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Word problems can be classified according to what they
are about.
Integer/Number Problems
Area Problems
Age Problems
Double Situation Problems
Integer/Number Problems
Example 1:
Find two non-negative integers whose difference is 18 and
whose product is 544.
We have two pieces of
We are looking for 2 integers.
information.
Difference is 18
Product is 544
1st Integer = x
2nd Integer = x – 18
By defining the second
integer in this way, I
am assuming that the
first integer is larger.
x(x – 18) = 544
x2 – 18x – 544 = 0
(x + 16)(x – 34) = 0
Difference is 18. This means that
there is a separation of 18
between the integers (one integer
is 18 more than the other).
Now we can write the
equation, but we must use
the other piece of info.
Product is 544. This means that
when you multiply the integers
by each other, you get 544.
(x + 16) = 0 OR (x – 34) = 0
x = -16
OR x = 34
– 544
1
2
4
8
16
17
-544
-272
-136
-68
-34
-32
-18
x = -16
OR x = 34
We must eliminate –16 as an answer because we are asked
to find non-negative integers only in the word problem.
This means that the non-negative integer that represents
‘x’ (the 1st integer) is 34. Now we must determine the
integer that represents ‘x – 18’ (the 2nd integer).
If x = 34, then x – 18 = 34 – 18
= 16
Now, we can address the problem with the answer:
The two non-negative integers whose product is 544 are
34 and 16.
Example 2:
The product of two consecutive numbers is 156. What are
these numbers?
We have two pieces of
We are looking for 2 integers.
information.
Numbers are consecutive
1st Integer = x
2nd Integer = x + 1
For the second number
to be consecutive to
the first, we must add
1.
Product is 156
Consecutive means one after
another, such as 4, 5 or 38, 39.
Notice that the second of the
consecutive numbers is 1 more
than the first.
We are ready to write the equation using the fact that
the product is 156. This means that when you multiply the
integers by each other, you get 156.
– 156
x(x + 1) = 156
-1 156
-2 78
+1
2
x + 1x – 156 = 0
-3 52
-4 39
(x - 12)(x + 13) = 0
-6 26
-12 13
(x - 12) = 0 OR (x + 13) = 0
x = 12
OR x = -13
Both of these results are acceptable, which means
that there are 2 possible values for ‘x’. This also
means that there will be 2 possible values for ‘x + 1’.
If x = 12, then x + 1 = 12 + 1 If x = -13, then x + 1 = -13 + 1
= 13
= -12
There are two possible pairs of numbers that fit the
description:
Either the numbers could be 12 and 13 OR –13 and –12.
Example 3:
Ten times a number and its square is 119. What is the
number?
We are looking for only 1 number.
Number = x
We must translate the information into an
algebraic equation. The word that
identifies the ‘equal sign’ is the word ‘is’.
Left Side
Right Side
10x + x2 = 119
– 119
x2 + 10x – 119 = 0
(x - 7)(x + 17) = 0
-7 17
(x - 7) = 0 OR (x + 17) = 0
x=7
OR x = -17
(7)2 + 10(7) = 119
(-17)2 + 10(-17) = 119
49 + 70 = 119
289 + -170 = 119
119 = 119
119 = 119
+10
Area Problems: Example 4
The area of a rectangular field is 10 800m2. Given that its
length is three times its width, find its dimensions?
The dimensions of a shape are particular to the type
of shape. We are working with a rectangle and the
dimensions here refer to the length and width.
Width = x
Length = 3x
Formula: Area = length  width
length
width
10800 = (3x) (x)
10800 = 3x2
-60 doesn’t make sense as the
0 = 3x2 - 10800
width so we disqualify it.
0 = x2 - 3600
0 = (x - 60)(x + 60)
The width is 60 m and the
(x - 60) = 0 OR (x + 60) = 0 length is 180 m.
x = 60 OR x = -60
Example 5:
Find the dimensions of a triangle whose area is 35 cm2 if
its base measures 3 cm more than its height.
We are working with a triangle and the dimensions
here refer to the base and height.
height = x
base = x + 3
base  height
2
xx  3
35 
2
Formula: Area 
70 = x2 + 3x
0 = x2 + 3x - 70
0 = (x - 7)(x + 10)
(x - 7) = 0 OR (x + 10) = 0
x=7
OR x = -10
height
– 70
-1
-2
-5
-7
70
35
14
10
3
Base
-10 doesn’t make sense as the
height so we disqualify it.
The height is 7 cm and the
base is 10 cm.
Age Problems: Example 6
The sum of a father’s age and his son’s age is 35. In 5
years, the product of the ages will be 350. How old is each
person?
NOW
THEN (In 5 years)
Father’s
Age
Son’s
Age
x
35 - x
x+5
35 – x + 5 =
40 - x
(x + 5)(40 - x) = 350
+150
-1 -150
40x – x2 + 200 - 5x = 350
-2 -75
-35
2
– x + 35x + 200 = 350
-3 -50
-5 -30
0 = x2 - 35x - 200 + 350
-10 -15
0 = x2 - 35x + 150
5 doesn’t make sense as the
0 = (x - 5)(x - 30)
father’s age so we disqualify it.
(x - 5) = 0 OR (x - 30) = 0
That means the father is 30
x=5
OR x = 30
years old and his son is 5.
Example 7:
A sister is 6 years older than her brother. 8 years ago, the
product of their ages was 135. How old are they now?
Sister’s
Age
Brother’s
Age
NOW
THEN (8 years ago)
x
x-6
x-8
x–6-8=
x - 14
(x - 8)(x - 14) = 135
x2 – 14x - 8x + 112 = 135
– 23
x2 – 22x + 112 - 135 = 0
x2 – 22x - 23 = 0
1 -23
-22
0 = (x - 23)(x + 1)
-1 doesn’t make sense as the
(x - 23) = 0 OR (x + 1) = 0 sister’s age so we disqualify it.
x = 23
OR x = -1
That means the sister is 23 years
old and her brother is 17 years old.
Double Situation Problems: Example 8
The price of a certain number of balls is $60. If each ball
cost $1 less, we would have 5 more balls for the same price.
What is the price of one ball?
SITUATION 1 SITUATION 2
60
60

x1
x
Price of a
ball
# of
balls
x
x-1
60
x
60
x1
If each ball is cheaper as it is in situation 2, you can
buy more balls. This means that the number of balls
in situation 2 is more than those in situation 1.
To balance the inequality, we can add 5 to the right
side so that it becomes an equation. 60
60
x1

x
5
60
60
C.D. = x(x – 1)

5
x1
x
60x
60(x  1) 5x(x  1)


x(x  1)
x(x  1)
x(x  1)
60x = 60x – 60 + 5x2 – 5x
60x = 60x – 60 + 5x2 – 5x
0 = 5x2 – 5x – 60
0 = x2 – x – 12
(x - 4)(x + 3) = 0
(x - 4) = 0 OR (x + 3) = 0
x=4
OR x = -3
-3 doesn’t make sense as answer, so we
disqualify it. That means each ball costs $4.
– 12
1 -12
2 -6
3 -4
-1
Example 9:
Two contestants at a banana eating contest must eat 90
bananas as fast as possible. By eating 3 more per minute,
the first contestant beats the second contestant by 5
minutes. How long did it take the winner to eat the
CONTESTANT 1
CONTESTANT
bananas?
90
90

x
x5
Time to eat
Bananas
x
x+5
Rate of
consumption
90
x
90
x5
Contestant 1 consumes more quickly than contestant 2.
To balance the inequality, we can add 3 to the right
side so that it becomes an equation.
90
90

3
x
x5
2
90
90
C.D. = x(x + 5)

3
x
x5
90(x  5)
90x
3x(x  5)


x(x  5)
x(x  5)
x(x  5)
90x + 450 = 90x + 3x2 + 15x
90x + 450 = 90x + 3x2 + 15x
0 = 3x2 + 15x – 450
0 = x2 + 5x – 150
(x - 10)(x + 15) = 0
(x - 10) = 0 OR (x + 15) = 0
x = 10
OR x = -15
-15 doesn’t make sense as answer, so we
disqualify it. This means the winner ate the
bananas in 10 minutes.
– 150
-1
-2
-3
-5
-6
-10
150
75
50
30
25
15
+5