Different Problem Solving Strategies.pptx

Different Problem Solving Strategies

Guess and Check
Is a problem solving strategy that students can use to
solve mathematical problem by guessing the answer and
then checking that the guess fits the conditions of the
problem.

Problem 1
The sum of 3 consecutive integers is 258. Find the integers.
Step 1: Understand the problem
 3 consecutive integers
 Sum of 3 consecutive integers is 258
Step 2:Plan
Guess and Check

Step 3:Carry out a Plan
Step 4:Look back
258/3 = 86
86, 87, 88 = 261
258/3 = 86
85, 86, 87 = 258
85, 86, 87 are three consecutive integers
There sum is 258
Thus, the 3 consecutive integers are 85, 86, and 87.

Problem 2
The ages of 3 sisters are 3 consecutive even integers. If the
sum of twice the 1st even integer, 3 times the 2nd even
integer, and the 3rd even integer is 34. Find each age.
 3 consecutive even integers
 1st even integer = x2
 2nd even integer = x3
 Sum is 34
Step 2:Plan
Guess and Check

Step 4:Look back
2 x 2 = 4
4 x 3 = 12
6
4 + 12 + 6 = 22 is not equal to 34
4 x 2 = 8
6 x 3 = 18
8
8 +18+ 8 = 34
4, 6 and 8 are 3 consecutive integers.
Thus , the ages of 3 sisters are 4 years old, 6 years old, and
8 years old.

Problem 3
Suppose decorated bookmarks cost P5.50 and the plain ones cost P3.50.You bought
equal number of bookmarks and paid exactly P72.00. How much of each kind did you
buy?
 Decorated bookmark = P5.50
 Plain bookmark = P3.50
 Bought equal number of bookmarks
 Paid P72.00
Step 2:Plan
Guess and Check
5.50 x ________
3.50 x ________
P72.00

Step 4:Look back
Therefore, there are 8 decorated bookmarks and 8 plain
bookmarks.
5.50 x 2 = P 11.00
3.50 X 2 = P7.00
P18.00 is not equal to P72.00
5.50 X 8 = P44.00
3.50 x 8 = P28.00
P44.00 + P28.00 = P72.00

Draw a Table
Is a problem solving strategy that students can use to
solve mathematical word problem by writing information in
a more organized format.

Problem 1
Mary ran a lemonade stand for 5 days. On the 1st day, he made $5.Every day after
that he made $2 more than the previous day. How much money did Mary make in
all after 5 days.
 Mary ran a lemonade for 5 days.
 1st day = $5
 Every day $2 more than the previous
Step 2:Plan Draw a Table
Day Income

Step 4:Look back
After 5 days, Mary made $ 45 in total.
Day Income
1 $5
2 $7
3 $9
4 $11
5 $13
Total $54

Problem 2
How many hours will a car travelling at 65 miles per hour to take to
catch up a car traveling at 55 miles per hour if the slower car starts
one hour before the faster car?
Step 2:Plan
 Slower car going 55 miles per hour
 Faster car going 65 miles per hour
 Slower car starts 1 hour before the faster car
Draw a Table
Hour 1 2 3 4 5 6
Slower Car
Faster Car

Step 4:Look back
Thus, in 6 hours a faster car catches up with the slower car.
Hour 1 2 3 4 5 6 7 8
Slower Car 55 110 165 220 275 330 385 440
Faster Car 0 65 130 195 260 325 390 455
In 7th hour:
Slower car = 385
Faster car = 390
(7-1= 6)

Problem 3
Jen was able to save P5.25 on the first day of school in his piggy bank. He doubles
the amount each day thereafter. He then had a total of P330.75. How many days did
he save this amount?
 Total saved: P330.75
 1st day = P5.25
 Doubles the amount each day
Step 2:Plan Draw a Table
Day Amount
1 5.25
2
Total 330.75

Step 4:Look back
Therefore, he saved the amount in 6 days.
Day Amount
1 5.25
2 10.50
3 21.00
4 42.00
5 84.00
6 168.00
Total 330.75

Finding a Pattern
Is a strategy in which students look for a patterns in the
data to solve the problem.

Problem 1
The 1st week of chess club had 3 students. The 2nd had 5 students.The
3rd had 8 students and 4th week had 12. If this pattern continues, how
many students will show up for the 8th week?
 1st week= 3 students
 2nd week = 5 students
 3rd week = 8 students
 4th week = 12 students
 8th week = ?
Step 2:Plan Finding a Pattern
1st 2nd 3rd 4th 8th
3 5 8 12 ?

Step 4:Look back
Thus, there are 38 students showed up for the eight week.
Week 1 2 3 4 5 6 7 8
No. of
students
3 5 8 12 17 23 30 38
+2 +3 +4 +5 +6 +7 +8

Problem 2
Tomas looks through a window and sees the top 3 of a 7 row display of digital
cameras. He sees 4, 6 and 8 cameras in these rows. How many cameras are in there in
the whole display.
Step 2:Plan Finding a Pattern
Sees the top 3 rows of a 7-rows display of digital
cameras
4, 6, 8 cameras on these rows
7th 6th 5th 4th 3rd 2nd 1st
4 6 8 ? ? ? ?

Step 4:Look back
Row 7 6 5 4 3 2 1 Total
No. of
Cameras
4 6 8 10 12 14 16 70
+2 +2 +2 +2 +2 +2
There are 70 cameras on the whole display.

Problem 3
A stack of firewood has 28 pieces of wood at the bottom, 24 on top of these, then
20, so on. If there are 108 pieces of wood, how many rows are there?
Step 2:Plan
• There are 108 pieces of wood.
• Bottom= 226 pieces
• On top of 28 pieces = 24 pieces
• Next = 20 pieces
Finding a Pattern

Step 4:Look back
8
12
16
20
24
28
-4
-4
-4
-4
-4
8 + 12+16+20+24+28 = 108
Thus, there are 6 rows of stack of firewood.

Draw a Picture/Diagram
Is a problem solving strategy in which students make a
visual representation of the problem.

Problem 1
In a class, there are 24 students, 16 play badminton and 12 play
volleyball. How many students only play volleyball?
Step 2:Plan Draw a picture or diagram
 No. of students= 24
 No. of players = 28
 Play badminton = 16
 Play volleyball = 12
 How many students play volleyball?
 No. of students= 24
 No. of players = 28
 Play badminton = 16
 Play volleyball = 12
 How many students play volleyball?

Step 4:Look back
8 4 12
Volleyball Badminton
8+4+12= 24 students
8+4+12+4 = 28 players
Thus, there are only 8 students who only play volleyball.

Problem 2
How many diagonals are there in a hexagon?
 Number of diagonals in a hexagon?

Step 4:Look back
 First side = 3 diagonals
 Second side = 3 diagonals
 Third side = 2 diagonals
 Fourth side = 1 diagonal
 Fifth side = 0
 Sixth side = 0
In a hexagon, there are 9 diagonals.

Problem 3
I am planning to make a square-shaped garden. How many posts 2 meters apart are
needed to fence it when one side of my garden measures 12 meters?
 Square-shaped garden
 2 meters apart needed to fence the garden
 One-side of the garden measures 12 meters

Step 4:Look back
 There are 20 posts needed to fence the square-shaped garden.

Using a Formula
Is a problem solving strategy that students can use to find
answers to math problems involving geometry, percents,
measurements, or Aljebra.

Problem 1
A small house is located inside a bigger square. The length of one side of the
small square is 3 inches and the length of one side of the big square is 7 inches.
What is the area of the region located outside the small square, but inside the big
square?
Step 2:Plan Using a formula
 Small square located inside bigger square
 Length of one side of small square is 3 inches
 Length of one side of bigger square is 7 inches
 Area of the region located outside the region of the small
square but inside the big square?

Step 4:Look back
 A = S2
 A =3 squared
 A =9 inches squared
 A = S2
 A =7 squared
 A =49 inches squared
 Area of the bigger square – Area of the smaller square
 49 inches squared – 9 inches squared = 40 inches squared
 Therefore, the area of the region located outside the small
square but inside the big square is 40 inches squared.

Problem 2
Carlos is surveying a plot of land in the shape of a right triangle. The area of the land
is 45,000 square meters. If one leg of the triangular plot is 180 meters long, what is
the other leg of the triangle.

Step 4:Look back
A = 45, 000 meters squared
180 m
A = BH/2
45, 000 meter squared
1 =
180 m (h)
1 2
95, 000 meters squared/180 m
h = 500 m
Therefore, the measure of the other leg of the triangle is 500m

Problem 3
How many 5 cm. square tiles are needed to tile the floor of a bathroom 2.4 long and
1.5 m wide?

Step 4:Look back
Area of the square= S2
Area of the rectangle = LXW
1m = 100cm
S = 5 cm
A = 5cm squared = 25cm squared
Length = 2.4 m Width = 1.5m = 1.5 x 100 = 150cm
Area of rectangle = L X W
A = 240 cm x 150 cm
A = 36, 000 cm squared
36 000 cm squared/25 cm squared = 1, 440
 Therefore, there are 1, 440 square tiles needed to tile a floor
of a bathroom.

Using an Equation and Variables
Is a problem solving strategy that uses algebraic
expressions and variables.

Problem 1
Combined, Megan and Kelly worked 60 hours. Kelly worked twice as
many hours as Megan. How many hours did they each work?
Step 2:Plan Using an equation or variables

Step 4:Look back
M + 2m =60
3m/2 = 60/3
m =20
Megan = 20 hours
Kelly = 2m
= 2 (20)
Kelly = 40 hours
Therefore, Kelly worked 40 hours and worked 20 hours

Problem 2
You went to the store and bought x bags of carrots and y bananas. Each bag of carrot
costs 1.50 and each bananas is 0.25. You spent 6.50. The total number of items you
purchased is 11. How many bags of carrots did you buy? How many bananas did you
buy?
 5x + 0.25 y = 6.5
 X + y = 11

Step 4:Look back
 5x + 0.25 y = 6.5
 X + y = 11
1.5x + 0.25 (11-x)= 6.5
1.5x +2.75-0.25x = 6.5
1.25x =6.25-2.75
1.25x =3.75
1.25x/1.25 = 3.75/1.25
X = 3 bags of carrots
X + Y =11
3 + y = 11
Y = 8 number of bananas
Therefore, there are 3 bags of carrots and 8 bananas

Problem 3
A coin purse contains 10 coins. There are 5 more P1.00 coins than P10.00 coins and
thrice as many P5.00 coins as the P10.00 coins. How many of each denominations are
there? How much does the purse contain?
X + (x+5) + 3x = 10

Step 4:Look back
X + (x+5) + 3x = 10
2x +3x +5 =10-5
2x +3x = 5
5x/5 = 5/5
X = 1 no. of P10.00 coin
X+5 = 1 +5 = 6 no. of P1.00 coin
3x = 3 no. of P5.00 coin
1+6+3 = 10 coins
Amount of money in the purse:
1 (P10.00) + 6 (P1.00) + 3 (P5.00)
P10 +P6 + P15 = P31.00
Therefore, the purse contains P31.00

Make a List
It is writing down all the combination for possibilities in an
organized list.

Problem 1
She is taking pictures of Jerick, Kanah, and April. How many different
ways could the 3 stand in a line?
Step 2:Plan Make a List
K, J, A

Step 4:Look back
1. Jerick, Kanah, April
2. Jerick, April, Kanah
3. Kanah, April, Jerick
4. Kanah, Jerick, April
5. April, Jerick, Kanah
6. April, Kanah, Jerick
Therefore, there are 6 different ways that the 3
can stand on a line.

Problem 2
There will be 5 teams playing in the Pine cone regional softball tournament. Each
team is scheduled to play other team once. How many games are they scheduled for
the tournament?
5 teams ( A, B, C, D, E

Step 4:Look back
A, B, C, D, E
AB, AC, AD, AE = 4
BC, BD, BE = 3
CD, CE = 2
DE = 1
Therefore, there are 10 games scheduled for the tournament.

Problem 3
Jazmine has 4 pairs of shoes; white, pink, blue and black. She has 6 pairs of socks of
different colors. How many different pair of shoes and socks she can wear?

Step 4:Look back
Shoes
white
Pink
blue
black
Sock
1
2
3
4
5
6
White = 6
Pink = 6
Blue = 6
Black = 6
Therefore, there are 24 different pair of shoes and sock she can wear.

Eliminating Possibilities
It is a problem solving strategy in which students remove
impossible answers until the correct answer remains.

Problem 1
The product of an unknown number multiplied by 4 is greater than 20
but less than 35. The unknown number is divisible by 4. What is the
unknown number could be?
Step 2:Plan Eliminating possibilities
• Product of an unknown number multiplied by 4 is
greater than 20 but less than 35
• Unknown number is divisible by 4
• What is the unknown number?
4, 8, 12, 16, 20, 24, 28, 32

Step 4:Look back
Therefore, the unknown number is 8.
>20 <35
4x4 = 16 X /
8x4 =32 / /
12x4 =48 / X
16x4 =64 / X
20x4=80 / X
24x4 =96 / X
28x4 =112 / X
32x4 =128 / x

Problem 2
In the game of football, a team can score either a touchdown for 6 points or field goal
for 3 points. If a team only scores touchdowns or field goals but does not get any
extra points (no points for an extra kick) what scores cannot be achieved if the team
scored under 30 points?
Step 2:Plan
Eliminating possibilities

Step 4:Look back
Divisible by 6
1, 2, 3,4,5,6,7,8,9, 10,11,12,13,14,15,16,17,18,19,20
21,22,23,24,25,26,27,28,29
Divisible by 3
1, 2, 3,4,5,6,7,8,9, 10,11,12,13,14,15,16,17,18,19,20
21,22,23,24,25,26,27,28,29
Therefore, the scores that cannot be achieved are
1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25,26,28,29

Problem 3
Tom, Tanya, and Josh live on Main Street. Two of them live on the right side of the street. The other
one is across the street. One house is painted red, another has a circular driveway, and a third house is
made of brick. The brick house is on the left side of the street. Tom has a pickup truck which is parked
in his circular driveway. Tanya lives across the street from Tom. Which house does Josh live in?
Step 2:Plan Eliminating possibilities

Step 4:Look back
TOM TANYA JOSH
Painted red X X /
Has circular driveway / X X
Made of brick x / x
Therefore, Josh lives in the painted red house located at the right side
of the street.

Work backwards
It resolve problems which do not have beginning.
They usually begin with final answer and requires to work
backwards through the steps to find the beginning.

Problem 1
After Bill gave 17 trading cards to his friend Jim, he has still 68 cards
on his own. How many trading cards Bill have before he gave some
away?
Step 2:Plan Work backward
- 17 =68

Step 4:Look back
- 17 =68
68 +17 = 85
Therefore, there are 85 cards Bill have before he gave it
away.

Problem 2
I think of a number and add 3 to it, multiply the result by 2, subtract 4 and divide by
7. The number I ended up is 2. What was the number I first think of?

Step 4:Look back
Therefore, the first number is 6.
2 x 7 + 4 /2 – 3
14 +4 =18/2=9-3 = 6
+ 3 x 2 -4 / 7 =2

Problem 3
Juan had P80.00. He spent P40.00 for his snack and lost some amount on the way
home. If he had P20.00 when he reaches home, how much did he lose?
+ 20 + 40=80

Step 4:Look back
+ 20 + 40=80
80 – 40 -20
40 – 20 = 20
Therefore, he lost P20.00 on the way home.

Design a Model
A problem solving strategy that needs to create a
representation of something.

Problem 1
Dr. Santos bought a DLP set and a laptop. She spends 200, 000
altogether. If the Laptop cost 65, 000, how much did she pay for the
DLP set?
Step 2:Plan Design a Model

Step 4:Look back
P200,00.00
P65,000.00 ?
P200,00.00 – P65,000.00 = P135,000.00
Therefore, she paid P135,000.00 for the DLP set.

Problem 2
A vegetable seller packed 153 tomatoes into bags of 6 each. How many bags were
there? How many tomatoes were left over?

Step 4:Look back
153 tomatoes 6 tomatoes( 1 bag)
25 bags and 3 pieces
Therefore, there are 25 bags of tomatoes and 3 pieces remaining.

Problem 3
Jen owns a bag store. She is buying a new keychains to put on her bags. She has 6 blue bags that
needs 3 blue keychains each and 5 red bags that needs 2 red keychains each. Each keychain cost
P10.00. How many keychain does Jen needs? How much will all these keychains cost?
Each bag needs 3 blue
keychains
Each bag needs 2 red
keychains

Step 4:Look back
Each bag needs 3 blue
keychains
Each bag needs 2 red
keychains
6 blue bags x 3 blue
keychains = 18 x 10 =
P180
5 red bags x 2 red
keychains = 10 x 10 =
P100
Therefore, Jen needs 28 keychains and it costs P280.00

Solving A Simpler Problem
It is used in solving problems that deals with large numbers.

Problem 1
Tony’s restaurant has 30 small tables to be used for a banquet. Each table can
seat only one person in each side. If the tables are pushed together to make one
long table, how many people can sit at the table?
Step 2:Plan Solving a simpler problem

Step 4:Look back
X X
X
X
X
X
X
X
X X X
X X
X
2 tables = 6 people
Therefore, there are 62 people who can sit at the long table composed of 30
small tables.
2 tables = 6 people 3 tables = 8 people 4 tables = 10 people
No. of tables x 2 people = +2 no. of outside seat
2 tables = 6 people
30 tables x 2 = 60 + 2 = 62 people

Problem 2
How many squares(of any size) can be found on 8x8 square grid similar to the
chessboard below?

Step 4:Look back
Grid sizes Number of squares
1x1 1
2x2 5
3x3 14
4x4 30
5x5 55
6x6 91
7x7 140
8x8 204
+4
+9
+16
+25
+36
+49
+64
8X8 = 64+49+36+25+16+9+4+1 = 204
There are 204 squares in a 8x8 square grid.

Problem 3
In a certain supermarket, in every 2 oranges you purchased there is a free one
orange and it costs P40.00. When I have P400.00, how many oranges can I have?

Step 4:Look back
2 oranges (3) + 1 free = P40.00
4 oranges (6) + 2 free = P80.00
2 oranges =P40
1 orange + P20
Thus, (amount of money)/20 =ans +ans/2
Budget Money = P400
P400/2O=20
20 +1O =30 oranges
Therefore, there are 30 oranges for a P400.00

Logical Reasoning
It is the process of using rational, systematic steps based
on mathematical procedure to arrive at a conclusion about
a problem.

Problem 1
Rees walks 10 m towards East and then turns left and walk 12m.
Again, he turns right and walks 18m. Finally, he turns to his right and
walks 12 m. How far is Rees from starting point?
Step 2:Plan Logical Reasoning
18m ®
12m ®
12m l
10m E

Step 4:Look back
Therefore, Rees is 28 m from the starting point.
18m ®
12m ®
1Om + 18m = 28m
12m l
10m E

Problem 2
In a certain code INACTIVE is written as VITCANIE. How is computer written in that
code?

Step 4:Look back
I N A C T I V E
V I T C A N I E
 4 and 8 stays on its position
 Interchange 7 and 1
C O M P U T E R
E T U P M O C R
Therefore,computer is written in the code etupmocr.

Problem 3
There are 2 equilateral triangles. The larger one has a Perimeter that is thrice than
the smaller one. About how much smaller are the side lengths in the smaller triangle?
The perimeter of the smaller one is 15 cm.

Step 4:Look back
P=15 cm
P=45 cm
15 x3 =45 cm
P=S1 + S2 + S3
P= 15/3 =5CM
The side length of the smaller
triangle is 5 cm.
P=S1 + S2 + S3
P= 45/3 =15CM
The side of the larger triangle is 15
cm.
Therefore, the side length of the smaller triangle is
10 cm smaller than the side length og bigger triangle.
Side of bigger triangle – Side of smaller triangle
15 cm -5c m = 10 cm

Act it Out
 A strategy in which students physically act out which is
taking place in a word problem.

Problem 1
The window cleaner was standing on the middle rung of the ladder cleaning the
outside windows of the office block. He climbed up to three rungs to clean some
windows then saw a spot he had missed below him. He climbed down seven rungs
to clean it and then climbed up the remaining ten rungs. He was now at the top
of the ladder. How many rungs were there altogether on the ladder?
Step 2:Plan Act it Out

Problem 2
Place 14 blocks in three piles. The first pile should have one less than the third. The
third pile should have twice as many as the second. How many blocks are there in
each pile?
Step 2:Plan
Act it out

Step 4:Look back

Problem 3
Some students were in line at the cashier to pay for their tuition fee. There was a
student in front of 2 students, a student between 2 students, and a student behind 4
students. What is the least number of students that could be in the line?
Step 2:Plan Act it out

Symmetric Strategy
This suggests a problem solving principle, in that if we
know that a problem has a symmetric solution, then that
solution can be easier to solve.

Problem 1
Matt and five other boys had 18 km. to run in a relay race. How far
should each boy run so they all cover an equal distance?
Step 2:Plan Symmetric Strategy

Step 4:Look back
Therefore, each boy should run 3 km so they can cover an
equal distance.
18km/6 boys = 3km

Problem 2
Jen and Alwyna have combined weight of 101 kg. Jen and Hazel have a combined
weight of 109kg. Alwyna and Hazel have combined weight of 112kg. How heavy does
Jen weight?
Step 2:Plan Symmetric Strategy
J + A = 101
J + H = 109
A + H =112

Step 4:Look back
J + A = 101
J + H = 109
A + H =112
2(J+A+H)=322
J +A+H =161
J+112 = 161
J=49kg =Jen’s weight
Thus, Jen is 49kg.

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