Direct variation :
Two variables x and y show direct variation when
y = ax
for some nonzero constant a.
Another type of variation is called inverse variation.
Inverse variation :
Two variables x and y show inverse variation when they are related as follows:
y = a/x, a ≠ 0
The constant a is the constant of variation, and y is said to vary inversely with x.
While dealing with word problems on direct and inverse variation, we have to follow the steps given below.
Step 1 :
Understanding the problem and create the equation representing the situation.
Step 2 :
Find the constant of variation.
Step 3 :
In order to receive equation only in two terms, we have to apply the constant of variation in the equation that we have created initially.
Step 4 :
Apply the given value to find the unknown.
Determine whether each situation is an example of a direct variation or inverse variation. Write and equations of variation to represent the situation and solve for the indicated information.
Problem 1 :
The volume V of a gas kept at a constant - temperature varies inversely with the pressure p. If the pressure is 24 pounds per square inch, the volume is 15 cubic feet. What will the volume be when the pressure is 30 pounds per square inch?
Solution :
V - volume of a gas
p - temperature
Step 1 :
Writing the proportional relationship between the variables V and p.
V ∝ p
V = kp -----(1)
k is constant of variation.
Step 2 :
Evaluating the constant of variation.
When V = 15 cubic feet, p = 24 pounds
15 = k (24)
k = 15/24
k = 5/8
Step 3 :
Applying the value of k in (1)
V = (5/8)p
Step 4 :
When p = 30 pounds, V = ?
V = (5/8)(30)
V = 18.75 cubic feet
Volume of gas is 18.75 cubic feet.
Problem 2 :
The amount of money spent at the gas station varies directly with the number of gallons purchased. When 11.5 gallons of gas were purchased the cost was $37.72. How much would 8 gallons of gas cost?
Solution :
The amount of money be M and number of gallons purchased is g.
M ∝ g
M = kg -----(1)
k is constant of variation.
Step 2 :
Evaluating the constant of variation.
When g = 11.5 gallons, M = $37.72
37.72 = k(11.5)
k = 37.72/11.5
k = 3.28
Step 3 :
Applying the value of k in (1)
M = 3.28g
Step 4 :
When g = 8 gallons, M = ?
M = 3.28(8)
M = $26.24
Cost of 8 gallons of gas is $26.24
Problem 3 :
The time to complete a project varies inversely with the number of employees. If 3 people can complete the project in 7 days, how long will it take 5 people?
Solution :
The time to complete a project be T and number of employees be E
T ∝ E
T = k/E -----(1)
k is constant of variation.
Step 2 :
Evaluating the constant of variation.
When T = 7, E = 3
7= k/3
k = 7(3)
k = 21
Step 3 :
Applying the value of k in (1)
T = 21/E
Step 4 :
When E = 5 people, M = ?
T = 21/5
T = 4.2 days
5 people can complete the project is 4.2 days.
Problem 4 :
The time needed to travel a certain distance varies inversely with the rate of speed. If it takes 8 hours to travel a certain distance at 36 miles per hour, how long will it take to travel the same distance at 60 miles per hour?
Solution :
Time varies inversely with the rate of speed.
Let T be the time taken and S be the speed.
Step 1 :
T ∝ S
T = k/S -----(1)
k is constant of variation.
Step 2 :
Evaluating the constant of variation.
When T = 8, D = 36
8 = k/36
k = 8(36)
k = 288
Step 3 :
Applying the value of k in (1)
T = 288/S
Step 4 :
When T = ? people, S = 60
T = 288/60
T = 4.8
So, it is will 4.8 hours to travel 60 miles
Problem 5 :
The number of centimeters y in a linear measurement varies directly with the number of inches x in the measurement. Pablo's height is 152.4 centimeters or 60 inches. What is Maria's height in centimeters if she is 64 inches tall?
Solution :
Let y be the linear measurement and x be Pablo's height in inches
Step 1 :
y ∝ x
y = kx -----(1)
Step 2 :
When y = 152.4 centimeter, x = 60 inches
152.4 = k(60)
k = 152.4/60
k = 2.54
Applying the value of k in (1)
Step 3 :
y = 2.54x
Step 4 :
When x = 64, y = ?
y = 2.54(64)
y = 162.56
So, Maria's height is 162.56 centimeter.
Problem 6 :
The number of revolutions made by a tire traveling over a fixed distance varies inversely with the radius of the tire. A 12-inch radius tire makes 100 revolutions to travel a certain distance. How many revolutions would a 16-inch radius tire require to travel the same distance?
Solution :
Let number of revolutions be R and radius be r.
R ∝ r
R = k/r ----(1)
When R = 100, r = 12
100 = k/12
k = 100(12)
k = 1200
Applying the value of k in (1), we get
When r = 16, R = ?
R = 1200/16
R = 75
So, the required number of revolution is 75.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM