building. The apartments
are numbered consecutively starting at 1. At the hardware store the digits cost $1
apiece. The bill for the digits came to $282.
How many apartments will Tabitha be numbering?
Be sure to explain how you found your answer.
Extra: On Mondays the store sells 1s for half price. How much could Tabitha save by
buying the digits on Monday?
38. Lunch Bunch Budgets
I am a lunch bunch teacher at the preschool where my children are enrolled. Since
Aidan is not yet three, he is too young to attend; but T.J. always comes with me.
Lunch bunch teachers earn $19 for each session they teach, and their children may
attend without paying the $8 fee. There are no other operating fees, as the children
bring their own lunches and the space is already available.
On Tuesdays and Thursdays there are two lunch bunch teachers, and each brings one
child with her. On Wednesdays there are three teachers, and again each brings one
child.
How many kids must attend on a Tuesday or Thursday for the program to break
even? How many kids must attend on a Wednesday for the program to break even?
39. Zoo Train
The zoo has a train that carries people between exhibits. One morning the first
passengers got on at Monkey House. At Alligator Pond the number of people who got
on was 3 more than got on at Monkey House.
The train made 4 more stops: Tiger Thicket, Panda Playground, Giraffe Savannah, and
Big Cats. At each of these stops, 3 more passengers boarded the train than at the
previous stop. At Big Cats 20 people got on the train. How many passengers in all
boarded the train?
Extra: What is the minimum (fewest) number of train cars that it would take to hold
all those passengers at once, if each car holds 12 passengers?
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
40. Feathers and Fur
Xiao, who lives in New York, is a lover of homing pigeons. She also trains seeing-eye
dogs. When I spoke to her recently, she told me in her quirky way that she was
currently hosting 36 heads and 80 feet. I was left to figure out how many puppies
and pigeons she had. Can you help me?
Don't forget to include a complete explanation of how you found your solution. If you
try something that doesn't work but gives you another idea, include that information,
too.
Solutions
1. $75
2. $334
3. $231
4. $268
5. 2 hours
6. 40
7. 16 days
8. 12 days
9. 30
10. 100
11. 155
12. The total length of the wire in two cables would be 128,596 km. The answer to the
extra is yes, it will wind around the earth 3 times. Since I have to find the total
length of all the wire used in the cables to the nearest km, I need to find the length
of one wire first. I know that 1 km equals 1000 meters. First I will divide 2332
meters by 1000 to get the length of one wire in kms.
Math: 2332 / 1000 = 2.332 km
Second I need to multiply 2.332 km by the number of wires in one cable.
Math: 2.332 km x 27572 wires = 64297.904 km total
Now, I will multiply the total wire length by 2 since there are two cables.
Math: 64297.904 km x 2 cables = 128595.808 km
This rounds to 128,596 km.
There are 128,595 km of wire in the cables
Extra: Since the earth's equator is about 40,000 km around, I could wrap this wire
around the equator three times and still have some left over.
Math: 40,000 X 3 = 120,000 km
I rechecked my math and the math is correct and my thinking seems to make sense.
13. Tristan rounded down to the nearest dollar and Emily rounded up to the nearest
dollar.
The three girls are going to buy three t-shirts, a cap, and two stickers. Their dad is
going to give them some money to buy what they want.
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
In Emily's estimation, their dad would give them $65.00. We figured that Emily got
her estimation by first estimating $15.95 into $16.00. She took the $16.00 on the
t-shirts and multiplied it by 3 (meaning the three girls) and got $48.00. She
estimated $12.45 into $13.00 and added the $48.00 to it. She got $61.00. Emily &
Tristan each wanted a sticker that cost $1.60. She rounded it to two dollars and
since Emily and Tristan both wanted one, the total would be $4.00. The $61.00 plus
$4.00 would give her $65.00.That's how she got her answer.
In Tristan's estimation, their dad would give them $59.00. Tristan figured it out by
estimating $15.95 to $15.00, multiplying it by three, and getting $45.00. Since
Meryem wanted a cap, Tristan rounded the $12.45 to $12.00 and added it to
$45.00.The answer was $57.00.She rounded the two stickers that cost $1.60 each
to $1.00.They both wanted a sticker, so it would be $2.00. She added the $2.00 to
the $57.00 and got $59.00.That is how Tristan got her answer.
If we were estimating we would round the $15.95 to $16.00 and then we would
multiply the $16.00 by 3.Then we would get $48.00.Then we would round $12.45 to
$12.00.Then we added $12.00 to the $48.00, and we got $60.00.Then we estimated
$1.60 to $2.00.Since Emily and Tristan both wanted a sticker it would be $4.00 for
the two stickers. Then add the $60.00 with the $4.00 which will equal to
$64.00.Then if you wanted to leave Meryem some money for candy, there will be
$1.00 change because we would use Emily's estimation which is $65.00
14. If Ms. Decartes picked a 3, Sylvia would have 6003 points. Sylvia would hope Ms.
Descartes would pick 4 next week so that she will have the highest posible score.
Extra: Carina would hope Ms. Descartes would pick the numbers 9,7,and 2.
Question 1: If Ms. Descartes pulled out 3, Sylvia would have 6003 because Sylvia's
one card (83483) has 3003 points and another card (43281) has 3000 points, so the
total points would be
3003 + 3000 = 6003.
Question 2:Sylvia would hope Ms. Descartes will pick a 4 next week because 4 is the
highest digit for 2 of the cards and the third card has a 4 in the hundreds which
would be 40000 + 40000 + 400 = 80400. If Ms. Descartes picks 8 next week, Sylvia
would have 80080+80+8=80168 points which is less than 80400.
All other numbers will not give a number higher than 80400 points. For details, see
the explanation in the Extra section below.
Extra:
Carina would hope Ms. Descartes will choose numbers 9, 7, or 2, so that she will have
a higher score than Sylvia.
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Solution:
Number
Sylvia's score
Carina's score
Carina is higher
1
2
3
4
5
6
7
8
9
1
200 + 200 = 400
3003 + 3000 = 6003
40000 + 40000 + 400 = 80400
0
0
0
8 + 80080 + 80 = 80168
9090
0
2020
300 + 300 + 300 = 630
40000 + 40004 = 80004
0
0
70 + 7 + 700 = 777
80008
9000 + 9000 = 18000
no
yes
no
no
no
no
yes
no
yes
15. The population at the end of 2008 for Heidi's hometown is 19,087. Extra: The 4
numbers that could be the population at the end of 2008 are 19,870, 19,807, 19,708,
19,780.
First I saw that the answer to the problem would have to start with a 19,000.
Therefore I only have 6 combinations to choose from. The answer would have to
between 19,079 and 19,597. The 6 combinations to choose from are; 19,870 which
would be no because it is too high. 19,807 would be no because it is too high. 19,708
would be no because it is too high. 19,780 would be no because it is too high. 19,078
which would be no because it is too low. 19,087 which would be yes because it is the
only number between 19,079 and 19,579. So the population at the end of 2008 for
Heidi's hometown is 19,087.
Extra: First I figured out the numbers made up of the combination of digits in
19,087 and were between 19,087 and 20,000. Then I subtracted 19,087 from each of
those 4 combinations. I came up with these 4 combinations and subtracted amounts:
19,870 - 19,087= 783
19,807 - 19,087= 720
19,708 - 19,087= 621
19,780 - 19,087= 693
All of these 4 combinations are correct answers to the question because they're less
than 20,000 and are also multiples of 9. So then the 4 possible populations at the end
of 2008 are 19,870, 19,807, 19,708, 19, 780.
16. Zach got 14 multiplication facts correct and 10 facts incorrect. I used the strategy
trial and error to get my answer. First I tried to see how many points Zach would get
if he answered 12 facts correctly and 12 incorrectly. To do this I multiplied 12 by 5
because he might answer 12 correctly and 5 because each correct answer is worth
five points and I got 60 points. Then I subtracted the quantity of 12 by 7 because 12
problems might be answered incorrectly and you lose 7 points for every incorrect
answer. I got 24 because 12 x 5 = 60 and 12 x 7 = 84 and 60 – 84 = -24. Because that
didn’t work I tried the amount of 13 answers correctly and 11 incorrectly where I
got -12. And then I tried the amount of 14 answers correctly and 10 incorrect.
14 x 5 = 70 (representing how many answers Zach got correct) and 10 x 7 = 70
(representing how many questions Zach got incorrect) and 70 – 70 = 0, which is how
many points Zach should get.
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Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Extra: Jorge and Zach played 20 rounds. I know this because Zach got the same
amount of answers correct and his final score was 28. So I thought 14 x 5 = 70 and
what multiple of 7 subtracted from 70 gets a difference of 28? And I got it, the
answer was 42 (6 x 7). And I added 6 and 14 because those were the amounts of
answers correct and incorrect and the sum was 20, which is how many rounds they
played.
17. Angela ate 8 grapes on Monday.
To solve this problem, I did a guess
number. 5 + 11 + 17 + 23 + 29 = 85.
everyone, the sum would be 5 more.
my first number would have been 8.
32 = 100
and check. I started and guessed 5 as the first
I knew if I went up one in the beginning and for
100 – 85 = 15. 15 divided by 5 = 3 5 + 3 = 8. So
She ate 8 grapes on Monday. 8 + 14 + 20 + 26 +
18. Dustin has 42 baseball cards, Kevin has 39 cards, and Mike has 54 cards.
Extra: Dustin needs 5 pages, Kevin needs 5 pages, and Mike needs 6 pages. I made a
D column, a K column, and an M column (standing for Dustin, Kevin, and Mike). Then I
wrote a K + M column. I also wrote to equations of how many card people had
(D + K = 81, D + M = 96, and K + M = 96)
If I subtracted whatever number of cards Dustin had from 81 I would get how many
Kevin had. Then, since Dustin's cards + Mike's cards = 96 cards I could subtract
Dustin's cards from 96 and I could get Mike's number of cards. Then I would check
it by adding Kevin's cards and Mike's cards together to see if it equaled 93.
First I guessed that Dustin had 27 cards and then I subtracted 27 from 81 and got
54. So then I subtracted 27 from 96 and got 69 so that's how many cards Mike had.
Then I had to add Kevin's cards + Mike's cards and I got 123 cards so I knew that
wasn't it because K + M had to equal 93. I tried Dustin having 47 cards, Kevin having
34 cards, and Mike having 49 cards and that didn't work either because K + M = 83.
Then I noticed when Dustin has a big amount of cards the K + M column has fewer
cards. So I tried Dustin having 46 cards, Kevin having 35, and Mike having 50. The
K + M column's total was 85. Then I figured out that when the D column goes down 1
the K + M goes up 2. So I counted by 2s from 85 all the way up to 93 and got 4 sets
of 2s. I subtracted 4 from 46 and got 42. So I tried Dustin having 42 cards, Kevin
having 39 cards, and Mike having 54 cards and I added K + M together and got 93.
Extra:
Kevin needs 5 pages, because 39 ÷ 9 = 4 r3. 4 pages are not enough because he needs
an extra page to hold the remaining 3
Dustin needs 5 pages, because 42 ÷ 9 = 4 r6. 4 pages are not enough because he
needs an extra page to hold the remaining 6
Mike needs 6 pages, because 54 ÷ 9 = 6. Mike has no remainder.
19. Damien and Carla won by 7 points.
The question is:
Which team won the basketball game and by how many points?
17
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
The problem is:
I need to keep my information straight and not get mixed up.
I know:
1) To figure out how many points the winning team won by; you subtract the losing
team’s points from the winning team’s points.
2) Abby and Bill are on a team and Damien and Carla are on a team.
3) Damien’s favorite number backwards is 13.
First I multiplied Damien’s points (13) by 3 then added 3 since Carla made 3 more
than 3 times as much as Damien.
( 13 points x 3 ) + 3 = 42 points
Second I divided Carla’s 42 points by 2 since Bill has half as many points as Carla.
42 points / 2 = 21 points
Third I added Abby’s 6 more points to Bill’s 21 since Abby made 6 more points than
Bill.
21 points + 6 =27 points
Forth I added Abby’s 27 points to Bill’s 21 to find out how many points they made.
27 points + 21 points = 48 points
Fifth I added Carla’s points to Damien’s to find out how many points they made.
42 points + 13 points = 55 points
Sixth I subtracted Abby’s and Bill’s points from Carla’s and Damien’s points to find
out how many points Damien and Carla won by.
55 points - 48 points = 7 points left
The answer is:
Damien and Carla won by 7 points.
Verification
(13 x 3) + 3 = 42
42 / 2 = 21
21 + 6 = 7
42 + 13 = 55
55 - 48 = 7
20. 1. There were 17 people able to sit at 5 tables. 2. There were 62 people able to sit at
20 tables.
First we started out drawing the tables. We figured out that 17 people could sit at 5
tables. After that we drew 10 tables and doubled it to get 64 people at 20 tables.
But we figured out that you can't have 4 end tables when there are only 2 ends.
So we subtracted the two people and came up with 62 people at 20 tables.
We had 18 middle tables that seat 3 so its 18 x 3 = 54 people. Then we added 8
people for the two end tables. 54 + 8 = 62 people.
That also worked for the first problem. 3*3=9+8=17 people so we know we are right.
18
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Extra 1: The rule is number of middle tables * 3= people at middle tables + 8 people
on both end tables = total people.
Extra 2 There are 28 tables for 85 people.
We used the rule from extra number 1 to help us find the 28 tables.
85 – 8 = 77 people at middle tables.
25 x 3 = 75 people at middle tables. We knew we needed one more table in the
middle and made it 26 tables. Then we added the end tables on 26+2=28 tables. We
would have one extra space.
21. 42 guests can sit at 10 tables, 102 guests can sit at 25 tables, and 402 can sit at 100
tables.
We took 2 tables away from the total of the tables. The middle tables had 4 places
each.
Then we multiplied by 4 and added 10. For the first problem we used real pattern
blocks.
Then we just used mental math, with middle tables being 4 and end tables being 5.
Problem 1. 10 - 2 = 8
8 x 4 = 32
32 + 10 = 42
Problem 2. 25 - 2 = 23
23 x 4 = 92
92 + 10 = 102
Problem 3. 100 - 2 = 98
98 x 4 = 392 392 + 10 = 402
Extra 1: Subtract 2 from the tables. Multiply the rest by 4. Add 10.
Extra 2: 100 guests will need 25 tables. 104 will need 26 tables. 108 will need 27
tables.
112 will need 28 tables. 112 + 8 will need 30 tables (because 28 tables + 2 tables is
30 tables) and we'll have 2 spots left over.
22. There are 3 girls and 4 boys.
Alex has the same number of brothers as he has sisters. Since Alex himself doesn't
count as one of his brothers but counts as a boy there will be one more boy then girl
because if Alex has the same number of brothers and sisters if you add Alex that's
one more boy then girl.
Since Megan has twice as many brothers then sisters there has to be an even number
of boys or else she would have half a sister which is impossible.
2 boys wouldn't work because then Megan wouldn't have any sisters. 4 boys works
because then there would be 2 girls (not including Megan), and since Alex would have
one more brother then sister without Megan he would have 3 brothers, so with Alex
there would be 4 boys, and 4 is 2 doubled so Megan would still have double the
amount of brothers then sisters since Megan doesn't count as her own sister. if you
add Megan then that makes 4 boys and 3 girls. Since 4 is one higher then 3 it works
for Alex too.
The main thing you need to remember about this problem is that the child doesn't
count as his/her own sibling.
EXTRA:
Since there are seven kids in the family and they make up 1/4 of the
kids at the reunion you would have to multiply 7 by 4 which is 28
19
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
23. It is possible for Alana's mother to be 3 times Alana's age. Alana would be 12 years
old and her mother would be 36 years old. Extra: It is possible for Alana's mother's
age to be twice Alana's age. Alana would be 24 years old and her mother would be 48
years old.
To figure out the answer I first made a chart to show Mom's age compared to
Alana's age. I then entered Alana's age and Mom's age from the problem. Then I
added 1 to Alana's age and Mom's age and multiplied Alana's age by 3 to determine if
Alana's age x 3 = Mom's age.
I found that when I got to Alana's age of 12, her mom's age was 36 and 12 x 3 = 36,
so when Alana was 12 years old Alana's mom would be 3 times Alana's age.
Alana's age
8
9
10
11
12
Mom's age
32
33
34
35
36
Alana's Age X 3
24
27
30
33
36
Extra:
I noticed that there was an age difference between Alana and her mom of 24 years.
So 24 years added to Alana's age will equal her mom's age. And since I was looking
for Alana's age X 2 to equal her mom's age, and I knew that 24 X 2 = 48, Alana must
be 24 years old to have her mother be 2 times her age.
24. 1) It would take 9 one-way trips. 2) It would take 17 one-way trips. 3) It would take
187 one-way trips.
1) I drew a picture and found out the following:
First trip: One pirate takes a chest and brings it over to the other side.
Second trip: The pirate sails back.
Third trip: The pirate takes the second chest and brings it over to the other side.
Fourth trip: The pirate sails back.
Fifth trip: The pirate takes the third chest and brings it over to the other side.
Sixth trip: The pirate sails back.
Seventh trip: The pirate takes the last chest and brings it over to the other side.
Eighth trip: The pirate sails back.
Ninth trip: The two pirates sail to the other side.
Therefore, it would take 9 one-way trips.
2) I realized that the rule to solving this equation is 2x+1, where x = the number of
chests there are. Since there are 8 chests, then 2(8)+1, or 16+1, or 17 trips, is our
answer.
Extra: As mentioned before, the rule to solving this equation is 2x + 1, where x = the
number of chests there are. The 2x in the equation stands for two trips for each
20
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
chest the pirates have to carry over, since there is one trip to bring the chest to the
other side, and one return trip. The +1 at the end of the equation stands for the final
trip that the pirate has to make in order to bring himself and the other pirate over
to the other side. Therefore, since there are 93 chests in this problem, then
2(93) + 1, or 187, is our answer.
25. If there was 6 ogres and 2 gnomes then it would take 25 times to get everyone
across. If there was 9 ogres and 2 gnomes then it would take 37 times to get
everyone across. EXTRA:0 x 4 + 1 is the rule I used to find the answer to the
question.
Six ogres and two gnomes wanted to cross a river they found a boat tied to a tree.
Only two gnomes or one ogre could fit in the boat and the boat must always be rowed.
How many times must the boat cross the river to get everyone on the other side?
How many crossings will it take to get 9 ogres and 2 gnomes on the other side?
I got the answer to the story problem of how many trips it would take to get 6 ogres
and 2 gnomes across by. . . . . . . . .
First I drew a picture. Here is what it looked like.
2 gnomes ---------------->
1 gnome <--------------1 ogre ---------------->
1 gnome <-------------2 gnomes -------------->
1 gnome <----------1 ogre --------------->
1 gnome <------------I continued the pattern 2 gnomes, 1 gnome, 1 ogre, 1 gnome. While keeping track of
how many ogres I brought across by tallying it. Then at the end when I brought all
the ogres across I had to bring one of the gnomes back so I rowed the gnome on the
other sided across and rowed the two gnomes back. Next I counted the number of
times the creatures crossed and got the varible 25.
Now I had to figure out how many trips it would take to bring nine ogres and two
gnomes across. I didn't want to draw the picture over again so I looked at the
picture I already drew. Since I already spotted the pattern I used it to help me.
First I divided 25 by 4. I chose the two variables because 25 was the variable that
stood for the number of trips it took to get everyone across. I chose 4 because it
was the number of variables it took before the pattern started again. It divided into
the varible 25 6 times with a remainder of 1. Since there was six ogres I knew I
found the way to solve the problems faster. I multiplied 4 and 9 which I then got 36
Then I added 1 and got 37. That is how I got the answer to the problem.
EXTRA: I got extra by using the rule I found out. I multiplied 4 and 50 in which I
got 200. Then I added 1 and got 201. Here is my algebra expression 0 x 4 + 1.
21
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
26. 1. The smallest number of flowers Jana has to work with is 51
2. Using 51 flowers, she could make 12 bouquets of 4 flowers and still have 3 flowers
left over. Using 51 flowers, she could make 7 bouquets of 7 flowers and still have 2
flowers left over. Using 51 flowers, she could make 6 bouquets of 8 flowers and still
have 3 flowers left over.
To find the total number of flowers, we made a chart
Bouquets
4
7
8
2
11
16
19
3
15
23
27
4
19
30
35
5
23
37
43
6
27
44
51
7
31
51
8
35
9
39
10
43
11
47
12
51
Using this chart, we could see that the number of flowers needed to make these
bouquets is 51.
Then we tried to use a chart to see the number she would need
bouquets of 7, 8 and 9 flowers. The chart got so long and we weren't
stopped! Then we came back to the problem and decided that if we
three numbers together, the answer would be divisible by 7, 8 and 9.
these and got 504.
to make even
through so we
multiplied the
We multiplied
27. A girl would be picked 47 out of 100.
This is how I solved the problem. First of all, I looked at the problem. Then, I
worked out a picture in my head. It looked like this:
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Fifty on each side. Then I saw that there had to be 6 more boys than girls. So I saw
that I had to even it out to the number requested. Then I thought of my basic
addition facts. I found 7 + 3 = 10. I tried 47 + 53 = 100. Now I knew that the
probability would be: A girl would be picked 47 times out of a hundred.
.....................................+ ...........................................= 100
28. The trip to Aunt Mazie's house is 870 miles. I need to find out how far is the trip to
Aunt Mazie's house. I will do a rectangle diagram to solve the problem.
I solved the problem by cutting a rectangle in half since they drove half way on the
first day. Then I cut the remainder into thirds since they drove two thirds on the
second day. A third of a half is one sixth of the whole, so the remaining sixth is 145
miles. Then I multiplied 145 time six and that equaled 870 miles--the answer!
145 x 6 = 870
22
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Extra: I need to find out if the family drives 50 mph how long will it take for them to
get home. I did a division equation to solve this problem. I divided 870 by 50 and
that equaled 17.4--the answer.
870 ÷ 50 = 17.4
The 17 is the amount of hours it will take them and the .4 is 4/10 of a hour. 4/10 is
the same as 2/5 so:
60 x 2/5 = 24 minutes
So, yes, they can make the trip in one day because 870 miles divided by 50mph equals
17 hours 24 minutes.
The first 5 hours one parent would drive and the next 5 hours the other parent can
drive. Then take a rest and after that one of the parents can drive the 7 hours and
24 min.
29. The stump was 185 feet long, and the part that was cut off was 37 feet.
I solved the problem by dividing 222 feet by 6. I divided by 6 because when the
fallen part was still on the tree, it was a sixth of the tree. This can be seen in the
following diagram:
stump broken piece
----|1|
|6|
----|2|
--|3|
--|4|
--|5|
--We can see that the broken piece is 1/5 of the stump, as the problem explains. But if
you put that broken piece back on the tree, you see that it is 1/6 of the complete
tree.
The result of dividing 222 feet by 6 was 37. Next I subtracted 37 from 222, and
the result was 185. That was the height of the stump. The height of the fallen part
was 37 feet.
30. Uncle Paddy is 51 years old. EXTRA: Sean was 14 years old.
To find the age of Uncle Paddy, I had to find the total age of the people when Uncle
Paddy was there and then subtract it by the total age of the people when Uncle
Paddy wasn't there.
I knew that to figure the total age of the 13 people at the dinner party I had to
multiply the number of people times the average age. I had to multiply 13 times 27.
23
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
13 x 27 = 351
The answer was 351, so the total age of the people when Uncle Paddy was there was
351. Now I had to find the total age of the people when Uncle Paddy wasn’t there.
When Uncle Paddy had left, the average age of the people dropped to 25 and the
number of people would also drop, there would be 12 people in the dinner party. Now
I had to multiply 12 times 25 to figure out the total age of the people at the dinner
party when Uncle Paddy left.
12 x 25 = 300
The answer was 300, so the total age of the people at the dinner party when Uncle
Paddy had left would be 300. Now I had to subtract the total age of the people at
the dinner party when Uncle Paddy was there minus the total age of the people at
the dinner party when Uncle Paddy had left to find the age of Uncle Paddy. I had to
subtract 300 from 351.
351 – 300 = 51
The answer was 51, so the age of uncle Paddy was 51 years old.
Extra: The problem told me that when uncle Paddy left, the ages of the remaining
people were consecutive even numbers and Sean was the youngest. Using this
information, I had to figure out the age of Sean.
I knew that the average age of the people when Uncle Paddy left was 25 so I first
drew a number line and I put 25 in the middle. I put the six even numbers on each
side. I put the 6 even numbers greater than 25 on the right side and the 6 even
numbers less than 25 on the left side. This is what my number line looked like:
36 34 32 30 28 26
25
24 22 20 18 16 14
Since Sean was the youngest at the party, the least number on the line had to be
Sean’s age. The least number on the line was 14, so Sean was 14 years old.
31. Jason spent $15.00 on his first trip. Extra: On his 8th trip he had spent 2/3 or 400
dollars.
We used guess and check as a problem solving strategy. We started at 10 dollars as
his first trip and we found that
$10 + $20 + $30 + $40 + $50 + $60 + $70 + $80 + $90 + $100 = $550
That was off by $50 dollars. Then we divided $50 (the leftover) by 10 (trips) and we
got $5. We then added $5 dollars to the first trip giving us $15 dollars as the first
trip. Next we tested the problem,
15 + 25 + 35 + 45 + 55 + 65 + 75 + 85 + 95 + 105 = 600 dollars in total, so he started
out spending $15 dollars on his first trip.
24
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Extra: 2/3 of 600 dollars is 400 dollars we know this because we divided it by 2/3,
so next we made a table.
15 1st trip
25 2nd trip
35 3rd trip
45 4th trip
55 5th trip
65 6th trip
75 7th trip
85 8th trip
__________
400 dollars on his 8th trip
32. The answer is he got 14 correct answers. It is not possible to answer every problem
and score 0 on the test.
I started off by guessing that he got 12 correct answers. So I multiplied 12 x 5 as
Mr. Garcia was giving 5 points for every correct answer.
12 x 5 = 60.
Since I knew he had 20 problems, I added 12 + 8 = 20.
Then I multiplied 8 x 2 because he was subtracting 2 points for every incorrect
answer.
8 x 2 = 16.
Then I subtracted 16 from 60 and got 44.
This was not correct because he scored 58 and since it was too low I decided to try
16 correct answers.
So I did the same calculations:
16 x 5 = 80, 20 – 16 = 4, 4 x 2 = 8 and 80 – 8 = 72.
72 is too high so I tried 14.
I calculated: 14 x 5 = 70, 20 – 14 = 6, 6 x 2 = 12
and then I subtracted 70-12=58 correct answers.
Extra: Since I knew that 12 correct answers would give a test score of 44, I knew
that I had to go lower than 12 to get a zero on the test. I tried 10 correct answers
x 5 points which gives him 50, and 10 wrong answers x 2 points gives him 20.
Then 50 – 20 = 30 for a score.
Next I tried 5 x 5 = 25.
This would leave him with 15 wrong answers.
15 x 2 = 30.
25 – 30 = -5 which is below 0.
25
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
So next I tried 6 x 5 = 30 and that would leave him with 14 wrong answers.
14 x 2 = 28.
So 30 – 28 = 2 which is more than 0.
So I concluded that you cannot answer all the questions and score 0 on this test.
33. Greta received $16 from peppers, $16 from squash, $64 from tomatoes, and $104
from corn. Extra: The percent of profit made by Greta from each vegetable are:
Squash 8% ; Peppers 8% ; Tomatoes 32% ; Corn 52%.
After reading the problem carefully, I figured out what the problem was and what
information it gave me. I used letters to help me, so that P (peppers) = S (Squash);
T = 2(P + S); C (corn) = $8 + T + P + S. So the total was $200 = C + T + P + S.
I made a chart and started to estimate money amounts, starting with peppers and
squash. I started with $25 dollars for both peppers and squash because they had to
be the same and because $25 didn't seem too big or too small.
T P S C
Total
100 25 25 158 308 $308 was too high, so I tried
$10 for both peppers and squash.
40 10 10
68
60 15
98
15
128
188
$128 was too low, so I
tried different higher numbers.
$188 was only $12 dollars away!
I first thought to add $6 to both peppers and to squash because it is half of 12. But
then I realized that I would have to change tomatoes too and that would make my
answer too high. So I decided to add $3 to both peppers and squash making them
each $18.
T
72
P
18
S
C
18
Total
116 224
This time I was way too high.
So I tried adding $1 to each peppers and squash instead. Making them each bring in
$16.
T
P
64
16
S
16
C
104
Total
200
By using a chart and "guess and check" I found out the answer.
Extra: To find the percent for each vegetable we put the dollar amount divided by
200 and pressed the percent key. Peppers and squash were 8% each. Tomatoes were
32% and corn was 52%. We know this is correct because 8 + 8 + 32 + 52 = 100%
34. Monique contributed $7.00 and Misty contributed $3.50. Extra: Dad ate 9 pieces and
mom ate 27 pieces of candy.
We figured this out by taking $10.50 divided by 3 because Misty has 1 amount of
money and Monique has twice that which is 2. 1 + 2 = 3. We got $3.50 as our answer
26
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
to $10.50 divided by 3. If you double that, you get $7.00. Then, we did $7.00 +
$3.50 = $10.50. We then knew that our answer was right.
Extra: We needed to think of 2 numbers that would equal 36 when you add then
together. We made a list since we know that mom ate 3 times as much as dad. Dad's
number times 3 should equal mom's number. We looked at our list and found out that
9 + 27 = 36 and 9 x 3 = 27. So mom ate 27 pieces and dad ate 9 pieces of candy.
35. It will take 45 swaps (in 9 weeks) for all 10 people in the 4-H club to have looked at
everyone’s video that they taped at the fair.
Each member will have a total of nine videos each that they would have to watch
(This doesn’t include their own video). This tells us that there must have been 9
weeks of trading.
Each time the clubs meet there are 10 videos. This means that 5 swaps had taken
place (10 divided by 2).
To work out the number of trades taken the nine week period, I multiplied by 5
(being the number of swaps taken a week) by 9 (being the number of swaps each
player must have and being the number of weeks that swapping takes place).
The answer for this problem is that it will take 45 swaps for all 10 people to have
looked at everyone’s video.
EXAMPLE ? All the swaps the take place
1st person = A
2nd person = B
3rd person = C
4th person = D
5th person = E
6th person = F
7th person = G
8th person = H
9th person = I
10th person = J
AB AC AD AE AF AG AH AI AJ
BC BD BE BF BG BH BI BJ
CD CE CF CG CH CI CJ
DE DF DG DH DI DJ
EF EG EH EI EJ
FG FH FI FJ
GH GI GJ
HI HJ
IJ
TOTAL: 9
TOTAL: 8
TOTAL: 7
TOTAL: 6
TOTAL: 5
TOTAL: 4
TOTAL: 3
TOTAL: 2
TOTAL: 1
GRAND TOTAL: 45
27
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
NOTE: Although the number of swaps seems to decrease each time a new person
(letter) starts to swap the answer is still correct because if you look at the letter C
for example, it has seven swaps instead of nine, but if you look at the A and B line, C
has also swapped with them. You can’t swap the video twice.
36. The mystery numbers are: A = 15, B = 9, C = 6, D = 5, E = 3, F = 3, G = 7, H = 27, and I
= 56.
To find the solution, I looked for patterns. I looked for patterns of odd and even
numbers, but that didn't help me. I saw that adding the first and second column
equalled the third column.
7 + 5 = 12
5+4=9
2+1=3
I tried to continue that pattern, but adding the numbers in the last row of the 7th
and 8th column did not equal the number in the last row of the 9th column.
3 + 14 NOT = 8
I saw that the diagonal from the upper left number 7 was 7 4 3 2. 7 = 4 + 3. Also a
diagonal from the 4 in the top row was 4 2
2. 4 = 2 + 2. But a lot of the other diagonals didn't work the same way. I noticed that
the diagonal from the lower left number 10 was 10 1 9
3. 10 = 1 + 9. A diagonal from the 4 in the bottom row was 4 3 1 A. 4 = 3 + 1.
But some of the other diagonals didn't work the same way.
After a long time, I saw that adding numbers in the second and third rows gave the
numbers in the top row. From the first column, 5 + 2 = 7. It worked all the way
across.
In the 5th column, 8 + 7 = A = 15.
In the 7th column, E + 1 = 4, so E = 3.
In the 9th column, 4 + 2 = C = 6.
That was all I could get for a while. I repeated some stuff that didn't work.
Finally, I saw that multiplying numbers in the second and third rows gave the
numbers in the bottom row. From the first column, 5 x 2 = 10. It worked all the way
across.
In the 3rd column, 9 x 3 = H = 27.
In the 5th column, 8 x 7 = I = 56.
In the 8th column, 2 x G = 14, so G = 7.
With G = 7, I could solve for 2 + 7 = B = 9 in the top row.
Then I had the 6th column left.
D+F=8
D x F = 15
I knew that 3 x 5 = 15. Then adding: 3 + 5 = 8. So D and F are 3 and 5. I tried to
decide whether D = 3, F = 5 or D = 5, F = 3. I looked at the numbers in the rows. No
numbers are repeated side by side in the rows. I decided D = 5, F = 3 so that D and E
wouldn't be the same number side by side.
28
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
37. Tabitha can number 130 apartments.
Step 1:
To determine how many apartments Tabitha will be numbering, we need to identify
the cost of each number. Numbers 1 - 9 will cost $1 a piece because there is 1 digit
used to form one of these numbers and 1 digit cost $1. Therefore, 1 digit x $1 = $1.
Numbers 10 - 99 will cost $2 because there are 2 digits used to form one of these
numbers and 1 digit cost $1. Therefore, 2 digits x $1 = $2.
Thus, in using this same idea, we can determine that numbers 100-999 will cost $3
per number because there are 3 digits used to form one of these numbers.
Step 2:
The bill for the digits was $282. Using this information I can begin to calculate how
many apartments Tabitha can number using a step-by-step process.
I know that Tabitha will begin numbering the apartments starting at 1 and she will
proceed in numerical order.
If Tabitha is able to number apartments 1-9 it will cost her:
9 (the number of apartments) x $1 (the cost to number a one digit number) = $9.
If it cost Tabitha $9 to number the first 9 apartments, there is $273($282 - $9)
with of digits left to number the remaining apartments.
If Tabitha is able to number apartments 10-99, it would cost her:
90 (the number of apartments) x $2 (the cost of a two digit number) = $180.
It would cost Tabitha an additional $180 to number apartments 10-99.
To determine if Tabitha has purchased enough numbers to do this add this total to
the cost it would be to number apartments 1-9. This total needs to be less than
$282.
$9 (cost to number apartments 1-9) + $180 (cost to number apartments 10 - 99) =
$189.
$282 (cost of digits bought) - $189 (cost of digits used to number apartments 1-99)
= $93
It would cost Tabitha $189 to number apartments 1-99. Therefore, Tabitha has $93
worth of digits left to number the remaining apartments. Since Tabitha is working
with 3 digit numbers now (100-999), the cost to number one apartment is $3 ($1
(cost of one digit) x 3 (number of digits used to form the number) = $3).Therefore,
$93 (cost of remaining digits) / $3 (cost to number a three digit number) = 31.
Tabitha can number 31 more apartments.
Therefore, 99 (apartments already numbered) + 31 (additional apartments that can
be numbered) = 130. Tabitha can number 130 apartments.
Verification:
To verify my solution I will check to make sure the total cost of the digits used
equals $282 (the price for all the digits purchased).
29
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Apartments 1 - 9 cost: (9 x $1(the cost for one digit)) = $9
Apartments 10 - 99 cost: (90 x $2(the cost for two digits)) = $180
Apartments 100 - 130: (31 x $3(the cost for three digits)) = $93
$9 + $180 + $93 = $282
My solution was verified. Tabitha can number 130 apartments.
Extra: On Monday the store sells 1s for half price. How much money could Tabitha
save by buying the digits on Monday?
Step 1:
If the store was selling 1s for $1 originally, then half off would be $0.50. Therefore,
we need to calculate how many numbers used the digit 1 to form the number.
I know that digits 10-19, used one “1” to form the number, with the exception of 11
which used two 1s. Also numbers: 1, 21, 31, 41, 51, 61, 71, 81, and 91 used one “1” to
form the number. So far this is a total of 20 numbers that used one “1” in the
formation of the number.
Next, I know that numbers 100-109 also have one “1” used with the exception of the
number 101 which has two ‘1s” used. Therefore, 11 more “1s” need to be accounted
for.
Furthermore, numbers 110-119 have two “1s” used with the exception of the number
111 which has three “1s” used. Therefore, we need to account for 11 more ones.
Lastly, numbers 120-130 have one “1” used with the exception of the number 121
which has two “1s” used. Therefore, we need to account for 12 more ones.
Thus, 20 + 11 + 11 + 12 = 64 ones were used to number apartments 1-130.
Step 2:
To calculate the how much Tabitha would save if she bought the digits on Monday,
multiply the cost for the digit “1” on Monday by the number of 1s needed to number
the apartments.
$0.50 x 64 = $32.00
Tabitha could have saved $32.00 if she purchased the numbers on Monday.
38. 7 Kids would attend on Tuesday or Thursday and 11 kids will attend on Wednesday.
On Tuesday and Thursday, since there are 2 lunch bunch teachers, they both need to
be paid $19 each. So the total amount that needs to be paid to them per day :
$19 x 2 = $38.
Each child pays $8 for a day.
One kid will pay
Two kids will pay
Three kids will pay
Four kids will pay
Five kids will pay
: $8
: $8 X 2 = $16
: $8 X 3 = $24
: $8 X 4 = $32
: $8 X 5 = $40.
30
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
On 5 kids, it will break even because amount kids will pay is more than teacher pay.
Total number of kids who will attend school on Tuesday or Thursday to break-even:
: 5 +2 = 7.
2 has been added to above number for kids of Lunch-Bunch teachers because each
Lunch-Bunch teacher brings one kid along.
On Wednesday there are three teachers, total amount to be paid to teachers :
$19 x 3 = $57.
One kid will pay : $8 X 1 = $8
Two kids will pay : $8 X 2 = $16
7 kids will pay
8 kids will pay
: $8 X 7 = $56
: $8 X 8 = $64
So the program on Wednesday will break-even with 8 kids.
Total number of kids in school on Wednesday : 8 + 3 = 11
This 3 has been added for one kid for each of Lunch-Bunch teachers.
39. 75 passengers boarded the train that day. Bonus: The fewest number of train cars it
would take to hold all passengers is 7 train cars.
To solve the problem I set up a chart using the graphing data and numbers method to
solve the problem starting at 20 and subtracting 3 each time.
Big cat: 20 got on.
Giraffe Savannah: 20-3=17
Panda Playground: 17-3=14
Tiger Ticket: 14-3=11
Alligator Pond: 11-3=8
monkey house: 8-3=5
Since Monkey House is where the train started according to the question so that
means I am done subtracting 3 from the previous number. The last thing I did which
gave me my answer of 75 was add all the numbers together, so I did 20 + 17 + 14 + 11
+ 8 + 5 which is 75.
Bonus: To get my answer I simply did 75 divided by 12 since there are 75 total
passengers boarding and 12 seats on each cart. When I did that I got 6.25 and since
you can’t have half a cart I knew that if there were 6 carts it would not hold
everyone so there has to be 7 or more.
That is why the minimum amount of carts there could be is 7.
40. She has 4 puppies and 32 pigeons.
I made a chart with numbers of each animal and the number of feet they have. I
started with one puppy because if I started with 1 pigeon, there would have to be 35
puppies which would give too many feet (35 x 4 = 140).
31
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Puppies
Pigeons
Total Heads
Total feet
Number Feet
Number Feet
1
(1x4)= 4
35 (35x2)= 70
1+35=36
4+70= 74
2
(2x4)= 8
34 (34x2)= 68
2+34=36
8+68= 76
3
(3x4)=12
33 (33x2)= 66
3+33=36
12+66= 78
4
(4x4)=16
32 (32x2)= 64
4+32=36
16+64=80
Therefore, she has 4 puppies and 32 pigeons.
32
Unit 1 – Number Sense and Algebraic Thinking
Lesson 1 & 2 – Whole Number Problem Solving
Bibliography Information
Teachers attempted to cite the sources for the problems included in this problem set. In some cases,
sources may not have been known.
Problems
Bibliography Information
All examples and
1 – 8.
Lee, Joseph D. Primary Mathematics 5:
Challenging Word Problems. Singapore: EPB
Pan Pacific, 2006.
9 – 11
Lee, Joseph D. Primary Mathematics 6:
Challenging Word Problems. Singapore: EPB
Pan Pacific, 2006.
12 – 40
The Math Forum @ Drexel
(http://mathforum.org/)
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