What is a word problem in mathematics

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This article is about algorithmic word problems in mathematics and computer science. For other uses, see Word problem.

In computational mathematics, a word problem is the problem of deciding whether two given expressions are equivalent with respect to a set of rewriting identities. A prototypical example is the word problem for groups, but there are many other instances as well. A deep result of computational theory is that answering this question is in many important cases undecidable.^[1]

Background and motivation[edit]

In computer algebra one often wishes to encode mathematical expressions using an expression tree. But there are often multiple equivalent expression trees. The question naturally arises of whether there is an algorithm which, given as input two expressions, decides whether they represent the same element. Such an algorithm is called a solution to the word problem. For example, imagine that are symbols representing real numbers — then a relevant solution to the word problem would, given the input , produce the output EQUAL, and similarly produce NOT_EQUAL from .

The most direct solution to a word problem takes the form of a normal form theorem and algorithm which maps every element in an equivalence class of expressions to a single encoding known as the normal form — the word problem is then solved by comparing these normal forms via syntactic equality.^[1] For example one might decide that is the normal form of , , and , and devise a transformation system to rewrite those expressions to that form, in the process proving that all equivalent expressions will be rewritten to the same normal form.^[2] But not all solutions to the word problem use a normal form theorem — there are algebraic properties which indirectly imply the existence of an algorithm.^[1]

While the word problem asks whether two terms containing constants are equal, a proper extension of the word problem known as the unification problem asks whether two terms containing variables have instances that are equal, or in other words whether the equation has any solutions. As a common example, is a word problem in the integer group ℤ,
while is a unification problem in the same group; since the former terms happen to be equal in ℤ, the latter problem has the substitution as a solution.

History[edit]

One of the most deeply studied cases of the word problem is in the theory of semigroups and groups. A timeline of papers relevant to the Novikov-Boone theorem is as follows:^[3][4]

• 1910: Axel Thue poses a general problem of term rewriting on tree-like structures. He states «A solution of this problem in the most general case may perhaps be connected with unsurmountable difficulties».^[5][6]
• 1911: Max Dehn poses the word problem for finitely presented groups.^[7]
• 1912: Dehn presents Dehn’s algorithm, and proves it solves the word problem for the fundamental groups of closed orientable two-dimensional manifolds of genus greater than or equal to 2.^[8] Subsequent authors have greatly extended it to a wide range of group-theoretic decision problems.^[9][10][11]
• 1914: Axel Thue poses the word problem for finitely presented semigroups.^[12]
• 1930 – 1938: The Church-Turing thesis emerges, defining formal notions of computability and undecidability.^[13]
• 1947: Emil Post and Andrey Markov Jr. independently construct finitely presented semigroups with unsolvable word problem.^[14][15] Post’s construction is built on Turing machines while Markov’s uses Post’s normal systems.^[3]
• 1950: Alan Turing shows the word problem for cancellation semigroups is unsolvable,^[16] by furthering Post’s construction. The proof is difficult to follow but marks a turning point in the word problem for groups.^[3]: 342
• 1955: Pyotr Novikov gives the first published proof that the word problem for groups is unsolvable, using Turing’s cancellation semigroup result.^{[17][3]: 354} The proof contains a «Principal Lemma» equivalent to Britton’s Lemma.^[3]: 355
• 1954 – 1957: William Boone independently shows the word problem for groups is unsolvable, using Post’s semigroup construction.^[18][19]
• 1957 – 1958: John Britton gives another proof that the word problem for groups is unsolvable, based on Turing’s cancellation semigroups result and some of Britton’s earlier work.^[20] An early version of Britton’s Lemma appears.^[3]: 355
• 1958 – 1959: Boone publishes a simplified version of his construction.^[21][22]
• 1961: Graham Higman characterises the subgroups of finitely presented groups with Higman’s embedding theorem,^[23] connecting recursion theory with group theory in an unexpected way and giving a very different proof of the unsolvability of the word problem.^[3]
• 1961 – 1963: Britton presents a greatly simplified version of Boone’s 1959 proof that the word problem for groups is unsolvable.^[24] It uses a group-theoretic approach, in particular Britton’s Lemma. This proof has been used in a graduate course, although more modern and condensed proofs exist.^[25]
• 1977: Gennady Makanin proves that the existential theory of equations over free monoids is solvable.^[26]

The word problem for semi-Thue systems[edit]

The accessibility problem for string rewriting systems (semi-Thue systems or semigroups) can be stated as follows: Given a semi-Thue system and two words (strings) , can be transformed into by applying rules from ? Note that the rewriting here is one-way. The word problem is the accessibility problem for symmetric rewrite relations, i.e. Thue systems.^[27]

The accessibility and word problems are undecidable, i.e. there is no general algorithm for solving this problem.^[28] This even holds if we limit the systems to have finite presentations, i.e. a finite set of symbols and a finite set of relations on those symbols.^[27] Even the word problem restricted to ground terms is not decidable for certain finitely presented semigroups.^[29][30]

The word problem for groups[edit]

Given a presentation for a group G, the word problem is the algorithmic problem of deciding, given as input two words in S, whether they represent the same element of G. The word problem is one of three algorithmic problems for groups proposed by Max Dehn in 1911. It was shown by Pyotr Novikov in 1955 that there exists a finitely presented group G such that the word problem for G is undecidable.^[31]

The word problem in combinatorial calculus and lambda calculus[edit]

One of the earliest proofs that a word problem is undecidable was for combinatory logic: when are two strings of combinators equivalent? Because combinators encode all possible Turing machines, and the equivalence of two Turing machines is undecidable, it follows that the equivalence of two strings of combinators is undecidable. Alonzo Church observed this in 1936.^[32]

Likewise, one has essentially the same problem in (untyped) lambda calculus: given two distinct lambda expressions, there is no algorithm which can discern whether they are equivalent or not; equivalence is undecidable. For several typed variants of the lambda calculus, equivalence is decidable by comparison of normal forms.

The word problem for abstract rewriting systems[edit]

The word problem for an abstract rewriting system (ARS) is quite succinct: given objects x and y are they equivalent under ?^[29] The word problem for an ARS is undecidable in general. However, there is a computable solution for the word problem in the specific case where every object reduces to a unique normal form in a finite number of steps (i.e. the system is convergent): two objects are equivalent under if and only if they reduce to the same normal form.^[33]
The Knuth-Bendix completion algorithm can be used to transform a set of equations into a convergent term rewriting system.

The word problem in universal algebra[edit]

In universal algebra one studies algebraic structures consisting of a generating set A, a collection of operations on A of finite arity, and a finite set of identities that these operations must satisfy. The word problem for an algebra is then to determine, given two expressions (words) involving the generators and operations, whether they represent the same element of the algebra modulo the identities. The word problems for groups and semigroups can be phrased as word problems for algebras.^[1]

The word problem on free Heyting algebras is difficult.^[34]
The only known results are that the free Heyting algebra on one generator is infinite, and that the free complete Heyting algebra on one generator exists (and has one more element than the free Heyting algebra).

The word problem for free lattices[edit]

Example computation of x∧z ~ x∧z∧(x∨y)

x∧z∧(x∨y) ≤~ x∧z by 5. since x∧z ≤~ x∧z by 1. since x∧z = x∧z

x∧z ≤~ x∧z∧(x∨y) by 7. since x∧z ≤~ x∧z and x∧z ≤~ x∨y by 1. since x∧z = x∧z by 6. since x∧z ≤~ x by 5. since x ≤~ x by 1. since x = x

The word problem on free lattices and more generally free bounded lattices has a decidable solution. Bounded lattices are algebraic structures with the two binary operations ∨ and ∧ and the two constants (nullary operations) 0 and 1. The set of all well-formed expressions that can be formulated using these operations on elements from a given set of generators X will be called W(X). This set of words contains many expressions that turn out to denote equal values in every lattice. For example, if a is some element of X, then a ∨ 1 = 1 and a ∧ 1 = a. The word problem for free bounded lattices is the problem of determining which of these elements of W(X) denote the same element in the free bounded lattice FX, and hence in every bounded lattice.

The word problem may be resolved as follows. A relation ≤_~ on W(X) may be defined inductively by setting w ≤_~ v if and only if one of the following holds:

1.   w = v (this can be restricted to the case where w and v are elements of X),
2.   w = 0,
3.   v = 1,
4.   w = w₁ ∨ w₂ and both w₁ ≤_~ v and w₂ ≤_~ v hold,
5.   w = w₁ ∧ w₂ and either w₁ ≤_~ v or w₂ ≤_~ v holds,
6.   v = v₁ ∨ v₂ and either w ≤_~ v₁ or w ≤_~ v₂ holds,
7.   v = v₁ ∧ v₂ and both w ≤_~ v₁ and w ≤_~ v₂ hold.

This defines a preorder ≤_~ on W(X), so an equivalence relation can be defined by w ~ v when w ≤_~ v and v ≤_~ w. One may then show that the partially ordered quotient set W(X)/~ is the free bounded lattice FX.^[35][36] The equivalence classes of W(X)/~ are the sets of all words w and v with w ≤_~ v and v ≤_~ w. Two well-formed words v and w in W(X) denote the same value in every bounded lattice if and only if w ≤_~ v and v ≤_~ w; the latter conditions can be effectively decided using the above inductive definition. The table shows an example computation to show that the words x∧z and x∧z∧(x∨y) denote the same value in every bounded lattice. The case of lattices that are not bounded is treated similarly, omitting rules 2 and 3 in the above construction of ≤_~.

Example: A term rewriting system to decide the word problem in the free group[edit]

Bläsius and Bürckert
^[37]
demonstrate the Knuth–Bendix algorithm on an axiom set for groups.
The algorithm yields a confluent and noetherian term rewrite system that transforms every term into a unique normal form.^[38]
The rewrite rules are numbered incontiguous since some rules became redundant and were deleted during the algorithm run.
The equality of two terms follows from the axioms if and only if both terms are transformed into literally the same normal form term. For example, the terms

, and

share the same normal form, viz. ; therefore both terms are equal in every group.
As another example, the term and has the normal form and , respectively. Since the normal forms are literally different, the original terms cannot be equal in every group. In fact, they are usually different in non-abelian groups.

Group axioms used in Knuth–Bendix completion

A1
A2
A3

Term rewrite system obtained from Knuth–Bendix completion

R1
R2
R3
R4
R8
R11
R12
R13
R14
R17

See also[edit]

• Conjugacy problem
• Group isomorphism problem

References[edit]

Solving Word Problems in Mathematics

     Learn what word problems are and how to solve them in 7 easy steps.

     Real life math problems don’t usually look as simple as 3 + 5 = ?. Instead, things are a bit more complex. To show this, sometimes, math curriculum creators use word problems to help students see what happens in the real world. Word problems often show math happening in a more natural way in real life circumstances. 
     As a teacher, you can share some tips with your students to show that in everyday life they actually solve such problems all the time, and it’s not as scary as it may seem. 

     As you know, word problems can involve just about any operation: from addition to subtraction and division, or even multiple operations simultaneously.

If you’re a teacher, you may sometimes wonder how to teach students to solve word problems. It may be helpful to introduce some basic steps of working through a word problem in order to guide students’ experience. So, what steps do students need for solving a word problem in math?

Steps of Solving a Word Problem

     To work through any word problem, students should do the following:

     1. Read the problem: first, students should read through the problem once.
     2. Highlight facts: then, students should read through the problem again and highlight or underline important facts such as numbers or words that indicate an operation.
     3. Visualize the problem: drawing a picture or creating a diagram can be helpful. 

     Drawing the picture can be a big help in figuring this out. However, they can also look for the clues in the words such as:

          – Addition: add, more, total, altogether, and, plus, combine, in all;

          – Subtraction: fewer, than, take away, subtract, left, difference;

          – Multiplication: times, twice, triple, in all, total, groups;

     These key words may be very helpful when learning how to determine the operation students need to perform, but we should still pay attention to the fact that in the end it all depends on the context of the wording. The same word can have different meanings in different word problems. 

     Here they should learn to identify the steps they need to perform first to solve the problem, whether it’s a simple or a complex sentence. 
     6. Solve the problem: then, students can solve the number sentence and determine the solution. For example, 3 + 8 = 11.
     7. Check the answer: finally, students should check their work to make sure that the answer is correct.

     These 7 steps will help students get closer to mastering the skill of solving word problems. Of course, they still need plenty of practice. So, make sure to create enough opportunities for that! 

     At Happy Numbers, we gradually include word problems throughout the curriculum to ensure math flexibility and application of skills. Check out how easy it is to learn how to solve word problems with our visual exercises! 

     Word problems can be introduced in Kindergarten and be used through all grades as an important part of an educational process connecting mathematics to real life experience. 

     Happy Numbers introduces young students to the first math symbols by first building conceptual understanding of the operation through simple yet engaging visuals and key words. Once they understand the connection between these keywords and the actions they represent, they begin to substitute them with math symbols and translate word problems into number sentences. In this way, students gradually advance to the more abstract representations of these concepts.

     From the beginning, visualization helps the youngest students to understand the concepts of addition, subtraction, and even more complex operations. Even if they don’t draw the representations by themselves yet, students learn the connection between operations they need to perform in the problem and the real-world process this problem describes. 

     Next, students organize data from the word problem and pictures into a number sentence. To diversify the activity, you can ask students to match a word problem with the number sentence it represents.  

     Solving measurement problems is also a good way of mastering practical math skills. This is an example where students can see that math problems are closely related to real-world situations. Happy Numbers applies this by introducing more complicated forms of word problems as we help students advance to the next skill. By solving measurement word problems, students upgrade their vocabulary, learning such new terms as “difference” and “sum,” and continue mastering the connection between math operations and their word problem representations. 

     Later, students move to the next step, in which they learn how to create drawings and diagrams by themselves. They start by distributing light bulbs equally into boxes, which helps them to understand basic properties of division and multiplication. Eventually, with the help of Dino, they master tape diagrams! 

To see the full exercise, follow this link.

     The importance of working with diagrams and models becomes even more apparent when students move to more complex word problems. Pictorial representations help students master conceptual understanding by representing a challenging multi-step word problem in a visually simple and logical form. The ability to interact with a model helps students better understand logical patterns and motivates them to complete the task. 

     Having mentioned complex word problems, we have to show some of the examples that Happy Numbers uses in its curriculum. As the last step of mastering word problems, it is not the least important part of the journey. It’s crucial for students to learn how to solve the most challenging math problems without being intimidated by them. This only happens when their logical and algorithmic thinking skills are mastered perfectly, so they easily start talking in “math” language. 

Solving word problems at KS1 and KS2 is an essential part of the new maths curriculum. Here you can find expert guidance on how to solve maths word problems as well as examples of the many different types of word problems primary school children will encounter with links to hundreds more.

What is a word problem?

A word problem in maths is a maths question written as one sentence or more that requires children to apply their maths knowledge to a ‘real-life’ scenario. 

This means that children must be familiar with the vocabulary associated with the mathematical symbols they are used to, in order to make sense of the word problem. 

For example:

Importance of word problems within the national curriculum

The National Curriculum states that its mathematics curriculum “aims to ensure that all pupils:

• become fluent in the fundamentals of mathematics, including through varied and frequent practice with increasingly complex problems over time, so that pupils develop conceptual understanding and the ability to recall and apply knowledge rapidly and accurately;
• reason mathematically by following a line of enquiry, conjecturing relationships and generalisations, and developing an argument, justification or proof using mathematical language;
• can solve problems by applying their mathematics to a variety of routine and non-routine problems with increasing sophistication, including breaking down problems into a series of simpler steps and persevering in seeking solutions.”

To support this schools are adopting a ‘mastery’ approach to maths

The National Centre for Excellence in the Teaching of Mathematics (NCETM) have defined “teaching for mastery”, with some aspects of this definition being: 

• Maths teaching for mastery rejects the idea that a large proportion of people ‘just can’t do maths’.  
• All pupils are encouraged by the belief that by working hard at maths they can succeed. 
• Procedural fluency and conceptual understanding are developed in tandem because each supports the development of the other. 
• Significant time is spent developing deep knowledge of the key ideas that are needed to underpin future learning. The structure and connections within the mathematics are emphasised, so that pupils develop deep learning that can be sustained.

(The Essence of Maths Teaching for Mastery, 2016)

Mastery helps children to explore maths in greater depth

One of NCETM’s Five Big Ideas in Teaching for Mastery (2017) is “Mathematical Thinking: if taught ideas are to be understood deeply, they must not merely be passively received but must be worked on by the student: thought about, reasoned with and discussed with others”.

In other words – yes, fluency in arithmetic is important; however, with this often lies the common misconception that once a child has learnt the number skills appropriate to their level/age, they should be progressed to the next level/age of number skills. 

The mastery approach encourages exploring the breadth and depth of these concepts (once fluency is secure) through reasoning and problem solving. 

See the following example:

What sort of word problems might my child encounter at school?

In Key Stage 2, there are nine ‘strands’ of maths – these are then further split into ‘sub-strands’.  For example, ‘number and place value’ is the first strand: a Year 3 sub-strand of this is to “find 10 or 100 more or less than a given number”; a Year 6 sub-strand of this is to “determine the value of each digit in numbers up to 10 million”. The table below shows how the ‘sub-strands’ are distributed across each strand and year group in KS2.

How to teach children to solve word problems? 

Here are two simple strategies that can be applied to many word problems before solving them.

1. What do you already know?
2. How can this problem be drawn/represented pictorially?

Let’s see how this can be applied to a word problem to help achieve the answer.

Solving a simple word problem

There are 28 pupils in a class.
The teacher has 8 litres of orange juice.
She pours 225 millilitres of orange juice for every pupil.
How much orange juice is left over?

1. What do you already know?

• There are 1,000ml in 1 litre
• Pours = liquid leaving the bottle = subtraction
• For every = multiply
• Left over = requires subtraction at some point

2. How can this problem be drawn/represented pictorially?

The bar model is always a brilliant way of representing problems, but if you are not familiar with this, there are always other ways of drawing it out. 

Read more: What is a bar model

For example, for this question, you could draw 28 pupils (or stick man x 28) with ‘225 ml’ above each one and then a half-empty bottle with ‘8 litres’ marked at the top.

Now to put the maths to work. This is a Year 6 multi-step problem, so we need to use what we already know and what we’ve drawn to break down the steps.

Solving a more complex word problem

Mara is in a bookshop.
She buys one book for £6.99 and another that costs £3.40 more than the first book.
She pays using a £20 notes.
What change does Mara get?

1. What do you already know?

• More than = add
• Using decimals means I will have to line up the decimal points correctly in calculations
• Change from money = subtract

2. How can this problem be drawn/represented pictorially?

See this example of bar modelling for this question:

Now to put the maths to work using what we already know and what we’ve drawn to break down the steps.

Mara is in a bookshop. 

She buys one book for £6.99 and another that costs £3.40 more than the first book. 1) £6.99 + (£6.99 + £3.40) = £17.38

She pays using a £20 note.
What change does Mara get? 2) £20 – £17.38 = £2.62

Maths word problems for years 1 to 6

The more children learn about maths as the go through primary school, the trickier the word problems they face will become.

Below you will find some information about the types of word problems your child will be coming up against on a year by year basis, and how word problems apply to each primary year group

Word problems in Year 1

Throughout Year 1 a child is likely to be introduced to word problems with the help of concrete resources (pieces of physical apparatus like coins, cards, counters or number lines) to help them understand the problem.

An example of a word problem for Year 1 would be:

Chris is going to buy a cake for his mum which costs 80p. How many 20p coins would he need to do this? 

Word problems in Year 2

Year 2 is a continuation of Year 1 when it comes to word problems, with children still using concrete maths resources to help them understand and visualise the problems they are working on

An example of a word problem for Year 2 would be:

A class of 10 children each have 5 pencils in their pencil cases. How many pencils are there in total?

Word problems in Year 3 

With word problems for year 3, children will move away from using concrete resources when solving word problems, and move towards using written methods. Teachers will begin to demonstrate the four operations such as addition and subtraction word problems, multiplication and division problems too.

This is also the year in which 2-step word problems will be introduced. This is a problem which requires two individual calculations to be completed.

Year 3 word problem: Geometry properties of shape

Shaun is making 3-D shapes out of plastic straws.

At the vertices where the straws meet, he uses blobs of modelling clay to fix them together

Here are some of the shapes he makes:

One of Sean’s shapes is a cuboid. Which is it? Explain your answer.

Answer: shape B as a cuboid has 12 edges (straws) and 8 vertices (clay)

Year 3 word problem: Statistics

Year 3 are collecting pebbles. This pictogram shows the different numbers of pebbles each group finds.

Answer: a) 9   b) 3 pebbles drawn

Top tip
By the time children are in Year 3 many of the word problems, even one-step story problems tend to be a variation on a multiplication problem. For this reason learning times tables becomes increasingly essential at this stage. One of the best things you can do to help with Year 3 maths at home is support your child to do this.

Word problems in Year 4

At this stage of their primary school career, children should feel confident using the written method for each of the four operations. 

Word problems for year 4 will include a variety of problems, including 2-step problems and be children will be expected to work out the appropriate method required to solve each one. 

Year 4 word problem: Number and place value

My number has four digits and has a 7 in the hundreds place.
The digit which has the highest value in my number is 2.
The digit which has the lowest value in my number is 6.
My number has 3 fewer tens than hundreds.
What is my number?

Answer: 2,746

Word problems in Year 5

One and 2-step word problems continue with word problems for year 5, but this is also the year that children will be introduced to word problems containing decimals.

These are some examples of Year 5 maths word problems.

Year 5 word problem: Fractions, decimals and percentages

Answer: Stan (he has washed 0.5 whereas Frank has only washed 0.2 and Norm 0.05)

Word problems in Year 6

Word problems for year 6 shift from 2-step word problems to multi-step word problems. These will include fractions, decimals, percentages and time word problems. 

Here are some examples of the types of maths word problems Year 6 will have to solve.

Year 6 word problem – Ratio and proportion

This question is from the 2018 key stage 2 SATs paper. It is worth 1 mark.

The Angel of the North is a large statue in England. It is 20 metres tall and 54 metres wide. 

Ally makes a scale model of the Angel of the North. Her model is 40 centimetres tall. How wide is her model?

Answer: 108cm

Year 6 word problem – Algebra

This question is from the 2018 KS2 SATs paper. It is worth 2 marks as there are 2 parts to the answer.

Amina is making designs with two different shapes.

She gives each shape a value.

Calculate the value of each shape.

Answer: 36 (hexagon) and 25. 

Year 6 word problem: Measurement

This question is from the 2018 KS2 SATs paper. It is worth 3 marks as it is a multi-step problem.

There are 28 pupils in a class.
The teacher has 8 litres of orange juice.
She pours 225 millilitres of orange juice for every pupil.
How much orange juice is left over?

Answer: 1.7 litres or 1,700ml

Topic based word problems

The following examples give you an idea of the kinds of maths word problems your child will encounter for each of the 9 strands of maths in KS2.

Place value word problems

Place value word problem Year 5

This machine subtracts one hundredth each time the button is pressed. The starting number is 8.43. What number will the machine show if the button is pressed six times? Answer: 8.37

Download free number and place value word problems for Years 3, 4, 5 and 6

Addition and subtraction word problems

Addition and subtraction word problem Year 3

In Year 3 pupils will solve addition word problems and subtraction word problems with 2 and 3 digits.

Sam has 364 sweets. He gets given 142 more. He then gives 277 away. How many sweets is he left with? Answer: 229

Download free addition and subtraction word problems for Years 3, 4, 5 and 6

Addition word problem Year 3

Alfie thinks of a number. He subtracts 70. His new number is 12. What was the number Alfie thought of? Answer: 82

Subtraction word problem Year 6

The temperature at 7pm was 4oC. By midnight, it had dropped by 9 degrees. What was the temperature at midnight? Answer: -5oC

More here: 25 addition and subtraction word problems

Multiplication and division word problems

Multiplication and division word problem Year 3

A baker is baking chocolate cupcakes. She melts 16 chocolate buttons to make the icing for 9 cakes. How many chocolate buttons will she need to melt to make the icing for 18 cakes? Answer: 32

Multiplication word problem Year 4

Eggs are sold in boxes of 12. The egg boxes are delivered to stores in crates. Each crate holds 9 boxes. How many eggs are in a crate? Answer: 108

Download free multiplication word problems for Years 3, 4, 5 and 6

Division word problem Year 6

A factory produces 1,692 paintbrushes every day. They are packaged into boxes of 9. How many boxes does the factory produce every day? Answer: 188

Download our free division word problems worksheets for Years 3, 4, 5 and 6.

More here: 20 multiplication word problems
More here: 25 division word problems

Free resource: Use these four operations word problems to practise addition, subtraction, multiplication and division all together.

Fraction word problems

Fraction word problem Year 5

At the end of every day, a chocolate factory has 1 and 2/6 boxes of chocolates left over. How many boxes of chocolates are left over by the end of a week? Answer: 9 and 2/6 or 9 and 1/3

Download free fractions and decimals word problems worksheets for Years 3, 4, 5 and 6

More here: 28 fraction word problems

Decimals word problem Year 4 (crossover with subtraction)

Which two decimals that have a difference of 0.5? 0.2, 0.25, 0.4, 0.45, 0.6, 0.75. Answer: 0.25 and 0.75

Download free decimals and percentages word problems resources for Years 3, 4, 5 and 6

Percentage word problem Year 5

There are 350 children in a school. 50% are boys. How many boys are there? Answer: 175

Measurement word problem Year 3 (crossover with subtraction)

Lucy and Ffion both have bottles of strawberry smoothie. Each bottle contains 1 litre. Lucy drinks ½ of her bottle. Ffion drinks 300ml of her bottle. How much does each person have left in both bottles? Answer: Lucy = 500ml, Ffion = 300ml

More here: 25 percentage word problems

Money word problem Year 3

James and Lauren have different amounts of money. James has twelve 2p coins. Lauren has seven 5p coins. Who has the most money and by how much? Answer: Lauren by 11p.

More here: 25 money word problems

Area word problem Year 4

A rectangle measures 6cm by 5cm.

What is its area? Answer: 30cm2

Perimeter word problem Year 4

The swimming pool at the Sunshine Inn hotel is 20m long and 7m wide. Mary swims around the edge of the pool twice. How many metres has she swum? Answer: 108m

Ratio word problem Year 6 (crossover with measurement)

A local council has spent the day painting double yellow lines. They use 1 pot of yellow paint for every 100m of road they paint. How many pots of paint will they need to paint a 2km stretch of road? Answer: 20 pots

More here: 24 ratio word problems

Bodmas word problem Year 6

Draw a pair of brackets in one of these calculations so that they make two different answers. What are the answers?

50 – 10 × 5 =

50 – 10 × 5 =

Volume word problem Year 6

This large cuboid has been made by stacking shipping containers on a boat. Each individual shipping container has a length of 6m, a width of 4m and a height of 3m. What is the volume of the large cuboid? Answer: 864m3

How important are word problems when it comes to the SATs? 

In the KS1 SATs, 58% (35/60 marks) of the test is comprised of maths ‘reasoning’ (word problems). 

In KS2, this increases to 64% (70/110 marks) spread over two reasoning papers, each worth 35 marks. Considering children have, in the past, needed approximately 55-60% to reach the ‘expected standard’, it’s clear that children need regular exposure to and a solid understanding of how to solve a variety of word problems.

Children have the opportunity to practice SATs style word problems in Third Space Learning’s online one-to-one SATs revision programme. Personalised to meet the needs of each student, our programme helps to fill gaps and give students more confidence going in to the SATs exams.

Remember: The word problems can change but the maths won’t 

It can be easy for children to get overwhelmed when they first come across word problems in KS2, but it is important that you remind them that whilst the context of the problem may be presented in a different way, the maths behind it remains the same. 

Word problems are a good way to bring maths into the real world and make maths more relevant for your child, so help them practise, or even ask them to turn the tables and make up some word problems for you to solve. 

This article while written by a teacher for teachers is also suitable for those at home supporting children with home learning. More free home learning resources are also available.

In computational mathematics, a word problem is the problem of deciding whether two given expressions are equivalent with respect to a set of rewriting identities. A prototypical example is the word problem for groups, but there are many other instances as well. A deep result of computational theory is that answering this question is in many important cases undecidable.^[1]

Background and motivation

In computer algebra one often wishes to encode mathematical expressions using an expression tree. But there are often multiple equivalent expression trees. The question naturally arises of whether there is an algorithm which, given as input two expressions, decides whether they represent the same element. Such an algorithm is called a solution to the word problem. For example, imagine that [math]displaystyle{ x,y,z }[/math] are symbols representing real numbers — then a relevant solution to the word problem would, given the input [math]displaystyle{ (x cdot y)/z mathrel{overset{?}{=}} (x/z)cdot y }[/math], produce the output EQUAL, and similarly produce NOT_EQUAL from [math]displaystyle{ (x cdot y)/z mathrel{overset{?}{=}} (x/x)cdot y }[/math].

The most direct solution to a word problem takes the form of a normal form theorem and algorithm which maps every element in a equivalence class of expressions to a single encoding known as the normal form — the word problem is then solved by comparing these normal forms via syntactic equality.^[1] For example one might decide that [math]displaystyle{ x cdot y cdot z^{-1} }[/math] is the normal form of [math]displaystyle{ (x cdot y)/z }[/math], [math]displaystyle{ (x/z)cdot y }[/math], and [math]displaystyle{ (y/z)cdot x }[/math], and devise a transformation system to rewrite those expressions to that form, in the process proving that all equivalent expressions will be rewritten to the same normal form.^[2] But not all solutions to the word problem use a normal form theorem — there are algebraic properties which indirectly imply the existence of an algorithm.^[1]

While the word problem asks whether two terms containing constants are equal, a proper extension of the word problem known as the unification problem asks whether two terms [math]displaystyle{ t_1,t_2 }[/math] containing variables have instances that are equal, or in other words whether the equation [math]displaystyle{ t_1 = t_2 }[/math] has any solutions. As a common example, [math]displaystyle{ 2 + 3 stackrel{?}{=} 8 + (-3) }[/math] is a word problem in the integer group ℤ,
while [math]displaystyle{ 2 + x stackrel{?}{=} 8 + (-x) }[/math] is a unification problem in the same group; since the former terms happen to be equal in ℤ, the latter problem has the substitution [math]displaystyle{ {x mapsto 3} }[/math] as a solution.

History

One of the most deeply studied cases of the word problem is in the theory of semigroups and groups. A timeline of papers relevant to the Novikov-Boone theorem is as follows:^[3][4]

• 1910: Axel Thue poses a general problem of term rewriting on tree-like structures. He states «A solution of this problem in the most general case may perhaps be connected with unsurmountable difficulties».^[5][6]
• 1911: Max Dehn poses the word problem for finitely presented groups.^[7]
• 1912: Dehn presents Dehn’s algorithm, and proves it solves the word problem for the fundamental groups of closed orientable two-dimensional manifolds of genus greater than or equal to 2.^[8] Subsequent authors have greatly extended it to a wide range of group theoretic decision problems.^[9][10][11]
• 1914: Axel Thue poses the word problem for finitely presented semigroups.^[12]
• 1930 – 1938: The Church-Turing thesis emerges, defining formal notions of computability and undecidability.^[13]
• 1947: Emil Post and Andrey Markov Jr. independently construct finitely presented semigroups with unsolvable word problem.^[14][15] Post’s construction is built on Turing machines while Markov’s uses Post’s normal systems.^[3]
• 1950: Alan Turing shows the word problem for cancellation semigroups is unsolvable,^[16] by furthering Post’s construction. The proof is difficult to follow but marks a turning point in the word problem for groups.^[3]:342
• 1955: Pyotr Novikov gives the first published proof that the word problem for groups is unsolvable, using Turing’s cancellation semigroup result.^[17][3]:354 The proof contains a «Principal Lemma» equivalent to Britton’s Lemma.^[3]:355
• 1954 – 1957: William Boone independently shows the word problem for groups is unsolvable, using Post’s semigroup construction.^[18][19]
• 1957 – 1958: John Britton gives another proof that the word problem for groups is unsolvable, based on Turing’s cancellation semigroups result and some of Britton’s earlier work.^[20] An early version of Britton’s Lemma appears.^[3]:355
• 1958 – 1959: Boone publishes a simplified version of his construction.^[21][22]
• 1961: Graham Higman characterises the subgroups of finitely presented groups with Higman’s embedding theorem,^[23] connecting recursion theory with group theory in an unexpected way and giving a very different proof of the unsolvability of the word problem.^[3]
• 1961 – 1963: Britton presents a greatly simplified version of Boone’s 1959 proof that the word problem for groups is unsolvable.^[24] It uses a group-theoretic approach, in particular Britton’s Lemma. This proof has been used in a graduate course, although more modern and condensed proofs exist.^[25]
• 1977: Gennady Makanin proves that the existential theory of equations over free monoids is solvable.^[26]

The word problem for semi-Thue systems

The accessibility problem for string rewriting systems (semi-Thue systems or semigroups) can be stated as follows: Given a semi-Thue system [math]displaystyle{ T:=(Sigma, R) }[/math] and two words (strings) [math]displaystyle{ u, v in Sigma^* }[/math], can [math]displaystyle{ u }[/math] be transformed into [math]displaystyle{ v }[/math] by applying rules from [math]displaystyle{ R }[/math]? Note that the rewriting here is one-way. The word problem is the accessibility problem for symmetric rewrite relations, i.e. Thue systems.^[27]

The accessibility and word problems are undecidable, i.e. there is no general algorithm for solving this problem.^[28] This even holds if we limit the systems to have finite presentations, i.e. a finite set of symbols and a finite set of relations on those symbols.^[27] Even the word problem restricted to ground terms is not decidable for certain finally presented semigroups.^[29][30]

The word problem for groups

Main page: Word problem for groups

Given a presentation [math]displaystyle{ langle Smid mathcal{R} rangle }[/math] for a group G, the word problem is the algorithmic problem of deciding, given as input two words in S, whether they represent the same element of G. The word problem is one of three algorithmic problems for groups proposed by Max Dehn in 1911. It was shown by Pyotr Novikov in 1955 that there exists a finitely presented group G such that the word problem for G is undecidable.^[31]

The word problem in combinatorial calculus and lambda calculus

One of the earliest proofs that a word problem is undecidable was for combinatory logic: when are two strings of combinators equivalent? Because combinators encode all possible Turing machines, and the equivalence of two Turing machines is undecidable, it follows that the equivalence of two strings of combinators is undecidable. Alonzo Church observed this in 1936.^[32]

Likewise, one has essentially the same problem in (untyped) lambda calculus: given two distinct lambda expressions, there is no algorithm which can discern whether they are equivalent or not; equivalence is undecidable. For several typed variants of the lambda calculus, equivalence is decidable by comparison of normal forms.

The word problem for abstract rewriting systems

The word problem for an abstract rewriting system (ARS) is quite succinct: given objects x and y are they equivalent under [math]displaystyle{ stackrel{*}{leftrightarrow} }[/math]?^[29] The word problem for an ARS is undecidable in general. However, there is a computable solution for the word problem in the specific case where every object reduces to a unique normal form in a finite number of steps (i.e. the system is convergent): two objects are equivalent under [math]displaystyle{ stackrel{*}{leftrightarrow} }[/math] if and only if they reduce to the same normal form.^[33]
The Knuth-Bendix completion algorithm can be used to transform a set of equations into a convergent term rewriting system.

The word problem in universal algebra

In universal algebra one studies algebraic structures consisting of a generating set A, a collection of operations on A of finite arity (typically binary operations), and a finite set of identities that these operations must satisfy. The word problem for an algebra is then to determine, given two expressions (words) involving the generators and operations, whether they represent the same element of the algebra modulo the identities. The word problems for groups and semigroups can be phrased as word problems for algebras.^[1]

The word problem on free Heyting algebras is difficult.^[34]
The only known results are that the free Heyting algebra on one generator is infinite, and that the free complete Heyting algebra on one generator exists (and has one more element than the free Heyting algebra).

The word problem for free lattices

Example computation of x∧z ~ x∧z∧(x∨y)

x∧z∧(x∨y) ≤~ x∧z by 5. since x∧z ≤~ x∧z by 1. since x∧z = x∧z

x∧z ≤~ x∧z∧(x∨y) by 7. since x∧z ≤~ x∧z and x∧z ≤~ x∨y by 1. since x∧z = x∧z by 6. since x∧z ≤~ x by 5. since x ≤~ x by 1. since x = x

The word problem on free lattices and more generally free bounded lattices has a decidable solution. Bounded lattices are algebraic structures with the two binary operations ∨ and ∧ and the two constants (nullary operations) 0 and 1. The set of all well-formed expressions that can be formulated using these operations on elements from a given set of generators X will be called W(X). This set of words contains many expressions that turn out to denote equal values in every lattice. For example, if a is some element of X, then a ∨ 1 = 1 and a ∧ 1 =a. The word problem for free bounded lattices is the problem of determining which of these elements of W(X) denote the same element in the free bounded lattice FX, and hence in every bounded lattice.

The word problem may be resolved as follows. A relation ≤_~ on W(X) may be defined inductively by setting w ≤_~ v if and only if one of the following holds:

1.   w = v (this can be restricted to the case where w and v are elements of X),
2.   w = 0,
3.   v = 1,
4.   w = w₁ ∨ w₂ and both w₁ ≤_~ v and w₂ ≤_~ v hold,
5.   w = w₁ ∧ w₂ and either w₁ ≤_~ v or w₂ ≤_~ v holds,
6.   v = v₁ ∨ v₂ and either w ≤_~ v₁ or w ≤_~ v₂ holds,
7.   v = v₁ ∧ v₂ and both w ≤_~ v₁ and w ≤_~ v₂ hold.

This defines a preorder ≤_~ on W(X), so an equivalence relation can be defined by w ~ v when w ≤_~ v and v ≤_~ w. One may then show that the partially ordered quotient space W(X)/~ is the free bounded lattice FX.^[35][36] The equivalence classes of W(X)/~ are the sets of all words w and v with w ≤_~ v and v ≤_~ w. Two well-formed words v and w in W(X) denote the same value in every bounded lattice if and only if w ≤_~ v and v ≤_~ w; the latter conditions can be effectively decided using the above inductive definition. The table shows an example computation to show that the words x∧z and x∧z∧(x∨y) denote the same value in every bounded lattice. The case of lattices that are not bounded is treated similarly, omitting rules 2 and 3 in the above construction of ≤_~.

Example: A term rewriting system to decide the word problem in the free group

Bläsius and Bürckert
^[37]
demonstrate the Knuth–Bendix algorithm on an axiom set for groups.
The algorithm yields a confluent and noetherian term rewrite system that transforms every term into a unique normal form.^[38]
The rewrite rules are numbered incontiguous since some rules became redundant and were deleted during the algorithm run.
The equality of two terms follows from the axioms if and only if both terms are transformed into literally the same normal form term. For example, the terms

[math]displaystyle{ ((a^{-1} cdot a) cdot (b cdot b^{-1}))^{-1} stackrel{R2}{rightsquigarrow} (1 cdot (b cdot b^{-1}))^{-1} stackrel{R13}{rightsquigarrow} (1 cdot 1)^{-1} stackrel{R1}{rightsquigarrow} 1 ^{-1} stackrel{R8}{rightsquigarrow} 1 }[/math], and

[math]displaystyle{ b cdot ((a cdot b)^{-1} cdot a) stackrel{R17}{rightsquigarrow} b cdot ((b^{-1} cdot a^{-1}) cdot a) stackrel{R3}{rightsquigarrow} b cdot (b^{-1} cdot (a^{-1} cdot a)) stackrel{R2}{rightsquigarrow} b cdot (b^{-1} cdot 1) stackrel{R11}{rightsquigarrow} b cdot b^{-1} stackrel{R13}{rightsquigarrow} 1 }[/math]

share the same normal form, viz. [math]displaystyle{ 1 }[/math]; therefore both terms are equal in every group.
As another example, the term [math]displaystyle{ 1 cdot (a cdot b) }[/math] and [math]displaystyle{ b cdot (1 cdot a) }[/math] has the normal form [math]displaystyle{ a cdot b }[/math] and [math]displaystyle{ b cdot a }[/math], respectively. Since the normal forms are literally different, the original terms cannot be equal in every group. In fact, they are usually different in non-abelian groups.

Group axioms used in Knuth–Bendix completion

A1	[math]displaystyle{ 1 cdot x }[/math]	[math]displaystyle{ = x }[/math]
A2	[math]displaystyle{ x^{-1} cdot x }[/math]	[math]displaystyle{ = 1 }[/math]
A3	[math]displaystyle{ (x cdot y) cdot z }[/math]	[math]displaystyle{ = x cdot (y cdot z) }[/math]

Term rewrite system obtained from Knuth–Bendix completion

R1	[math]displaystyle{ 1 cdot x }[/math]	[math]displaystyle{ rightsquigarrow x }[/math]
R2	[math]displaystyle{ x^{-1} cdot x }[/math]	[math]displaystyle{ rightsquigarrow 1 }[/math]
R3	[math]displaystyle{ (x cdot y) cdot z }[/math]	[math]displaystyle{ rightsquigarrow x cdot (y cdot z) }[/math]
R4	[math]displaystyle{ x^{-1} cdot (x cdot y) }[/math]	[math]displaystyle{ rightsquigarrow y }[/math]
R8	[math]displaystyle{ 1^{-1} }[/math]	[math]displaystyle{ rightsquigarrow 1 }[/math]
R11	[math]displaystyle{ x cdot 1 }[/math]	[math]displaystyle{ rightsquigarrow x }[/math]
R12	[math]displaystyle{ (x^{-1})^{-1} }[/math]	[math]displaystyle{ rightsquigarrow x }[/math]
R13	[math]displaystyle{ x cdot x^{-1} }[/math]	[math]displaystyle{ rightsquigarrow 1 }[/math]
R14	[math]displaystyle{ x cdot (x^{-1} cdot y) }[/math]	[math]displaystyle{ rightsquigarrow y }[/math]
R17	[math]displaystyle{ (x cdot y)^{-1} }[/math]	[math]displaystyle{ rightsquigarrow y^{-1} cdot x^{-1} }[/math]

See also

• Conjugacy problem
• Group isomorphism problem

References

/en/algebra-topics/solving-equations/content/

What are word problems?

A word problem is a math problem written out as a short story or scenario. Basically, it describes a realistic problem and asks you to imagine how you would solve it using math. If you’ve ever taken a math class, you’ve probably solved a word problem. For instance, does this sound familiar?

Johnny has 12 apples. If he gives four to Susie, how many will he have left?

You could solve this problem by looking at the numbers and figuring out what the problem is asking you to do. In this case, you’re supposed to find out how many apples Johnny has left at the end of the problem. By reading the problem, you know Johnny starts out with 12 apples. By the end, he has 4 less because he gave them away. You could write this as:

12 — 4

12 — 4 = 8, so you know Johnny has 8 apples left.

Word problems in algebra

If you were able to solve this problem, you should also be able to solve algebra word problems. Yes, they involve more complicated math, but they use the same basic problem-solving skills as simpler word problems.

You can tackle any word problem by following these five steps:

1. Read through the problem carefully, and figure out what it’s about.
2. Represent unknown numbers with variables.
3. Translate the rest of the problem into a mathematical expression.
4. Solve the problem.
5. Check your work.

We’ll work through an algebra word problem using these steps. Here’s a typical problem:

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took two days, and the van cost $360. How many miles did she drive?

It might seem complicated at first glance, but we already have all of the information we need to solve it. Let’s go through it step by step.

Step 1: Read through the problem carefully.

With any problem, start by reading through the problem. As you’re reading, consider:

• What question is the problem asking?
• What information do you already have?

Let’s take a look at our problem again. What question is the problem asking? In other words, what are you trying to find out?

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

There’s only one question here. We’re trying to find out how many miles Jada drove. Now we need to locate any information that will help us answer this question.

There are a few important things we know that will help us figure out the total mileage Jada drove:

• The van cost $30 per day.
• In addition to paying a daily charge, Jada paid $0.50 per mile.
• Jada had the van for 2 days.
• The total cost was $360.

Step 2: Represent unknown numbers with variables.

In algebra, you represent unknown numbers with letters called variables. (To learn more about variables, see our lesson on reading algebraic expressions.) You can use a variable in the place of any amount you don’t know. Looking at our problem, do you see a quantity we should represent with a variable? It’s often the number we’re trying to find out.

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

Since we’re trying to find the total number of miles Jada drove, we’ll represent that amount with a variable—at least until we know it. We’ll use the variable m for miles. Of course, we could use any variable, but m should be easy to remember.

Step 3: Translate the rest of the problem.

Let’s take another look at the problem, with the facts we’ll use to solve it highlighted.

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

We know the total cost of the van, and we know that it includes a fee for the number of days, plus another fee for the number of miles. It’s $30 per day, and $0.50 per mile. A simpler way to say this would be:

$30 per day plus $0.50 per mile is $360.

If you look at this sentence and the original problem, you can see that they basically say the same thing: It cost Jada $30 per day and $0.50 per mile, and her total cost was $360. The shorter version will be easier to translate into a mathematical expression.

Let’s start by translating $30 per day. To calculate the cost of something that costs a certain amount per day, you’d multiply the per-day cost by the number of days—in other words, 30 per day could be written as 30 ⋅days, or 30 times the number of days. (Not sure why you’d translate it this way? Check out our lesson on writing algebraic expressions.)

$30 per day and $.50 per mile is $360

$30 ⋅ day + $.50 ⋅ mile = $360

As you can see, there were a few other words we could translate into operators, so and $.50 became + $.50, $.50 per mile became $.50 ⋅ mile, and is became =.

Next, we’ll add in the numbers and variables we already know. We already know the number of days Jada drove, 2, so we can replace that. We’ve also already said we’ll use m to represent the number of miles, so we can replace that too. We should also take the dollar signs off of the money amounts to make them consistent with the other numbers.

$30 ⋅ day + $.50 ⋅ mile = $360

30 ⋅ 2 + .5 ⋅ m = 360

Now we have our expression. All that’s left to do is solve it.

Step 4: Solve the problem.

This problem will take a few steps to solve. (If you’re not sure how to do the math in this section, you might want to review our lesson on simplifying expressions.) First, let’s simplify the expression as much as possible. We can multiply 30 and 2, so let’s go ahead and do that. We can also write .5 ⋅ m as 0.5m.

30 ⋅ 2 + .5 ⋅ m = 360

60 + .5m = 360

Next, we need to do what we can to get the m alone on the left side of the equals sign. Once we do that, we’ll know what m is equal to—in other words, it will let us know the number of miles in our word problem.

We can start by getting rid of the 60 on the left side by subtracting it from both sides.

The only thing left to get rid of is .5. Since it’s being multiplied with m, we’ll do the reverse and divide both sides of the equation with it.

.5m / .5 is m and 300 / 0.50 is 600, so m = 600. In other words, the answer to our problem is 600—we now know Jada drove 600 miles.

Step 5: Check the problem.

To make sure we solved the problem correctly, we should check our work. To do this, we can use the answer we just got—600—and calculate backward to find another of the quantities in our problem. In other words, if our answer for Jada’s distance is correct, we should be able to use it to work backward and find another value, like the total cost. Let’s take another look at the problem.

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

According to the problem, the van costs $30 per day and $0.50 per mile. If Jada really did drive 600 miles in 2 days, she could calculate the cost like this:

$30 per day and $0.50 per mile

30 ⋅ day + .5 ⋅ mile

30 ⋅ 2 + .5 ⋅ 600

60 + 300

360

According to our math, the van would cost $360, which is exactly what the problem says. This means our solution was correct. We’re done!

While some word problems will be more complicated than others, you can use these basic steps to approach any word problem. On the next page, you can try it for yourself.

Practice!

Let’s practice with a couple more problems. You can solve these problems the same way we solved the first one—just follow the problem-solving steps we covered earlier. For your reference, these steps are:

1. Read through the problem carefully, and figure out what it’s about.
2. Represent unknown numbers with variables.
3. Translate the rest of the problem into a mathematical expression.
4. Solve the problem.
5. Check your work.

If you get stuck, you might want to review the problem on page 1. You can also take a look at our lesson on writing algebraic expressions for some tips on translating written words into math.

Problem 1

Try completing this problem on your own. When you’re done, move on to the next page to check your answer and see an explanation of the steps.

A single ticket to the fair costs $8. A family pass costs $25 more than half of that. How much does a family pass cost?

Problem 2

Here’s another problem to do on your own. As with the last problem, you can find the answer and explanation to this one on the next page.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

Problem 1 Answer

Here’s Problem 1:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

Answer: $29

Let’s solve this problem step by step. We’ll solve it the same way we solved the problem on page 1.

Step 1: Read through the problem carefully

The first in solving any word problem is to find out what question the problem is asking you to solve and identify the information that will help you solve it. Let’s look at the problem again. The question is right there in plain sight:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

So is the information we’ll need to answer the question:

• A single ticket costs $8.
• The family pass costs $25 more than half the price of the single ticket.

Step 2: Represent the unknown numbers with variables

The unknown number in this problem is the cost of the family pass. We’ll represent it with the variable f.

Step 3: Translate the rest of the problem

Let’s look at the problem again. This time, the important facts are highlighted.

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

In other words, we could say that the cost of a family pass equals half of $8, plus $25. To turn this into a problem we can solve, we’ll have to translate it into math. Here’s how:

1. First, replace the cost of a family pass with our variable f.

f equals half of $8 plus $25

2. Next, take out the dollar signs and replace words like plus and equals with operators.

f = half of 8 + 25

3. Finally, translate the rest of the problem. Half of can be written as 1/2 times, or 1/2 ⋅ :

f = 1/2 ⋅ 8 + 25

Step 4: Solve the problem

Now all we have to do is solve our problem. Like with any problem, we can solve this one by following the order of operations.

1. f is already alone on the left side of the equation, so all we have to do is calculate the right side.

f = 1/2 ⋅ 8 + 25

2. First, multiply 1/2 by 8. 1/2 ⋅ 8 is 4.

f = 4 + 25

3. Next, add 4 and 25. 4 + 25 equals 29 .

f = 29

That’s it! f is equal to 29. In other words, the cost of a family pass is $29.

Step 5: Check your work

Finally, let’s check our work by working backward from our answer. In this case, we should be able to correctly calculate the cost of a single ticket by using the cost we calculated for the family pass. Let’s look at the original problem again.

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

We calculated that a family pass costs $29. Our problem says the pass costs $25 more than half the cost of a single ticket. In other words, half the cost of a single ticket will be $25 less than $29.

1. We could translate this into this equation, with s standing for the cost of a single ticket.

1/2s = 29 — 25

2. Let’s work on the right side first. 29 — 25 is 4.

1/2s = 4

3. To find the value of s, we have to get it alone on the left side of the equation. This means getting rid of 1/2. To do this, we’ll multiply each side by the inverse of 1/2: 2.

s = 8

According to our math, s = 8. In other words, if the family pass costs $29, the single ticket will cost $8. Looking at our original problem, that’s correct!

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

So now we’re sure about the answer to our problem: The cost of a family pass is $29.

Problem 2 Answer

Here’s Problem 2:

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

Answer: $70

Let’s go through this problem one step at a time.

Step 1: Read through the problem carefully

Start by asking what question the problem is asking you to solve and identifying the information that will help you solve it. What’s the question here?

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

To solve the problem, you’ll have to find out how much money Mo gave to charity. All the important information you need is in the problem:

• The amount Flor donated is three times as much the amount Mo donated
• Flor and Mo’s donations add up to $280 total

Step 2: Represent the unknown numbers with variables

The unknown number we’re trying to identify in this problem is Mo’s donation. We’ll represent it with the variable m.

Step 3: Translate the rest of the problem

Here’s the problem again. This time, the important facts are highlighted.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

The important facts of the problem could also be expressed this way:

Mo’s donation plus Flor’s donation equals $280

Because we know that Flor’s donation is three times as much as Mo’s donation, we could go even further and say:

Mo’s donation plus three times Mo’s donation equals $280

We can translate this into a math problem in only a few steps. Here’s how:

1. Because we’ve already said we’ll represent the amount of Mo’s donation with the variable m, let’s start by replacing Mo’s donation with m.

m plus three times m equals $280

2. Next, we can put in mathematical operators in place of certain words. We’ll also take out the dollar sign.

m + three times m = 280

3. Finally, let’s write three times mathematically. Three times m can also be written as 3 ⋅ m, or just 3m.

m + 3m = 280

Step 4: Solve the problem

It will only take a few steps to solve this problem.

1. To get the correct answer, we’ll have to get m alone on one side of the equation.

m + 3m = 280

2. To start, let’s add m and 3m. That’s 4m.

4m = 280

3. We can get rid of the 4 next to the m by dividing both sides by 4. 4m / 4 is m, and 280 / 4 is 70.

m = 70.

We’ve got our answer: m = 70. In other words, Mo donated $70.

Step 5: Check your work

The answer to our problem is $70, but we should check just to be sure. Let’s look at our problem again.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

If our answer is correct, $70 and three times $70 should add up to $280.

1. We can write our new equation like this:

70 + 3 ⋅ 70 = 280

2. The order of operations calls for us to multiply first. 3 ⋅ 70 is 210.

70 + 210 = 280

3. The last step is to add 70 and 210. 70 plus 210 equals 280.

280 = 280

280 is the combined cost of the tickets in our original problem. Our answer is correct: Mo gave $70 to charity.

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