We come across many such situations in our day-to-day life, where we need to see variation in one quantity bringing in variation in the other quantity.
For example:
(i) As the speed of a vehicle increases, the time taken to cover the same distance decreases
(ii) More apples cost more money
(iii) More interest earned for more money deposited.
(iv) More distance to travel, more petrol needed.
Direct Proportion
Two quantities x and y are said to be in direct proportion
if they increase (decrease) together in such a manner that the ratio of their corresponding values remains constant.
That is if x/y=k= [k is a positive number] = Constant
Then x and y are said to vary directly. In such a case if y1, y2 are the values of y corresponding to the values x1, x2 of x respectively then
$\frac {x_1}{y_1}= \frac {x_2}{y_2}$
How to Solve Direct Proportion Problems
Two method to solve the problem
(1) We know that in direct Proportion
$\frac {x_1}{y_1}= \frac {x_2}{y_2}=\frac {x_3}{y_3} =\frac {x_n}{y_n}$
In a problem, one ratio would be given. So using the above equation, we can easily find the unknown terms
Example
The cost of 5 kg of a particular quality of sugar is Rs 200. Tabulate the cost of 1,2, 4, 10 and 14 kg of sugar of the same type Solution
Let x kg of sugar cost y rupees
x
1
2
4
5
10
14
y
?
?
?
200
?
?
As the number of kg increases, cost of the sugar also increases in the same ratio. It is case of direct proportion.
Now let’s use the above expression
$\frac {x_1}{y_1}= \frac {x_2}{y_2}=\frac {x_3}{y_3} =\frac {x_n}{y_n}$
Now we have
(a)
$\frac {x_4}{y_4} = \frac {5}{200}$
$ \frac {1}{y} = \frac {5}{200}$
y=Rs 40
(b)
$\frac {x_2}{y_2} = \frac {x_4}{y_4}
2/y = 5/200
y=Rs 80
Similarly, other can be found
Complete table would be
x
1
2
4
5
10
14
y
40
80
160
200
400
560
(2) We know that in direct Proportion
$ \frac {x}{y}= k$
$x=ky$
So we can find value of k from known values, and then use the formula to calculate the unknown values
Inverse proportion
Two quantities x and y are said to be in inverse proportion
if an increase in x causes a proportional decrease in y (and vice-versa) in such a manner that the product of their corresponding values remains constant.
That is, if xy = k= Constant
Then x and y are said to vary inversely.
In this case if y1, y2 are the values of y corresponding to the values x1, x2 of x respectively then x1 y1 = x2 y2
How to Solve Inverse Proportion Problems
Two method to solve the problem
(1) We know that in inverse Proportion x1y1 = x2y2 = x2y2 = x2y2
In a problem, one pair would be given. So using the above equation, we can easily find the unknown terms
(2) We know that in inverse Proportion
xy= k
x=k/y
So we can find value of k from known values, and then use the formula to calculate the unknown values
Example:
If 20 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours? Solution:
Let the number of workers employed to build the wall in 30 hours be y.
We have the following table.
Number of Hours
48
30
Number of workers
20
y
Obviously more the number of workers, faster will they build the wall.
So, the number of hours and number of workers vary in inverse proportion.
So 48 × 20 = 30 × y y=32 workers
So to finish the work in 30 hours, 32 workers are required
Solved Example
Example
Following are the car parking charges near an Airport up to
2 hours Rs 60
6 hours Rs 100
12 hours Rs 140
24 hours Rs 180
Check if the parking charges are in direct proportion to the parking time. Answer
We know that two quantities are in direct proportion if whenever the values of one quantity increase, then the value of another quantity increase in such a way that ratio of the quantities remains same
Here The charges are not increasing in direct proportion to the parking time because
2/60 ≠ 6/100 ≠ 12/140 ≠ 24/180
Extra Zing
When two quantities x and y are in direct proportion (or vary directly) they are also written as x∝y.
When two quantities x and y are in inverse proportion (or vary inversely) they are also written as x∝ 1/y
