In math word problems what is and

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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 x,y,z are symbols representing real numbers — then a relevant solution to the word problem would, given the input {displaystyle (xcdot y)/zmathrel {overset {?}{=}} (x/z)cdot y}, produce the output EQUAL, and similarly produce NOT_EQUAL from {displaystyle (xcdot y)/zmathrel {overset {?}{=}} (x/x)cdot y}.

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 {displaystyle xcdot ycdot z^{-1}} is the normal form of {displaystyle (xcdot y)/z}, {displaystyle (x/z)cdot y}, and {displaystyle (y/z)cdot x}, 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 t_{1},t_{2} containing variables have instances that are equal, or in other words whether the equation t_{1}=t_{2} has any solutions. As a common example, {displaystyle 2+3mathrel {overset {?}{=}} 8+(-3)} is a word problem in the integer group ℤ,
while {displaystyle 2+xmathrel {overset {?}{=}} 8+(-x)} is a unification problem in the same group; since the former terms happen to be equal in ℤ, the latter problem has the substitution {xmapsto 3} 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 T:=(Sigma ,R) and two words (strings) u,vin Sigma ^{*}, can u be transformed into v by applying rules from R? 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 {displaystyle langle Smid {mathcal {R}}rangle } 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]

Solving the word problem: deciding if x{stackrel {*}{leftrightarrow }}y usually requires heuristic search (red, green), while deciding xdownarrow =ydownarrow is straightforward (grey).

The word problem for an abstract rewriting system (ARS) is quite succinct: given objects x and y are they equivalent under {stackrel {*}{leftrightarrow }}?[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 {stackrel {*}{leftrightarrow }} 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 xz ~ xz∧(xy)

xz∧(xy) ~ xz
by 5. since xz ~ xz
by 1. since xz = xz
 
 
xz ~ xz∧(xy)
by 7. since xz ~ xz and xz ~ xy
by 1. since xz = xz by 6. since xz ~ 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 = w1w2 and both w1~ v and w2~ v hold,
  5.   w = w1w2 and either w1~ v or w2~ v holds,
  6.   v = v1v2 and either w~ v1 or w~ v2 holds,
  7.   v = v1v2 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 xz and xz∧(xy) 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

{displaystyle ((a^{-1}cdot a)cdot (bcdot b^{-1}))^{-1}mathrel {overset {R2}{rightsquigarrow }} (1cdot (bcdot b^{-1}))^{-1}mathrel {overset {R13}{rightsquigarrow }} (1cdot 1)^{-1}mathrel {overset {R1}{rightsquigarrow }} 1^{-1}mathrel {overset {R8}{rightsquigarrow }} 1}, and
{displaystyle bcdot ((acdot b)^{-1}cdot a)mathrel {overset {R17}{rightsquigarrow }} bcdot ((b^{-1}cdot a^{-1})cdot a)mathrel {overset {R3}{rightsquigarrow }} bcdot (b^{-1}cdot (a^{-1}cdot a))mathrel {overset {R2}{rightsquigarrow }} bcdot (b^{-1}cdot 1)mathrel {overset {R11}{rightsquigarrow }} bcdot b^{-1}mathrel {overset {R13}{rightsquigarrow }} 1}

share the same normal form, viz. 1; therefore both terms are equal in every group.
As another example, the term 1cdot (acdot b) and bcdot (1cdot a) has the normal form acdot b and bcdot a, 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 1cdot x =x
A2 x^{{-1}}cdot x =1
A3     (xcdot y)cdot z =xcdot (ycdot z)
Term rewrite system obtained from Knuth–Bendix completion

R1 1cdot x {displaystyle rightsquigarrow x}
R2 x^{{-1}}cdot x {displaystyle rightsquigarrow 1}
R3 (xcdot y)cdot z rightsquigarrow xcdot (ycdot z)
R4 {displaystyle x^{-1}cdot (xcdot y)} {displaystyle rightsquigarrow y}
R8 {displaystyle 1^{-1}} {displaystyle rightsquigarrow 1}
R11 {displaystyle xcdot 1} {displaystyle rightsquigarrow x}
R12 {displaystyle (x^{-1})^{-1}} {displaystyle rightsquigarrow x}
R13 {displaystyle xcdot x^{-1}} {displaystyle rightsquigarrow 1}
R14 {displaystyle xcdot (x^{-1}cdot y)} {displaystyle rightsquigarrow y}
R17     {displaystyle (xcdot y)^{-1}} {displaystyle rightsquigarrow y^{-1}cdot x^{-1}}

See also[edit]

  • Conjugacy problem
  • Group isomorphism problem

References[edit]

  1. ^ a b c d Evans, Trevor (1978). «Word problems». Bulletin of the American Mathematical Society. 84 (5): 790. doi:10.1090/S0002-9904-1978-14516-9.
  2. ^ Cohen, Joel S. (2002). Computer algebra and symbolic computation: elementary algorithms. Natick, Mass.: A K Peters. pp. 90–92. ISBN 1568811586.
  3. ^ a b c d e f g Miller, Charles F. (2014). Downey, Rod (ed.). «Turing machines to word problems» (PDF). Turing’s Legacy: 330. doi:10.1017/CBO9781107338579.010. hdl:11343/51723. ISBN 9781107338579. Retrieved 6 December 2021.
  4. ^ Stillwell, John (1982). «The word problem and the isomorphism problem for groups». Bulletin of the American Mathematical Society. 6 (1): 33–56. doi:10.1090/S0273-0979-1982-14963-1.
  5. ^ Müller-Stach, Stefan (12 September 2021). «Max Dehn, Axel Thue, and the Undecidable». p. 13. arXiv:1703.09750 [math.HO].
  6. ^ Steinby, Magnus; Thomas, Wolfgang (2000). «Trees and term rewriting in 1910: on a paper by Axel Thue». Bulletin of the European Association for Theoretical Computer Science. 72: 256–269. CiteSeerX 10.1.1.32.8993. MR 1798015.
  7. ^ Dehn, Max (1911). «Über unendliche diskontinuierliche Gruppen». Mathematische Annalen. 71 (1): 116–144. doi:10.1007/BF01456932. ISSN 0025-5831. MR 1511645. S2CID 123478582.
  8. ^ Dehn, Max (1912). «Transformation der Kurven auf zweiseitigen Flächen». Mathematische Annalen. 72 (3): 413–421. doi:10.1007/BF01456725. ISSN 0025-5831. MR 1511705. S2CID 122988176.
  9. ^ Greendlinger, Martin (June 1959). «Dehn’s algorithm for the word problem». Communications on Pure and Applied Mathematics. 13 (1): 67–83. doi:10.1002/cpa.3160130108.
  10. ^ Lyndon, Roger C. (September 1966). «On Dehn’s algorithm». Mathematische Annalen. 166 (3): 208–228. doi:10.1007/BF01361168. hdl:2027.42/46211. S2CID 36469569.
  11. ^ Schupp, Paul E. (June 1968). «On Dehn’s algorithm and the conjugacy problem». Mathematische Annalen. 178 (2): 119–130. doi:10.1007/BF01350654. S2CID 120429853.
  12. ^ Power, James F. (27 August 2013). «Thue’s 1914 paper: a translation». arXiv:1308.5858 [cs.FL].
  13. ^ See History of the Church–Turing thesis. The dates are based on On Formally Undecidable Propositions of Principia Mathematica and Related Systems and Systems of Logic Based on Ordinals.
  14. ^ Post, Emil L. (March 1947). «Recursive Unsolvability of a problem of Thue» (PDF). Journal of Symbolic Logic. 12 (1): 1–11. doi:10.2307/2267170. JSTOR 2267170. S2CID 30320278. Retrieved 6 December 2021.
  15. ^ Mostowski, Andrzej (September 1951). «A. Markov. Névožmoinost’ nékotoryh algoritmov v téorii associativnyh sistém (Impossibility of certain algorithms in the theory of associative systems). Doklady Akadémii Nauk SSSR, vol. 77 (1951), pp. 19–20». Journal of Symbolic Logic. 16 (3): 215. doi:10.2307/2266407. JSTOR 2266407.
  16. ^ Turing, A. M. (September 1950). «The Word Problem in Semi-Groups With Cancellation». The Annals of Mathematics. 52 (2): 491–505. doi:10.2307/1969481. JSTOR 1969481.
  17. ^ Novikov, P. S. (1955). «On the algorithmic unsolvability of the word problem in group theory». Proceedings of the Steklov Institute of Mathematics (in Russian). 44: 1–143. Zbl 0068.01301.
  18. ^ Boone, William W. (1954). «Certain Simple, Unsolvable Problems of Group Theory. I». Indagationes Mathematicae (Proceedings). 57: 231–237. doi:10.1016/S1385-7258(54)50033-8.
  19. ^ Boone, William W. (1957). «Certain Simple, Unsolvable Problems of Group Theory. VI». Indagationes Mathematicae (Proceedings). 60: 227–232. doi:10.1016/S1385-7258(57)50030-9.
  20. ^ Britton, J. L. (October 1958). «The Word Problem for Groups». Proceedings of the London Mathematical Society. s3-8 (4): 493–506. doi:10.1112/plms/s3-8.4.493.
  21. ^ Boone, William W. (1958). «The word problem» (PDF). Proceedings of the National Academy of Sciences. 44 (10): 1061–1065. Bibcode:1958PNAS…44.1061B. doi:10.1073/pnas.44.10.1061. PMC 528693. PMID 16590307. Zbl 0086.24701.
  22. ^ Boone, William W. (September 1959). «The Word Problem». The Annals of Mathematics. 70 (2): 207–265. doi:10.2307/1970103. JSTOR 1970103.
  23. ^ Higman, G. (8 August 1961). «Subgroups of finitely presented groups». Proceedings of the Royal Society of London. Series A. Mathematical and Physical Sciences. 262 (1311): 455–475. Bibcode:1961RSPSA.262..455H. doi:10.1098/rspa.1961.0132. S2CID 120100270.
  24. ^ Britton, John L. (January 1963). «The Word Problem». The Annals of Mathematics. 77 (1): 16–32. doi:10.2307/1970200. JSTOR 1970200.
  25. ^ Simpson, Stephen G. (18 May 2005). «A Slick Proof of the Unsolvability of the Word Problem for Finitely Presented Groups» (PDF). Retrieved 6 December 2021.
  26. ^ «Subgroups of finitely presented groups». Mathematics of the USSR-Sbornik. 103 (145): 147–236. 13 February 1977. doi:10.1070/SM1977v032n02ABEH002376.
  27. ^ a b Matiyasevich, Yuri; Sénizergues, Géraud (January 2005). «Decision problems for semi-Thue systems with a few rules». Theoretical Computer Science. 330 (1): 145–169. doi:10.1016/j.tcs.2004.09.016.
  28. ^ Davis, Martin (1978). «What is a Computation?» (PDF). Mathematics Today Twelve Informal Essays: 257–259. doi:10.1007/978-1-4613-9435-8_10. ISBN 978-1-4613-9437-2. Retrieved 5 December 2021.
  29. ^ a b Baader, Franz; Nipkow, Tobias (5 August 1999). Term Rewriting and All That. Cambridge University Press. pp. 59–60. ISBN 978-0-521-77920-3.
  30. ^
    • Matiyasevich, Yu. V. (1967). «Простые примеры неразрешимых ассоциативных исчислений» [Simple examples of undecidable associative calculi]. Doklady Akademii Nauk SSSR (in Russian). 173 (6): 1264–1266. ISSN 0869-5652.
    • Matiyasevich, Yu. V. (1967). «Simple examples of undecidable associative calculi». Soviet Mathematics. 8 (2): 555–557. ISSN 0197-6788.

  31. ^ Novikov, P. S. (1955). «On the algorithmic unsolvability of the word problem in group theory». Trudy Mat. Inst. Steklov (in Russian). 44: 1–143.
  32. ^ Statman, Rick (2000). «On the Word Problem for Combinators». Rewriting Techniques and Applications. Lecture Notes in Computer Science. 1833: 203–213. doi:10.1007/10721975_14. ISBN 978-3-540-67778-9.
  33. ^ Beke, Tibor (May 2011). «Categorification, term rewriting and the Knuth–Bendix procedure». Journal of Pure and Applied Algebra. 215 (5): 730. doi:10.1016/j.jpaa.2010.06.019.
  34. ^ Peter T. Johnstone, Stone Spaces, (1982) Cambridge University Press, Cambridge, ISBN 0-521-23893-5. (See chapter 1, paragraph 4.11)
  35. ^ Whitman, Philip M. (January 1941). «Free Lattices». The Annals of Mathematics. 42 (1): 325–329. doi:10.2307/1969001. JSTOR 1969001.
  36. ^ Whitman, Philip M. (1942). «Free Lattices II». Annals of Mathematics. 43 (1): 104–115. doi:10.2307/1968883. JSTOR 1968883.
  37. ^ K. H. Bläsius and H.-J. Bürckert, ed. (1992). Deduktionsssysteme. Oldenbourg. p. 291.; here: p.126, 134
  38. ^ Apply rules in any order to a term, as long as possible; the result doesn’t depend on the order; it is the term’s normal form.

Solving Word Problems in Mathematics

     What Is a Word Problem? (And How to Solve It!)

     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. 

     Students can start visualizing simple or more complex problems by creating relevant images, from concrete (like drawings of putting away cookies from a jar) to more abstract (like tape diagrams). It can also help students clarify the operations they need to carry out. (next step!)
     4. Determine the operation(s): next, students should determine the operation or operations they need to perform. Is it addition, subtraction, multiplication, division? What needs to be done? 

     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;

          – Division: each, equal pieces, split, share, per, out of, average.

     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. 

     Another way to determine the operation is to search for certain situations, Jennifer Findley suggests. She has a great resource that lists various situations you might find in the most common word problems and the explanation of which operation applies to each situation. 

     5. Make a math sentence: next, students should try to translate the word problem and drawings into a math or number sentence. This means students might write a sentence such as 3 + 8 =. 

     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.

     For example, during the first steps, simple wording and animation help students realize what action the problem represents and find the connection between these actions and key words like “take away” and “left” that may signal them. 

     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. 


     These are the common steps that may help students overcome initial feelings of anxiety and fear of difficulty of the task they are given. Together with a teacher, they can master these foundational skills and build their confidence toward solving word problems. And Happy Numbers can facilitate this growth, providing varieties of engaging exercises and challenging word problems! 


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You can solve many real world problems with the help of math. In order to familiarize students with these kinds of problems, teachers include word problems in their math curriculum. However, word problems can present a real challenge if you don’t know how to break them down and find the numbers underneath the story. Solving word problems is an art of transforming the words and sentences into mathematical expressions and then applying conventional algebraic techniques to solve the problem.

  1. Image titled Solve Word Problems in Algebra Step 1

    1

    Read the problem carefully.[1]
    A common setback when trying to solve algebra word problems is assuming what the question is asking before you read the entire problem. In order to be successful in solving a word problem, you need to read the whole problem in order to assess what information is provided, and what information is missing.[2]

  2. Image titled Solve Word Problems in Algebra Step 2

    2

    Determine what you are asked to find. In many problems, what you are asked to find is presented in the last sentence. This is not always true, however, so you need to read the entire problem carefully.[3]
    Write down what you need to find, or else underline it in the problem, so that you do not forget what your final answer means.[4]
    In an algebra word problem, you will likely be asked to find a certain value, or you may be asked to find an equation that represents a value.

    • For example, you might have the following problem: Jane went to a book shop and bought a book. While at the store Jane found a second interesting book and bought it for $80. The price of the second book was $10 less than three times the price of he first book. What was the price of the first book?
    • In this problem, you are asked to find the price of the first book Jane purchased.

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    3

    Summarize what you know, and what you need to know. Likely, the information you need to know is the same as what information you are asked to find. You also need to assess what information you already know. Again, underline or write out this information, so you can keep track of all the parts of the problem. For problems involving geometry, it is often helpful to draw a sketch at this point.[5]

    • For example, you know that Jane bought two books. You know that the second book was $80. You also know that the second book cost $10 less than 3 times the price of the first book. You don’t know the price of the first book.
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    Assign variables to the unknown quantities. If you are being asked to find a certain value, you will likely only have one variable. If, however, you are asked to find an equation, you will likely have multiple variables. No matter how many variables you have, you should list each one, and indicate what they are equal to.[6]

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    5

    Look for keywords.[7]
    Word problems are full of keywords that give you clues about what operations to use. Locating and interpreting these keywords can help you translate the words into algebra.[8]

    • Multiplication keywords include times, of, and factor.[9]
    • Division keywords include per, out of, and percent.[10]
    • Addition keywords include some, more, and together.[11]
    • Subtraction keywords include difference, fewer, and decreased.[12]
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  1. Image titled Solve Word Problems in Algebra Step 6

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    Write an equation. Use the information you learn from the problem, including keywords, to write an algebraic description of the story.[13]

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    2

    Solve an equation for one variable. If you have only one unknown in your word problem, isolate the variable in your equation and find which number it is equal to. Use the normal rules of algebra to isolate the variable. Remember that you need to keep the equation balanced. This means that whatever you do to one side of the equation, you must also do to the other side.[14]

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    Solve an equation with multiple variables. If you have more than one unknown in your word problem, you need to make sure you combine like terms to simplify your equation.

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    4

    Interpret your answer. Look back to your list of variables and unknown information. This will remind you what you were trying to solve. Write a statement indicating what your answer means.[15]

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  1. Image titled Solve Word Problems in Algebra Step 10

    1

    Solve the following problem. This problem has more than one unknown value, so its equation will have multiple variables. This means you cannot solve for a specific numerical value of a variable. Instead, you will solve to find an equation that describes a variable.

    • Robyn and Billy run a lemonade stand. They are giving all the money that they make to a cat shelter. They will combine their profits from selling lemonade with their tips. They sell cups of lemonade for 75 cents. Their mom and dad have agreed to double whatever amount they receive in tips. Write an equation that describes the amount of money Robyn and Billy will give to the shelter.
  2. Image titled Solve Word Problems in Algebra Step 11

    2

    Read the problem carefully and determine what you are asked to find.[16]
    You are asked to find how much money Robyn and Billy will give to the cat shelter.

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    3

    Summarize what you know, and what you need to know. You know that Robyn and Billy will make money from selling cups of lemonade and from getting tips. You know that they will sell each cup for 75 cents. You also know that their mom and dad will double the amount they make in tips. You don’t know how many cups of lemonade they sell, or how much tip money they get.

  4. Image titled Solve Word Problems in Algebra Step 13

    4

    Assign variables to the unknown quantities. Since you have three unknowns, you will have three variables. Let x equal the amount of money they will give to the shelter. Let c equal the number of cups they sell. Let t equal the number of dollars they make in tips.

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    5

    Look for keywords. Since they will “combine” their profits and tips, you know addition will be involved. Since their mom and dad will “double” their tips, you know you need to multiply their tips by a factor of 2.

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    Write an equation. Since you are writing an equation that describes the amount of money they will give to the shelter, the variable x will be alone on one side of the equation.

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    Interpret your answer. The variable x equals the amount of money Robyn and Billy will donate to the cat shelter. So, the amount they donate can be found by multiplying the number of cups of lemonade they sell by .75, and adding this product to the product of their tip money and 2.

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Add New Question

  • Question

    How do you solve an algebra word problem?

    Daron Cam

    Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary’s College.

    Daron Cam

    Academic Tutor

    Expert Answer

    Carefully read the problem and figure out what information you’re given and what that information should be used for. Once you know what you need to do with the values they’ve given you, the problem should be a lot easier to solve.

  • Question

    If Deborah and Colin have $150 between them, and Deborah has $27 more than Colin, how much money does Deborah have?

    Donagan

    Let x = Deborah’s money. Then (x — 27) = Colin’s money. That means that (x) + (x — 27) = 150. Combining terms: 2x — 27 = 150. Adding 27 to both sides: 2x = 177. So x = 88.50, and (x — 27) = 61.50. Deborah has $88.50, and Colin has $61.50, which together add up to $150.

  • Question

    Karl is twice as old Bob. Nine years ago, Karl was three times as old as Bob. How old is each now?

    Donagan

    Let x be Bob’s current age. Then Karl’s current age is 2x. Nine years ago Bob’s age was x-9, and Karl’s age was 2x-9. We’re told that nine years ago Karl’s age (2x-9) was three times Bob’s age (x-9). Therefore, 2x-9 = 3(x-9) = 3x-27. Subtract 2x from both sides, and add 27 to both sides: 18 = x. So Bob’s current age is 18, and Karl’s current age is 36, twice Bob’s current age. (Nine years ago Bob would have been 9, and Karl would have been 27, or three times Bob’s age then.)

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  • Word problems can have more than one unknown and more the one variable.

  • The number of variables is always equal to the number of unknowns.

  • While solving word problems you should always read every sentence carefully and try to extract all the numerical information.

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About This Article

Article SummaryX

To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Next, summarize what information you know and what you need to know. Then, assign variables to the unknown quantities. For example, if you know that Jane bought 2 books, and the second book cost $80, which was $10 less than 3 times the price of the first book, assign x to the price of the 1st book. Use this information to write your equation, which is 80 = 3x — 10. To learn how to solve an equation with multiple variables, keep reading!

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This article is for parents who think about how to help with math and support their children. The math word problems below provide a gentle introduction to common math operations for schoolers of different grades.

What are math word problems?

During long-time education, kids face various hurdles that turn into real challenges. Parents shouldn’t leave their youngsters with their problems. They need an adult’s possible help, but what if the parents themselves aren’t good at mathematics? All’s not lost. You can provide your kid with different types of support. Not let a kid burn the midnight oil! Help him/ her to get over the challenges thanks to these captivating math word examples.

Math word problems are short math questions formulated into one or several sentences. They help schoolers to apply their knowledge to real-life scenarios. Besides, this kind of task helps kids to understand this subject better.

Addition for the first and second grades

math word problems for kids

These math examples are perfect for kids that just stepped into primary school. Here you find six easy math problems with answers:

1. Peter has eight apples. Dennis gives Peter three more. How many apples does Peter have in all?

Show answer

Answer: 8 apples + 3 apples = 11 apples.

2. Ann has seven candies. Lack gives her seven candies more. How many candies does Ann have in all?

Show answer

Answer: 7 candies + 7 candies = 14 candies.

3. Walter has two books. Matt has nine books. If Matt gives all his books to Walter, how many books will Walter have?

Show answer

Answer: 2 books + 9 books = 11 books.

4. There are three crayons on the table. Albert puts five more crayons on the table. How many crayons are on the table?

Show answer

Answer: 3 crayons + 5 crayons = 8 crayons.

5. Bill has nine oranges. His friend has one orange. If his friend gives his orange to Bill, how many oranges will Bill have?

Show answer

Answer: 9 oranges + 1 orange = 10 oranges.

6. Jassie has four leaves. Ben has two leaves. Ben gives her all his leaves. How many leaves does Jessie have in all?

Show answer

Answer: 4 leaves + 2 leaves = 6 leaves.

Subtraction for the first and second grades

1. There were three books in total at the book shop. A customer bought one book. How many books are left?

Show answer

Answer: 3 books – 1 book = 2 books.

2. There are five pizzas in total at the pizza shop. Andy bought one pizza. How many pizzas are left?

Show answer

Answer: 5 pizzas – 1 pizza = 4 pizzas.

3. Liza had eleven stickers. She gave one of her stickers to Sarah. How many stickers does Liza have?

Show answer

Answer: 11 stickers – 1 sticker = 10 stickers.

4. Adrianna had ten stones. But then she left two stones. How many stones does Adrianna have?

Show answer

Answer: 10 stones – 2 stones = 8 stones.

5. Mary bought a big bag of candy to share with her friends. There were 20 candies in the bag. Mary gave three candies to Marissa. She also gave three candies to Kayla. How many candies were left?

Show answer

Answer: 20 candies – 3 candies – 3 candies = 14 candies.

6. Betty had a pack of 25 pencil crayons. She gave five to her friend Theresa. She gave three to her friend Mary. How many pencil crayons does Betty have left?

Show answer

Answer: 25 crayons – 5 crayons – 3 crayons = 17 crayons.

Multiplication for the 2nd grade and 3rd grade

See the simple multiplication word problems. Make sure that the kid has a concrete understanding of the meaning of multiplication before.

Bill is having his friends over for the game night. He decided to prepare snacks and games.

1. He makes mini sandwiches. If he has five friends coming over and he made three sandwiches for each of them, how many sandwiches did he make?

Show answer

Answer: 5 x 3 = 15 sandwiches.

2. He also decided to get some juice from fresh oranges. If he used two oranges per glass of juice and made six glasses of juice, how many oranges did he use?

Show answer

Answer: 2 x 6 = 12 oranges.

3. Then Bill prepared the games for his five friends. If each game takes 7 minutes to prepare and he prepared a total of four games, how many minutes did it take for Bill to prepare all the games?

Show answer

Answer: 7 x 4 = 28 minutes.

4. Bill decided to have takeout food as well. If each friend and Bill eat three slices of pizza, how many slices of pizza do they have in total?

Show answer

Answer: 6 (5 friends and Bill) x 3 slices of pizza = 18 slices of pizza.

Mike is having a party at his house to celebrate his birthday. He invited some friends and family.

1. He and his mother prepared cupcakes for dessert. Each box had 8 cupcakes, and they prepared four boxes. How many cupcakes have they prepared in the total?

Show answer

Answer: 8 x 4 = 32 cupcakes.

2. They also baked some cookies. If they baked 6 pans of cookies, and there were 7 cookies per pan, how many cookies did they bake?

Show answer

Answer: 6 x 7 = 42 cookies.

3. Mike planned to serve some cold drinks as well. If they make 7 pitchers of drinks and each pitcher can fill 5 glasses, how many glasses of drinks are they preparing?

Show answer

Answer: 7 x 5 = 35 glasses.

4. At the end of the party, Mike wants to give away some souvenirs to his 6 closest friends. If he gives 2 souvenir items for each friend, how many souvenirs does Mike prepare?

Show answer

Answer: 6 x 2 = 12 souvenirs.

Division: best for 3rd and 4th grades

1. If you have 10 books split evenly into 2 bags, how many books are in each bag?

Show answer

Answer: 10 : 2 = 5 books.

2. You have 40 tickets for the fair. Each ride costs 2 tickets. How many rides can you go on?

3. The school has $20,000 to buy new equipment. If each piece of equipment costs $100, how many pieces can the school buy in total?

Show answer

Answer: $20,000 : $100= 200.

4. Melissa has 2 packs of tennis balls for $10 in total. How much does 1 pack of tennis balls cost?

5. Jack has 25 books. He has a bookshelf with 5 shelves on it. If Jack puts the same number of books on each shelf, how many books will be on each shelf?

6. Matt is having a picnic for his family. He has 36 cookies. There are 6 people in his family. If each person gets the same number of cookies, how many cookies will each person get?

Division with remainders for fourth and fifth grades

1. Sarah sold 35 boxes of cookies. How many cases of ten boxes, plus extra boxes does Sarah need to deliver?

Show answer

Answer: 35 boxes divided by 10 boxes per case = 3 cases and 5 boxes.

2. Candies come in packages of 16. Mat ate 46 candies. How many whole packages of candies did he eat, and how many candies did he leave? 46 candies divided by 16 candies = 2 packages and 2 candies left over.

3. Mary sold 24 boxes of chocolate biscuits. How many cases of ten boxes, plus extra boxes does she need to deliver?

Show answer

Answer: 24 boxes divided by 10 boxes per case = 2 cases and four boxes.

4. Gummy bears come in packages of 25. Suzie and Tom ate 30 gummy bears. How many whole packages did they eat? How many gummy bears did they leave?

Show answer

Answer: 30 divided by 25 = 1 package they have eaten and 20 gummy bears left over.

5. Darel sold 55 ice-creams. How many cases of ten boxes, plus extra boxes does he need to deliver?

Show answer

Answer: 55 boxes divided by 10 boxes per case = 5 cases and 5 boxes.

6. Crackers come in packages of 8. Mat ate 20 crackers. How many whole packages of crackers did he eat, and how many crackers did he leave?

Show answer

Answer: 20 divided by 8 = 2 packages eaten and 4 crackers are left.

Mixed operations for the fifth grade

simple math word problems

These math word problems involve four basic operations: addition, multiplication, subtraction, and division. They suit best for the fifth-grade schoolers.

200 planes are taking off from the airport daily. During the Christmas holidays, the airport is busier — 240 planes are taking off every day from the airport.

1. During the Christmas holidays, how many planes take off from the airport in each hour if the airport opens 12 hours daily?

Show answer

Answer: 240÷12=20 planes take off from this airport each hour during the Christmas holidays.

2. Each plane takes 220 passengers. How many passengers depart from the airport every hour during the Christmas holidays? 20 x 220 = 4400.

Show answer

Answer: 4400 passengers depart from the airport every hour.

3. Compared with a normal day, how many more passengers are departing from the airport in a day during the Christmas holidays?

Show answer

Answer: (240-200) x 220 = 8800 more passengers departing from the airport in a day during the Christmas holidays.

4. During normal days on average 650 passengers are late for their plane daily. During the Christmas holidays, 1300 passengers are late for their plane. That’s why 14 planes couldn’t take off and are delayed. How many more passengers are late for their planes during Christmas week?

Show answer

Answer: 1300 – 650 = 650 more passengers are late for their planes each day during the Christmas holidays.

5. According to the administration’s study, an additional 5 minutes of delay in the overall operation of the airport is caused for every 27 passengers that are late for their flights. What is the delay in the overall operation if there are 732 passengers late for their flights?

Show answer

Answer: 732 ÷ 27 × 5 = 136. There will be a delay of 136 minutes in the overall operation of the airport.

Extra info math problems for the fifth grade

1. Ann has 7 pairs of red socks and 8 pairs of pink socks. Her sister has 12 pairs of white socks. How many pairs of socks does Ann have?

2. Kurt spent 17 minutes doing home tasks. He took a 3-minute snack break. Then he studied for 10 more minutes. How long did Kurt study altogether?

Show answer

Answer: 17 + 10 = 27 minutes.

3. There were 15 spelling words on the test. The first schooler spelled 9 words correctly. Miguel spelled 8 words correctly. How many words did Miguel spell incorrectly?

4. In the morning, Jack gave his friend 2 gummies. His friend ate 1 of them. Later Jack gave his friend 7 more gummies. How many gummies did Jack give his friend in all?

5. Peter wants to buy 2 candy bars. They cost 8 cents, and the gum costs 5 cents. How much will Peter pay?

Finding averages for 5th grade

We need to find averages in many situations in everyday life.

1. The dog slept 8 hours on Monday, 10 hours on Tuesday, and 900 minutes on Wednesday. What was the
average number of hours the dog slept per day?

Show answer

Answer: (8+10+(900:60)) : 3 = 11 hours.

2. Jakarta can get a lot of rain in the rainy season. The rainfall during 6 days was 90 mm, 74 mm, 112 mm, 30 mm, 100 mm, and 44 mm. What was the average daily rainfall during this period?

Show answer

Answer: (90+74+112+30+100+44) : 6 = 75 mm.

3. Mary bought 4 books. The prices of the first 3 books were $30, $15, and $18. The average price she paid for the 4 books was $25 per books. How much did she pay for the 4th books?

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Ordering and number sense for the 5th grade

1. There are 135 pencils, 200 pens, 167 crayons, and 555 books in the bookshop. How would you write these numbers in ascending order?

Show answer

Answer: 135, 167, 200, 555

2. There are five carrots, one cabbage, eleven eggs, and 15 apples in the fridge. How would you write these numbers in descending order?

3. Peter has completed exercises on pages 279, 256, 264, 259, and 192. How would you write these numbers in ascending order?

Show answer

Answer: 192, 256, 259, 264, 279.

4. Mary picked 32 pants, 15 dresses, 26 pairs of socks, 10 purses. Put all these numbers in order.

5. The family bought 12 cans of tuna, 23 potatoes, 11 onions, and 33 pears. Put all these numbers in order.

Fractions for the 6th-8th grades

1. Jannet cooked 12 lemon biscuits for her daughter, Jill. She ate up 4 biscuits. What fraction of lemon biscuits did Jill eat?

Show answer

Answer: 1/3 of the lemon biscuits.

2. Guinet travels a distance of 7 miles to reach her school. The bus covers only 5 miles. Then she has to walk 2 miles to reach the school. What fraction of the distance does Guinet travel by bus?

Show answer

Answer: 5/7 of the distance

3. Bob has 24 pencils in a box. Eighteen pencils have #2 marked on them, and the 6 are marked #3. What fraction of pencils are marked #3?

Show answer

Answer: 1/4 of the pencils.

4. My mother places 15 tulips in a glass vase. It holds 6 yellow tulips and 9 red tulips. What fraction of tulips are red?

Show answer

Answer: 3/5 of the tulips.

5. Bill owns 14 pairs of socks, of which 7 pairs are white, and the rest are brown. What fraction of pairs of socks are brown?

Show answer

Answer: 1/2 of the pairs of socks.

6. Bred spotted a total of 39 birds in an aviary at the Zoo. He counted 18 macaws and 21 cockatoos. What fraction of macaws did Bred spot at the aviary?

Show answer

Answer: 6/13 of the birds.

Decimals for the 6th grade

Write in words the following decimals:

  • 0,004
  • 0,07
  • 2,1
  • 0,725
  • 46,36
  • 2000,19

Show answer

Answer:

  • 0,004 = four thousandths.
  • 0,07 = seven hundredths.
  • 2,1 = two and one tenth.
  • 0,725 = seven hundred twenty five thousandths.
  • 46,36 = foury six and thirty six hundredths.
  • 2000,19 = two thousand and nineteen hundredths.

Comparing and sequencing for the 6th grade

1. The older brother picked 42 apples at the orchard. The younger brother picked only 22 apples. How many more apples did the older brother pick?

Show answer

Answer: 42 – 22 = 20 apples more.

2. There were 16 oranges in a basket and 66 oranges in a barrel. How many fewer oranges were in the basket than were in the barrel?

Show answer

Answer: 66 – 16 = 50 fewer oranges.

3. There were 40 parrots in the flock. Some of them flew away. Then there were 25 parrots in the flock. How many parrots flew away?

Show answer

Answer: 40 – 25 = 15 parrots flew away.

4. One hundred fifty is how much greater than fifty-three?

5. On Monday, the temperature was 13°C. The next day, the temperature dropped by 8 degrees. What was the temperature on Tuesday?

6. Zoie picked 15 dandelions. Her sister picked 22 ones. How many more dandelions did her sister pick than Zoie?

Show answer

Answer: 22-15 = 7 dandelions more.

Time for the 4th grade

1. The bus was scheduled to arrive at 7:10 p.m. However, it was delayed for 45 minutes. What time was it when the bus arrived?

2. My mother starts her 7-hour work at 9:15 a.m. What time does she get off from work?

3. Jack’s walk started at 6:45 p.m. and ended at 7:25 p.m. How long did his walk last?

4. The school closes at 9:00 p.m. Today, the school’s principal left 15 minutes after the office closed, and his secretary left the office 25 minutes after he left. When did the secretary leave work?

5. Suzie arrives at school at 8:20 a.m. How much time does she need to wait before the school opens? The school opens at 8:35 a.m.

6. The class starts at 9:15 a.m.. The first bell will ring 20 minutes before the class starts. When will the first bell ring?

Money word problems for the fourth grade

kids math word problems

1. James had $20. He bought a chocolate bar for $2.30 and a coffee cup for $5.50. How much money did he have left?

Show answer

Answer: $20.00 – $2.30 – $5.50 = $12.20. James had $12.20 left.

2. Coffee mugs cost $1.50 each. How much do 7 coffee mugs cost?

Show answer

Answer: $1.5 x 7 = $10.5.

3. The father gives $32 to his four children to share equally. How much will each of his children get?

4. Each donut costs $1.20. How much do 6 donuts cost?

Show answer

Answer: $1.20 * 6 = $7,2.

5. Bill and Bob went out for takeout food. They bought 4 hamburgers for $10. Fries cost $2 each. How much does one hamburger with fries cost?

Show answer

Answer: $10 ÷ 4 = $2.50. One hamburger costs $2.50. $2.50 + $2.00 = $4.50. One hamburger with fries costs $4.50.

6. A bottle of juice costs $2.80, and a can is $1.50. What would it cost to buy two cans of soft drinks and a bottle of juice?

Show answer

Answer: $1.50 x 2 + $2.80 = $5.80.

Measurement word problems for the 6th grade

The task is to convert the given measures to new units. It best suits the sixth-grade schoolers.

  • 55 yd = ____ in.
  • 43 ft = ____ yd.
  • 31 in = ____ ft.
  • 29 ft = ____ in.
  • 72 in = ____ ft.
  • 13 ft = ____ yd.
  • 54 lb = ____ t.
  • 26 t = ____ lb.
  • 77 t = ____ lb.
  • 98 lb = ____ t.
  • 25 lb = ____ t.
  • 30 t = ____ lb.

Show answer

Answer:

  • 55 yd = 1.980 in
  • 43 ft = 14 yd 1 ft
  • 31 in = 2 ft 7 in
  • 29 ft = 348 in
  • 72 in = 6 ft
  • 13 ft = 4 yd 1 ft.
  • 54 lb = 0,027 t
  • 26 t = 52.000 lb
  • 77 t = 154.000 lb
  • 98 lb = 0,049 t
  • 25 lb = 0?0125 t
  • 30 t = 60.000 lb.

Ratios and percentages for the 6th-8th grades

It is another area that children can find quite difficult. Let’s look at simple examples of how to find percentages and ratios.

1. A chess club has 25 members, of which 13 are males, and the rest are females. What is the ratio of males to all club members?

2. A group has 8 boys and 24 girls. What is the ratio of girls to all children?

3. A pattern has 4 red triangles for every 12 yellow triangles. What is the ratio of red triangles to all triangles?

4. An English club has 21 members, of which 13 are males, and the rest are females. What is the ratio of females to all club members?

5. Dan drew 1 heart, 1 star, and 26 circles. What is the ratio of circles to hearts?

6. Percentages of whole numbers:

  • 50% of 60 = …
  • 100% of 70 = …
  • 90% of 70 = …
  • 20% of 30 = …
  • 40% of 10 = …
  • 70% of 60 = …
  • 100% of 20 = …
  • 80% of 90 = …

Show answer

Answer:

  • 50% of 60 = 30
  • 100% of 70 = 70
  • 90% of 70 = 63
  • 20% of 30 = 6
  • 40% of 10 = 4
  • 70% of 60 = 42
  • 100% of 20 = 20
  • 80% of 90 = 72.

Probability and data relationships for the 8th grade

1. John ‘s probability of winning the game is 60%. What is the probability of John not winning the game?

2. The probability that it will rain is 70%. What is the probability that it won’t rain?

3. There is a pack of 13 cards with numbers from 1 to 13. What is the probability of picking a number 9 from the pack?

4. A bag had 4 red toy cars, 6 white cars, and 7 blue cars. When a car is picked from this bag, what is the probability of it being red or blue?

5. In a class, 22 students like orange juice, and 18 students like milk. What is the probability that a schooler likes juice?

Geometry for the 7th grade

The following task is to write out equations and find the angles. Complementary angles are two angles that sum up to 90 degrees, and supplementary angles are two angles that sum up to 180 degrees.

1. The complement of a 32° angle = …

2. The supplement of a 10° angle = …

3. The complement of a 12° angle = …

4. The supplement of a 104° angle = …

Variables/ equation word problems for the 5th grades

1. The park is 𝑥 miles away from Jack’s home. Jack had to drive to and from the beach with a total distance of 36 miles. How many miles is Jack’s home away from the park?

Show answer

Answer: 2𝑥 = 36 → 𝑥 = 18 miles.

2. Larry bought some biscuits which cost $24. He paid $x and got back $6 of change. Find x.

Show answer

Answer: x = 24 + 6 = $30.

3. Mike played with his children on the beach for 90 minutes. After they played for x minutes, he had to remind them that they would be leaving in 15 minutes. Find x.

Show answer

Answer: x = 90 – 15 = 75 minutes.

4. At 8 a.m., there were x people at the orchard. Later at noon, 27 of the people left the orchard, and there were 30 people left in the orchard. Find x.

Show answer

Answer: x = 30 + 27 = 57 people

Travel time word problems for the 5th-7th grades

1. Tony sprinted 22 miles at 4 miles per hour. How long did Tony sprint?

Show answer

Answer: 22 miles divided by 4 miles per hour = 5.5 hours.

2. Danny walked 15 miles at 3 miles per hour. How long did Danny walk?

Show answer

Answer: 15 miles divided by 3 miles per hour = 5 hours.

3. Roy sprinted 30 miles at 6 miles per hour. How long did Roy sprint?

Show answer

Answer: 30 miles divided by 6 miles per hour = 5 hours.

4. Harry wandered 5 hours to get Pam’s house. It is 20 miles from his house to hers. How fast did Harry go?

Show answer

Answer: 20 miles divided by 5 hours = 4 miles per hour.

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When most people hear “word problems,” they often think of the popular example of trains traveling at different speeds, or unrealistic applications of math. However, word problems represent how most people use math in everyday life, and the SAT includes these problems to test students’ ability to reason logically and solve problems.

For many students, the hardest part of a word problem is figuring out exactly what the word problem is asking of you. Not all word problems are created the same though, so we’re going to break down what you need to know so that you feel confident solving them on the SAT.

Types of Math Word Problems on the SAT

You could divide math word problem in a few different ways. You could do it by topic, for example, such as using the SAT subscores as your guide. Or you could do it by the type of solution the problem requires of you. We’re going to focus on the second way, simply because trying to divide word problems by topic would mean we’d have to cover virtually every math topic the SAT covers, which is way more than we could fit in a single post.

Creating an equation or a formula. In these problems, you’ll be given some information about a scenario and asked to come up with an equation or formula that represents that scenario. In fact, looking for the word “represents” in the question might tip you off that you’re dealing with a problem of this type. These are almost exclusively found in the multiple-choice questions, with different equations or formulas listed as the answer choices.

Solving for a value. This is the most common form of word problem. Based on the information in the word problem, you’ll need to come up with at least one specific value that satisfies the requirements of the problem. Examples include finding “how much” or “how many” of something in the problem, or finding minimums and maximum values.

Defining or interpreting. For these problems, you are often asked to interpret what a variable or a constant means in the situation described, or you might be asked what kinds of conclusions can be drawn from a survey. These types of problems often have no “math” involved, and instead ask you to think logically about a situation.

Tip: Being able to identify problems by type on the SAT is a useful prep exercise and can help you on the day you take the test. You may want to use the Math section from a free practice SAT test and circle all of the word problems. Then, read each word problem and assign it to one of the types. Doing this will help you see the commonalities between different types of questions and will allow you to spend less time on the actual test wondering what type of question you’re dealing with!

How to Approach SAT Math Word Problems

For every SAT Math word problem, there are two things to consider. The first is what we discussed above, which is to figure out what type of problem you’re dealing with to best identify what you need to do. Once you know whether you need to provide a definition, write an equation, or solve for a value, you’ve taken the first step.

The next step will be to do the math that you’ve identified you need to do based on the type of word problem. Let’s take a look at each one:

Write an equation/formula. For problems of this type and for problems that ask you to solve for a specific value, you will often need to understand how to “translate” a word problem into an equation, diagram, or graph that you can work with. Luckily, the SAT is very particular about how it uses words in its word problems, so you can always be sure that certain words are meant to relate to specific mathematical concepts. Here are some of the common phrases used by test makers to clue you in to what kind of math you’re looking for:

Word/phrase Mathematical meaning/symbol
Is, will be, is equal to, is the same as Equals, =
Plus, sum, increased by, added to, received Addition, +
Minus, fewer, difference, decreased by Subtraction, –
Times, product, multiplied by, of Multiplication, x
Divided by, quotient, per, for Division, ÷
More than, greater than Inequality, >
At least, minimum Inequality, ≥
Fewer than, less than Inequality, <
At most, maximum Inequality, ≤
What, how many Variable, x (or any other letter)

Solve for a value. With these problems, you might be given an equation as part of your information, but most likely you have to write an equation or formula and solve for the answer. You may benefit from drawing a diagram on some of these questions, especially if the word problem refers to geometric shapes but doesn’t already include one. This can help you visualize the information you’re looking for (the measure of an angle, for example) and help you focus on the right concepts to find the solution.

When translating a word problem into a diagram, make sure you follow the set-up as described in the word problem to the best of your ability, and don’t be fooled by your own diagram. For example, a problem may mention a right triangle, so you draw one. You may end up drawing a diagram that looks like a 45-45-90 triangle, but the problem is actually dealing with a 30-60-90 triangle. By labeling all of the information on your diagram and making a mental note that your drawings aren’t to scale, you can quickly make sense of most problems.

Defining and interpreting. With these types of problems, you often have the equation or formula given to you, and you just need to explain how a part of it works. If you’re not sure, consider whether you’re being asked about a constant or a variable, and eliminate any choices that don’t correspond to the type of value you’re looking at. You can also test the equation with different numbers to see how the value in question influences the equation.

Some problems involve your knowledge of statistical topics, such as understanding how to draw conclusions from surveys, or what factors affect standard deviation. For many of these problems, you’ll need to review these concepts in order to form the correct conclusion.

Examples of SAT Math Word Problems

The following example problems come from the College Board’s free SAT practice tests.

Example of Creating an Equation/Formula

How to solve: We can tell we need to come up with our own formula based on the answer choices and the word “represents” in the problem’s question. We need to find the formula that represents what the original price of a laptop was in terms of p, what Alma paid for it.

Let’s make x the original price of the laptop. In this case, x would be multiplied by .8 to get to the sales price (representing the 20% discount). The sales prices would then be multiplied by 1.08 to account for sales tax. This means that p=(1.08)(0.8)(x). We need to isolate x, so we need to divide both sides by 1.08 and 0.8. That gives us D as the correct answer.

Alternate method: Many problems requiring you to create your own equation are not too difficult, but they can get tricky when there are several steps involved or several variables. You can substitute a number for the variable you’re solving for to find the solution that way. For example, you can set $100 to be the original price of the laptop. Multiply it by .8 to find the sales price, and then by 1.08 to find the price that Alma paid. Then plug the price Alma paid into the answer choices to find the correct answer.

Example of Solving for a Value

How to solve: The word “maximum” tips us off that we are looking for a specific value and that we’ll be dealing with an inequality (see above). Given the information in the problem, we can write this inequality to start: (weight of driver) + (weight of truck) + (14x, or the weight of the boxes) ≤ 6000.

Because we’re told that the weight of the driver and the truck equals 4,500, we can replace that part of the equation: 4500 + 14x ≤ 6000. Now, all we need to do is solve for x. Subtract 4500 from both sides to get 14x ≤ 1500. Then divide both sides by 14 to get x ≤ 107.14… The answer is thus 107.

Note: Pay close attention to how you need to round when working with inequalities. Although this particular decimal probably wouldn’t have made you think to round up, if the result of the division had been 107.9999 you should still round down in this case because the number you’re looking for needs to be less than that value.

Example of Defining/Interpreting

How to solve: Again, the use of the word “interpretation” lets us know what type of problem we’re dealing with. In this case, 12 is a coefficient for the variables n and h, which are the number of landscapers and the number of hours respectively. We can thus rule out choices B and D, since those answers define 12 as a constant in place of these variables.

Since this equation is used to calculate the price of a job, we can be sure 12 represents a dollar amount as reflected in choices A and C. To figure out how things are affected, we can substitute simple numbers for n and h. When both are equal to 1 (there is one landscaper who works for one hour), then 12nh=12(1)(1)=12. When we increase both of these numbers to two, then 12nh=12(2)(2)=48. We can thus rule out C, because the price increased by more than 12 when we added another hour. Choice A is the best interpretation.

How to solve: This is an interpretation problem dealing with statistical concepts rather than interpreting an equation. To be able to “solve” you’d need to know what factors contribute to a reliable conclusion and see what was missing from this survey set-up. In this case, the correct answer is D.

In order for reliable conclusions to be drawn, a sample must be selected at random from the population to ensure that the sample is representative of the population, which in this case is the entire town. By choosing people at the same restaurant on one day, the researcher is possibly biasing their results by narrowing their sample to “people who like this one restaurant.” Because the sample is not representative, choice D identifies the factor that makes the conclusion unreliable. The other choices are factors that, given what we know from the problem, don’t negatively impact the conclusion.

SAT Math Word Problems Tips

As you may have noticed in the last example, you’ll still need to be familiar with math concepts and understand how they might show up in different situations to really feel confident taking any problem on. Here are a couple of other tips worth mentioning:

For most problems, there is more than one way to solve it. We showed this with the first example, where we provided an alternate method for solving a problem. If you find yourself stumped on a problem (we’ve all been there!), then move on to other problems and return to the one that tripped you up later. You may find that you see what you need to do more clearly.

Read carefully. The SAT test makers put a lot of care into how they write math word problems. Be sure to carefully read the question so that you don’t misread something and solve for the wrong thing. (The test makers may even include the answers that result from these common misreadings in the choices!). It’s difficult to read a question correctly once you’ve misread it. If you’re a fast reader who tends to skip over things, try mouthing the words silently as you read. It will force you to slow down just enough that you won’t miss anything, but not so much that you won’t have time to finish the test.

In general, preparing for the SAT Math section requires a multi-faceted approach. You need to be able to identify the types of questions you’re dealing with, know mathematical concepts and how to use them, and deal with other testing issues like rushing through the end or dealing with test anxiety. To help you get started, check out our free SAT Guide that includes 8 of our best test prep tips!

For more information about preparing for the SAT, check out these posts below:

30 SAT Math Formulas You Need to Know

25 Tips and Tricks for the SAT

5 Common SAT Math Mistakes to Avoid

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