6
6.1
Proportion: Fractions, Direct and Inverse
Variation, and Percent
Fractions
Every rational number can be written as a fraction, that is a quotient of
two integers, where the divisor of course cannot be zero. However, this
representation is not unique. We need to know how to tell when two fractions
represent the same rational number (which we think of as being a unique
point on the real number line). We also need to know how to perform the
arithmetic operations of addition, subtraction, multiplication, and division
on fractions. Subsequently we will deal with fractional expressions where the
numerator and/or denominator are made up of algebraic expressions instead
of integers, but the same rules for determining when two are equal, and for
performing arithmetic operations on them, will apply.
The general method for telling when two fractions are equal is the rule
for cross-multiplying: Let m, n, r, s be integers and assume that n != 0, s != 0.
= rs if and only if ms = rn. Let’s look at why this makes sense: we
Then m
n
know that m
= mp
for any integer p. If we think about this in terms of the
n
np
number line representation, this is clear. We can think of m
, where m, n are
n
positive, as dividing each unit segment up into n subunits, each of length 1/n,
and now we take m of those subunits, to have something of length m/n. If
we further subdivide each of those subunits p times, now each little segment
will have length 1/(np), and in taking m of the old subunits, we will have
. It should be the case
taken mp of the new ones, so the length is also mp
np
m
r
that n = s if and only if there is some way to “subdivide”each of them so
that we get a fraction mp
and a fraction rt
where the numerator of each and
np
st
the denominator of each are exactly the same, that is, mp = rt, np = st. But
then, mp · st = rt · np, or mspt = nrpt, and canceling pt, we have ms = nr.
The approach just given justifies the criterion for when two fractions are
equal, but we can check it more simply if we accept some other facts about
fractions. We have that ms = rn if and only if ms ÷ sn = rn ÷ sn. Now
if we accept that dividing by an integer is the same as multiplying by its
= rn
, and then if we accept as
reciprocal, we have this equivalent to ms
sn
sn
above that we can cancel common factors of numerators and denominators,
we have m
= rs .
n
is represented in lowest terms if the only comWe say that a fraction m
n
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mon divisor of m and n is 1. (In a separate section we discuss the use of
the Euclidean algorithm to determine the greatest common divisor of two
integers.)
Recall from §1 the discussion of how fractions are added, subtracted,
and multiplied. There we also stated without proof that ab = dc if and only
a
c
if a+b
= c+d
. We show this simple fact now, using the remarks above on
a
c
determining when two fractions are equal. We have a+b
= c+d
if and only if
a · (c + d) = (a + b) · c. Applying the distributive law, we see this is equivalent
to ac + ad = ac + bc, and subtracting ac from each side of the equation shows
us this is the same as ad = bc. Then since we are assuming neither b nor d
is 0, we have that this is equivalent to ab = dc .
We also revisit the idea of dividing with fractions. There are many ways
to look at this. We can accept as “obvious” the fact that if we take any
number and divide it by itself, the quotient will be 1. Thus for rational
÷m
= 1. In terms of our properties of real numbers, and the
numbers, m
n
n
idea that multiplying any number by its multiplicative inverse gives you 1,
has the same effect as multiplying by ( m
)−1 . But
we see that dividing by m
n
n
m −1
what is ( n ) as a rational number, that is as a quotient of two integers?
It is a number which when multiplied together with m
gives us a fraction
n
m
n
mn
that is equivalent to 1. Now observe that n × m = mn = 1, so we have that
n
)−1 = m
. Thus dividing by a fraction is the same as multiplying by its
(m
n
reciprocal.
This also makes sense when we think of division as the inverse operation
to multiplication. Working with integers, if we are trying to solve a ÷ b =
x, we are asking to find the number x such that b times x will give us a.
= x, we are looking for
Working with fractions, if we ar trying to solve pq ÷ m
n
p
m
that number x such that q = n · x. Using our understanding of equivalent
pn
works, thus verifying that dividing
fractions, it is easy to check that x = qm
m
n
by n is the same as multiplying by m .
When our fractions involve algebraic expressions, exactly the same rules
x+1
+ x+2
we will first have to find a common
will apply. For example, to add x−1
x+3
denominator, which here will be the product of the two individual denominators. We need to convert each of the summands into an equivalent fraction
have the common denominator (x − 1)(x + 3). This will be done by multiplying the first term by x+3
and the second term by x−1
. This gives
x+3
x−1
(x2 + 4x + 3) + (x2 + x − 2)
2x2 + 5x − 1
(x + 1)(x + 3) (x − 1)(x + 2)
+
=
= 2
.
(x − 1)(x + 3) (x − 1)(x + 3)
(x − 1)(x + 3)
x + 2x − 3
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We summarize the rules for fractions below. Make sure you understand
why each one is true! And never forget that we cannot divide by 0, so the
denominator of a fraction can never be 0.
a
b
1. Equivalent Fractions:
=
ax
bx
∀ x != 0
2. Cross-Multiplying Rule for Equivalence:
a
b
3. Multiplication Rule:
4. Division Rule:
a
b
÷
c
d
×
=
a
b
c
d
=
ac
bd
d
c
=
×
5. Addition/Subtraction Rule:
a
b
±
6.2
−a
−b
=
c
d
if and only if ad = bc
ad
bc
c
d
6. Simplifying Compound Fractions:
7. Working with Negatives:
a
b
= ab ;
=
a/b
c/d
−a
b
ad±bc
bd
=
=
a
b
a
−b
÷
c
d
=
ad
bc
= − ab
Direct and Inverse Variation Problems
Two values are said to be directly proportional to each other if one value is
a fixed numerical multiple of the other. Thus y is directly proportional to
x if there exists a constant c such that y = cx. What this means is that
the ratio xy is always equal to the constant c. Two values are said to be
inversely proportional to each other if one value is directly proportional to
the multiplicative inverse (reciprocal) of the other number. In this case, there
would be a constant c such that y = c· x1 . Equivalently, this would say yx = c.
Direct and inverse proportion problems are easy to solve algebraically and
provide a powerful tool. Direct proportion is what is used in order to convert
units of one kind to units of another. For example, suppose we know that
on a map 1/8 of an inch equals 10miles, and we measure on the map that
two towns are 3.5 inches apart. How far are they apart in real life? We know
that the ratio of miles to inches is a constant, and we know that constant is
12
. Then we also know that the ratio of the distance the two towns
equal to 1/8
are apart to their distance on the map is also equal to this constant. To solve
the problem, set the two ratios equal to each other. Then we have two equal
fractions, so we know if we cross-multiply, they must be equal. We can cross
multiply and solve for the unknown distance. In this case we have
x miles
12 miles
=
1/8 inches
3.5 inches
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which gives 12 · 3.5 = 18 x, or 12 · 3.5 · 8 = x, so x = 336 miles.
The general method for solving proportion problems is to set up the correct
ratios, cross-multiply, and solve. Often, thinking about the units involved
will help you set up the ratio in the correct order. Let’s look at a few more
examples:
A man is in a field with a charging bull, which is 363 feet away from
him. If the man can run 12 feet per second, and the bull can run 45 feet per
second, how long until the bull catches up to the man? If there is a sturdy
fence the man can climb over and be safe, which is 150 feet away, will the
man make it? We can figure this out as follows: Every second, the bull runs
45 − 12 = 33 feet farther than the man. So we can set up the ratio 33 feet / 1
second = 363 feet / x seconds, and solve for x. If we cross-muultiply, we have
33x = 363, so x = 11. Now in 11 seconds, the man will travel 11 · 12 = 132
feet, so unfortunately he will not reach the fence.
Ratios and direct or inverse proportions come up in many real world
problems. The distance an object travels is equal to its speed times the time
it travels. Thus distance is directly proportional to speed, and distance is
directly proportional to time, and time and speed are inversely proportional
to each other. Likewise, the amount of work done by a person who works at a
certain rate is the product of the rate at which the person works, multiplied by
the amount of time the person works. Let’s start with some simple examples
and move up to more complicated ones.
Example 1: Susie walks at a rate of 3.5 miles per hour. She is walking
around a lake path that is 2.25 miles long. If she starts around the lake
at 9:30, what time (to the nearest minute) will she get back to where she
started? If we let t stand for how long Susie walks, we have 3.5 miles / 1
hour = 2.25 miles / t hours. We cross multiply to get 3.5t = 2.25, or t =
2.25/3.5 = 9/14, so it will take her 9/14 of an hour. Now we know we have
m
= 9/14
, or m = 9 · 60/14 = 270/7 minutes,
60 minutes / 1 hour, so we have 60
1
which is 38.57... minutes, so to the nearest minute it will take her 39 minutes,
and she will be back at 10:09.
Example 2: It took John two hours to ride his bike to his grandmother’s
house. On the way home he was tired, and it took three hours. If John was
biking 5 mph slower on the way home than he was on the way there, what
speed was he biking on the way to his grandmother’s, and how far away did
she live? Let’s let s be the speed (in miles per hour) John biked on the way
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to his grandmother’s, and let’s let d be the distance he traveled. Then we
know d = 2s, because distance is rate × time. On the way back, the distance
he traveled is still d, but the rate is s − 5, and the time is 3, so we have
d = 3(s − 5). Thus 2s = 3(s − 5), so 2s = 3s − 15, and s = 15. He biked at
a rate of 15 miles per hour on the way there and 10 miles per hour on the
way home, and his grandmother lived 30 miles away.
Example 3: Bill can paint four houses in 28 days. How long will it take
him to paint five houses? John can paint nine houses in 45 days. How long
will it take him to paint seven houses? If the two of them work together,
how long will it take them to paint two houses? To answer this question, we
have to determine the rate at which each of them paint houses. Let b denote
the rate at which Bill paints houses, measured in houses per day, and let j
denote the rate at which John paints houses, also measured in houses per
day. Then b = 4/28 = 1/7, that is, Bill can paint 1/7 house per day. Or,
taking reciprocals, it takes him 7 days per house. To paint five houses will
take 5 · 7 = 35 days. And j = 9/45 = 1/5, that is, John can paint 1/5 house
per day, or it takes him 5 days per house. To paint seven houses will also take
him 35 days. If they both work together, the first thing we need to figure out
is the rate at which they paint. In one day, they can paint 1/7 + 1/5 houses
working together, so they paint 5/35 + 7/35 = 12/35 houses per day. That
means it takes them 35/12 days to paint one house, and 70/12 = 35/6 days
to paint two houses.
6.3
Solving Problems Involving Percents
There are basically three types of problems involving percents: You may be
asked to find the percent, you may be asked to find the part, or you may be
asked to find the whole. We’ll look at three examples that illustrate this:
a) 60 is what percent of 75? (Find the percent.)
b) What is 80% of 75? (Find the part.)
c) 60 is 80% of what number? (Find the whole.)
All three of these problems can be solved by setting up the following ratio:
part
percent
=
100
whole
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In each problem, you are given two of the three values “percent”, “part”,
and “whole.” By setting up the ratio and cross-multiplying, you can solve
x
80
x
for the remaining value. Thus (a) becomes 100
= 60
, (b) becomes 100
= 75
,
75
80
60
and (c) becomes 100 = x . The task in solving each problem then becomes
identifying which of the three, percent, part, or whole, you need to find to
solve the problem, and which have been given.
Percent problems may also involve finding the percent increase, or the percent decrease. This is given by solving for the “percent” where the “whole”
is the original amount, and the “part” is the amount by which the original
value went up (in the case of increase) or went down (in the case of decrease).
Suppose the price of gas one day is $1.60, and the next day it has jumped
to $1.80. The following day it is back at $1.60. What is the percent increase
from the first day to the second? What is the percent decrease from the second day to the third? Notice that although the price goes up $.20 and then
back down $.20, the percent decrease will be less than the percent increase,
because $.20 is a smaller percent of $1.80 than it is of $1.60. Let’s do the
calculations:
.20
x
= 1.60
to obtain x = 12.5, so
To find the percent increase, we solve 100
x
.20
the price increased 12.5%. To find the percent decrease, we solve 100
= 1.80
to obtain x = 11.1̄, so the percent decrease is approximately 11%.
Summarizing, percent problems are not that difficult if you remember to
set up the ratio percent/100 = part/whole, and cross-multiply to solve.
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