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Math Problem Solving Strategies

In these lessons, we will learn some math problem solving strategies for example, Verbal Model (or Logical Reasoning), Algebraic Model, Block Model (or Singapore Math), Guess & Check Model and Find a Pattern Model.

Related Pages Solving Word Problems Using Block Models Heuristic Approach to Problem-Solving Algebra Lessons

Problem Solving Strategies

The strategies used in solving word problems:

• What do you know?
• What do you need to know?
• Draw a diagram/picture

Solution Strategies Label Variables Verbal Model or Logical Reasoning Algebraic Model - Translate Verbal Model to Algebraic Model Solve and Check.

Solving Word Problems

Step 1: Identify (What is being asked?) Step 2: Strategize Step 3: Write the equation(s) Step 4: Answer the question Step 5: Check

Problem Solving Strategy: Guess And Check

Using the guess and check problem solving strategy to help solve math word problems.

Example: Jamie spent $40 for an outfit. She paid for the items using $10, $5 and $1 bills. If she gave the clerk 10 bills in all, how many of each bill did she use?

Problem Solving : Make A Table And Look For A Pattern

• Identify - What is the question?
• Plan - What strategy will I use to solve the problem?
• Solve - Carry out your plan.
• Verify - Does my answer make sense?

Example: Marcus ran a lemonade stand for 5 days. On the first day, he made $5. Every day after that he made $2 more than the previous day. How much money did Marcus made in all after 5 days?

Find A Pattern Model (Intermediate)

In this lesson, we will look at some intermediate examples of Find a Pattern method of problem-solving strategy.

Example: The figure shows a series of rectangles where each rectangle is bounded by 10 dots. a) How many dots are required for 7 rectangles? b) If the figure has 73 dots, how many rectangles would there be?

a) The number of dots required for 7 rectangles is 52.

b) If the figure has 73 dots, there would be 10 rectangles.

Example: Each triangle in the figure below has 3 dots. Study the pattern and find the number of dots for 7 layers of triangles.

The number of dots for 7 layers of triangles is 36.

Example: The table below shows numbers placed into groups I, II, III, IV, V and VI. In which groups would the following numbers belong? a) 25 b) 46 c) 269

Solution: The pattern is: The remainder when the number is divided by 6 determines the group. a) 25 ÷ 6 = 4 remainder 1 (Group I) b) 46 ÷ 6 = 7 remainder 4 (Group IV) c) 269 ÷ 6 = 44 remainder 5 (Group V)

Example: The following figures were formed using matchsticks.

a) Based on the above series of figures, complete the table below.

b) How many triangles are there if the figure in the series has 9 squares?

c) How many matchsticks would be used in the figure in the series with 11 squares?

b) The pattern is +2 for each additional square.   18 + 2 = 20   If the figure in the series has 9 squares, there would be 20 triangles.

c) The pattern is + 7 for each additional square   61 + (3 x 7) = 82   If the figure in the series has 11 squares, there would be 82 matchsticks.

Example: Seven ex-schoolmates had a gathering. Each one of them shook hands with all others once. How many handshakes were there?

Total = 6 + 5 + 4 + 3 + 2 + 1 = 21 handshakes.

The following video shows more examples of using problem solving strategies and models. Question 1: Approximate your average speed given some information Question 2: The table shows the number of seats in each of the first four rows in an auditorium. The remaining ten rows follow the same pattern. Find the number of seats in the last row. Question 3: You are hanging three pictures in the wall of your home that is 16 feet wide. The width of your pictures are 2, 3 and 4 feet. You want space between your pictures to be the same and the space to the left and right to be 6 inches more than between the pictures. How would you place the pictures?

The following are some other examples of problem solving strategies.

Explore it/Act it/Try it (EAT) Method (Basic) Explore it/Act it/Try it (EAT) Method (Intermediate) Explore it/Act it/Try it (EAT) Method (Advanced)

Finding A Pattern (Basic) Finding A Pattern (Intermediate) Finding A Pattern (Advanced)

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Module 1: Problem Solving Strategies

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Unlike exercises, there is never a simple recipe for solving a problem. You can get better and better at solving problems, both by building up your background knowledge and by simply practicing. As you solve more problems (and learn how other people solved them), you learn strategies and techniques that can be useful. But no single strategy works every time.

Pólya’s How to Solve It

George Pólya was a great champion in the field of teaching effective problem solving skills. He was born in Hungary in 1887, received his Ph.D. at the University of Budapest, and was a professor at Stanford University (among other universities). He wrote many mathematical papers along with three books, most famously, “How to Solve it.” Pólya died at the age 98 in 1985.1

1. Image of Pólya by Thane Plambeck from Palo Alto, California (Flickr) [CC BY

In 1945, Pólya published the short book How to Solve It , which gave a four-step method for solving mathematical problems:

First, you have to understand the problem.

After understanding, then make a plan.

Carry out the plan.

Look back on your work. How could it be better?

This is all well and good, but how do you actually do these steps?!?! Steps 1. and 2. are particularly mysterious! How do you “make a plan?” That is where you need some tools in your toolbox, and some experience to draw upon.

Much has been written since 1945 to explain these steps in more detail, but the truth is that they are more art than science. This is where math becomes a creative endeavor (and where it becomes so much fun). We will articulate some useful problem solving strategies, but no such list will ever be complete. This is really just a start to help you on your way. The best way to become a skilled problem solver is to learn the background material well, and then to solve a lot of problems!

Problem Solving Strategy 1 (Guess and Test)

Make a guess and test to see if it satisfies the demands of the problem. If it doesn't, alter the guess appropriately and check again. Keep doing this until you find a solution.

Mr. Jones has a total of 25 chickens and cows on his farm. How many of each does he have if all together there are 76 feet?

Step 1: Understanding the problem

We are given in the problem that there are 25 chickens and cows.

All together there are 76 feet.

Chickens have 2 feet and cows have 4 feet.

We are trying to determine how many cows and how many chickens Mr. Jones has on his farm.

Step 2: Devise a plan

Going to use Guess and test along with making a tab

Many times the strategy below is used with guess and test.

Make a table and look for a pattern:

Procedure: Make a table reflecting the data in the problem. If done in an orderly way, such a table will often reveal patterns and relationships that suggest how the problem can be solved.

Step 3: Carry out the plan:

Notice we are going in the wrong direction! The total number of feet is decreasing!

Better! The total number of feet are increasing!

Step 4: Looking back:

Check: 12 + 13 = 25 heads

24 + 52 = 76 feet.

We have found the solution to this problem. I could use this strategy when there are a limited number of possible answers and when two items are the same but they have one characteristic that is different.

Videos to watch:

1. Click on this link to see an example of “Guess and Test”

http://www.mathstories.com/strategies.htm

2. Click on this link to see another example of Guess and Test.

http://www.mathinaction.org/problem-solving-strategies.html

Check in question 1:

Place the digits 8, 10, 11, 12, and 13 in the circles to make the sums across and vertically equal 31. (5 points)

Check in question 2:

Old McDonald has 250 chickens and goats in the barnyard. Altogether there are 760 feet . How many of each animal does he have? Make sure you use Polya’s 4 problem solving steps. (12 points)

Problem Solving Strategy 2 (Draw a Picture). Some problems are obviously about a geometric situation, and it is clear you want to draw a picture and mark down all of the given information before you try to solve it. But even for a problem that is not geometric thinking visually can help!

Videos to watch demonstrating how to use "Draw a Picture".

1. Click on this link to see an example of “Draw a Picture”

2. Click on this link to see another example of Draw a Picture.

Problem Solving Strategy 3 ( Using a variable to find the sum of a sequence.)

Gauss's strategy for sequences.

last term = fixed number ( n -1) + first term

The fix number is the the amount each term is increasing or decreasing by. "n" is the number of terms you have. You can use this formula to find the last term in the sequence or the number of terms you have in a sequence.

Ex: 2, 5, 8, ... Find the 200th term.

Last term = 3(200-1) +2

Last term is 599.

To find the sum of a sequence: sum = [(first term + last term) (number of terms)]/ 2

Sum = (2 + 599) (200) then divide by 2

Sum = 60,100

Check in question 3: (10 points)

Find the 320 th term of 7, 10, 13, 16 …

Then find the sum of the first 320 terms.

Problem Solving Strategy 4 (Working Backwards)

This is considered a strategy in many schools. If you are given an answer, and the steps that were taken to arrive at that answer, you should be able to determine the starting point.

Videos to watch demonstrating of “Working Backwards”

https://www.youtube.com/watch?v=5FFWTsMEeJw

Karen is thinking of a number. If you double it, and subtract 7, you obtain 11. What is Karen’s number?

1. We start with 11 and work backwards.

2. The opposite of subtraction is addition. We will add 7 to 11. We are now at 18.

3. The opposite of doubling something is dividing by 2. 18/2 = 9

4. This should be our answer. Looking back:

9 x 2 = 18 -7 = 11

5. We have the right answer.

Check in question 4:

Christina is thinking of a number.

If you multiply her number by 93, add 6, and divide by 3, you obtain 436. What is her number? Solve this problem by working backwards. (5 points)

Problem Solving Strategy 5 (Looking for a Pattern)

Definition: A sequence is a pattern involving an ordered arrangement of numbers.

We first need to find a pattern.

Ask yourself as you search for a pattern – are the numbers growing steadily larger? Steadily smaller? How is each number related?

Example 1: 1, 4, 7, 10, 13…

Find the next 2 numbers. The pattern is each number is increasing by 3. The next two numbers would be 16 and 19.

Example 2: 1, 4, 9, 16 … find the next 2 numbers. It looks like each successive number is increase by the next odd number. 1 + 3 = 4.

So the next number would be

25 + 11 = 36

Example 3: 10, 7, 4, 1, -2… find the next 2 numbers.

In this sequence, the numbers are decreasing by 3. So the next 2 numbers would be -2 -3 = -5

-5 – 3 = -8

Example 4: 1, 2, 4, 8 … find the next two numbers.

This example is a little bit harder. The numbers are increasing but not by a constant. Maybe a factor?

So each number is being multiplied by 2.

16 x 2 = 32

1. Click on this link to see an example of “Looking for a Pattern”

2. Click on this link to see another example of Looking for a Pattern.

Problem Solving Strategy 6 (Make a List)

Example 1 : Can perfect squares end in a 2 or a 3?

List all the squares of the numbers 1 to 20.

1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400.

Now look at the number in the ones digits. Notice they are 0, 1, 4, 5, 6, or 9. Notice none of the perfect squares end in 2, 3, 7, or 8. This list suggests that perfect squares cannot end in a 2, 3, 7 or 8.

How many different amounts of money can you have in your pocket if you have only three coins including only dimes and quarters?

Quarter’s dimes

0 3 30 cents

1 2 45 cents

2 1 60 cents

3 0 75 cents

Videos demonstrating "Make a List"

Check in question 5:

How many ways can you make change for 23 cents using only pennies, nickels, and dimes? (10 points)

Problem Solving Strategy 7 (Solve a Simpler Problem)

Geometric Sequences:

How would we find the nth term?

Solve a simpler problem:

1, 3, 9, 27.

1. To get from 1 to 3 what did we do?

2. To get from 3 to 9 what did we do?

Let’s set up a table:

Term Number what did we do

Looking back: How would you find the nth term?

Find the 10 th term of the above sequence.

Let L = the tenth term

Problem Solving Strategy 8 (Process of Elimination)

This strategy can be used when there is only one possible solution.

I’m thinking of a number.

The number is odd.

It is more than 1 but less than 100.

It is greater than 20.

It is less than 5 times 7.

The sum of the digits is 7.

It is evenly divisible by 5.

a. We know it is an odd number between 1 and 100.

b. It is greater than 20 but less than 35

21, 23, 25, 27, 29, 31, 33, 35. These are the possibilities.

c. The sum of the digits is 7

21 (2+1=3) No 23 (2+3 = 5) No 25 (2 + 5= 7) Yes Using the same process we see there are no other numbers that meet this criteria. Also we notice 25 is divisible by 5. By using the strategy elimination, we have found our answer.

Check in question 6: (8 points)

Jose is thinking of a number.

The number is not odd.

The sum of the digits is divisible by 2.

The number is a multiple of 11.

It is greater than 5 times 4.

It is a multiple of 6

It is less than 7 times 8 +23

What is the number?

Click on this link for a quick review of the problem solving strategies.

https://garyhall.org.uk/maths-problem-solving-strategies.html

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Math Word Problems and Solutions - Distance, Speed, Time

Problem 1 A salesman sold twice as much pears in the afternoon than in the morning. If he sold 360 kilograms of pears that day, how many kilograms did he sell in the morning and how many in the afternoon? Click to see solution Solution: Let $x$ be the number of kilograms he sold in the morning.Then in the afternoon he sold $2x$ kilograms. So, the total is $x + 2x = 3x$. This must be equal to 360. $3x = 360$ $x = \frac{360}{3}$ $x = 120$ Therefore, the salesman sold 120 kg in the morning and $2\cdot 120 = 240$ kg in the afternoon.

Problem 2 Mary, Peter, and Lucy were picking chestnuts. Mary picked twice as much chestnuts than Peter. Lucy picked 2 kg more than Peter. Together the three of them picked 26 kg of chestnuts. How many kilograms did each of them pick? Click to see solution Solution: Let $x$ be the amount Peter picked. Then Mary and Lucy picked $2x$ and $x+2$, respectively. So $x+2x+x+2=26$ $4x=24$ $x=6$ Therefore, Peter, Mary, and Lucy picked 6, 12, and 8 kg, respectively.

Problem 3 Sophia finished $\frac{2}{3}$ of a book. She calculated that she finished 90 more pages than she has yet to read. How long is her book? Click to see solution Solution: Let $x$ be the total number of pages in the book, then she finished $\frac{2}{3}\cdot x$ pages. Then she has $x-\frac{2}{3}\cdot x=\frac{1}{3}\cdot x$ pages left. $\frac{2}{3}\cdot x-\frac{1}{3}\cdot x=90$ $\frac{1}{3}\cdot x=90$ $x=270$ So the book is 270 pages long.

Problem 4 A farming field can be ploughed by 6 tractors in 4 days. When 6 tractors work together, each of them ploughs 120 hectares a day. If two of the tractors were moved to another field, then the remaining 4 tractors could plough the same field in 5 days. How many hectares a day would one tractor plough then? Click to see solution Solution: If each of $6$ tractors ploughed $120$ hectares a day and they finished the work in $4$ days, then the whole field is: $120\cdot 6 \cdot 4 = 720 \cdot 4 = 2880$ hectares. Let's suppose that each of the four tractors ploughed $x$ hectares a day. Therefore in 5 days they ploughed $5 \cdot 4 \cdot x = 20 \cdot x$ hectares, which equals the area of the whole field, 2880 hectares. So, we get $20x = 2880$ $ x = \frac{2880}{20} = 144$. Hence, each of the four tractors would plough 144 hectares a day.

Problem 5 A student chose a number, multiplied it by 2, then subtracted 138 from the result and got 102. What was the number he chose? Click to see solution Solution: Let $x$ be the number he chose, then $2\cdot x - 138 = 102$ $2x = 240$ $x = 120$

Problem 6 I chose a number and divide it by 5. Then I subtracted 154 from the result and got 6. What was the number I chose? Click to see solution Solution: Let $x$ be the number I chose, then $\frac{x}{5}-154=6$ $\frac{x}{5}=160$ $x=800$

Problem 8 One side of a rectangle is 3 cm shorter than the other side. If we increase the length of each side by 1 cm, then the area of the rectangle will increase by 18 cm 2 . Find the lengths of all sides. Click to see solution Solution: Let $x$ be the length of the longer side $x \gt 3$, then the other side's length is $x-3$ cm. Then the area is S 1 = x(x - 3) cm 2 . After we increase the lengths of the sides they will become $(x +1)$ and $(x - 3 + 1) = (x - 2)$ cm long. Hence the area of the new rectangle will be $A_2 = (x + 1)\cdot(x - 2)$ cm 2 , which is 18 cm 2 more than the first area. Therefore $A_1 +18 = A_2$ $x(x - 3) + 18 = (x + 1)(x - 2)$ $x^2 - 3x + 18 = x^2 + x - 2x - 2$ $2x = 20$ $x = 10$. So, the sides of the rectangle are $10$ cm and $(10 - 3) = 7$ cm long.

Problem 9 The first year, two cows produced 8100 litres of milk. The second year their production increased by 15% and 10% respectively, and the total amount of milk increased to 9100 litres a year. How many litres were milked from each cow each year? Click to see solution Solution: Let x be the amount of milk the first cow produced during the first year. Then the second cow produced $(8100 - x)$ litres of milk that year. The second year, each cow produced the same amount of milk as they did the first year plus the increase of $15\%$ or $10\%$. So $8100 + \frac{15}{100}\cdot x + \frac{10}{100} \cdot (8100 - x) = 9100$ Therefore $8100 + \frac{3}{20}x + \frac{1}{10}(8100 - x) = 9100$ $\frac{1}{20}x = 190$ $x = 3800$ Therefore, the cows produced 3800 and 4300 litres of milk the first year, and $4370$ and $4730$ litres of milk the second year, respectively.

Problem 10 The distance between stations A and B is 148 km. An express train left station A towards station B with the speed of 80 km/hr. At the same time, a freight train left station B towards station A with the speed of 36 km/hr. They met at station C at 12 pm, and by that time the express train stopped at at intermediate station for 10 min and the freight train stopped for 5 min. Find: a) The distance between stations C and B. b) The time when the freight train left station B. Click to see solution Solution a) Let x be the distance between stations B and C. Then the distance from station C to station A is $(148 - x)$ km. By the time of the meeting at station C, the express train travelled for $\frac{148-x}{80}+\frac{10}{60}$ hours and the freight train travelled for $\frac{x}{36}+\frac{5}{60}$ hours. The trains left at the same time, so: $\frac{148 - x}{80} + \frac{1}{6} = \frac{x}{36} + \frac{1}{12}$. The common denominator for 6, 12, 36, 80 is 720. Then $9(148 - x) +120 = 20x +60$ $1332 - 9x + 120 = 20x + 60$ $29x = 1392$ $x = 48$. Therefore the distance between stations B and C is 48 km. b) By the time of the meeting at station C the freight train rode for $\frac{48}{36} + \frac{5}{60}$ hours, i.e. $1$ hour and $25$ min. Therefore it left station B at $12 - (1 + \frac{25}{60}) = 10 + \frac{35}{60}$ hours, i.e. at 10:35 am.

Problem 11 Susan drives from city A to city B. After two hours of driving she noticed that she covered 80 km and calculated that, if she continued driving at the same speed, she would end up been 15 minutes late. So she increased her speed by 10 km/hr and she arrived at city B 36 minutes earlier than she planned. Find the distance between cities A and B. Click to see solution Solution: Let $x$ be the distance between A and B. Since Susan covered 80 km in 2 hours, her speed was $V = \frac{80}{2} = 40$ km/hr. If she continued at the same speed she would be $15$ minutes late, i.e. the planned time on the road is $\frac{x}{40} - \frac{15}{60}$ hr. The rest of the distance is $(x - 80)$ km. $V = 40 + 10 = 50$ km/hr. So, she covered the distance between A and B in $2 +\frac{x - 80}{50}$ hr, and it was 36 min less than planned. Therefore, the planned time was $2 + \frac{x -80}{50} + \frac{36}{60}$. When we equalize the expressions for the scheduled time, we get the equation: $\frac{x}{40} - \frac{15}{60} = 2 + \frac{x -80}{50} + \frac{36}{60}$ $\frac{x - 10}{40} = \frac{100 + x - 80 + 30}{50}$ $\frac{x - 10}{4} = \frac{x +50}{5}$ $5x - 50 = 4x + 200$ $x = 250$ So, the distance between cities A and B is 250 km.

Problem 12 To deliver an order on time, a company has to make 25 parts a day. After making 25 parts per day for 3 days, the company started to produce 5 more parts per day, and by the last day of work 100 more parts than planned were produced. Find how many parts the company made and how many days this took. Click to see solution Solution: Let $x$ be the number of days the company worked. Then 25x is the number of parts they planned to make. At the new production rate they made: $3\cdot 25 + (x - 3)\cdot 30 = 75 + 30(x - 3)$ Therefore: $25 x = 75 + 30(x -3) - 100$ $25x = 75 +30x -90 - 100$ $190 -75 = 30x -25$ $115 = 5x$ $x = 23$ So the company worked 23 days and they made $23\cdot 25+100 = 675$ pieces.

Problem 13 There are 24 students in a seventh grade class. They decided to plant birches and roses at the school's backyard. While each girl planted 3 roses, every three boys planted 1 birch. By the end of the day they planted $24$ plants. How many birches and roses were planted? Click to see solution Solution: Let $x$ be the number of roses. Then the number of birches is $24 - x$, and the number of boys is $3\times (24-x)$. If each girl planted 3 roses, there are $\frac{x}{3}$ girls in the class. We know that there are 24 students in the class. Therefore $\frac{x}{3} + 3(24 - x) = 24$ $x + 9(24 - x) = 3\cdot 24$ $x +216 - 9x = 72$ $216 - 72 = 8x$ $\frac{144}{8} = x$ $x = 18$ So, students planted 18 roses and 24 - x = 24 - 18 = 6 birches.

Problem 14 A car left town A towards town B driving at a speed of V = 32 km/hr. After 3 hours on the road the driver stopped for 15 min in town C. Because of a closed road he had to change his route, making the trip 28 km longer. He increased his speed to V = 40 km/hr but still he was 30 min late. Find: a) The distance the car has covered. b) The time that took it to get from C to B. Click to see solution Solution: From the statement of the problem we don't know if the 15 min stop in town C was planned or it was unexpected. So we have to consider both cases. A The stop was planned. Let us consider only the trip from C to B, and let $x$ be the number of hours the driver spent on this trip. Then the distance from C to B is $S = 40\cdot x$ km. If the driver could use the initial route, it would take him $x - \frac{30}{60} = x - \frac{1}{2}$ hours to drive from C to B. The distance from C to B according to the initially itinerary was $(x - \frac{1}{2})\cdot 32$ km, and this distance is $28$ km shorter than $40\cdot x$ km. Then we have the equation $(x - 1/2)\cdot 32 + 28 = 40x$ $32x -16 +28 = 40x$ $-8x = -12$ $8x = 12$ $x = \frac{12}{8}$ $x = 1 \frac{4}{8} = 1 \frac{1}{2} = 1 \frac{30}{60} =$ 1 hr 30 min. So, the car covered the distance between C and B in 1 hour and 30 min. The distance from A to B is $3\cdot 32 + \frac{12}{8}\cdot 40 = 96 + 60 = 156$ km. B Suppose it took $x$ hours for him to get from C to B. Then the distance is $S = 40\cdot x$ km. The driver did not plan the stop at C. Let we accept that he stopped because he had to change the route. It took $x - \frac{30}{60} + \frac{15}{60} = x - \frac{15}{60} = x - \frac{1}{4}$ h to drive from C to B. The distance from C to B is $32(x - \frac{1}{4})$ km, which is $28$ km shorter than $40\cdot x$, i.e. $32(x - \frac{1}{4}) + 28 = 40x$ $32x - 8 +28 = 40x$ $20= 8x$ $x = \frac{20}{8} = \frac{5}{2} = 2 \text{hr } 30 \text{min}.$ The distance covered equals $ 40 \times 2.5 = 100 km$.

Problem 15 If a farmer wants to plough a farm field on time, he must plough 120 hectares a day. For technical reasons he ploughed only 85 hectares a day, hence he had to plough 2 more days than he planned and he still has 40 hectares left. What is the area of the farm field and how many days the farmer planned to work initially? Click to see solution Solution: Let $x$ be the number of days in the initial plan. Therefore, the whole field is $120\cdot x$ hectares. The farmer had to work for $x + 2$ days, and he ploughed $85(x + 2)$ hectares, leaving $40$ hectares unploughed. Then we have the equation: $120x = 85(x + 2) + 40$ $35x = 210$ $x = 6$ So the farmer planned to have the work done in 6 days, and the area of the farm field is $120\cdot 6 = 720$ hectares.

Problem 16 A woodworker normally makes a certain number of parts in 24 days. But he was able to increase his productivity by 5 parts per day, and so he not only finished the job in only 22 days but also he made 80 extra parts. How many parts does the woodworker normally makes per day and how many pieces does he make in 24 days? Click to see solution Solution: Let $x$ be the number of parts the woodworker normally makes daily. In 24 days he makes $24\cdot x$ pieces. His new daily production rate is $x + 5$ pieces and in $22$ days he made $22 \cdot (x + 5)$ parts. This is 80 more than $24\cdot x$. Therefore the equation is: $24\cdot x + 80 = 22(x +5)$ $30 = 2x$ $x = 15$ Normally he makes 15 parts a day and in 24 days he makes $15 \cdot 24 = 360$ parts.

Problem 17 A biker covered half the distance between two towns in 2 hr 30 min. After that he increased his speed by 2 km/hr. He covered the second half of the distance in 2 hr 20 min. Find the distance between the two towns and the initial speed of the biker. Click to see solution Solution: Let x km/hr be the initial speed of the biker, then his speed during the second part of the trip is x + 2 km/hr. Half the distance between two cities equals $2\frac{30}{60} \cdot x$ km and $2\frac{20}{60} \cdot (x + 2)$ km. From the equation: $2\frac{30}{60} \cdot x = 2\frac{20}{60} \cdot (x+2)$ we get $x = 28$ km/hr. The intial speed of the biker is 28 km/h. Half the distance between the two towns is $2 h 30 min \times 28 = 2.5 \times 28 = 70$. So the distance is $2 \times 70 = 140$ km.

Problem 18 A train covered half of the distance between stations A and B at the speed of 48 km/hr, but then it had to stop for 15 min. To make up for the delay, it increased its speed by $\frac{5}{3}$ m/sec and it arrived to station B on time. Find the distance between the two stations and the speed of the train after the stop. Click to see solution Solution: First let us determine the speed of the train after the stop. The speed was increased by $\frac{5}{3}$ m/sec $= \frac{5\cdot 60\cdot 60}{\frac{3}{1000}}$ km/hr = $6$ km/hr. Therefore, the new speed is $48 + 6 = 54$ km/hr. If it took $x$ hours to cover the first half of the distance, then it took $x - \frac{15}{60} = x - 0.25$ hr to cover the second part. So the equation is: $48 \cdot x = 54 \cdot (x - 0.25)$ $48 \cdot x = 54 \cdot x - 54\cdot 0.25$ $48 \cdot x - 54 \cdot x = - 13.5$ $-6x = - 13.5$ $x = 2.25$ h. The whole distance is $2 \times 48 \times 2.25 = 216$ km.

Problem 19 Elizabeth can get a certain job done in 15 days, and Tony can finish only 75% of that job within the same time. Tony worked alone for several days and then Elizabeth joined him, so they finished the rest of the job in 6 days, working together. For how many days have each of them worked and what percentage of the job have each of them completed? Click to see solution Solution: First we will find the daily productivity of every worker. If we consider the whole job as unit (1), Elizabeth does $\frac{1}{15}$ of the job per day and Tony does $75\%$ of $\frac{1}{15}$, i.e. $\frac{75}{100}\cdot \frac{1}{15} = \frac{1}{20}$. Suppose that Tony worked alone for $x$ days. Then he finished $\frac{x}{20}$ of the total job alone. Working together for 6 days, the two workers finished $6\cdot (\frac{1}{15}+\frac{1}{20}) = 6\cdot \frac{7}{60} = \frac{7}{10}$ of the job. The sum of $\frac{x}{20}$ and $\frac{7}{10}$ gives us the whole job, i.e. $1$. So we get the equation: $\frac{x}{20}+\frac{7}{10}=1$ $\frac{x}{20} = \frac{3}{10}$ $x = 6$. Tony worked for 6 + 6 = 12 days and Elizabeth worked for $6$ days. The part of job done is $12\cdot \frac{1}{20} = \frac{60}{100} = 60\%$ for Tony, and $6\cdot \frac{1}{15} = \frac{40}{100} = 40\%$ for Elizabeth.

Problem 20 A farmer planned to plough a field by doing 120 hectares a day. After two days of work he increased his daily productivity by 25% and he finished the job two days ahead of schedule. a) What is the area of the field? b) In how many days did the farmer get the job done? c) In how many days did the farmer plan to get the job done? Click to see solution Solution: First of all we will find the new daily productivity of the farmer in hectares per day: 25% of 120 hectares is $\frac{25}{100} \cdot 120 = 30$ hectares, therefore $120 + 30 = 150$ hectares is the new daily productivity. Lets x be the planned number of days allotted for the job. Then the farm is $120\cdot x$ hectares. On the other hand, we get the same area if we add $120 \cdot 2$ hectares to $150(x -4)$ hectares. Then we get the equation $120x = 120\cdot 2 + 150(x -4)$ $x = 12$ So, the job was initially supposed to take 12 days, but actually the field was ploughed in 12 - 2 =10 days. The field's area is $120 \cdot 12 = 1440$ hectares.

Problem 21 To mow a grass field a team of mowers planned to cover 15 hectares a day. After 4 working days they increased the daily productivity by $33 \times \frac{1}{3}\%$, and finished the work 1 day earlier than it was planned. A) What is the area of the grass field? B) How many days did it take to mow the whole field? C) How many days were scheduled initially for this job? Hint : See problem 20 and solve by yourself. Answer: A) 120 hectares; B) 7 days; C) 8 days.

Problem 22 A train travels from station A to station B. If the train leaves station A and makes 75 km/hr, it arrives at station B 48 minutes ahead of scheduled. If it made 50 km/hr, then by the scheduled time of arrival it would still have 40 km more to go to station B. Find: A) The distance between the two stations; B) The time it takes the train to travel from A to B according to the schedule; C) The speed of the train when it's on schedule. Click to see solution Solution: Let $x$ be the scheduled time for the trip from A to B. Then the distance between A and B can be found in two ways. On one hand, this distance equals $75(x - \frac{48}{60})$ km. On the other hand, it is $50x + 40$ km. So we get the equation: $75(x - \frac{48}{60}) = 50x + 40$ $x = 4$ hr is the scheduled travel time. The distance between the two stations is $50\cdot 4 +40 = 240$ km. Then the speed the train must keep to be on schedule is $\frac{240}{4} = 60$ km/hr.

Problem 23 The distance between towns A and B is 300 km. One train departs from town A and another train departs from town B, both leaving at the same moment of time and heading towards each other. We know that one of them is 10 km/hr faster than the other. Find the speeds of both trains if 2 hours after their departure the distance between them is 40 km. Click to see solution Solution: Let the speed of the slower train be $x$ km/hr. Then the speed of the faster train is $(x + 10)$ km/hr. In 2 hours they cover $2x$ km and $2(x +10)$km, respectively. Therefore if they didn't meet yet, the whole distance from A to B is $2x + 2(x +10) +40 = 4x +60$ km. However, if they already met and continued to move, the distance would be $2x + 2(x + 10) - 40 = 4x - 20$km. So we get the following equations: $4x + 60 = 300$ $4x = 240$ $x = 60$ or $4x - 20 = 300$ $4x = 320$ $x = 80$ Hence the speed of the slower train is $60$ km/hr or $80$ km/hr and the speed of the faster train is $70$ km/hr or $90$ km/hr.

Problem 24 A bus travels from town A to town B. If the bus's speed is 50 km/hr, it will arrive in town B 42 min later than scheduled. If the bus increases its speed by $\frac{50}{9}$ m/sec, it will arrive in town B 30 min earlier than scheduled. Find: A) The distance between the two towns; B) The bus's scheduled time of arrival in B; C) The speed of the bus when it's on schedule. Click to see solution Solution: First we will determine the speed of the bus following its increase. The speed is increased by $\frac{50}{9}$ m/sec $= \frac{50\cdot60\cdot60}{\frac{9}{1000}}$ km/hr $= 20$ km/hr. Therefore, the new speed is $V = 50 + 20 = 70$ km/hr. If $x$ is the number of hours according to the schedule, then at the speed of 50 km/hr the bus travels from A to B within $(x +\frac{42}{60})$ hr. When the speed of the bus is $V = 70$ km/hr, the travel time is $x - \frac{30}{60}$ hr. Then $50(x +\frac{42}{60}) = 70(x-\frac{30}{60})$ $5(x+\frac{7}{10}) = 7(x-\frac{1}{2})$ $\frac{7}{2} + \frac{7}{2} = 7x -5x$ $2x = 7$ $x = \frac{7}{2}$ hr. So, the bus is scheduled to make the trip in $3$ hr $30$ min. The distance between the two towns is $70(\frac{7}{2} - \frac{1}{2}) = 70\cdot 3 = 210$ km and the scheduled speed is $\frac{210}{\frac{7}{2}} = 60$ km/hr.

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20 Effective Math Strategies To Approach Problem-Solving 

Katie Keeton

Math strategies for problem-solving help students use a range of approaches to solve many different types of problems. It involves identifying the problem and carrying out a plan of action to find the answer to mathematical problems.  

Problem-solving skills are essential to math in the general classroom and real-life. They require logical reasoning and critical thinking skills. Students must be equipped with strategies to help them find solutions to problems.

This article explores mathematical problem solving strategies, logical reasoning and critical thinking skills to help learners with solving math word problems independently in real-life situations. 

What are problem-solving strategies?

Problem-solving strategies in math are methods students can use to figure out solutions to math problems. Some problem-solving strategies: 

• Draw a model
• Use different approaches
• Check the inverse to make sure the answer is correct

Students need to have a toolkit of math problem-solving strategies at their disposal to provide different ways to approach math problems. This makes it easier to find solutions and understand math better. 

Strategies can help guide students to the solution when it is difficult ot know when to start.

The ultimate guide to problem solving techniques

Download these ready-to-go problem solving techniques that every student should know. Includes printable tasks for students including challenges, short explanations for teachers with questioning prompts.

20 Math Strategies For Problem-Solving

Different problem-solving math strategies are required for different parts of the problem. It is unlikely that students will use the same strategy to understand and solve the problem. 

Here are 20 strategies to help students develop their problem-solving skills. 

Strategies to understand the problem

Strategies that help students understand the problem before solving it helps ensure they understand: 

• The context
• What the key information is
• How to form a plan to solve it

Following these steps leads students to the correct solution and makes the math word problem easier .

Here are five strategies to help students understand the content of the problem and identify key information. 

1. Read the problem aloud

Read a word problem aloud to help understand it. Hearing the words engages auditory processing. This can make it easier to process and comprehend the context of the situation.

2. Highlight keywords 

When keywords are highlighted in a word problem, it helps the student focus on the essential information needed to solve it. Some important keywords help determine which operation is needed.  For example, if the word problem asks how many are left, the problem likely requires subtraction.  Ensure students highlight the keywords carefully and do not highlight every number or keyword. There is likely irrelevant information in the word problem.

3. Summarize the information

Read the problem aloud, highlight the key information and then summarize the information. Students can do this in their heads or write down a quick summary.  Summaries should include only the important information and be in simple terms that help contextualize the problem.

4. Determine the unknown

A common problem that students have when solving a word problem is misunderstanding what they are solving. Determine what the unknown information is before finding the answer.  Often, a word problem contains a question where you can find the unknown information you need to solve. For example, in the question ‘How many apples are left?’ students need to find the number of apples left over.

5. Make a plan

Once students understand the context of the word problem, have dentified the important information and determined the unknown, they can make a plan to solve it.  The plan will depend on the type of problem. Some problems involve more than one step to solve them as some require more than one answer.  Encourage students to make a list of each step they need to take to solve the problem before getting started.

Strategies for solving the problem 

1. draw a model or diagram.

Students may find it useful to draw a model, picture, diagram, or other visual aid to help with the problem solving process.  It can help to visualize the problem to understand the relationships between the numbers in the problem. In turn, this helps students see the solution.

Similarly, you could draw a model to represent the objects in the problem:

2. Act it out

This particular strategy is applicable at any grade level but is especially helpful in math investigation in elementary school . It involves a physical demonstration or students acting out the problem using movements, concrete resources and math manipulatives .  When students act out a problem, they can visualize and contectualize the word problem in another way and secure an understanding of the math concepts.  The examples below show how 1st-grade students could “act out” an addition and subtraction problem:

3. Work backwards

Working backwards is a popular problem-solving strategy. It involves starting with a possible solution and deciding what steps to take to arrive at that solution.  This strategy can be particularly helpful when students solve math word problems involving multiple steps. They can start at the end and think carefully about each step taken as opposed to jumping to the end of the problem and missing steps in between.

For example,

To solve this problem working backwards, start with the final condition, which is Sam’s grandmother’s age (71) and work backwards to find Sam’s age. Subtract 20 from the grandmother’s age, which is 71.  Then, divide the result by 3 to get Sam’s age. 71 – 20 = 51 51 ÷ 3 = 17 Sam is 17 years old.

4. Write a number sentence

When faced with a word problem, encourage students to write a number sentence based on the information. This helps translate the information in the word problem into a math equation or expression, which is more easily solved.  It is important to fully understand the context of the word problem and what students need to solve before writing an equation to represent it.

5. Use a formula

Specific formulas help solve many math problems. For example, if a problem asks students to find the area of a rug, they would use the area formula (area = length × width) to solve.   Make sure students know the important mathematical formulas they will need in tests and real-life. It can help to display these around the classroom or, for those who need more support, on students’ desks.

Strategies for checking the solution 

Once the problem is solved using an appropriate strategy, it is equally important to check the solution to ensure it is correct and makes sense. 

There are many strategies to check the solution. The strategy for a specific problem is dependent on the problem type and math content involved.

Here are five strategies to help students check their solutions. 

1. Use the Inverse Operation

For simpler problems, a quick and easy problem solving strategy is to use the inverse operation. For example, if the operation to solve a word problem is 56 ÷ 8 = 7 students can check the answer is correct by multiplying 8 × 7. As good practice, encourage students to use the inverse operation routinely to check their work. 

2. Estimate to check for reasonableness

Once students reach an answer, they can use estimation or rounding to see if the answer is reasonable.  Round each number in the equation to a number that’s close and easy to work with, usually a multiple of ten.  For example, if the question was 216 ÷ 18 and the quotient was 12, students might round 216 to 200 and round 18 to 20. Then use mental math to solve 200 ÷ 20, which is 10.  When the estimate is clear the two numbers are close. This means your answer is reasonable. 

3. Plug-In Method

This method is particularly useful for algebraic equations. Specifically when working with variables.  To use the plug-in method, students solve the problem as asked and arrive at an answer. They can then plug the answer into the original equation to see if it works. If it does, the answer is correct.

If students use the equation 20m+80=300 to solve this problem and find that m = 11, they can plug that value back into the equation to see if it is correct. 20m + 80 = 300 20 (11) + 80 = 300 220 + 80 = 300 300 = 300 ✓

4. Peer Review

Peer review is a great tool to use at any grade level as it promotes critical thinking and collaboration between students. The reviewers can look at the problem from a different view as they check to see if the problem was solved correctly.   Problem solvers receive immediate feedback and the opportunity to discuss their thinking with their peers. This strategy is effective with mixed-ability partners or similar-ability partners. In mixed-ability groups, the partner with stronger skills provides guidance and support to the partner with weaker skills, while reinforcing their own understanding of the content and communication skills.  If partners have comparable ability levels and problem-solving skills, they may find that they approach problems differently or have unique insights to offer each other about the problem-solving process.

5. Use a Calculator

A calculator can be introduced at any grade level but may be best for older students who already have a foundational understanding of basic math operations. Provide students with a calculator to allow them to check their solutions independently, accurately, and quickly. Since calculators are so readily available on smartphones and tablets, they allow students to develop practical skills that apply to real-world situations.  

Step-by-step problem-solving processes for your classroom

In his book, How to Solve It , published in 1945, mathematician George Polya introduced a 4-step process to solve problems. 

Polya’s 4 steps include:

• Understand the problem
• Devise a plan
• Carry out the plan

Today, in the style of George Polya, many problem-solving strategies use various acronyms and steps to help students recall. 

Many teachers create posters and anchor charts of their chosen process to display in their classrooms. They can be implemented in any elementary, middle school or high school classroom. 

Here are 5 problem-solving strategies to introduce to students and use in the classroom.

How Third Space Learning improves problem-solving 

Resources .

Third Space Learning offers a free resource library is filled with hundreds of high-quality resources. A team of experienced math experts carefully created each resource to develop students mental arithmetic, problem solving and critical thinking. 

Explore the range of problem solving resources for 2nd to 8th grade students. 

One-on-one tutoring 

Third Space Learning offers one-on-one math tutoring to help students improve their math skills. Highly qualified tutors deliver high-quality lessons aligned to state standards. 

Former teachers and math experts write all of Third Space Learning’s tutoring lessons. Expertly designed lessons follow a “my turn, follow me, your turn” pedagogy to help students move from guided instruction and problem-solving to independent practice. 

Throughout each lesson, tutors ask higher-level thinking questions to promote critical thinking and ensure students are developing a deep understanding of the content and problem-solving skills.

Problem-solving

Educators can use many different strategies to teach problem-solving and help students develop and carry out a plan when solving math problems. Incorporate these math strategies into any math program and use them with a variety of math concepts, from whole numbers and fractions to algebra. 

Teaching students how to choose and implement problem-solving strategies helps them develop mathematical reasoning skills and critical thinking they can apply to real-life problem-solving.

READ MORE : 8 Common Core math examples

There are many different strategies for problem-solving; Here are 5 problem-solving strategies: • draw a model  • act it out  • work backwards  • write a number sentence • use a formula

Here are 10 strategies of problem-solving: • Read the problem aloud • Highlight keywords • Summarize the information • Determine the unknown • Make a plan • Draw a model  • Act it out  • Work backwards  • Write a number sentence • Use a formula

1. Understand the problem 2. Devise a plan 3. Carry out the plan 4. Look back

Some strategies you can use to solve challenging math problems are: breaking the problem into smaller parts, using diagrams or models, applying logical reasoning, and trying different approaches.

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Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

• For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
• For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

• Add or Subtract the same value from both sides
• Clear out any fractions by Multiplying every term by the bottom parts
• Divide every term by the same nonzero value
• Combine Like Terms
• Expanding (the opposite of factoring) may also help
• Recognizing a pattern, such as the difference of squares
• Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

• solve Quadratic Equations
• solve Radical Equations
• solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

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Unit 1: Intro to multiplication

Unit 2: 1-digit multiplication, unit 3: addition, subtraction, and estimation, unit 4: intro to division, unit 5: understand fractions, unit 6: equivalent fractions and comparing fractions, unit 7: more with multiplication and division, unit 8: arithmetic patterns and problem solving, unit 9: quadrilaterals, unit 10: area, unit 11: perimeter, unit 12: time, unit 13: measurement, unit 14: represent and interpret data.