Business Mathematics: Ratio, Proportion, and Percent - Proportion

In algebra, unknown quantities are represented by letters, such as x and y, called variables. The value of one variable is often related to the values of others.

Two very special relationships occur often:
y is said to be directly proportional to x if their relationship is expressed in the form y = kx.
y is said to be inversely proportional to x if their relationship is expressed in the form y = k/x.
In each case, k is a number whose value is constant; k is called the constant of proportionality. Note that in a direct proportion y gets larger as x gets larger while in an inverse proportion, as x gets larger y gets smaller.
The constant of proportionality can be easily determined if values of x and y are given.

Example 6
(a)   Determine the value of k assuming that x and y are directly proportional and y = 15 when x = 3.
 Since, for direct proportions, y = kx, we simply substitute the appropriate values of x and y:
15 = k × 3
To solve, divide both sides of the equation by 3 to get k = 5.
(b)   Determine the value of k given that x and y are inversely proportional and y = 15 when x = 3.
For inverse proportions, y = k/x. Substitution yields 15 = k/3. This time we multiply both sides by 3 to get k = 45.
Once we have determined the value of k, we can use the equation to solve problems.

Example 7
The amount of money Jim earns is directly proportional to the number of hours he works. If he works 40 hours, he will make $200. How much will he earn if he works 55 hours?
Let A represent the amount of money earned and h the number of hours worked. Then A = kh represents the relationship between A and h. Since A = 200 when h = 40, the equation becomes 200 = k × 40 so, upon division by 40, k = 5. The equation now becomes A = 5h. Letting h = 55, we see that A = $275.A direct proportion can often be solved by setting up a ratio in the form of a fraction. If y is directly proportional to x, and y1 and y2 correspond, respectively, to x1 and x2, then

Note that the numerator (denominator) of the fraction on the left corresponds to the numerator (denominator) of the fraction on the right.

Example 8
Solve example 7 using a ratio expressed as a fraction.
Let A represent the amount of money earned if Jim works 55 hours:

If we cross-multiply, we get 40A = 11,000 from which it follows (by dividing by 40) that A = $275.

Example 9
It takes 72 hours for five workers to paint an office building. If the number of hours worked is inversely proportional to the number of workers, how many hours should it take six workers to paint the building?
Let H be the number of hours to paint the building and n the number of workers. Since H is inversely proportional to n, H = kin. Since H = 72 when n = 5, 72 = k/5, and multiplication by 5 yields k = 360. We now have H = 360/n. If n = 6, H = 360/6 = 60 hours.
An inverse proportion can also be solved by setting up a ratio in the form of a fraction. If y is inversely proportional to x, and y1 and y2 correspond, respectively, to x1 and x2, then

Note that the numerator of the fraction on the left corresponds to the denominator of the fraction on the right.

Example 10
Solve example 9 using a ratio expressed as a fraction.
Let H represent the number of hours it would take six workers to complete the job.

Cross-multiplying, we get 6H = 360, from which it follows that H = 60 hours.

Solved Problems

2.13   Determine the constant of proportionality, k, given y directly proportional to x and y = 200 when x = 40.
Solution
Since y is inversely proportional to x, y = kx. Since y = 200 when x = 40, we have, on substitution, 200 = k × 40. Therefore k = 5.


2.14   Determine the constant of proportionality, k, given y inversely proportional to x and y = 20 when x = 3.
Solution
Since y is inversely proportional to x, y = k/x. Since y = 20 when × = 3, 20 = k/3. Multiplication by 3 gives k = 60.

2.15   The number of gallons of gasoline consumed by an automobile is directly proportional to the number of miles driven. If the car can travel 133 mi on 7 gal of gasoline, how many gallons will be consumed on a 228 mi trip?
Solution
Let x represent the number of miles driven and let y represent the number of gallons of gasoline used. Since y is directly proportional to x, y = kx. When × = 133, y = 7 so 7 = k× 133. Divide both sides of the equation by 133 and k = 7/133 = 1/19. Now we have y = (1/19)x. When x = 228, y = (1/19) × 228 = 12 gal of gasoline.

2.16   Solve problem 2.15 using a ratio expressed as a fraction.
Solution
Let y be the number of gallons of gasoline needed. Since the number of gallons is directly proportional to the number of miles y = 12 gal.

2.17   It takes four workers a total of 15 days to frame a house. How many days would it take six workers to complete the project?
Solution
Let x represent the number of days to frame the house and n the number of workers. Since x is inversely proportional to n, x = k/n. Since x = 15 when n = 4,15 = k/4, so k = 60. The equation of inverse proportionality now becomes x = 60/n, and when n = 6, x = 10. It would take six workers 10 days to frame the house.

2.18   Solve problem 2.17 using a ratio expressed as a fraction.
Solution
Let x represent the number of days to frame the house. x is inversely proportional to the number of workers on the job. Therefore x/15 = 4/6. Note that the 15 in the denominator on the left, which represents the number of hours to complete the job, corresponds to the 4 on the left, representing the number of workers. It follows that 6x = 60, and x = 10.