A compact guide for solving word problems in Algebra

1. Mathematics Reviewer
WORD PROBLEMS IN ALGEBRA
Solving Word Problems
George Polya’s Four-step Problem-Solving 2. Check each step of the plan as you proceed. 5. If the first number is x and another
Process This may be intuitive checking or a formal number is k less than the first, then the
As part of his work on problem solving, Polya proof of each step. other number is x  k .
developed a four-step problem-solving process 3. Keep an accurate record of your work. 6. Consecutive integers: If x is the first
similar to the following: D. Looking Back integer, then x  1 is the 2nd, x  2 is the
A. Understanding the Problem 1. Check the results in the original problem. In 3rd, etc.
1. Can you state the problem in your own some cases, this will require a proof. 7. Consecutive even/odd integers: If x is the
words? 2. Interpret the solution in terms of the original first even/odd integer, then x  2 is the
2. What are you trying to find or do? problem. Does your answer make sense? Is 2nd even/odd integer, x  4 is the 2nd
3. What are the unknowns? it reasonable? even/odd integer, etc.
4. What information do you obtain from the 3. Determine whether there is another method 8. Digits. A two digit number can be written
problem? of finding the solution. in the from 10T  U , where T is the ten’s
5. What information, if any, is missing or not 4. If possible, determine other related or more digit and U is the unit’s digit. A three-
needed? general problems for which the techniques digit number can be written in the form
B. Devising a Plan will work.
100H  10T  U , where H is the
The following list of strategies, although not (Source:
hundred’s digit.
exhaustive, is very useful: http://www.drkhamsi.com/classe/polya.html)
N.B. If the digit of a two-digit number
1. Look for a pattern.
10T  U is REVERSED, the new number
2. Examine related problems and determine if Solution Strategies & Tips for Particular Word
the same technique can be applied. becomes 10U  T . Reversing a three-digit
Problems
3. Examine a simpler or special case of the number means reading the number backwards;
problem to gain insight into the solution of A. Representations for Number Problems i.e., 100H  10T  U becomes 100U  10T  H .
the original problem. 1. If the sum of two numbers is s and x is one
4. Make a table. number, then the other number is s  x . B. Age Problems
5. Make a diagram. 2. If the difference between two numbers is d If a is the present age of person A, then a  y
6. Write an equation. and x is the smaller number, then the larger is the age of A y years ago, while a  y is the
7. Use a guess and check. number is d  x . age of A y years from now (or hence).
8. Work backward. 3. If the first number is x and another is k times Age table for age problems:
9. Identify a sub goal. the first, then the other number is kx . Age some
C. Carrying out the Plan 4. If the first number is x and another number Persons Age now years ago or
1. Implement the strategy in Step 2 and is k more than the first, then the other from now
perform any necessary actions or number is x  k .
computations.

2. C. Mixture/Collection Problems meeting point Note: If B undoes what A does, the “+” becomes a
A dA B
For mixtures, the general equation is “” between 1/a and 1/b in #s 2, 3, and 4.
 amount of   % concen-   amount of  dB
 × =  dapart  d A  d B The situations described above can be extended for
 solution   tration   substance  3 or more persons doing a job.
while for those involving prices or money,
4. For bodies traveling in air/current:
 price or cost   totol cost 
 quantity  ×  =   rate against the   rate in still   rate of wind 
F. Some Formulas in Geometry
 
 per unit   or price  1. Perimeter Formulas
 
 wind/current   wind/water   or current  a. square: P  4s, s  length of one side
Types of No. of Amount/price/percent Total  rate with the   rate in still   rate of wind  b. rectangle: P  2l  2w (l = length, w
quantities units per unit amount     = width
 wind/current   wind/water   or current 
c. triangle: P  a  b  c (a, b, and c are
E. Work Problems the sides of the triangle)
TOTAL = sum of all entries at last column If person A can finish a job alone in a time units d. circle (circumference): C  2r (r =
and B can finish the same job alone in b time radius)
D. Motion Problems units, then 2. Area formulas
 distance    rate  time  1 a. square: A  s 2
General Formula: 1. A can finish of the job in 1 time unit,
d  rt a b. rectangle: A  lw
Situations: 1 c. triangle: A  1 bh
2
1. Overtaking = equal distances covered while B can finish of the job in 1 time
b d. circle: A  r 2
Starting point unit. So after k time units, A and B can 3. Volume Formulas
dA A
1 1 a. cube: V  s3
finish k   and k   of the job,
a b b. rectangular solid/prism: V  lwh
d B  d A  rAt0
B respectively. c. sphere: V  3 r 3
4
d A  dB
2. Together, A and B can finish   of
1 1 1 d. right circular cylinder: V  r 2 h
2. Bodies moving in opposite directions = the a b x e. right circular cone: V  1 r 2 h
3
distance apart is the sum of the distances the job in 1 time unit, where x is the no. of
traveled by each body time units that A and B can finish the job
together. Anything you can solve in
Starting point 3. Suppose that A started working the job five minutes should not
A dA B
along in the first p time units then B joined
be considered a problem.
dB for x more time units until they finished the
-- George Polya
dapart  d A  d B 1 1 1
job, then p    x     1 where 1
 a a b
3. Bodies moving toward each other = the sum of represents the whole job.
the distance between the origin of 1st body and 4. If A and B were doing the job together for
2nd body to the meeting point is equal to the sum the first q time units until B left, letting A
of the distances between the origin of each point finish the job alone in x more time units, © gjnabueg 2 August 2003
1 1 1 Revised 19 July 2009
then q     x    1 . All rights reserved 
a b a