Math problem solving strategies pdf

Pick three children to represent Marcus, Nina, and Ophelia. Start by giving Ophelia 15 chips. This is 5 more than Nina, so give Nina 10 chips. Nina found 7 fewer than Marcus, so Marcus must have 17 chips.

He claims that he can make the sums of the numbers on all three stacks equal the same amount by moving just one block. How can Max do it? Solution: Get 9 blocks and paste a slip of paper on each with the number as shown.

Students can now physically move the blocks around to see when the sums are the same in all three stacks. A little bit of logical rea- soning can help. The middle stack already has a sum of 15, so the block to be moved must come from the other two stacks.

Furthermore, because the first stack only adds up to 6, the block must come from the third stack. In fact, moving the blocks from stack 3 one at a time to stack 1 will reveal the correct answer.

Now all three stacks have a sum of She wants to put them on display in the school showcase. She arranges them so that there are the same number of rocks in each row. Solution: Take 16 chips and arrange them in all possible ways in equal rows see Figure 5. Teaching Notes: This might be a good time to discuss the fact that 8 rows of 2 are different from 2 rows of 8—even though they are companions. This could be used to develop the concept of a perfect square. T his strategy can be difficult for the students to master.

For most of their mathematical lives, they have been taught to start at the begin- ning of a problem and carry the action through, on a step-by-step basis. The students begin with the end result of the problem, and carry the action backwards to find conditions at the beginning. The mathematical opera- tions are reversed; so, for example, what was subtraction now becomes the inverse operation, namely, addition. Once the answer has been found, the results can be checked by start- ing with this answer and carrying the action through from start to finish.

Although, on the surface, the procedure may seem unnatural, it is used in everyday decision making without much fanfare. Take, for example, the task of finding the best route to an unfamiliar place on a map. Typically, we first try to locate the destination point and then gradually work back- wards through a network of roads until we get to familiar surroundings.

However, when it comes to mathematical applications of this technique, we have to encourage students to include this procedure in their arsenal of problem-solving tools, even where it may not be in an obvious problem- solving technique. Evelyn, Henry, and Al play a certain game. The player who loses each round must give each of the other players as much money as each of them has at that time.

In Round 1, Evelyn loses and gives Henry and Al as much money as they each have. In Round 2, Henry loses, and gives Evelyn and Al as much money as they each then have. Al loses in Round 3 and gives Evelyn and Henry as much money as they each have. How much money did they each start with? Solution: You may have begun this problem by setting up a system of three equations in three variables.

Can it be done? Of course! However, because the problem requires a great deal of subtraction and simplification of parenthetical expressions, the final set of equations stands the chance of being incorrect.

Even if the correct set of equations is obtained, it must then be solved simultaneously:. These are the same answers we arrived at by solving the problem algebraically.

This is fundamental to using this strategy. Students must have a reasonable grasp of the structure of a problem prior to being able to trace the problem from the ending conditions back to the starting point. In particular, it is essential that students are able, once they determine a set of starting conditions, to check their answer.

Such discussion might usefully include mention of analogous compu- tational exercises they have previously mastered. You might focus on sim- ilarities and ask students to look for clues. As always, students should be encouraged to plan their mathematics efforts and look for elegant and effi- cient solutions. Working Backwards Problem 6. Last week, they traded stamps. When the meeting was over and they had finished trading, Ilana had 28 stamps.

She had given 10 to Jennie, and she received 12 from Karl and 7 from Luis. How many stamps did Ilana start with? Solution: Because we know how many stamps Ilana finished with the end condition and want to know what she started with the beginning condition , we can use the working backwards strategy. Ilana finished with She had She got 7 from Luis. She must have previously had She got 12 from Karl.

She must have previously had 9. She gave Jennie She must have started with Teaching Notes: To check the answer, begin at the beginning with 19 stamps and follow the action from start to finish. You should end up with Ilana having 28 stamps.

If each song he plays averages 3 minutes, how many songs can he plan for his Saturday evening program? Solution: Because we know the end of the problem 1 hour and want to find the earlier information, we can use the working backwards strategy. Answer: He can play 7 songs in one hour. Teaching Notes: Again, work forward as a check. Begin with 21 minutes of songs. Add 27 minutes for commercials, 3 minutes for local announce- ments, 4 minutes for weather, and 5 minutes for news.

The total should be 60 minutes. Stan has 9 more cards than Tara. Tara has 17 cards. How many baseball cards do the three friends have together? Solution: We know the end of the problem: Tara has 17 cards.

We can begin there and work backwards as follows. Ron has two times the number of cards that Stan has, so he has 52 cards. Answer: They have a total of 95 cards. ThenVictor came along and noticed the cookies. He ate a third of what was left in the jar. Shawanna, who was waiting around nearby, decided to take a fourth of the cookies left in the jar.

Then Tiffany came rushing up and took one cookie to munch on in her class. When Valerie looked at the cookie jar, she saw there were two cookies left. Can you answer Valerie? Solution: Because we know the ending of the problem 2 cookies left and we want to find the starting situation how many cookies were in the jar , the working backwards strategy is one we might find useful.

Therefore, 3 cookies must have been in the jar when she arrived. Therefore, 4 cookies must have been in the jar when she arrived—that is, she took one cookie.

Therefore, 6 cookies must have been in the jar when he arrived, so when he took 1—3 of the cookies 2 , he left 4 in the jar. Therefore, there must have been double the number of cookies in the jar when they arrived, or 12 cookies.

Answer: There were 12 cookies in the cookie jar at the start. They sent 1—3 of their baseballs to be rubbed with special mud to take the gloss off. They gave 15 more baseballs to their star outfielder to autograph. The batboy took 20 baseballs for batting practice.

They had only 15 baseballs left. How many baseballs were in the box at the start? This is a working backwards problem. The problem is, however, you only have a 7-minute timer and an minute timer. How can you boil the egg for exactly 15 minutes using only these two timers? Solution: The strategy here is to work backwards.

Neither egg timer alone can time the egg for 15 minutes. Running the 7-minute timer twice consec- utively will only give you 14 minutes of timing. The minute timer is still missing 4 minutes to reach the desired 15 minutes. How might we measure or time 4 minutes that could be added on to the minute timer?

Turn both timers over at the same time and let them run. When the 7-minute timer is finished, there will be exactly 4 minutes left on the minute timer.

This is when you start your timing of the egg boiling. When the minute timer runs out, turn it over and continue boiling the egg. Answer: See above.

As he is walking home, he meets his friend Yehor. He gives Yehor half of what he has and then gives him one extra. He walks on a little further and meets Rochelle. He gives her half of what he now has, and then one extra.

Walking a little further, he comes upon a child crying. He gives him half of what he has, and one extra. When he gets home, he looks in the bag and sees that he has five pieces of candy left.

How many did he start with? Solution: Because we know the ending of the problem and wish to find out how many pieces of candy he had at the beginning, we can use the working backwards strategy. Roberto ends up with 5 pieces of candy. However he gave the crying child half of what he had and one extra. So, at that point, he had Now, he gave half of what he had plus one extra when he met Yehor. Thus, Roberto started with 54 pieces of candy in the bag.

Teaching Notes: We can check the answer by starting with 54 pieces of candy and working forward. Roberto starts with 54 pieces of candy. He meets Yehor, gives him half 27 plus one extra or Thus, 54 — 28 leaves him with He meets Rochelle and gives her half 13, plus an extra or 14 , leaving him with He meets the crying child and gives him half 6 plus an extra, or 7 leaving him with 5.

Hence, our answer is correct. Each of them ate one food pack for breakfast, one for lunch, and one for dinner on each of the first two days. That evening, by mistake, someone dumped half of what food was left into space.

When they woke up the next morning, they each had one pack for breakfast. They counted and found only 5 food packs left. How many food packs had they started with? His teacher decided that each student could drop his or her lowest test score and then the teacher would compute their averages over again. Jack dropped the 25 he had scored on one test.

What would be his new average? To compute an aver- age or mean, we add all the scores and divide by the number of scores. The new total is with only 10 tests. Answer: His new average would be He checked the price on January 1. How much would he have saved if he had bought it in January? By mistake, he multiplies by 10 then divides by 5 instead of multiplying by 5 and then dividing by If the incorrect answer he got was 3, what is the correct answer?

Solution: We must first find the original number with which Doug started. The working backwards method will be useful here. Doug ended up with 3 after dividing by 5, so he must have had 15 at that point. He got 15 when he multiplied by 10, so he must have begun with 1.

Now, we will work forward from 1. Answer: The correct answer is 0. She left an envelope with the money in it on the kitchen table.

On his way to work, Stu took half the money. An hour later, Rich needed some money so he took half the money that was in the envelope. Later, Ed took half the money that was in the envelope when he left the house. How much money had been in the envelope at the start? To share the costs equally, how much money should each child give to the others? Solution: We can work backwards by first determining what each person had to pay by finding the total amount required and then dividing by the number of participants.

Now we can examine what each child spent and decide who owes money to whom. Teaching Notes: This might be an excellent time to discuss with the students how to make a problem simpler by breaking it up into its parts and solving each part separately.

A famous mathematician once said that mathematics is a search for patterns. Patterns occur in many situations. Students need practice in examining data to see if a pattern exists. Other problems may require a table or list to organize the data and see if a pattern emerges. However, a very powerful problem-solving strategy for problems that do not directly call for finding a pattern is, in fact, to search for a pattern and then use it to solve the problem.

In everyday life situations, we are often called on to find a pattern to solve a problem, but not asked to do so directly. Take, for example, search- ing for a particular address in a neighborhood with which you are not familiar. If you are looking for Main Street, you first determine on which side are the odd-numbered addresses, and then in which order the numbers are ascending or descending. This involves finding the pattern and then continuing it to your goal.

Finding a pattern can sometimes be quite challenging, whereas at other times, it is almost directly presented to you. The best way for students to learn to discover patterns is to practice finding patterns in different problem situations. Thus, if we input 3, the machine can only operate on 3s. The machine uses the four fundamental operations of arithmetic addition, subtraction, multiplication, and division either alone or in combination.

Solution: You may have begun this problem by attempting to guess the function rule. This is a very difficult and time-consuming task. However, the problem can be solved by using the finding a pattern strategy together with some reasoning to determine what the function machine is doing when we input a number.

The output appears to be close to the cube of the input number. That is,. Understand the purpose for and context of the pattern. Example: In the above problem, understanding that we are looking for a pattern that describes the mathematical relationship between input and output. Identify the repeating elements in the pattern.

Example: In the above problem, noticing that the output values are near the cubes of the input. Extend the identified pattern.

Although such skills have their roots in the primary grades, they are continuing to be developed in grades 3 through 5. In addition to continued and varied exploration of patterns—such as offered in this chapter—your students need to learn to first notice and to then systematically organize their observations. Much of this can be addressed in class discussions prior to and subsequent to students beginning independent work.

Multiple student descriptions will lay the groundwork for exploring how a pattern might be continued. Multiple student solutions will allow you to address both efficiency and elegance. Keep in mind, however, that noticing and extending only provide an outline for solving the problem. Your students must still use their conceptual understandings and computational skills to create a solution.

Problem 7. Solution: The pattern for each term after the first in this sequence is to double the previous term and add 1. Some students may recognize a pattern of add 6, add 12, add 24, and so on.

Either pattern rule is fine to solve the problem. Remember, many mathematics problems have more than one method of solution. It is good to provide students with alternate solutions, because each one increases their problem-solving fluency. Answer: The next two terms are 95 and The diameter of the first circle is 6 cm.

The diameter of the next circle is 5. He continues in this manner, cutting circles, and his final cir- cle has a diameter of 2. How many circles did he cut out from the cardboard? Solution: We examine the data, looking for a pattern.

The pattern rule tells us that each circle is 0. Circle 1 2 3 4 5 6 7 8 Diameter Size 6 5. He reaches a circle of diameter 2. Answer: He cut 8 circles from the cardboard.

Finding a Pattern Evans is designing square swimming pools. Each pool has a square center that is the area of the water. Evans uses blue tiles to show the water. Around the square pool, he puts a border of white tiles. Figure 7. Each time the number of blue tiles increases to the next perfect square, the number of white tiles increases by 4.

Continue the table. This leads to our answer. Some students may notice that the number of white tiles is the same as the perimeter of the square made up of the blue tiles plus the 4 tiles needed for each corner. Answer: There are 24 white squares in the pool that has 25 blue tiles. Two hexagons placed side by side as shown in Figure 7. Three hexa- gons placed side by side have a perimeter of 14" the darkened line in Figure 7. What would be the perimeter of the figure formed by 6 hexagons placed side by side in the same manner?

If there is such a pattern we can use it to solve the problem for any number of hexagons. Number of Hexagons Perimeter 1 6" 2 10" 3 14".

The perimeter seems to increase by 4 as the number of hexagons increases by 1. Number of Hexagons Perimeter 1 6" 2 10" 3 14" 4 18" 5 22" 6 26". We now have the number of tiles for 6 hexagons. Teaching Notes: The problem could be made more visual using hexagonal pattern blocks. Students might also notice a pattern: There are four sides on each of the hexagons when they are placed end to end and always two sides on the ends. Solution: If we look at these numbers we should be able to recognize a pattern of a different sort; namely, these are consecutive prime numbers— that is, numbers that can be divided only by themselves and 1.

Therefore, the next two prime numbers are 19 and What color is the 13th light? Solution: The sequence of lights is green, yellow, red, green, yellow. The 13th light would be green. Notice that the sequence of green lights is 1, 4, 7, 10, 13,. Answer: The 13th light is green. When we cut it with a single vertical line, we get 3 pieces.

When we cut it with two vertical lines, we get 5 pieces. When we cut it with three vertical lines, we get 7 pieces. Into how many pieces will 7 vertical lines divide the C?

There is a pattern. Each time we add a cut, the number of pieces increases by 2. Thus, for 4 cuts, we have 9 pieces; for 5 cuts, we have 11 pieces; for 6 cuts, we have 13 pieces; and for 7 cuts, we have 15 pieces. Sam can hop up one step at a time or two steps at a time. Sam hops up the flight a different way each time. How many different ways can Sam hop up the flight of seven steps? Solution: We will begin with a simpler situation, building up to the required one, and searching for a pattern that might allow us to get the answer more quickly than if we were to exhaust all possibilities.

Each step appears to require the number of hops equal to the sum of the two previous steps i. Number of Steps Ways Sam Can Hop Number of Ways 1 1 1 2 1—1, 2 2 3 1—1—1, 2—1, 1—2 3 4 1—1—1—1, 2—1—1, 2—2 5 1—2—1 1—1—2 5 1—1—1—1—1, 2—1—1—1, 1—2—2 8 1—2—1—1, 2—1—2 1—1—2—1, 2—2—1 1—1—1—2. Our pattern rule is right, and the sequence is 1, 2, 3, 5, 8, 13, 21, Rather than write the sequence, some students might simply continue the table for seven steps.

This is perfectly correct and will lead to the same answer, Notice, too, that this sequence of numbers is known as the Fibonacci numbers, where each is the sum of the previous two numbers. This sequence occurs in this problem because, in order to reach a given step, Sam must hop from either the step one below or the step two below. Answer: Sam can hop up the staircase in 34 different ways. He is designing a large wall covering for a client.

The entire design is made up of 50 concentric squares squares with the same center and sides parallel. David is going to outline the perimeter of each square with wool.

How many feet of wool does he need to outline all 50 squares? Solution: We examine the data and see if we can find a pattern to help us. Because a square has four equal sides, we can find the sum of one side of each square and then multiply by four. This will let us use smaller numbers. Starting from the smallest square and moving outward, the lengths of the sides form a sequence of the first 50 odd numbers: 1, 3, 5, 7,.

Notice that the sums are perfect squares. In fact, each sum is the square of the number of terms being added. Using this pattern, the sum of all 50 terms will be , or 2, Now we multiply by 4 to find the perimeter and arrive at 10, feet of wool. You can also look at this problem from a different point of view, using a different pattern or, if you wish, organizing data. As before, we then multiply by 4 to get 10, Answer: David will need 10, feet of wool for the wall covering.

In some problems, it is often helpful to make tree diagrams similar to the one in Figure 7. This drawing shows four layers and 15 branches in all. If we continued this draw- ing for 8 layers, how many branches would there be?

Solution: We could actually continue drawing the tree diagram until we had shown 8 layers. However, this would be complicated to draw and might not be very accurate. The difference between successive numbers of branches is 2, 4, 8,. This would make sense, because each branch emits two new ones in the next layer. Thus, we continue the table as follows:. Answer: There would be branches in the eighth layer. She has 17 interlocking blocks and wants to make the largest shelf possible.

If she uses 5 blocks, she can make a shelf that holds 3 dolls. Six blocks will hold 4 dolls. Seven blocks make a 5-doll shelf, and 8 blocks make an 8-doll shelf as shown in Figure 7. How many dolls will fit on the shelf if she con- tinues in this manner and uses all 17 interlocking blocks? The number of dolls seems to be 2 less than the number of blocks. Blocks 5 6 7 Teaching Notes: Some students may reason the problem out deductively. Thus, for 17 blocks, 2 form the legs and 15 dolls can be accommodated.

This is an excellent solution with sound logical reasoning and should be discussed with the class. The pattern was two fastballs followed by two sliders followed by a change-of-pace pitch.

If the pattern continues, what kind of pitch will the 18th pitch be? Solution: Because the pitcher pitches according to a set pattern, we can try to discover the pattern rule. Thus, the 18th pitch will be a slider pitch, because there will be three groups of 5 when 15 pitches have been thrown.

Therefore, three pitches later, he will have thrown his 18th pitch. The actual sequence of pitches would be:. Teaching Notes: The emphasis in this problem is on the pattern that repeats. This happens in groups of 5. Thus, the students want to know the 3rd pitch in the 4th set of 5, which is the 18th pitch. A lthough solving any problem requires logical thinking or reasoning, some problems depend upon logical reasoning as the primary strat- egy for solving them.

These can range from the simple logic of what size product yields the best price per object i. An inference is made, which leads to a second inference, and so on. One inference leads to another. The logical process continues until the problem has been resolved.

Solution: One of the first solutions that comes to mind is algebraic. The solution strategy, as has been noted, is to draw logical conclusions from such data; that is, to use inference. However, this requires that your students learn how to sensibly and systematically organize the data. This includes learning how to analyze clues—for instance, how to use the process of elimination, lists, Venn diagrams, or matrix logic.

It is also critical that your students discuss their thinking with you and their classmates. When students are asked to solve a problem requiring logi- cal reasoning, it is not just a matter of jumping from one fact to another. It often requires reading between the lines. This will be a new experience for many of your students, and they will need to discuss and explore a number of problems as they acquire the taste of effi- ciency and elegance that logical reasoning brings to problem solving.

Problem 8. How many socks must he pull from the drawer to be sure he has a matching pair? Solution: A logical approach would be to assume the worst-case scenario, namely, that he pulls out one black sock and one blue sock on his first two pulls. The third sock must guarantee that he has a matching pair. Ron lost to Ursula in the first-round game. Tom won one game and lost one game.

Tom played Ursula in the second round. Who won the tournament? Solution: Here we can only use logical reasoning. He must have won his first- round game in order to get into the second round. Ron lost to Ursula in the first round.

This enabled Ursula to be in the second round. But Tom must have lost to Ursula in the second round, because he won only one game. Thus, Ursula was the tournament winner. Answer: Ursula won the tournament. Logical Reasoning The logical reasoning strategy must be used by students in an open- thinking mode, and as long as students can justify their steps logically, these steps should be accepted.

This can be a nice group activity. Adams planted three kinds of berry plants in her garden. Of these, 1—2 are blueberries, 1—4 are strawberries, and the rest are raspberries. She planted 10 raspberry plants. How many berry plants did she plant altogether? Therefore, the number of strawberry plants, which also represents 1—4 of the plants, must also be Then 1—2 of the plants are blueberry, which must be 20 in number.

Solution: A little logical reasoning should help. If we simply draw two intersecting lines to form four regions, we will be bunching the small numbers together and the large numbers together. This will be unbalanced from the start. The earlier the students could master the skills of four basic operations, the more the students could explore many possibilities of word problem computation problems. With this in mind, how does the very popular Math Kangaroo Contest test the grade 1 and grade 2 students?

How is it different from other math contests? The Math Kangaroo grades 1 and 2 Contest almost does not include the direct math computation problems which are very different from the math contests in China where direct computation problems could include skillful computation problems. I analyzed the most recent years of Canadian Math Kangaroo Contest grade 1 and 2 problems and they start to emerge some characteristics and categories, so I include here to help students prepare for it.

The lower grade math contest tends to skew to the more visual operation type of problems. Figures and Tables from this paper. Citation Type. Has PDF. Publication Type. More Filters. Scandinavian Journal of Educational Research.

ABSTRACT The aim of this study is to measure the effects of teaching problem-solving strategies as an integrated part of the mathematical content prescribed by the curriculum. By implementing the … Expand. View 7 excerpts, cites background and methods. Learning to solve problems that you have not learned to solve: Strategies in mathematical problem solving. Highly Influenced. View 4 excerpts, cites background.

A comparative study on problem solving strategies of gifted 4th grade students and their high-achieving counterparts. Calismanin amaci ustun yetenekli ilkokul 4.