Download Solving Word/Story Problems

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Solving Word Problems
Students (and others) frequently have an aversion to story/word problems; that is, to
problems that are expressed in story form rather than as definite mathematical
equations and expressions that need to be solved. However, most math problems that
are presented to us in the real world environment come to us in the form of prose, or in
the form of a story or verbal description of some kind. Seldom are we given "a straight
up math problem" to solve, where the equation (or equations) is already written out for
us so that all we have to do is apply the mechanical methods of mathematics that we
have learned to solve the problem. To solve problems of this type, we need to learn to
apply some logical thinking to the problem, converting some of the words that describe
the problem into mathematical symbols and/or operations that can be manipulated to
come up with an answer.
To solve problems of this kind, we need to learn to convert the words of the problem
into mathematical symbols, expressions, and operations that represent the problem
description. For example, the word(s) 'is/are' generally translate(s) to the mathematical
'=' sign and logically represent the concept of equality. The words ‘and’ and ‘sum’ imply
or translate to the operation of addition (+). The words ‘double’ or ‘triple’ imply the
operation of multiplying by 2 or 3, respectively. Something that is ‘squared’ or ‘cubed’ is
raised to the second or third power, respectively. Use symbols/letters such as 'x' and ‘y’
to represent quantities in the problem that are not known. And so on. In this way we
can translate the words into one or more mathematical expressions that we can solve.
The following Web sites often addition insights to solving word/story problems:
http://www.purplemath.com/modules/translat.htm
The following are examples of using the approaches described above to solving
word/story problems.
Example 1
Problem:
The sum of two numbers is 15, and the ratio of the first number to the second number is
two-thirds. What are the two numbers?
Solution:
1. We are trying to find two numbers, but we don't know the values of those numbers.
Let x represent the first number and y represent the second number.
2. We are told that the sum of the two numbers is 15. The word 'sum' translates to the
mathematical operation of addition (represented by the  sign), and the word 'is' translates
to the mathematical operation of equality (represented by the  sign). Thus, the first part of
the statement of the problem can be represented mathematiclly as:
x 
y  15  'the sum of two numbers is 15'
3. Next, we are told that the ratio of the first number to the second number is two-thirds.
The word 'ratio' implies the division of the first number by the second number, and the word 'is'
translates to the mathematical operation of equality (represented by the  sign). Thus, the
second part of the statement of the problem can be represented mathematiclly as:
x
y

2
 'the ratio of the first number to the second number is two-thirds'
3
4. So, we have the problem represented by the two equations
x 
y  15 and
x
y

2
3
Thus, we have two equations in two unknowns which we can now solve simultaneously for the
two numbers.
From the second equation, we can write that y 
3
x. Substituting this value for y into
2
the first equation yields:
x 
3
x  15
2
which can be solved for x :
x 
3
x 
2
5
x  15
2
2
x  15    6
5
Then
y 
3
x 
2
3
 6  9
2
So, the first number is 6 and the second number is 9.
Note that the sum of 6 and 9 is 15, and the ratio of 6 to 9 is two-thirds
as required by the problem statement.
Example 2
Problem:
To earn an A in a course, Mary must have an average score of at least 90 on five tests.
If her first four test scores are 92, 86, 79, and 96, what score does she need on the fifth
test to obtain a 90 average?
Solution:
1. The average score is calculated by adding the individual test scores and dividing by the number of test scores.
Let x represent the score Mary needs on the fifth test.
2. Mary's average for the five test is then the sum of the five test scores divided by five, or
Mary's average 
92  86  79  96 
5
x

353 
5
3. To get an A in the course, Mary's average must be 90 or more, so that
Mary's average 
353 
5
x
 90
4. So, we have the problem represented by the inequality
353 
5
x
 90
5. Solving, we have the following sequence of steps:
353 
5
x
353 
x 
353 
x  450
 90
 5 90 
x  450  353
x  97
So, Mary must score 97 or more on the fifth test to get an A in the course.
Note the if Mary scores 97 on the fifth test, her average will be exactly 90:
Mary's average 
Example 3
Problem:
353 
5
x

353  97
5

450
5
 90
x
All edges of a cube are expanding at a rate of 8 centimeters per second. How fast is
the surface area changing when each edge is 6.5 centimeters? How fast is the volume
changing when each edge is 6.5 centimeters?
Solution:
1. Note that for a cube, all edges are of the same length.
Let l represent the length of each edge in centimeters.
2. If l is the length of each edge of the cube, the cube's surface area and volume are given by the following equations:
A  6l 2
V
 l3
- the cube has six sides each of area l 2
- the cube has a volume of l 3
where A is the area of the cube in square centimeters and V is the volume of the cube in cubic centimeters.
3. We are given the rate of change of the length of each edge with respect to time, t,
and the length of each edge; that is, we are given the derivative of l with respect to t ,
dl
,
dt
and the length of each edge, l.
4. To get the rate at which the area is changing, we differentiate the area equation with respect to time:
A  6l 2
dA
dt
 6
d 2
l  
dt
 6  2  l
When l  6.5 centimeters and
dA
dt
 12l
dl
dt

dl
dt
 12l
dl
dt
dl
 8 centimeters per second, then the rate of change of the area is:
dt
12  6.58
 624 square centimeters per second
5. Similarly, to get the rate at which the volume is changing, we differentiate the volume equation with respect to time:
V
dV
dt
 l3

d 3
 l   3l 2 dldt
dt
When l  6.5 centimeters and
dV
dt
 3l 2
4/30/2016
dl
dt

dl
 8 centimeters per second, then the rate of change of the volume is:
dt
 3 6.5 8
2
 1014 cubic centimeters per second