In the previous sections we completed the discussion on 'Direct proportions'. In this section we will discuss about 'Inverse proportions'.
Consider the speed of a student when he walks to school. Let it be 3 km per hour. Let us convert this speed to a more basic form:
• 3 km/hr means he can walk a distance of 3 km in one hour.
• That means he can walk a distance of 3000 m in 60 minutes. (∵ since 1 km = 1000 m & 1 hr = 60 minutes).
• So he can walk a distance of 3000/60 = 50 m in 1 minute. So his speed is 50 m / minute.
Suppose the distance from his home to the school is 400 m. Then the time that he would take can be calculated as follows:
• In the 1st minute he travels 50 m
• In the 2nd minute he travels another 50 m. So the total distance at the end of the 2nd minute is 100m
• In the 3rd minute he travels another 50m. So the total distance at the end of the 3rd minute is 150 m
• In the 4th minute he travels another 50m. So the total distance at the end of the 4th minute is 200 m
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• In the 8th minute he travels another 50m. So the total distance at the end of the 8th minute is 400 m
So he takes 8 minutes to travel 400 m from home to school. We do not have to write the 8 steps above. The result, which is '8 minutes' can be obtained simply by dividing the total distance by speed. That is., Time = Distance ÷ speed = 400 m ÷ 50 m / minute = 8 minutes.
For both the methods, the speed must be uniform. That is., he must travel the same 50 m in every minute. If he travels 50 m in one minute and 52 or 45 m in another minute, the speed is not uniform, and our calculations will not work. For our present discussion, whenever we say speed, uniform speed is implied.
So we have calculated the time required for the travel from home to school. Suppose, instead of walking, he rides a cycle. Let the speed of cycle be 80 m/ minute. Then the time required for the travel will be 400/80 = 5 minutes.
Suppose he travels by a car. Let the speed of the car be 200 m/minute. Then the time required for the travel will be 400/200 = 2 minutes.
Let us arrange the above results in a table. It is shown below:
Two quantities are tabulated: 1. Speed (denoted as x) and 2. Time of travel (denoted as y).
These two quantities vary.
• When speed increases, time decreases.
• When speed decreases, time increases.
That is., When one quantity increases, the other decreases and vice versa. In the lower most row of the table, the product xy of the two quantities are written. We see that it is a constant (equal to 400) in all the columns.
Let us take another example. The example we saw above was the time taken to travel a certain distance. Now we will see the time taken to do a work. Let us dee the details:
A container holds a large number of bricks. The work to be done is the 'unloading all the bricks from the container'. From previous experience, the following information is already known:
• If 4 workers are employed, they will unload all the bricks in 2 days.
With this information, we want to calculate the number of days required when different number of workers are employed. Let us see the calculations:
• Let the ‘average quantity’ of work done by a single worker on a single day be equal to ‘u’.
[Note the use of the word 'average' in the above step. The reason for using it can be explained as follows:
All the workers will not be able to do the same exact quantity of work on a day. Some workers will do more work, and some may do less work. So it is appropriate to take the 'average quantity'. This average is applied to all the workers. That is., it is assumed that all the workers do an average of 'u' on a day]
• So the total work done by 2 workers in 1 day = 2u
• So the total work done by 4 workers in 2 days = 4 x 2u = 8u
• 8u is the total work (unloading all the bricks) to be done. It will not change, and hence is a constant.
Now we want to know the different possibilities:
(1) 2 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 2 workers in 1 day = 2u
• No. of days = Total work / Work done by 2 workers in one day = 8u / 2u = 4 days
(2) 8 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 8 workers in 1 day = 8u
• No. of days = Total work / Work done by 8 workers in one day = 8u / 8u = 1 day
(3) 6 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 6 workers in 1 day = 6u
• No. of days = Total work / Work done by 6 workers in one day = 8u / 6u = 4⁄3 days
4⁄3 days is one day and 1⁄3 days. But 1⁄3 day = (8 × 1⁄3) hours = 8⁄3 = 2 2⁄3 hours (assuming one day = 8 hours of work). So 4⁄3 days = One day plus 2 2⁄3 hours. But for our present discussion, it is sufficient to write 4/3 days.
(4) 5 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 5 workers in 1 day = 5u
• No. of days = Total work / Work done by 5 workers in one day = 8u / 5u = 8⁄5 days
(5) 3 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 3 workers in 1 day = 3u
• No. of days = Total work / Work done by 3 workers in one day = 8u / 3u = 8⁄3 days. Note that this is double the number of days when 6 workers are employed.
Let us now tabulate the above results. It is shown below:
Two quantities are tabulated: 1. Number of days (denoted as x) 2. Number of workers (denoted as y). These two quantities vary.
• When No. of workers increases, time decreases.
• When No. of workers decreases, time increases. In such a situation, we say: When one quantity increases, the other decreases and vice versa.
In the lower most row of the table, the product xy of the two quantities are written. We see that it is a constant (equal to 8.00) in all the columns.
In the next section, we will see some solved examples based on the above discussion.
Consider the speed of a student when he walks to school. Let it be 3 km per hour. Let us convert this speed to a more basic form:
• 3 km/hr means he can walk a distance of 3 km in one hour.
• That means he can walk a distance of 3000 m in 60 minutes. (∵ since 1 km = 1000 m & 1 hr = 60 minutes).
• So he can walk a distance of 3000/60 = 50 m in 1 minute. So his speed is 50 m / minute.
Suppose the distance from his home to the school is 400 m. Then the time that he would take can be calculated as follows:
• In the 1st minute he travels 50 m
• In the 2nd minute he travels another 50 m. So the total distance at the end of the 2nd minute is 100m
• In the 3rd minute he travels another 50m. So the total distance at the end of the 3rd minute is 150 m
• In the 4th minute he travels another 50m. So the total distance at the end of the 4th minute is 200 m
- - -
- - -
• In the 8th minute he travels another 50m. So the total distance at the end of the 8th minute is 400 m
So he takes 8 minutes to travel 400 m from home to school. We do not have to write the 8 steps above. The result, which is '8 minutes' can be obtained simply by dividing the total distance by speed. That is., Time = Distance ÷ speed = 400 m ÷ 50 m / minute = 8 minutes.
For both the methods, the speed must be uniform. That is., he must travel the same 50 m in every minute. If he travels 50 m in one minute and 52 or 45 m in another minute, the speed is not uniform, and our calculations will not work. For our present discussion, whenever we say speed, uniform speed is implied.
So we have calculated the time required for the travel from home to school. Suppose, instead of walking, he rides a cycle. Let the speed of cycle be 80 m/ minute. Then the time required for the travel will be 400/80 = 5 minutes.
Suppose he travels by a car. Let the speed of the car be 200 m/minute. Then the time required for the travel will be 400/200 = 2 minutes.
Let us arrange the above results in a table. It is shown below:
Two quantities are tabulated: 1. Speed (denoted as x) and 2. Time of travel (denoted as y).
These two quantities vary.
• When speed increases, time decreases.
• When speed decreases, time increases.
That is., When one quantity increases, the other decreases and vice versa. In the lower most row of the table, the product xy of the two quantities are written. We see that it is a constant (equal to 400) in all the columns.
Let us take another example. The example we saw above was the time taken to travel a certain distance. Now we will see the time taken to do a work. Let us dee the details:
A container holds a large number of bricks. The work to be done is the 'unloading all the bricks from the container'. From previous experience, the following information is already known:
• If 4 workers are employed, they will unload all the bricks in 2 days.
With this information, we want to calculate the number of days required when different number of workers are employed. Let us see the calculations:
• Let the ‘average quantity’ of work done by a single worker on a single day be equal to ‘u’.
[Note the use of the word 'average' in the above step. The reason for using it can be explained as follows:
All the workers will not be able to do the same exact quantity of work on a day. Some workers will do more work, and some may do less work. So it is appropriate to take the 'average quantity'. This average is applied to all the workers. That is., it is assumed that all the workers do an average of 'u' on a day]
• So the total work done by 2 workers in 1 day = 2u
• So the total work done by 4 workers in 2 days = 4 x 2u = 8u
• 8u is the total work (unloading all the bricks) to be done. It will not change, and hence is a constant.
Now we want to know the different possibilities:
(1) 2 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 2 workers in 1 day = 2u
• No. of days = Total work / Work done by 2 workers in one day = 8u / 2u = 4 days
(2) 8 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 8 workers in 1 day = 8u
• No. of days = Total work / Work done by 8 workers in one day = 8u / 8u = 1 day
(3) 6 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 6 workers in 1 day = 6u
• No. of days = Total work / Work done by 6 workers in one day = 8u / 6u = 4⁄3 days
4⁄3 days is one day and 1⁄3 days. But 1⁄3 day = (8 × 1⁄3) hours = 8⁄3 = 2 2⁄3 hours (assuming one day = 8 hours of work). So 4⁄3 days = One day plus 2 2⁄3 hours. But for our present discussion, it is sufficient to write 4/3 days.
(4) 5 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 5 workers in 1 day = 5u
• No. of days = Total work / Work done by 5 workers in one day = 8u / 5u = 8⁄5 days
(5) 3 workers are employed. How many days are required ?
We have:
• Work done by one worker in 1 day = u
• So the work done by 3 workers in 1 day = 3u
• No. of days = Total work / Work done by 3 workers in one day = 8u / 3u = 8⁄3 days. Note that this is double the number of days when 6 workers are employed.
Let us now tabulate the above results. It is shown below:
Two quantities are tabulated: 1. Number of days (denoted as x) 2. Number of workers (denoted as y). These two quantities vary.
• When No. of workers increases, time decreases.
• When No. of workers decreases, time increases. In such a situation, we say: When one quantity increases, the other decreases and vice versa.
In the lower most row of the table, the product xy of the two quantities are written. We see that it is a constant (equal to 8.00) in all the columns.
In the next section, we will see some solved examples based on the above discussion.
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