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30 Problem Solving Maths Questions And Answers For GCSE

Martin Noon

Problem solving maths questions can be challenging for GCSE students as there is no ‘one size fits all’ approach. In this article, we’ve compiled tips for problem solving, example questions, solutions and problem solving strategies for GCSE students. 

Since the current GCSE specification began, there have been many maths problem solving exam questions which take elements of different areas of maths and combine them to form new maths problems which haven’t been seen before. 

While learners can be taught to approach simply structured problems by following a process, questions often require students to make sense of lots of new information before they even move on to trying to solve the problem. This is where many learners get stuck.

How to teach problem solving

6 tips to tackling problem solving maths questions, 10 problem solving maths questions (foundation tier), 10 problem solving maths questions (foundation & higher tier crossover), 10 problem solving maths questions (higher tier).

In the Ofsted maths review , published in May 2021, Ofsted set out their findings from the research literature regarding the sort of curriculum and teaching that best supports all pupils to make good progress in maths throughout their time in school.

Regarding the teaching of problem solving skills, these were their recommendations:

• Teachers could use a curricular approach that better engineers success in problem-solving by teaching the useful combinations of facts and methods, how to recognise the problem types and the deep structures that these strategies pair to.
• Strategies for problem-solving should be topic specific and can therefore be planned into the sequence of lessons as part of the wider curriculum. Pupils who are already confident with the foundational skills may benefit from a more generalised process involving identifying relationships and weighing up features of the problem to process the information. 
• Worked examples, careful questioning and constructing visual representations can help pupils to convert information embedded in a problem into mathematical notation.
• Open-ended problem solving tasks do not necessarily mean that the activity is the ‘ideal means of acquiring proficiency’. While enjoyable, open ended problem-solving activities may not necessarily lead to improved results.  

Are you a KS2 teacher needing more support teaching reasoning, problem solving & planning for depth ? See this article for FREE downloadable CPD

There is no ‘one size fits all’ approach to successfully tackling problem solving maths questions however, here are 6 general tips for students facing a problem solving question:

• Read the whole question, underline important mathematical words, phrases or values.
• Annotate any diagrams, graphs or charts with any missing information that is easy to fill in.
• Think of what a sensible answer may look like. E.g. Will the angle be acute or obtuse? Is £30,000 likely to be the price of a coat?
• Tick off information as you use it.
• Draw extra diagrams if needed.
• Look at the final sentence of the question. Make sure you refer back to that at the end to ensure you have answered the question fully.

There are many online sources of mathematical puzzles and questions that can help learners improve their problem-solving skills. Websites such as NRICH and our blog on SSDD problems have some great examples of KS2, KS3 and KS4 mathematical problems.

Read more: KS2 problem solving and KS3 maths problem solving

In this article, we’ve focussed on GCSE questions and compiled 30 problem solving maths questions and solutions suitable for Foundation and Higher tier students. Additionally, we have provided problem solving strategies to support your students for some questions to encourage critical mathematical thinking . For the full set of questions, solutions and strategies in a printable format, please download our 30 Problem Solving Maths Questions, Solutions & Strategies.

30 Problem Solving Maths Questions, Solutions & Strategies

Help your students prepare for their maths GCSE with these free problem solving maths questions, solutions and strategies

These first 10 questions and solutions are similar to Foundation questions. For the first three, we’ve provided some additional strategies.

In our downloadable resource, you can find strategies for all 10 Foundation questions .

1) L-shape perimeter 

Here is a shape:

Sarah says, “There is not enough information to find the perimeter.”

Is she correct? What about finding the area?

• Try adding more information – giving some missing sides measurements that are valid. 
• Change these measurements to see if the answer changes.
• Imagine walking around the shape if the edges were paths. Could any of those paths be moved to another position but still give the same total distance?

The perimeter of the shape does not depend on the lengths of the unlabelled edges.

Edge A and edge B can be moved to form a rectangle, meaning the perimeter will be 22 cm. Therefore, Sarah is wrong.

The area, however, will depend on those missing side length measurements, so we would need more information to be able to calculate it.

2) Find the missing point

Here is a coordinate grid with three points plotted. A fourth point is to be plotted to form a parallelogram. Find all possible coordinates of the fourth point.

• What are the properties of a parallelogram?
• Can we count squares to see how we can get from one vertex of the parallelogram to another? Can we use this to find the fourth vertex?

There are 3 possible positions.

3) That rating was a bit mean!

The vertical line graph shows the ratings a product received on an online shopping website. The vertical line for 4 stars is missing.

If the mean rating is 2.65, use the information to complete the vertical line graph.

Strategies 

• Can the information be put into a different format, either a list or a table?
• Would it help to give the missing frequency an algebraic label, x ?
• If we had the data in a frequency table, how would we calculate the mean?
• Is there an equation we could form?

Letting the frequency of 4 star ratings be x , we can form the equation \frac{45+4x}{18+x} =2.65

Giving x=2 

4) Changing angles

The diagram shows two angles around a point. The sum of the two angles around a point is 360°.

Peter says “If we increase the small angle by 10% and decrease the reflex angle by 10%, they will still add to 360°.”

Explain why Peter might be wrong.

Are there two angles where he would be correct?

Peter is wrong, for example, if the two angles are 40° and 320°, increasing 40° by 10% gives 44°, decreasing 320° by 10% gives 288°. These sum to 332°.

10% of the larger angle will be more than 10% of the smaller angle so the sum will only ever be 360° if the two original angles are the same, therefore, 180°.

5) Base and power

The integers 1, 2, 3, 4, 5, 6, 7, 8 and 9 can be used to fill in the boxes. 

How many different solutions can be found so that no digit is used more than once?

There are 8 solutions.

6) Just an average problem 

Place six single digit numbers into the boxes to satisfy the rules.

The mean in maths is 5  \frac{1}{3}

The median is 5

The mode is 3.

How many different solutions are possible?

There are 4 solutions.

2, 3, 3, 7, 8, 9

3, 3, 4, 6, 7, 9

3, 3, 3, 7, 7, 9

3, 3, 3, 7, 8, 8

7) Square and rectangle  

The square has an area of 81 cm 2 . The rectangle has the same perimeter as the square.

Its length and width are in the ratio 2:1.

Find the area of the rectangle.

The sides of the square are 9 cm giving a perimeter of 36 cm. 

We can then either form an equation using a length 2x and width x . 

Or, we could use the fact that the length and width add to half of the perimeter and share 18 in the ratio 2:1. 

The length is 12 cm and the width is 6 cm, giving an area of 72 cm 2 .

8) It’s all prime

The sum of three prime numbers is equal to another prime number.

If the sum is less than 30, how many different solutions are possible?

There are 6 solutions. 

2 can never be used as it would force two more odd primes into the sum to make the total even.

9) Unequal share

Bob and Jane have £10 altogether. Jane has £1.60 more than Bob. Bob spends one third of his money. How much money have Bob and Jane now got in total?

Initially Bob has £4.20 and Jane has £5.80. Bob spends £1.40, meaning the total £10 has been reduced by £1.40, leaving £8.60 after the subtraction.

10) Somewhere between

Fred says, “An easy way to find any fraction which is between two other fractions is to just add the numerators and add the denominators.” Is Fred correct?

Solution 

Fred is correct. His method does work and can be shown algebraically which could be a good problem for higher tier learners to try.

If we use these two fractions \frac{3}{8} and \frac{5}{12} , Fred’s method gives us \frac{8}{20} = \frac{2}{5}

\frac{3}{8} = \frac{45}{120} , \frac{2}{5} = \frac{48}{120} , \frac{5}{12} = \frac{50}{120} . So \frac{3}{8} < \frac{2}{5} < \frac{5}{12}

The next 10 questions are crossover questions which could appear on both Foundation and Higher tier exam papers. We have provided solutions for each and, for the first three questions, problem solving strategies to support learners.

11) What’s the difference?

An arithmetic sequence has an nth term in the form an+b .

4 is in the sequence.

16 is in the sequence.

8 is not in the sequence.

-2 is the first term of the sequence.

What are the possible values of a and b ?

• We know that the first number in the sequence is -2 and 4 is in the sequence. Can we try making a sequence to fit? Would using a number line help?
• Try looking at the difference between the numbers we know are in the sequence.

If we try forming a sequence from the information, we get this:

We can now try to fill in the missing numbers, making sure 8 is not in the sequence. Going up by 2 would give us 8, so that won’t work.

The only solutions are 6 n -8 and 3 n -5.

12) Equation of the hypotenuse

The diagram shows a straight line passing through the axes at point P and Q .

Q has coordinate (8, 0). M is the midpoint of PQ and MQ has a length of 5 units.

Find the equation of the line PQ .

• We know MQ is 5 units, what is PQ and OQ ?
• What type of triangle is OPQ ?
• Can we find OP if we know PQ and OQ ?
• A line has an equation in the form y=mx+c . How can we find m ? Do we already know c ?

PQ is 10 units. Using Pythagoras’ Theorem OP = 6

The gradient of the line will be \frac{-6}{8} = -\frac{3}{4} and P gives the intercept as 6.

13) What a waste

Harry wants to cut a sector of radius 30 cm from a piece of paper measuring 30 cm by 20 cm. 

What percentage of the paper will be wasted?

• What information do we need to calculate the area of a sector? Do we have it all?
• Would drawing another line on the diagram help find the angle of the sector?

The angle of the sector can be found using right angle triangle trigonometry.

The angle is 41.81°.

This gives us the area of the sector as 328.37 cm 2 .

The area of the paper is 600 cm 2 .

The area of paper wasted would be 600 – 328.37 = 271.62 cm 2 .

The wasted area is 45.27% of the paper.

14) Tri-polygonometry

The diagram shows part of a regular polygon and a right angled triangle. ABC is a straight line. Find the sum of the interior angles of the polygon.

Finding the angle in the triangle at point B gives 30°. This is the exterior angle of the polygon. Dividing 360° by 30° tells us the polygon has 12 sides. Therefore, the sum of the interior angles is 1800°.

15) That’s a lot of Pi

A block of ready made pastry is a cuboid measuring 3 cm by 10 cm by 15 cm. 

Anne is making 12 pies for a charity event. For each pie, she needs to cut a circle of pastry with a diameter of 18 cm from a sheet of pastry 0.5 cm thick.

How many blocks of pastry will Anne need to buy?

The volume of one block of pastry is 450 cm 3 . 

The volume of one cylinder of pastry is 127.23 cm 3 .

12 pies will require 1526.81 cm 3 .

Dividing the volume needed by 450 gives 3.39(…). 

Rounding this up tells us that 4 pastry blocks will be needed.

16) Is it right?

A triangle has sides of (x+4) cm, (2x+6) cm and (3x-2) cm. Its perimeter is 80 cm.

Show that the triangle is right angled and find its area.

Forming an equation gives 6x+8=80

This gives us x=12 and side lengths of 16 cm, 30 cm and 34 cm.

Using Pythagoras’ Theorem

16 2 +30 2 =1156 

Therefore, the triangle is right angled.

The area of the triangle is (16 x 30) ÷ 2 = 240 cm 2 .

17) Pie chart ratio

The pie chart shows sectors for red, blue and green. 

The ratio of the angles of the red sector to the blue sector is 2:7. 

The ratio of the angles of the red sector to the green sector is 1:3. 

Find the angles of each sector of the pie chart.

Multiplying the ratio of red : green by 2, it can be written as 2:6. 

Now the colour each ratio has in common, red, has equal parts in each ratio.

The ratio of red:blue is 2:7, this means red:blue:green = 2:7:6.

Sharing 360° in this ratio gives red:blue:green = 48°:168°:144°.

18) DIY Simultaneously

Mr Jones buys 5 tins of paint and 4 rolls of decorating tape. The total cost was £167.

The next day he returns 1 unused tin of paint and 1 unused roll of tape. The refund amount is exactly the amount needed to buy a fan heater that has been reduced by 10% in a sale. The fan heater normally costs £37.50.

Find the cost of 1 tin of paint.    

The sale price of the fan heater is £33.75. This gives the simultaneous equations

p+t = 33.75 and 5 p +4 t = 167.

We only need the price of a tin of paint so multiplying the first equation by 4 and then subtracting from the second equation gives p =32. Therefore, 1 tin of paint costs £32. 

19) Triathlon pace

Jodie is competing in a Triathlon. 

A triathlon consists of a 5 km swim, a 40 km cycle and a 10 km run. 

Jodie wants to complete the triathlon in 5 hours. 

She knows she can swim at an average speed of 2.5 km/h and cycle at an average speed of 25 km/h. There are also two transition stages, in between events, which normally take 4 minutes each.

What speed must Jodie average on the final run to finish the triathlon in 5 hours?

Dividing the distances by the average speeds for each section gives times of 2 hours for the swim and 1.6 hours for the cycle, 216 minutes in total. Adding 8 minutes for the transition stages gives 224 minutes. To complete the triathlon in 5 hours, that would be 300 minutes. 300 – 224 = 76 minutes. Jodie needs to complete her 10 km run in 76 minutes, or \frac{19}{15} hours. This gives an average speed of 7.89 km/h.

20) Indices

a 2x × a y =a 3

(a 3 ) x ÷ a 4y =a 32

Find x and y .

Forming the simultaneous equations

Solving these gives

This final set of 10 questions would appear on the Higher tier only. Here we have just provided the solutions. Try asking your learners to discuss their strategies for each question.  

21) Angles in a polygon

The diagram shows part of a regular polygon.

A , B and C are vertices of the polygon. 

The size of the reflex angle ABC is 360° minus the interior angle.

Show that the sum of all of these reflex angles of the polygon will be 720° more than the sum of its interior angles.

Each of the reflex angles is 180 degrees more than the exterior angle: 180 + \frac{360}{n}

The sum of all of these angles is n (180 + \frac{360}{n} ). 

This simplifies to 180 n + 360

The sum of the interior angles is 180( n – 2) = 180 n – 360

The difference is 180 n + 360 – (180 n -360) = 720°

22) Prism and force (Non-calculator)

The diagram shows a prism with an equilateral triangle cross-section.

When the prism is placed so that its triangular face touches the surface, the prism applies a force of 12 Newtons resulting in a pressure of \frac{ \sqrt{3} }{4} N/m^{2}

Given that the prism has a volume of 384 m 3 , find the length of the prism.

Pressure = \frac{Force}{Area}

Area = 12÷ \frac{ \sqrt{3} }{4} = 16\sqrt{3} m 2

Therefore, the length of the prism is 384 ÷ 16\sqrt{3} = 8\sqrt{3} m

23) Geometric sequences (Non-calculator)

A geometric sequence has a third term of 6 and a sixth term of 14 \frac{2}{9}

Find the first term of the sequence.

The third term is ar 2 = 6

The sixth term is ar 5 = \frac{128}{9}

Diving these terms gives r 3 = \frac{64}{27}

Giving r = \frac{4}{3}

Dividing the third term twice by \frac{4}{3} gives the first term a = \frac{27}{8}

24) Printing factory

A printing factory is producing exam papers. When all 10 of its printers are working, it can produce all of the exam papers in 12 days.

For the first two days of printing, 3 of the printers are broken.

At the beginning of the third day it is discovered that 2 more printers have broken down, so the factory continues to print with the reduced amount of printers for 3 days. The broken printers are repaired and now all printers are available to print the remaining exams.

How many days in total does it take the factory to produce all of the exam papers?

If we assume one printer prints 1 exam paper per day, 10 printers would print 120 exam papers in 12 days. Listing the number printed each day for the first 5 days gives:

Day 5: 5 

This is a total of 29 exam papers.

91 exam papers are remaining with 10 printers now able to produce a total of 10 exam papers each day. 10 more days would be required to complete the job.

Therefore, 15 days in total are required.

25) Circles

The diagram shows a circle with equation x^{2}+{y}^{2}=13 .

A tangent touches the circle at point P when x=3 and y is negative.

The tangent intercepts the coordinate axes at A and B .

Find the length AB .

Using the equation  x^{2}+y^{2}=13 to find the y value for P gives y=-2 .

The gradient of the radius at this point is - \frac{2}{3} , giving a tangent gradient of \frac{3}{2} .

Using the point (3,-2) in y = \frac {3}{2} x+c gives the equation of the tangent as y = \frac {3}{2} x – \frac{13}{2}

Substituting x=0 and y=0 gives A and B as (0 , -\frac {13}{2}) and ( \frac{13}{3} , 0)

Using Pythagoras’ Theorem gives the length of AB as ( \frac{ 13\sqrt{13} }{6} ) = 7.812.

26) Circle theorems

The diagram shows a circle with centre O . Points A, B, C and D are on the circumference of the circle. 

EF is a tangent to the circle at A . 

Angle EAD = 46°

Angle FAB = 48°

Angle ADC = 78°

Find the area of ABCD to the nearest integer.

The Alternate Segment Theorem gives angle ACD as 46° and angle ACB as 48°.

Opposite angles in a cyclic quadrilateral summing to 180° gives angle ABC as 102°.

Using the sine rule to find AC will give a length of 5.899. Using the sine rule again to find BC will give a length of 3.016cm.

We can now use the area of a triangle formula to find the area of both triangles.

0.5 × 5 × 5.899 × sin (46) + 0.5 × 3.016 × 5.899 × sin (48) = 17 units 2 (to the nearest integer).

27) Quadratic function

The quadratic function f(x) = -2x^{2} + 8x +11 has a turning point at P .

Find the coordinate of the turning point after the transformation -f(x-3) .

There are two methods that could be used. We could apply the transformation to the function and then complete the square, or, we could complete the square and then apply the transformation.

Here we will do the latter.

This gives a turning point for f(x) as (2,19).

Applying -f(x-3) gives the new turning point as (5,-19).

28) Probability with fruit

A fruit bowl contains only 5 grapes and n strawberries.

A fruit is taken, eaten and then another is selected.

The probability of taking two strawberries is \frac{7}{22} .

Find the probability of taking one of each fruit. 

There are n+5 fruits altogether.

P(Strawberry then strawberry)= \frac{n}{n+5} × \frac{n-1}{n+4} = \frac{7}{22}

This gives the quadratic equation 15n^{2} - 85n - 140 = 0

This can be divided through by 5 to give 3n^{2} - 17n- 28 = 0

This factorises to (n-7)(3n + 4) = 0

n must be positive so n = 7.

The probability of taking one of each fruit is therefore, \frac{5}{12} × \frac{7}{11} + \frac {7}{12} × \frac {5}{11} = \frac {70}{132}

29) Ice cream tub volume

An ice cream tub in the shape of a prism with a trapezium cross-section has the dimensions shown. These measurements are accurate to the nearest cm.

An ice cream scoop has a diameter of 4.5 cm to the nearest millimetre and will be used to scoop out spheres of ice cream from the tub.

Using bounds find a suitable approximation to the number of ice cream scoops that can be removed from a tub that is full.

We need to find the upper and lower bounds of the two volumes. 

Upper bound tub volume = 5665.625 cm 3

Lower bound tub volume = 4729.375 cm 3

Upper bound scoop volume = 49.32 cm 3  

Lower bound scoop volume = 46.14 cm 3  

We can divide the upper bound of the ice cream tub by the lower bound of the scoop to get the maximum possible number of scoops. 

Maximum number of scoops = 122.79

Then divide the lower bound of the ice cream tub by the upper bound of the scoop to get the minimum possible number of scoops.

Minimum number of scoops  = 95.89

These both round to 100 to 1 significant figure, Therefore, 100 scoops is a suitable approximation the the number of scoops.

30) Translating graphs

 The diagram shows the graph of y = a+tan(x-b ).

The graph goes through the points (75, 3) and Q (60, q).

Find exact values of a , b and q .

The asymptote has been translated to the right by 30°. 

Therefore, b=30

So the point (45,1) has been translated to the point (75,3). 

Therefore, a=2

We hope these problem solving maths questions will support your GCSE teaching. To get all the solutions and strategies in a printable form, please download the complete resource .

Looking for additional support and resources at KS3? You are welcome to download any of the secondary maths resources from Third Space Learning’s resource library for free. There is a section devoted to GCSE maths revision with plenty of maths worksheets and GCSE maths questions . There are also maths tests for KS3, including a Year 7 maths test , a Year 8 maths test and a Year 9 maths test Other valuable maths practice and ideas particularly around reasoning and problem solving at secondary can be found in our KS3 and KS4 maths blog articles. Try these fun maths problems for KS2 and KS3, SSDD problems , KS3 maths games and 30 problem solving maths questions . For children who need more support, our maths intervention programmes for KS3 achieve outstanding results through a personalised one to one tuition approach.

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120 Math Word Problems To Challenge Students Grades 1 to 8

Written by Marcus Guido

Hey teachers! 👋

Use Prodigy to spark a love for math in your students – including when solving word problems!

• Teaching Tools
• Subtraction
• Multiplication
• Mixed operations
• Ordering and number sense
• Comparing and sequencing
• Physical measurement
• Ratios and percentages
• Probability and data relationships

You sit at your desk, ready to put a math quiz, test or activity together. The questions flow onto the document until you hit a section for word problems.

A jolt of creativity would help. But it doesn’t come.

Whether you’re a 3rd grade teacher or an 8th grade teacher preparing students for high school, translating math concepts into real world examples can certainly be a challenge.

This resource is your jolt of creativity. It provides examples and templates of math word problems for 1st to 8th grade classes.

There are 120 examples in total.

The list of examples is supplemented by tips to create engaging and challenging math word problems.

120 Math word problems, categorized by skill

Addition word problems.

Best for: 1st grade, 2nd grade

1. Adding to 10: Ariel was playing basketball. 1 of her shots went in the hoop. 2 of her shots did not go in the hoop. How many shots were there in total?

2. Adding to 20: Adrianna has 10 pieces of gum to share with her friends. There wasn’t enough gum for all her friends, so she went to the store to get 3 more pieces of gum. How many pieces of gum does Adrianna have now?

3. Adding to 100: Adrianna has 10 pieces of gum to share with her friends. There wasn’t enough gum for all her friends, so she went to the store and got 70 pieces of strawberry gum and 10 pieces of bubble gum. How many pieces of gum does Adrianna have now?

4. Adding Slightly over 100: The restaurant has 175 normal chairs and 20 chairs for babies. How many chairs does the restaurant have in total?

5. Adding to 1,000: How many cookies did you sell if you sold 320 chocolate cookies and 270 vanilla cookies?

6. Adding to and over 10,000: The hobby store normally sells 10,576 trading cards per month. In June, the hobby store sold 15,498 more trading cards than normal. In total, how many trading cards did the hobby store sell in June?

7. Adding 3 Numbers: Billy had 2 books at home. He went to the library to take out 2 more books. He then bought 1 book. How many books does Billy have now?

8. Adding 3 Numbers to and over 100: Ashley bought a big bag of candy. The bag had 102 blue candies, 100 red candies and 94 green candies. How many candies were there in total?

Subtraction word problems

Best for: 1st grade, second grade

9. Subtracting to 10: There were 3 pizzas in total at the pizza shop. A customer bought 1 pizza. How many pizzas are left?

10. Subtracting to 20: Your friend said she had 11 stickers. When you helped her clean her desk, she only had a total of 10 stickers. How many stickers are missing?

11. Subtracting to 100: Adrianna has 100 pieces of gum to share with her friends. When she went to the park, she shared 10 pieces of strawberry gum. When she left the park, Adrianna shared another 10 pieces of bubble gum. How many pieces of gum does Adrianna have now?

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12. Subtracting Slightly over 100: Your team scored a total of 123 points. 67 points were scored in the first half. How many were scored in the second half?

13. Subtracting to 1,000: Nathan has a big ant farm. He decided to sell some of his ants. He started with 965 ants. He sold 213. How many ants does he have now?

14. Subtracting to and over 10,000: The hobby store normally sells 10,576 trading cards per month. In July, the hobby store sold a total of 20,777 trading cards. How many more trading cards did the hobby store sell in July compared with a normal month?

15. Subtracting 3 Numbers: Charlene had a pack of 35 pencil crayons. She gave 6 to her friend Theresa. She gave 3 to her friend Mandy. How many pencil crayons does Charlene have left?

16. Subtracting 3 Numbers to and over 100: Ashley bought a big bag of candy to share with her friends. In total, there were 296 candies. She gave 105 candies to Marissa. She also gave 86 candies to Kayla. How many candies were left?

Multiplication word problems

Best for: 2nd grade, 3rd grade

17. Multiplying 1-Digit Integers: Adrianna needs to cut a pan of brownies into pieces. She cuts 6 even columns and 3 even rows into the pan. How many brownies does she have?

18. Multiplying 2-Digit Integers: A movie theatre has 25 rows of seats with 20 seats in each row. How many seats are there in total?

19. Multiplying Integers Ending with 0: A clothing company has 4 different kinds of sweatshirts. Each year, the company makes 60,000 of each kind of sweatshirt. How many sweatshirts does the company make each year?

20. Multiplying 3 Integers: A bricklayer stacks bricks in 2 rows, with 10 bricks in each row. On top of each row, there is a stack of 6 bricks. How many bricks are there in total?

21. Multiplying 4 Integers: Cayley earns $5 an hour by delivering newspapers. She delivers newspapers 3 days each week, for 4 hours at a time. After delivering newspapers for 8 weeks, how much money will Cayley earn?

Division word problems

Best for: 3rd grade, 4th grade, 5th grade

22. Dividing 1-Digit Integers: If you have 4 pieces of candy split evenly into 2 bags, how many pieces of candy are in each bag?

23. Dividing 2-Digit Integers: If you have 80 tickets for the fair and each ride costs 5 tickets, how many rides can you go on?

24. Dividing Numbers Ending with 0: The school has $20,000 to buy new computer equipment. If each piece of equipment costs $50, how many pieces can the school buy in total?

25. Dividing 3 Integers: Melissa buys 2 packs of tennis balls for $12 in total. All together, there are 6 tennis balls. How much does 1 pack of tennis balls cost? How much does 1 tennis ball cost?

26. Interpreting Remainders: An Italian restaurant receives a shipment of 86 veal cutlets. If it takes 3 cutlets to make a dish, how many cutlets will the restaurant have left over after making as many dishes as possible?

Mixed operations word problems

27. Mixing Addition and Subtraction: There are 235 books in a library. On Monday, 123 books are taken out. On Tuesday, 56 books are brought back. How many books are there now?

28. Mixing Multiplication and Division: There is a group of 10 people who are ordering pizza. If each person gets 2 slices and each pizza has 4 slices, how many pizzas should they order?

29. Mixing Multiplication, Addition and Subtraction: Lana has 2 bags with 2 marbles in each bag. Markus has 2 bags with 3 marbles in each bag. How many more marbles does Markus have?

30. Mixing Division, Addition and Subtraction: Lana has 3 bags with the same amount of marbles in them, totaling 12 marbles. Markus has 3 bags with the same amount of marbles in them, totaling 18 marbles. How many more marbles does Markus have in each bag?

Ordering and number sense word problems

31. Counting to Preview Multiplication: There are 2 chalkboards in your classroom. If each chalkboard needs 2 pieces of chalk, how many pieces do you need in total?

32. Counting to Preview Division: There are 3 chalkboards in your classroom. Each chalkboard has 2 pieces of chalk. This means there are 6 pieces of chalk in total. If you take 1 piece of chalk away from each chalkboard, how many will there be in total?

33. Composing Numbers: What number is 6 tens and 10 ones?

34. Guessing Numbers: I have a 7 in the tens place. I have an even number in the ones place. I am lower than 74. What number am I?

35. Finding the Order: In the hockey game, Mitchell scored more points than William but fewer points than Auston. Who scored the most points? Who scored the fewest points?

Fractions word problems

Best for: 3rd grade, 4th grade, 5th grade, 6th grade

36. Finding Fractions of a Group: Julia went to 10 houses on her street for Halloween. 5 of the houses gave her a chocolate bar. What fraction of houses on Julia’s street gave her a chocolate bar?

37. Finding Unit Fractions: Heather is painting a portrait of her best friend, Lisa. To make it easier, she divides the portrait into 6 equal parts. What fraction represents each part of the portrait?

38. Adding Fractions with Like Denominators: Noah walks ⅓ of a kilometre to school each day. He also walks ⅓ of a kilometre to get home after school. How many kilometres does he walk in total?

39. Subtracting Fractions with Like Denominators: Last week, Whitney counted the number of juice boxes she had for school lunches. She had ⅗ of a case. This week, it’s down to ⅕ of a case. How much of the case did Whitney drink?

40. Adding Whole Numbers and Fractions with Like Denominators: At lunchtime, an ice cream parlor served 6 ¼ scoops of chocolate ice cream, 5 ¾ scoops of vanilla and 2 ¾ scoops of strawberry. How many scoops of ice cream did the parlor serve in total?

41. Subtracting Whole Numbers and Fractions with Like Denominators: For a party, Jaime had 5 ⅓ bottles of cola for her friends to drink. She drank ⅓ of a bottle herself. Her friends drank 3 ⅓. How many bottles of cola does Jaime have left?

42. Adding Fractions with Unlike Denominators: Kevin completed ½ of an assignment at school. When he was home that evening, he completed ⅚ of another assignment. How many assignments did Kevin complete?

43. Subtracting Fractions with Unlike Denominators: Packing school lunches for her kids, Patty used ⅞ of a package of ham. She also used ½ of a package of turkey. How much more ham than turkey did Patty use?

44. Multiplying Fractions: During gym class on Wednesday, the students ran for ¼ of a kilometre. On Thursday, they ran ½ as many kilometres as on Wednesday. How many kilometres did the students run on Thursday? Write your answer as a fraction.

45. Dividing Fractions: A clothing manufacturer uses ⅕ of a bottle of colour dye to make one pair of pants. The manufacturer used ⅘ of a bottle yesterday. How many pairs of pants did the manufacturer make?

46. Multiplying Fractions with Whole Numbers: Mark drank ⅚ of a carton of milk this week. Frank drank 7 times more milk than Mark. How many cartons of milk did Frank drink? Write your answer as a fraction, or as a whole or mixed number.

Decimals word problems

Best for: 4th grade, 5th grade

47. Adding Decimals: You have 2.6 grams of yogurt in your bowl and you add another spoonful of 1.3 grams. How much yogurt do you have in total?

48. Subtracting Decimals: Gemma had 25.75 grams of frosting to make a cake. She decided to use only 15.5 grams of the frosting. How much frosting does Gemma have left?

49. Multiplying Decimals with Whole Numbers: Marshall walks a total of 0.9 kilometres to and from school each day. After 4 days, how many kilometres will he have walked?

50. Dividing Decimals by Whole Numbers: To make the Leaning Tower of Pisa from spaghetti, Mrs. Robinson bought 2.5 kilograms of spaghetti. Her students were able to make 10 leaning towers in total. How many kilograms of spaghetti does it take to make 1 leaning tower?

51. Mixing Addition and Subtraction of Decimals: Rocco has 1.5 litres of orange soda and 2.25 litres of grape soda in his fridge. Antonio has 1.15 litres of orange soda and 0.62 litres of grape soda. How much more soda does Rocco have than Angelo?

52. Mixing Multiplication and Division of Decimals: 4 days a week, Laura practices martial arts for 1.5 hours. Considering a week is 7 days, what is her average practice time per day each week?

Comparing and sequencing word problems

Best for: Kindergarten, 1st grade, 2nd grade

53. Comparing 1-Digit Integers: You have 3 apples and your friend has 5 apples. Who has more?

54. Comparing 2-Digit Integers: You have 50 candies and your friend has 75 candies. Who has more?

55. Comparing Different Variables: There are 5 basketballs on the playground. There are 7 footballs on the playground. Are there more basketballs or footballs?

56. Sequencing 1-Digit Integers: Erik has 0 stickers. Every day he gets 1 more sticker. How many days until he gets 3 stickers?

57. Skip-Counting by Odd Numbers: Natalie began at 5. She skip-counted by fives. Could she have said the number 20?

58. Skip-Counting by Even Numbers: Natasha began at 0. She skip-counted by eights. Could she have said the number 36?

59. Sequencing 2-Digit Numbers: Each month, Jeremy adds the same number of cards to his baseball card collection. In January, he had 36. 48 in February. 60 in March. How many baseball cards will Jeremy have in April?

Time word problems

66. Converting Hours into Minutes: Jeremy helped his mom for 1 hour. For how many minutes was he helping her?

69. Adding Time: If you wake up at 7:00 a.m. and it takes you 1 hour and 30 minutes to get ready and walk to school, at what time will you get to school?

70. Subtracting Time: If a train departs at 2:00 p.m. and arrives at 4:00 p.m., how long were passengers on the train for?

71. Finding Start and End Times: Rebecca left her dad’s store to go home at twenty to seven in the evening. Forty minutes later, she was home. What time was it when she arrived home?

Money word problems

Best for: 1st grade, 2nd grade, 3rd grade, 4th grade, 5th grade

60. Adding Money: Thomas and Matthew are saving up money to buy a video game together. Thomas has saved $30. Matthew has saved $35. How much money have they saved up together in total?

61. Subtracting Money: Thomas has $80 saved up. He uses his money to buy a video game. The video game costs $67. How much money does he have left?

62. Multiplying Money: Tim gets $5 for delivering the paper. How much money will he have after delivering the paper 3 times?

63. Dividing Money: Robert spent $184.59 to buy 3 hockey sticks. If each hockey stick was the same price, how much did 1 cost?

64. Adding Money with Decimals: You went to the store and bought gum for $1.25 and a sucker for $0.50. How much was your total?

65. Subtracting Money with Decimals: You went to the store with $5.50. You bought gum for $1.25, a chocolate bar for $1.15 and a sucker for $0.50. How much money do you have left?

67. Applying Proportional Relationships to Money: Jakob wants to invite 20 friends to his birthday, which will cost his parents $250. If he decides to invite 15 friends instead, how much money will it cost his parents? Assume the relationship is directly proportional.

68. Applying Percentages to Money: Retta put $100.00 in a bank account that gains 20% interest annually. How much interest will be accumulated in 1 year? And if she makes no withdrawals, how much money will be in the account after 1 year?

Physical measurement word problems

Best for: 1st grade, 2nd grade, 3rd grade, 4th grade

72. Comparing Measurements: Cassandra’s ruler is 22 centimetres long. April’s ruler is 30 centimetres long. How many centimetres longer is April’s ruler?

73. Contextualizing Measurements: Picture a school bus. Which unit of measurement would best describe the length of the bus? Centimetres, metres or kilometres?

74. Adding Measurements: Micha’s dad wants to try to save money on gas, so he has been tracking how much he uses. Last year, Micha’s dad used 100 litres of gas. This year, her dad used 90 litres of gas. How much gas did he use in total for the two years?

75. Subtracting Measurements: Micha’s dad wants to try to save money on gas, so he has been tracking how much he uses. Over the past two years, Micha’s dad used 200 litres of gas. This year, he used 100 litres of gas. How much gas did he use last year?

76. Multiplying Volume and Mass: Kiera wants to make sure she has strong bones, so she drinks 2 litres of milk every week. After 3 weeks, how many litres of milk will Kiera drink?

77. Dividing Volume and Mass: Lillian is doing some gardening, so she bought 1 kilogram of soil. She wants to spread the soil evenly between her 2 plants. How much will each plant get?

78. Converting Mass: Inger goes to the grocery store and buys 3 squashes that each weigh 500 grams. How many kilograms of squash did Inger buy?

79. Converting Volume: Shad has a lemonade stand and sold 20 cups of lemonade. Each cup was 500 millilitres. How many litres did Shad sell in total?

80. Converting Length: Stacy and Milda are comparing their heights. Stacy is 1.5 meters tall. Milda is 10 centimetres taller than Stacy. What is Milda’s height in centimetres?

81. Understanding Distance and Direction: A bus leaves the school to take students on a field trip. The bus travels 10 kilometres south, 10 kilometres west, another 5 kilometres south and 15 kilometres north. To return to the school, in which direction does the bus have to travel? How many kilometres must it travel in that direction?

Ratios and percentages word problems

Best for: 4th grade, 5th grade, 6th grade

82. Finding a Missing Number: The ratio of Jenny’s trophies to Meredith’s trophies is 7:4. Jenny has 28 trophies. How many does Meredith have?

83. Finding Missing Numbers: The ratio of Jenny’s trophies to Meredith’s trophies is 7:4. The difference between the numbers is 12. What are the numbers?

84. Comparing Ratios: The school’s junior band has 10 saxophone players and 20 trumpet players. The school’s senior band has 18 saxophone players and 29 trumpet players. Which band has the higher ratio of trumpet to saxophone players?

85. Determining Percentages: Mary surveyed students in her school to find out what their favourite sports were. Out of 1,200 students, 455 said hockey was their favourite sport. What percentage of students said hockey was their favourite sport?

86. Determining Percent of Change: A decade ago, Oakville’s population was 67,624 people. Now, it is 190% larger. What is Oakville’s current population?

87. Determining Percents of Numbers: At the ice skate rental stand, 60% of 120 skates are for boys. If the rest of the skates are for girls, how many are there?

88. Calculating Averages: For 4 weeks, William volunteered as a helper for swimming classes. The first week, he volunteered for 8 hours. He volunteered for 12 hours in the second week, and another 12 hours in the third week. The fourth week, he volunteered for 9 hours. For how many hours did he volunteer per week, on average?

Probability and data relationships word problems

Best for: 4th grade, 5th grade, 6th grade, 7th grade

89. Understanding the Premise of Probability: John wants to know his class’s favourite TV show, so he surveys all of the boys. Will the sample be representative or biased?

90. Understanding Tangible Probability: The faces on a fair number die are labelled 1, 2, 3, 4, 5 and 6. You roll the die 12 times. How many times should you expect to roll a 1?

91. Exploring Complementary Events: The numbers 1 to 50 are in a hat. If the probability of drawing an even number is 25/50, what is the probability of NOT drawing an even number? Express this probability as a fraction.

92. Exploring Experimental Probability: A pizza shop has recently sold 15 pizzas. 5 of those pizzas were pepperoni. Answering with a fraction, what is the experimental probability that he next pizza will be pepperoni?

93. Introducing Data Relationships: Maurita and Felice each take 4 tests. Here are the results of Maurita’s 4 tests: 4, 4, 4, 4. Here are the results for 3 of Felice’s 4 tests: 3, 3, 3. If Maurita’s mean for the 4 tests is 1 point higher than Felice’s, what’s the score of Felice’s 4th test?

94. Introducing Proportional Relationships: Store A is selling 7 pounds of bananas for $7.00. Store B is selling 3 pounds of bananas for $6.00. Which store has the better deal?

95. Writing Equations for Proportional Relationships: Lionel loves soccer, but has trouble motivating himself to practice. So, he incentivizes himself through video games. There is a proportional relationship between the amount of drills Lionel completes, in x , and for how many hours he plays video games, in y . When Lionel completes 10 drills, he plays video games for 30 minutes. Write the equation for the relationship between x and y .

Geometry word problems

Best for: 4th grade, 5th grade, 6th grade, 7th grade, 8th grade

96. Introducing Perimeter:  The theatre has 4 chairs in a row. There are 5 rows. Using rows as your unit of measurement, what is the perimeter?

97. Introducing Area: The theatre has 4 chairs in a row. There are 5 rows. How many chairs are there in total?

98. Introducing Volume: Aaron wants to know how much candy his container can hold. The container is 20 centimetres tall, 10 centimetres long and 10 centimetres wide. What is the container’s volume?

99. Understanding 2D Shapes: Kevin draws a shape with 4 equal sides. What shape did he draw?

100. Finding the Perimeter of 2D Shapes: Mitchell wrote his homework questions on a piece of square paper. Each side of the paper is 8 centimetres. What is the perimeter?

101. Determining the Area of 2D Shapes: A single trading card is 9 centimetres long by 6 centimetres wide. What is its area?

102. Understanding 3D Shapes: Martha draws a shape that has 6 square faces. What shape did she draw?

103. Determining the Surface Area of 3D Shapes: What is the surface area of a cube that has a width of 2cm, height of 2 cm and length of 2 cm?

104. Determining the Volume of 3D Shapes: Aaron’s candy container is 20 centimetres tall, 10 centimetres long and 10 centimetres wide. Bruce’s container is 25 centimetres tall, 9 centimetres long and 9 centimetres wide. Find the volume of each container. Based on volume, whose container can hold more candy?

105. Identifying Right-Angled Triangles: A triangle has the following side lengths: 3 cm, 4 cm and 5 cm. Is this triangle a right-angled triangle?

106. Identifying Equilateral Triangles: A triangle has the following side lengths: 4 cm, 4 cm and 4 cm. What kind of triangle is it?

107. Identifying Isosceles Triangles: A triangle has the following side lengths: 4 cm, 5 cm and 5 cm. What kind of triangle is it?

108. Identifying Scalene Triangles: A triangle has the following side lengths: 4 cm, 5 cm and 6 cm. What kind of triangle is it?

109. Finding the Perimeter of Triangles: Luigi built a tent in the shape of an equilateral triangle. The perimeter is 21 metres. What is the length of each of the tent’s sides?

110. Determining the Area of Triangles: What is the area of a triangle with a base of 2 units and a height of 3 units?

111. Applying Pythagorean Theorem: A right triangle has one non-hypotenuse side length of 3 inches and the hypotenuse measures 5 inches. What is the length of the other non-hypotenuse side?

112. Finding a Circle’s Diameter: Jasmin bought a new round backpack. Its area is 370 square centimetres. What is the round backpack’s diameter?

113. Finding a Circle's Area: Captain America’s circular shield has a diameter of 76.2 centimetres. What is the area of his shield?

114. Finding a Circle’s Radius: Skylar lives on a farm, where his dad keeps a circular corn maze. The corn maze has a diameter of 2 kilometres. What is the maze’s radius?

Variables word problems

Best for: 6th grade, 7th grade, 8th grade

115. Identifying Independent and Dependent Variables: Victoria is baking muffins for her class. The number of muffins she makes is based on how many classmates she has. For this equation, m is the number of muffins and c is the number of classmates. Which variable is independent and which variable is dependent?

116. Writing Variable Expressions for Addition: Last soccer season, Trish scored g goals. Alexa scored 4 more goals than Trish. Write an expression that shows how many goals Alexa scored.

117. Writing Variable Expressions for Subtraction: Elizabeth eats a healthy, balanced breakfast b times a week. Madison sometimes skips breakfast. In total, Madison eats 3 fewer breakfasts a week than Elizabeth. Write an expression that shows how many times a week Madison eats breakfast.

118. Writing Variable Expressions for Multiplication: Last hockey season, Jack scored g goals. Patrik scored twice as many goals than Jack. Write an expression that shows how many goals Patrik scored.

119. Writing Variable Expressions for Division: Amanda has c chocolate bars. She wants to distribute the chocolate bars evenly among 3 friends. Write an expression that shows how many chocolate bars 1 of her friends will receive.

120. Solving Two-Variable Equations: This equation shows how the amount Lucas earns from his after-school job depends on how many hours he works: e = 12h . The variable h represents how many hours he works. The variable e represents how much money he earns. How much money will Lucas earn after working for 6 hours?

How to easily make your own math word problems & word problems worksheets

Armed with 120 examples to spark ideas, making your own math word problems can engage your students and ensure alignment with lessons. Do:

• Link to Student Interests:  By framing your word problems with student interests, you’ll likely grab attention. For example, if most of your class loves American football, a measurement problem could involve the throwing distance of a famous quarterback.
• Make Questions Topical:  Writing a word problem that reflects current events or issues can engage students by giving them a clear, tangible way to apply their knowledge.
• Include Student Names:  Naming a question’s characters after your students is an easy way make subject matter relatable, helping them work through the problem.
• Be Explicit:  Repeating keywords distills the question, helping students focus on the core problem.
• Test Reading Comprehension:  Flowery word choice and long sentences can hide a question’s key elements. Instead, use concise phrasing and grade-level vocabulary.
• Focus on Similar Interests:  Framing too many questions with related interests -- such as football and basketball -- can alienate or disengage some students.
• Feature Red Herrings:  Including unnecessary information introduces another problem-solving element, overwhelming many elementary students.

A key to differentiated instruction , word problems that students can relate to and contextualize will capture interest more than generic and abstract ones.

Final thoughts about math word problems

You’ll likely get the most out of this resource by using the problems as templates, slightly modifying them by applying the above tips. In doing so, they’ll be more relevant to -- and engaging for -- your students.

Regardless, having 120 curriculum-aligned math word problems at your fingertips should help you deliver skill-building challenges and thought-provoking assessments.

The result?

A greater understanding of how your students process content and demonstrate understanding, informing your ongoing teaching approach.

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Our math problem solver that lets you input a wide variety of math math problems and it will provide a step by step answer. This math solver excels at math word problems as well as a wide range of math subjects.

• Math Word Problems
• Pre-Algebra
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• Trigonometry
• Precalculus
• Finite Math
• Linear Algebra

Here are example math problems within each subject that can be input into the calculator and solved. This list is constanstly growing as functionality is added to the calculator.

Basic Math Solutions

Below are examples of basic math problems that can be solved.

• Long Arithmetic
• Rational Numbers
• Operations with Fractions
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• Measurement, Area, and Volume
• Factors, Fractions, and Exponents
• Unit Conversions
• Data Measurement and Statistics
• Points and Line Segments

Math Word Problem Solutions

Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms.

Word Problem: Rachel has 17 apples. She gives some to Sarah. Sarah now has 8 apples. How many apples did Rachel give her?

Simplified Equation: 17 - x = 8

Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?

Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)

Simplified Equation: {r = d + 12, d = b + 6, r = 2 �� b}

Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?

Simplified: 40 - 10 - 5

Pre-Algebra Solutions

Below are examples of Pre-Algebra math problems that can be solved.

• Variables, Expressions, and Integers
• Simplifying and Evaluating Expressions
• Solving Equations
• Multi-Step Equations and Inequalities
• Ratios, Proportions, and Percents
• Linear Equations and Inequalities

Algebra Solutions

Below are examples of Algebra math problems that can be solved.

• Algebra Concepts and Expressions
• Points, Lines, and Line Segments
• Simplifying Polynomials
• Factoring Polynomials
• Linear Equations
• Absolute Value Expressions and Equations
• Radical Expressions and Equations
• Systems of Equations
• Quadratic Equations
• Inequalities
• Complex Numbers and Vector Analysis
• Logarithmic Expressions and Equations
• Exponential Expressions and Equations
• Conic Sections
• Vector Spaces
• 3d Coordinate System
• Eigenvalues and Eigenvectors
• Linear Transformations
• Number Sets
• Analytic Geometry

Trigonometry Solutions

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• Algebra Concepts and Expressions Review
• Right Triangle Trigonometry
• Radian Measure and Circular Functions
• Graphing Trigonometric Functions
• Simplifying Trigonometric Expressions
• Verifying Trigonometric Identities
• Solving Trigonometric Equations
• Complex Numbers
• Analytic Geometry in Polar Coordinates
• Exponential and Logarithmic Functions
• Vector Arithmetic

Precalculus Solutions

Below are examples of Precalculus math problems that can be solved.

• Operations on Functions
• Rational Expressions and Equations
• Polynomial and Rational Functions
• Analytic Trigonometry
• Sequences and Series
• Analytic Geometry in Rectangular Coordinates
• Limits and an Introduction to Calculus

Calculus Solutions

Below are examples of Calculus math problems that can be solved.

• Evaluating Limits
• Derivatives
• Applications of Differentiation
• Applications of Integration
• Techniques of Integration
• Parametric Equations and Polar Coordinates
• Differential Equations

Statistics Solutions

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• Algebra Review
• Average Descriptive Statistics
• Dispersion Statistics
• Probability
• Probability Distributions
• Frequency Distribution
• Normal Distributions
• t-Distributions
• Hypothesis Testing
• Estimation and Sample Size
• Correlation and Regression

Finite Math Solutions

Below are examples of Finite Math problems that can be solved.

• Polynomials and Expressions
• Equations and Inequalities
• Linear Functions and Points
• Systems of Linear Equations
• Mathematics of Finance
• Statistical Distributions

Linear Algebra Solutions

Below are examples of Linear Algebra math problems that can be solved.

• Introduction to Matrices
• Linear Independence and Combinations

Chemistry Solutions

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• Unit Conversion
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• Behavior of Gases
• Solutions and Concentrations

Physics Solutions

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• Static Equilibrium
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• Kinematics Equations
• Electricity
• Thermodymanics

Geometry Graphing Solutions

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• Step By Step Graphing
• Linear Equations and Functions
• Polar Equations

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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

Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

Example: Sam and Alex play tennis.

On the weekend Sam played 4 more games than Alex did, and together they played 12 games.

How many games did Alex play?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

• Read the whole thing first
• Do a sketch if possible
• Assign letters for the values
• Find or work out formulas

You should also write down what is actually being asked for , so you know where you are going and when you have arrived!

Also look for key words:

Thinking Clearly

Some wording can be tricky, making it hard to think "the right way around", such as:

Example: Sam has 2 dollars less than Alex. How do we write this as an equation?

• Let S = dollars Sam has
• Let A = dollars Alex has

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

( S − 2 = A is a common mistake, as the question is written "Sam ... 2 less ... Alex")

Example: on our street there are twice as many dogs as cats. How do we write this as an equation?

• Let D = number of dogs
• Let C = number of cats

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

( 2D = C is a common mistake, as the question is written "twice ... dogs ... cats")

Let's start with a really simple example so we see how it's done:

Example: A rectangular garden is 12m by 5m, what is its area ?

Turn the English into Algebra:

• Use w for width of rectangle: w = 12m
• Use h for height of rectangle: h = 5m

Formula for Area of a Rectangle : A = w × h

We are being asked for the Area.

A = w × h = 12 × 5 = 60 m 2

The area is 60 square meters .

Now let's try the example from the top of the page:

Example: Sam and Alex play Tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. How many games did Alex play?

• Use S for how many games Sam played
• Use A for how many games Alex played

We know that Sam played 4 more games than Alex, so: S = A + 4

And we know that together they played 12 games: S + A = 12

We are being asked for how many games Alex played: A

Which means that Alex played 4 games of tennis.

Check: Sam played 4 more games than Alex, so Sam played 8 games. Together they played 8 + 4 = 12 games. Yes!

A slightly harder example:

Example: Alex and Sam also build tables. Together they make 10 tables in 12 days. Alex working alone can make 10 in 30 days. How long would it take Sam working alone to make 10 tables?

• Use a for Alex's work rate
• Use s for Sam's work rate

12 days of Alex and Sam is 10 tables, so: 12a + 12s = 10

30 days of Alex alone is also 10 tables: 30a = 10

We are being asked how long it would take Sam to make 10 tables.

30a = 10 , so Alex's rate (tables per day) is: a = 10/30 = 1/3

Which means that Sam's rate is half a table a day (faster than Alex!)

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

Example: Jenna is training hard to qualify for the National Games. She has a regular weekly routine, training for five hours a day on some days and 3 hours a day on the other days. She trains altogether 27 hours in a seven day week. On how many days does she train for five hours?

• The number of "5 hour" days: d
• The number of "3 hour" days: e

We know there are seven days in the week, so: d + e = 7

And she trains 27 hours in a week, with d 5 hour days and e 3 hour days: 5d + 3e = 27

We are being asked for how many days she trains for 5 hours: d

The number of "5 hour" days is 3

Check : She trains for 5 hours on 3 days a week, so she must train for 3 hours a day on the other 4 days of the week.

3 × 5 hours = 15 hours, plus 4 × 3 hours = 12 hours gives a total of 27 hours

Some examples from Geometry:

Example: A circle has an area of 12 mm 2 , what is its radius?

• Use A for Area: A = 12 mm 2
• Use r for radius

And the formula for Area is: A = π r 2

We are being asked for the radius.

We need to rearrange the formula to find the area

Example: A cube has a volume of 125 mm 3 , what is its surface area?

Make a quick sketch:

• Use V for Volume
• Use A for Area
• Use s for side length of cube
• Volume of a cube: V = s 3
• Surface area of a cube: A = 6s 2

We are being asked for the surface area.

First work out s using the volume formula:

Now we can calculate surface area:

An example about Money:

Example: Joel works at the local pizza parlor. When he works overtime he earns 1¼ times the normal rate. One week Joel worked for 40 hours at the normal rate of pay and also worked 12 hours overtime. If Joel earned $660 altogether in that week, what is his normal rate of pay?

• Joel's normal rate of pay: $N per hour
• Joel works for 40 hours at $N per hour = $40N
• When Joel does overtime he earns 1¼ times the normal rate = $1.25N per hour
• Joel works for 12 hours at $1.25N per hour = $(12 × 1¼N) = $15N
• And together he earned $660, so:

$40N + $(12 × 1¼N) = $660

We are being asked for Joel's normal rate of pay $N.

So Joel’s normal rate of pay is $12 per hour

Joel’s normal rate of pay is $12 per hour, so his overtime rate is 1¼ × $12 per hour = $15 per hour. So his normal pay of 40 × $12 = $480, plus his overtime pay of 12 × $15 = $180 gives us a total of $660

More about Money, with these two examples involving Compound Interest

Example: Alex puts $2000 in the bank at an annual compound interest of 11%. How much will it be worth in 3 years?

This is the compound interest formula:

So we will use these letters:

• Present Value PV = $2,000
• Interest Rate (as a decimal): r = 0.11
• Number of Periods: n = 3
• Future Value (the value we want): FV

We are being asked for the Future Value: FV

Example: Roger deposited $1,000 into a savings account. The money earned interest compounded annually at the same rate. After nine years Roger's deposit has grown to $1,551.33 What was the annual rate of interest for the savings account?

The compound interest formula:

• Present Value PV = $1,000
• Interest Rate (the value we want): r
• Number of Periods: n = 9
• Future Value: FV = $1,551.33

We are being asked for the Interest Rate: r

So the annual rate of interest is 5%

Check : $1,000 × (1.05) 9 = $1,000 × 1.55133 = $1,551.33

And an example of a Ratio question:

Example: At the start of the year the ratio of boys to girls in a class is 2 : 1 But now, half a year later, four boys have left the class and there are two new girls. The ratio of boys to girls is now 4 : 3 How many students are there altogether now?

• Number of boys now: b
• Number of girls now: g

The current ratio is 4 : 3

Which can be rearranged to 3b = 4g

At the start of the year there was (b + 4) boys and (g − 2) girls, and the ratio was 2 : 1

b + 4 g − 2 = 2 1

Which can be rearranged to b + 4 = 2(g − 2)

We are being asked for how many students there are altogether now: b + g

There are 12 girls !

And 3b = 4g , so b = 4g/3 = 4 × 12 / 3 = 16 , so there are 16 boys

So there are now 12 girls and 16 boys in the class, making 28 students altogether .

There are now 16 boys and 12 girls, so the ratio of boys to girls is 16 : 12 = 4 : 3 At the start of the year there were 20 boys and 10 girls, so the ratio was 20 : 10 = 2 : 1

And now for some Quadratic Equations :

Example: The product of two consecutive even integers is 168. What are the integers?

Consecutive means one after the other. And they are even , so they could be 2 and 4, or 4 and 6, etc.

We will call the smaller integer n , and so the larger integer must be n+2

And we are told the product (what we get after multiplying) is 168, so we know:

n(n + 2) = 168

We are being asked for the integers

That is a Quadratic Equation , and there are many ways to solve it. Using the Quadratic Equation Solver we get −14 and 12.

Check −14: −14(−14 + 2) = (−14)×(−12) = 168 YES

Check 12: 12(12 + 2) = 12×14 = 168 YES

So there are two solutions: −14 and −12 is one, 12 and 14 is the other.

Note: we could have also tried "guess and check":

• We could try, say, n=10: 10(12) = 120 NO (too small)
• Next we could try n=12: 12(14) = 168 YES

But unless we remember that multiplying two negatives make a positive we might overlook the other solution of (−14)×(−12).

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• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Expanded Form Mean, Median & Mode
• Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems
• Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time Volume
• Pre Algebra
• One-Step Addition
• One-Step Subtraction
• One-Step Multiplication
• One-Step Division
• One-Step Decimals
• Two-Step Integers
• Two-Step Add/Subtract
• Two-Step Multiply/Divide
• Two-Step Fractions
• Two-Step Decimals
• Multi-Step Integers
• Multi-Step with Parentheses
• Multi-Step Rational
• Multi-Step Fractions
• Multi-Step Decimals
• Solve by Factoring
• Completing the Square
• Quadratic Formula
• Biquadratic
• Logarithmic
• Exponential
• Rational Roots
• Floor/Ceiling
• Equation Given Roots
• Newton Raphson
• Substitution
• Elimination
• Cramer's Rule
• Gaussian Elimination
• System of Inequalities
• Perfect Squares
• Difference of Squares
• Difference of Cubes
• Sum of Cubes
• Polynomials
• Distributive Property
• FOIL method
• Perfect Cubes
• Binomial Expansion
• Negative Rule
• Product Rule
• Quotient Rule
• Expand Power Rule
• Fraction Exponent
• Exponent Rules
• Exponential Form
• Logarithmic Form
• Absolute Value
• Rational Number
• Powers of i
• Complex Form
• Partial Fractions
• Is Polynomial
• Leading Coefficient
• Leading Term
• Standard Form
• Complete the Square
• Synthetic Division
• Linear Factors
• Rationalize Denominator
• Rationalize Numerator
• Identify Type
• Convergence
• Interval Notation
• Pi (Product) Notation
• Boolean Algebra
• Truth Table
• Mutual Exclusive
• Cardinality
• Caretesian Product
• Age Problems
• Distance Problems
• Cost Problems
• Investment Problems
• Number Problems
• Percent Problems
• Addition/Subtraction
• Multiplication/Division
• Dice Problems
• Coin Problems
• Card Problems
• Pre Calculus
• Linear Algebra
• Trigonometry
• Conversions

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• \mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
• \mathrm{Kira\:went\:for\:a\:drive\:in\:her\:new\:car.\:She\:drove\:for\:142.5\:miles\:at\:a\:speed\:of\:57\:mph.\:For\:how\:many\:hours\:did\:she\:drive?}
• \mathrm{The\:sum\:of\:two\:numbers\:is\:249\:.\:Twice\:the\:larger\:number\:plus\:three\:times\:the\:smaller\:number\:is\:591\:.\:Find\:the\:numbers.}
• \mathrm{If\:2\:tacos\:and\:3\:drinks\:cost\:12\:and\:3\:tacos\:and\:2\:drinks\:cost\:13\:how\:much\:does\:a\:taco\:cost?}
• \mathrm{You\:deposit\:3000\:in\:an\:account\:earning\:2\%\:interest\:compounded\:monthly.\:How\:much\:will\:you\:have\:in\:the\:account\:in\:15\:years?}
• How do you solve word problems?
• To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
• How do you identify word problems in math?
• Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
• Is there a calculator that can solve word problems?
• Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
• What is an age problem?
• An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

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• Middle School Math Solutions – Equation Calculator Welcome to our new "Getting Started" math solutions series. Over the next few weeks, we'll be showing how Symbolab...

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Age Problems

Age word problems.

Every now and then, we encounter word problems that require us to find the relationship between the ages of different people. Age word problems typically involve comparing two people’s ages at different points in time, i.e. at present, in the past, or in the future.

This lesson is divided into two parts. Part I involves age word problems that can be solved using a single variable while Part II contains age word problems that need to be solved using two variables .

Let’s get familiar with age word problems by working through some examples.

PART I: Age Word Problems Solvable with One Variable

Example 1: Tanya is 28 years older than Marcus. In 6 years, Tanya will be three times as old as Marcus. How old is Tanya now?

In this problem, we are only asked to find Tanya’s current age. However, the problem also gave us a lot of other information which can be overwhelming. To help us organize the important details, let’s create a table to list what we know so far.

Since we are only given details about their current ages and what they will be 6 years from now, we’ll go ahead and gray out the Past column.

You may notice that Tanya’s current age is defined using the age of Marcus. However, Marcus’s present age is currently unknown. So let’s express Marcus’s age using the variable [latex]x[/latex]. Since Tanya is 28 years older than Marcus , then Tanya’s present age must be [latex]x+28[/latex].

Next, let’s fill in the Future column which will consist of their ages in 6 years. All we have to do is add 6 to Tanya and Marcus’s present or current ages. Therefore, we have:

• Tanya: [latex]\left( {x + 28} \right) {\color{red}+ 6} = x + 34[/latex]
• Marcus: [latex]x {\color{red}+ 6}[/latex]

Now that our table is filled out, we can go ahead and create our equation based on the information provided. The problem states the following:

In 6 years , Tanya will be three times as old as Marcus.

Here we are trying to find the relationship between their ages in the future. We can simply say that,

Tanya’s age in 6 years = 3( Marcus’s age in 6 years )

With that in mind, we can easily construct our equation.

Our next step now is to solve for [latex]x[/latex]. But before that, remember that our problem is asking us to find Tanya’s current age. Since Tanya’s age is defined using Marcus’s current age (which is [latex]x[/latex]), we have to find his age first in order to determine what Tanya’s present age is.

Now that we have the value for [latex]x[/latex], let’s find out what Tanya and Marcus’s current ages are. We can do this by simply replacing the [latex]x[/latex]’s with [latex]8[/latex].

CURRENT AGES (present)

• Marcus: [latex]x = {\textbf{8}}[/latex] years old
• Tanya: [latex]x + 28 = {\color{red}8} + 28 = {\textbf{36}}[/latex] years old

Going back to the problem’s question, how old is Tanya now?

Answer: Tanya is 36 years old.

Answer Check:

At this point, we are confident that our answer is correct. But, how can we be 100% sure? Well, it’s always a good idea especially in math, to check our answers so we’re certain that we got the correct values.

For this problem, we can simply verify if our answer makes our future statement true. Do you remember this statement?

In 6 years, Tanya will be three times as old as Marcus.

We know the present ages of Marcus and Tanya which are [latex]8[/latex] and [latex]36[/latex], respectively. Hence in 6 years, Marcus will be [latex]14[/latex] years old while Tanya will be [latex]42[/latex] years old.

So, will Tanya be three times as old as Marcus in 6 years? The answer is Yes .

Example 2: Bruce is 4 years younger than Hector. Twenty years ago, Hector’s age was 13 years more than half the age of Bruce. How old are they now?

By just reading the problem, we can already tell that there is a great deal of information that we have to sort through and that this problem includes a fraction. Most students easily get lost in all the given information, let alone solving equations that involve fractions. But, don’t fret! As long as you stick with the basic principles and steps on how to solve age word problems, you’ll be fine.

Right now, we don’t know Bruce or Hector’s current age. But since Bruce’s age is expressed in relation to Hector’s age, then our unknown variable will be based on Hector’s age. In other words,

• Let [latex]{\textbf{\textit{h}}} =[/latex] Hector’s age
• [latex]{\textbf{\textit{h} – 4}} =[/latex] Bruce’s age, since he is 4 years younger than Hector

Let’s organize all these important data into a table. We’re only given details about their present and past (20 years ago) ages so we’ll gray out the Future column.

Twenty years ago, both Bruce and Hector were 20 years younger so we’ll subtract 20 from each of their present ages.

• Bruce: [latex]\left( {h – 4} \right) {\color{red}- 20} = h – 24[/latex]
• Hector: [latex]h {\color{red}- 20}[/latex]

Our table is now ready so we can proceed to create our equation. As you can see under the Past column, we were able to create algebraic expressions for Bruce and Hector’s ages 20 years ago. But our problem also told us that,

Twenty years ago , Hector’s age was 13 years more than half the age of Bruce.

Since Hector’s age 20 years ago is also 13 years more than half of Bruce’s age, we can take these two algebraic expressions and set them equal to each other, to create an equation.

Hector’s age 20 years ago = [latex]\Large{1 \over 2}[/latex]( Bruce’s age 20 years ago )[latex]+ 13[/latex]

We’re now ready to solve for the unknown variable, [latex]h[/latex].

Therefore, Hector’s present age is [latex]{\textbf{42}}[/latex] years old.

On the other hand, you may recall that Bruce’s current age is: [latex]h – 4[/latex]. Since [latex]h = 42[/latex], then Bruce’s current age is [latex]42 – 4 = {\textbf{38}}[/latex].

So, how old are they now?

Answer: Hector is 42 years old and Bruce is 38 years old .

The final step is to check our answers by substituting the unknown values into our original equation to verify if each side of the equation equals the other.

Great! Our answer checks. This just showed us that if we take Bruce’s age twenty years ago, which is 18, and divide it in half, we get 9. Adding 13 to that ([latex]9 + 13[/latex]), we get 22 which was Hector’s age twenty years ago.

Therefore, we are able to confirm that twenty years ago when Hector was 22 years old and Bruce was 18 years old, Hector’s age was 13 years more than half the age of Bruce.

Example 3: Stella is 13 years younger than Kwame. Nine years from now, the sum of their ages will be 43. Find the present age of each.

This problem is a little different from our previous two examples as we are given the sum of their ages in 9 years. But right off the bat, we can see that Stella’s age is defined in terms of Kwame’s age. Therefore, we’ll select a variable to represent Kwame’s current age. In this instance, let’s use “[latex]k[/latex]”.

• Let [latex]{\textbf{\textit{k}}} =[/latex] Kwame’s age
• [latex]{\textbf{\textit{k} – 13}} =[/latex] Stella’s age, since she is 13 years younger than Kwame

Nine years from now, both Kwame and Stella will be 9 years older. So we’ll simply add 9 to their present ages above to show their future ages.

• Kwame: [latex]k {\color{red}+ 9}[/latex]
• Stella: [latex]\left( {k – 13} \right) {\color{red}+ 9} = k – 4[/latex]

Let’s complete our table.

Now that we have the algebraic expressions for both their ages in 9 years, we can add these expressions to create our equation. We were given the following details:

Nine years from now , the sum of their ages will be 43 .

So we have,

Checking back at our table, [latex]k[/latex] stands for Kwame’s age. But since our problem asked us to find the current ages for both, let’s do a little bit more solving.

• Kwame: [latex]k = {\textbf{19}}[/latex] years old
• Stella: [latex]k – 13 = {\color{red}19} – 13 = {\textbf{6}}[/latex] years old

Answer: Kwame is 19 years old and Stella is 6 years old .

Let’s now verify if indeed the sum of Kwame and Stella’s ages in 9 years will be 43.

• Kwame’s age in 9 years: [latex]k + 9 = {\color{red}19} + 9 = {\textbf{28}}[/latex]
• Stella’s age in 9 years: [latex]k – 4 = {\color{red}19} – 4 = {\textbf{15}}[/latex]

Perfect! The total of their ages nine years from now is 43 so our answers are correct.

Example 4: Mr. Cook is 34 years old. His son is 22 years younger than him. In how many years will Mr. Cook’s age be 24 years less than three times as old as his son?

We already know their current ages, so before we delve any further, let’s start filling in our table.

Note that since the son is 22 years younger than Mr. Cook, we subtracted 22 from 34 to get his son’s current age, [latex]34 – {\color{red}22} = 12[/latex].

This problem is unique because it’s not asking us for their ages at a certain point in time like usual. Instead, it asks us to find out the number of years when Mr. Cook’s age will meet a certain relationship with his son’s age in the future.

But at this point, we don’t know how long it will take for Mr. Cook to be 24 years less than three times as old as his son. So, let’s assign the unknown variable “[latex]x[/latex]” to stand for the number of years then add [latex]x[/latex] to both of their current ages to create algebraic expressions that will represent how old they will be after [latex]x[/latex] years.

Since Mr. Cook’s age after [latex]x[/latex] number of years ([latex]x + 34[/latex]) will also be 24 years less than three times as old as his son , we can set these two algebraic expressions equal to each other, thus creating our equation.

Now that we have our equation, let’s solve for [latex]x[/latex].

As you may recall, [latex]x[/latex] stands for the number of years from now that will take for Mr. Cook to be 24 years less than three times as old as his son. Therefore,

Answer: In 11 years , Mr. Cook’s age will be 24 years less than three times as old as his son.

To check if our answer is correct, we must first find out how old will Mr. Cook and his son be in 11 years. Substituting the value of [latex]x[/latex] which is 11 into our algebraic expressions, we get:

• Mr. Cooks’s age in 11 years: [latex]x + 34 = {\color{red}11} + 34 = {\textbf{45}}[/latex]
• Son’s age in 11 years: [latex]x + 12 = {\color{red}11} + 12 = {\textbf{23}}[/latex]

So in 11 years, Mr. Cook will be 45 years old while his son will be 23 years old.

This time, I’ll leave it up to you to verify if indeed during that time, his age of 45 years old will be 24 years less than three times as old as his son. If it meets the condition, then our answer is correct.

Example 5: The sum of one-fifth of Annika’s age four years ago and half of her age in six years is 33. How old is she now?

Compared to our previous exercises, this problem only involves one person. Also, instead of comparing the ages of two people at a certain point in time, we will be comparing Annika’s ages at different points in time, i.e. 4 years ago and in 6 years.

We don’t know Annika’s current age so let’s select the variable [latex]{\textbf{\textit{a}}}[/latex] to represent this unknown value. We’ll use this variable as well to create algebraic expressions that will stand for her past and future ages.

• Let [latex]{\textbf{\textit{a}}} =[/latex] Annika’s current age
• [latex]{\textbf{\textit{a} – 4}} =[/latex] Annika’s age 4 years ago
• [latex]{\textbf{\textit{a} + 6}} =[/latex] Annika’s age 6 years from now

Our problem also told us that if we add [latex]\Large{1 \over 5}[/latex] of Annika’s age 4 years ago and [latex]\Large{1 \over 2}[/latex] of her age 6 years from now , the sum is 33 .

With this information, it’s easy for us to write our equation.

Our next step is to solve for the unknown variable, [latex]a[/latex].

So, how old is Annika now?

Answer: Annika is currently 44 years old.

As I mentioned before, it’s always a good practice to verify if you got the correct answer. To start, let’s find out what Annika’s past and future ages are.

• Annika’s age 4 years ago : [latex]a – 4 = {\color{red}44} – 4 = {\textbf{40}}[/latex]
• Annika’s age 6 years from now : [latex]a + 6 = {\color{red}44} + 6 = {\textbf{50}}[/latex]

Now that we know how old she was 4 years ago and how old she’ll be in 6 years, we’ll plug in these values into our original equation to see if both sides of the equation equal each other.

And they did! We were able to prove that the sum of [latex]\Large{1 \over 5}[/latex] of Annika’s age 4 years ago and [latex]\Large{1 \over 2}[/latex] of her age 6 years from now is indeed 33.

PART II: Age Word Problems Solvable with Two Variables

Example 6: The sum of Aaliyah and Harald’s ages is 28. Four years from now, Aaliyah will be three times as old as Harald. Find their present ages.

Neither Aaliyah nor Harald’s age is expressed in terms of the other. So for this problem, we will be using more than one variable to represent the unknown values. To start,

• Let [latex]{\textbf{\textit{a}}}[/latex] be Aaliyah’s age
• Let [latex]{\textbf{\textit{h}}}[/latex] be Harald’s age

Since they will be 4 years older in the next 4 years, we simply have to add 4 to their current ages to represent their future ages.

Looking back at our problem, there are two significant statements that can help us find our answers.

1) The sum of Aaliyah and Harald’s ages is 28.

From this statement, we can create the equation below:

2) Four years from now, Aaliyah will be three times as old as Harald.

Meanwhile, the statement above can be translated into the following equation:

We now have two equations to solve.

• Equation 1: [latex]a + h = 28[/latex]
• Equation 2: [latex]a + 4 = 3(h + 4)[/latex]

First, we’ll use equation 1 to solve for [latex]a[/latex].

Next, we’ll replace [latex]a[/latex] with [latex]28 – h[/latex] in equation 2 .

Perfect! We are able to find the values for both our unknown variables, [latex]a[/latex] and [latex]h[/latex], which also stand for the present ages for Aaliyah and Harald. So we have,

• Aaliyah’s present age: [latex]a = 28 – h = 28 – {\color{red}5} = {\textbf{23}}[/latex]
• Harald’s present age: [latex]h = {\textbf{5}}[/latex]

Answer: Currently, Aaliyah is 23 years old while Harald is 5 years old.

I’ll leave it up to you to check if our answers are correct. But as you can see, even with just using mental computation, we can already tell that the sum of Aaliyah and Harald’s ages is 28 ([latex]23 + 5 = 28[/latex]) which makes our first statement true. You may further check our answers by plugging in the values of [latex]a[/latex] and [latex]h[/latex] into equation 2 to verify if the left side of the equation equals the right, thus making our second statement true as well.

Example 7: The sum of the ages of Jaya and Nadia is three times Nadia’s age. Seven years ago, Jaya was three less than four times as old as Nadia. How old are they now?

This problem is similar to our previous example. However, for this one, we are not given the exact number for the sum. We first have to find out each of their current ages so we can determine what the sum is.

• Let [latex]{\textbf{\textit{y}}}[/latex] be Jaya’s age
• Let [latex]{\textbf{\textit{n}}}[/latex] be Nadia’s age

We then need to subtract 7 from their current ages to represent how old they were seven years ago.

Now that we’ve organized our data, let’s go through the significant statements given in our problem and translate each into an equation.

1) The sum of the ages of Jaya and Nadia is three times Nadia’s age.

2) Seven years ago, Jaya was three less than four times as old as Nadia.

Therefore, our two equations are:

• Equation 1: [latex]y + n = 3n[/latex]
• Equation 2: [latex]y – 7 = 4(n – 7) – 3[/latex]

Let’s first focus on equation 1 and solve for [latex]y[/latex].

Now we’ll solve for [latex]n[/latex] using the value of [latex]y[/latex] from equation 1. We’ll do this by replacing [latex]y[/latex] with [latex]2n[/latex] in equation 2 .

Taking the values of [latex]y[/latex] and [latex]n[/latex], we have:

• Jaya’s present age: [latex]y = 2n = 2({\color{red}12}) = {\textbf{24}}[/latex]
• Nadia’s present age: [latex]n = {\textbf{12}}[/latex]

So, going back to our problem. How old are they now?

Answer: Jaya is 24 years old and Nadia is 12 years old.

To check our answers, we’ll replace the values of [latex]y[/latex] and [latex]n[/latex] in equation 1 and equation 2. Again, I’ll leave it up to you to solve both equations and verify if each side of the equation equals the other. Once you’re done with your solutions, you’ll see that we are able to prove that both statements from our problem are true.

Example 8: The difference between the ages of Penelope and her son, Zack, is 34. In six years, Penelope will be four times as old as Zack’s age two years ago. How old are they now?

It’s easy to get lost in all the information given so we’ll focus first on assigning variables that will stand for the unknown values.

• Let [latex]{\textbf{\textit{p}}}[/latex] be Penelope’s current age
• Let [latex]{\textbf{\textit{z}}}[/latex] be Zack’s current age

One thing that’s unique about this problem is that it involves three different points in time. We are given not only the relationship between Penelope and her son’s age in the present time but also how their ages in 6 years are related to their ages two years ago.

To show this, we’ll subtract 2 from their ages now for their ages 2 years ago then add 6 to their current ages for their ages 6 years later .

Great! We now have variables and algebraic expressions to represent Penelope and Zack’s current ages as well as their ages in the past and in the future. Moving forward, let’s go through the important details given in the problem and create an equation from each statement.

1) The difference between the ages of Penelope and her son, Zack, is 34 .

Remember that Penelope is Zack’s mother so she’s definitely older than him. Therefore, we are subtracting Zack’s age from Penelope’s age to find the difference.

2) In six years, Penelope will be four times as old as Zack’s age two years ago.

Here are our two equations:

• Equation 1: [latex]p – z = 34[/latex]
• Equation 2: [latex]p + 6 = 4(z – 2)[/latex]

Let’s now work on equation 1 to solve for [latex]p[/latex].

Next, we’ll replace [latex]p[/latex] with [latex]34 + z[/latex] in equation 2 then solve for [latex]z[/latex].

• Penelope’s current age: [latex]p = 34 + z = 34 + ({\color{red}16}) = {\textbf{50}}[/latex]
• Zack’s current age: [latex]z = {\textbf{16}}[/latex]

How about we replace the unknown values in our table and also find out what their past and future ages are?

Going back to our original question, how old are they now?

Answer: Penelope is currently 50 years old while her son, Zack, is 16 years old.

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30 Fun Maths Questions with Answers

Table of Contents

Introduction

Mathematics can be fun if you treat it the right way. Maths is nothing less than a game, a game that polishes your intelligence and boosts your concentration. Compared to older times, people have a better and friendly approach to mathematics which makes it more appealing. The golden rule is to know that maths is a mindful activity rather than a task.

There is nothing like hard math problems or tricky maths questions, it’s just that you haven’t explored mathematics well enough to comprehend its easiness and relatability. Maths tricky questions and answers can be transformed into fun math problems if you look at it as if it is a brainstorming session. With the right attitude and friends and teachers, doing math can be most entertaining and delightful.

Math is interesting because a few equations and diagrams can communicate volumes of information. Treat math as a language, while moving to rigorous proof and using logical reason for performing a particular step in a proof or derivation.

Treating maths as a language totally eradicates the concept of hard math problems or tricky maths questions from your mind. Introducing children to fun maths questions can create a strong love and appreciation for maths at an early age. This way you are setting up the child’s successful future. Fun math problems will urge your child to choose to solve it over playing bingo or baking.

Apparently, there are innumerable methods to make easy maths tricky questions and answers. This includes the inception of the ideology that maths is simpler than their fear. This can be done by connecting maths with everyday life. Practising maths with the aid of dice, cards, puzzles and tables reassures that your child effectively approaches Maths.

If you wish to add some fun and excitement into educational activities, also check out

• Check out some mind-blowing Math Magic Tricks!
• Mental Maths: How to Improve it?

Cuemath is one of the world's leading math learning platforms that offers LIVE 1-to-1 online math classes for grades K-12 . Our mission is to transform the way children learn math, to help them excel in school and competitive exams. Our expert tutors conduct 2 or more live classes per week, at a pace that matches the child's learning needs.

Fun Maths Questions with answers - PDF

Here is the Downloadable PDF that consists of Fun Math questions. Click the Download button to view them.

Here are some fun, tricky and hard to solve maths problems that will challenge your thinking ability.

Answer: is 3, because ‘six’ has three letters

What is the number of parking space covered by the car?

This tricky math problem went viral a few years back after it appeared on an entrance exam in Hong Kong… for six-year-olds. Supposedly the students had just 20 seconds to solve the problem!

Believe it or not, this “math” question actually requires no math whatsoever. If you flip the image upside down, you’ll see that what you’re dealing with is a simple number sequence.

Replace the question mark in the above problem with the appropriate number.

Which number is equivalent to 3^(4)÷3^(2)

This problem comes straight from a standardized test given in New York in 2014.

There are 49 dogs signed up for a dog show. There are 36 more small dogs than large dogs. How many small dogs have signed up to compete? 

This question comes directly from a second grader's math homework.

To figure out how many small dogs are competing, you have to subtract 36 from 49 and then divide that answer, 13 by 2, to get 6.5 dogs, or the number of big dogs competing. But you’re not done yet! You then have to add 6.5 to 36 to get the number of small dogs competing, which is 42.5. Of course, it’s not actually possible for half a dog to compete in a dog show, but for the sake of this math problem let’s assume that it is.

Add 8.563 and 4.8292.

Adding two decimals together is easier than it looks. Don’t let the fact that 8.563 has fewer numbers than 4.8292 trip you up. All you have to do is add a 0 to the end of 8.563 and then add like you normally would.

I am an odd number. Take away one letter and I become even. What number am I?

Answer:  Seven (take away the ‘s’ and it becomes ‘even’).

Using only an addition, how do you add eight 8’s and get the number 1000?

Answer: 

888 + 88 + 8 + 8 + 8 = 1000

Sally is 54 years old and her mother is 80, how many years ago was Sally’s mother times her age?

41 years ago, when Sally was 13 and her mother was 39.

Which 3 numbers have the same answer whether they’re added or multiplied together?

There is a basket containing 5 apples, how do you divide the apples among 5 children so that each child has 1 apple while 1 apple remains in the basket?

4 children get 1 apple each while the fifth child gets the basket with the remaining apple still in it.

There is a three-digit number. The second digit is four times as big as the third digit, while the first digit is three less than the second digit. What is the number?

Fill in the question mark

Two girls were born to the same mother, at the same time, on the same day, in the same month and the same year and yet somehow they’re not twins. Why not?

Because there was a third girl, which makes them triplets!

A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each rung in 20cm and the bottom rung touches the water. The tide rises at a rate of 10cm an hour. When will the water reach the fifth rung?

The tide raises both the water and the boat so the water will never reach the fifth rung. 

The day before yesterday I was 25. The next year I will be 28. This is true only one day in a year. What day is my Birthday?  

You have a 3-litre bottle and a 5-litre bottle. How can you measure 4 litres of water by using 3L and 5L bottles? 

Solution 1 :

First, fill 3Lt bottle and pour 3 litres into 5Lt bottle.

Again fill the 3Lt bottle. Now pour 2 litres into the 5Lt bottle until it becomes full.

Now empty 5Lt bottle.

Pour remaining 1 litre in 3Lt bottle into 5Lt bottle.

Now again fill 3Lt bottle and pour 3 litres into 5Lt bottle.

Now you have 4 litres in the 5Lt bottle. That’s it.

Solution 2 :

First, fill the 5Lt bottle and pour 3 litres into 3Lt bottle.

Empty 3Lt bottle.

Pour remaining 2 litres in  5Lt bottle into 3Lt bottle.

Again fill the 5Lt bottle and pour 1 litre into 3 Lt bottle until it becomes full.

3 Friends went to a shop and purchased 3 toys. Each person paid Rs.10 which is the cost of one toy. So, they paid Rs.30 i.e. total amount. The shop owner gave a discount of Rs.5 on the total purchase of 3 toys for Rs.30. Then, among Rs.5, Each person has taken Rs.1 and remaining Rs.2 given to the beggar beside the shop. Now, the effective amount paid by each person is Rs.9 and the amount given to the beggar is Rs.2. So, the total effective amount paid is 9*3 = 27 and the amount given to beggar is Rs.2, thus the total is Rs.29. Where has the other Rs.1 gone from the original Rs.30?

The logic is payments should be equal to receipts. We cannot add the amount paid by persons and the amount given to the beggar and compare it to Rs.30.The total amount paid is ₹27. So, from ₹27, the shop owner received 25 rupees and beggar received ₹ 2. Thus, payments are equal to receipts.

How to get a number 100 by using four sevens (7’s) and a one (1)?

Answer 1:   177 – 77 = 100 ;

Answer 2: (7+7) * (7 + (1/7)) = 100 

Move any four matches to get 3 equilateral triangles only (don’t remove matches)

Find the area of the red triangle.

To solve this fun maths question, you need to understand how the area of a parallelogram works. If you already know how the area of a parallelogram and the area of a triangle are related, then adding 79 and 10 and subsequently subtracting 72 and 8 to get 9 should make sense.

 How many feet are in a mile? 

Solve  - 15+ (-5x) =0

What is 1.92÷3

A man is climbing up a mountain which is inclined. He has to travel 100 km to reach the top of the mountain. Every day He climbs up 2 km forward in the day time. Exhausted, he then takes rest there at night time. At night, while he is asleep, he slips down 1 km backwards because the mountain is inclined. Then how many days does it take him to reach the mountain top? 

 If 72 x 96 = 6927, 58 x 87 = 7885, then 79 x 86 = ?

Answer:  

Look at this series: 36, 34, 30, 28, 24, … What number should come next?

  Look at this series: 22, 21, 23, 22, 24, 23, … What number should come next?

If 13 x 12 = 651 & 41 x 23 = 448, then, 24 x 22 =?

Look at this series: 53, 53, 40, 40, 27, 27, … What number should come next?

The ultimate goals of mathematics instruction are students understanding the material presented, applying the skills, and recalling the concepts in the future. There's little benefit in students recalling a formula or procedure to prepare for an assessment tomorrow only to forget the core concept by next week.

Teachers must focus on making sure that the students understand the material and not just memorize the procedures. After you learn the answers to a fun maths question, you begin to ask yourself how you could have missed something so easy. The truth is, most trick questions are designed to trick your mind, which is why the answers to fun maths questions are logical and easy. 

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