In the previous sections we saw the type of variation in which when one quantity increases, the other decreases and vice versa. In this section, we will see some solved examples.
Solved example 3.17
A school has a total fund of ₹ 28000/- to buy chairs. There are 4 types of chairs available in the market: Types A, B, C and D. Their costs are shown in the table below:
How many chairs can be purchased if each of the four types are chosen? Form a table showing the variation of two quantities: (1) Cost of one chair (2) No. of chairs that can be purchased. What is the nature of the variation ?
Solution:
• If type A is chosen, the number that can be purchased is 28000 / 300 = 93.33
[Note that 0.33 numbers of chairs cannot be purchased. In such cases, we ignore the decimal part. Only 93 chairs will be purchased. The balance amount will be 0.33 x 300 = ₹ 99.00. This will remain in the school fund. For our discussion purpose, the number can be taken as 93.33]
• If type B is chosen, the number that can be purchased is 28000 / 250 = 112
• If type C is chosen, the number that can be purchased is 28000 / 375 = 74.67
• If type D is chosen, the number that can be purchased is 28000 / 224 = 125
The table showing the above results is given below:
We can see that, when the cost increases, no. of chairs decreases and vice versa. Also, the product xy is a constant in all the columns.
Solved example 3.18
Duration of a conference meeting is fixed as 4 hours. It is to be divided into equal sessions. The number of sessions should not be less than 3. Also it should not be greater than 7. Show that when the number of sessions increase, the duration of sessions decrease.
Solution:
The total duration of the conference is fixed as 4 hours. It is equal to 4 x 60 = 240 minutes.
Let us divide this available time into equal sessions:
• When the number of sessions is 3, duration of each session = 240/3 = 80 minutes.
• When the number of sessions is 4, duration of each session = 240/4 = 60 minutes.
• When the number of sessions is 5, duration of each session = 240/5 = 48 minutes.
• When the number of sessions is 6, duration of each session = 240/6 = 40 minutes.
• When the number of sessions is 7, duration of each session = 240/7 = 34.3 minutes.
Now we will tabulate the above results. The two quantities are: 1. Number of sessions and, 2. Duration of each session. It is shown below:
We can see that when the number of sessions increase, the duration of sessions decrease and vice versa. Also the product of the two quantities is always a constant.
In all the above examples,
Solved example 3.17
A school has a total fund of ₹ 28000/- to buy chairs. There are 4 types of chairs available in the market: Types A, B, C and D. Their costs are shown in the table below:
| Type | A | B | C | D |
|---|---|---|---|---|
| Cost (₹) | 300 | 250 | 375 | 224 |
Solution:
• If type A is chosen, the number that can be purchased is 28000 / 300 = 93.33
[Note that 0.33 numbers of chairs cannot be purchased. In such cases, we ignore the decimal part. Only 93 chairs will be purchased. The balance amount will be 0.33 x 300 = ₹ 99.00. This will remain in the school fund. For our discussion purpose, the number can be taken as 93.33]
• If type B is chosen, the number that can be purchased is 28000 / 250 = 112
• If type C is chosen, the number that can be purchased is 28000 / 375 = 74.67
• If type D is chosen, the number that can be purchased is 28000 / 224 = 125
The table showing the above results is given below:
We can see that, when the cost increases, no. of chairs decreases and vice versa. Also, the product xy is a constant in all the columns.
Solved example 3.18
Duration of a conference meeting is fixed as 4 hours. It is to be divided into equal sessions. The number of sessions should not be less than 3. Also it should not be greater than 7. Show that when the number of sessions increase, the duration of sessions decrease.
Solution:
The total duration of the conference is fixed as 4 hours. It is equal to 4 x 60 = 240 minutes.
Let us divide this available time into equal sessions:
• When the number of sessions is 3, duration of each session = 240/3 = 80 minutes.
• When the number of sessions is 4, duration of each session = 240/4 = 60 minutes.
• When the number of sessions is 5, duration of each session = 240/5 = 48 minutes.
• When the number of sessions is 6, duration of each session = 240/6 = 40 minutes.
• When the number of sessions is 7, duration of each session = 240/7 = 34.3 minutes.
Now we will tabulate the above results. The two quantities are: 1. Number of sessions and, 2. Duration of each session. It is shown below:
We can see that when the number of sessions increase, the duration of sessions decrease and vice versa. Also the product of the two quantities is always a constant.
In all the above examples,
• When one quantity changes, the other also changes. The change is proportional.
• But the change is such that, when one quantity increases, the other decreases and vice versa. So it is inverse.
When two quantities change according to the above rules, they are said to be in Inverse proportion. Also their products xy is always a constant. So such cases will satisfy the equation: xy =k. Where k is a constant.
• When two quantities x and y are in inverse proportion, it is some times written as : x ∝ 1⁄y
• It is read as: x proportional to 1⁄y.
In the equation xy = k, left side is a product of two quantities. The right side is a constant. So if one of the quantity increase, the other has to decrease. Then only the right side k will remain as a constant. This property can be effectively used to calculate unknown quantities if two quantities are known to be in inverse proportion. We will see a few such problems below:
• When two quantities x and y are in inverse proportion, it is some times written as : x ∝ 1⁄y
• It is read as: x proportional to 1⁄y.
In the equation xy = k, left side is a product of two quantities. The right side is a constant. So if one of the quantity increase, the other has to decrease. Then only the right side k will remain as a constant. This property can be effectively used to calculate unknown quantities if two quantities are known to be in inverse proportion. We will see a few such problems below:
Solved example 3.19
A camp has 80 participants. The food provisions in the camp will last for 15 days. 25 more participants join the camp. For how many days will the food provisions last?
Solution:
In this problem, the quantity of food provisions available is a constant. That is., it does not change. The organisers of the camp know that this available provision will last for 15 days if the no. of participants is 80. How do they know it? Let us analyse:
From previous camps, the organisers know that each participant will consume an average provision of ‘u’ on a single day. So the provision consumed by 80 participants on a single day will be 80u. The organisers divide the available provision by 80u. The result they got is 15.
Example:
Let u = 0.75 kg of rice.
For 80 participants rice required on a single day = 80u = 80 × 0.75 = 60 kg
If the total quantity of rice available is 900 kg, It will last for 900/60 = 15 days.
In our problem, 25 new participants arrived at the camp. The total no. of participants increase to 105. The new participants will also be consuming u per day. So the total consumption per day increases from 80u to 105u. It is a proportional increase. Obviously, the available provision will not last for 15 days. The no. of days will proportionately decrease. So it is a case of inverse proportion. We are required to find the new number of days. Let us form a table:
• Look at column (iii). 80u is multiplied with 15, to get 1200u, the total available provision.
• 105u, and the corresponding no. of days (denoted as y1) should be given appropriate places in the same table. And their product is also equal to 1200u. So we can write:
105u × y1 = 1200u ⇒ y1 = 1200u / 105u = 11.43 days (Eleven and a half days approximately). Note that the unknown 'u' cancels out from the numerator and denominator.
The above method involving an analysis with 'u' was shown just to give a basic understanding of the problem. We can solve it directly by using only the given data. The table so formed is given below:
The table is self explanatory. 105 and y1 should give the same constant 1200. So we can write:
105 × y1 = 1200 ⇒ y1 = 1200 / 105 = 11.43 days (same as before).
Note that for direct proportion, the constant was y⁄x. Here, for inverse proportion, the constant is xy
Solved example 3.20
14 workers can do a work in 42 days. How many days will it take if the number of workers is decreased to 12 ?
Solution:
First we will see the table with basic details:
The table is self explanatory.
• ‘u’ is the average work done by a single worker in a single day.
• So 14 workers will do a work of 14u on a single day.
• In 42 days, these 14 workers will do 42 x 14 u = 588u. This is the total work to be completed.
• If the number of workers increase, the number of days will decrease and vice versa. So this is a case of inverse proportion
• If there are 12 workers, the work done 12u by them on a single day, and the corresponding no. of days y1 should also be given appropriate places in the table.
• 12u and y1 will also give the same total work 588u. So we can write:
12u × y1 = 588u ⇒ y1 = 588u / 12u = 49 days
Once we understand the basics, we can write the table directly from the given data:
12 and y1 should give the same constant 588. So we can write:
12 × y1 = 588 ⇒ y1 = 588 / 12 = 49 days (same as before).
In the next section, we will see a few more solved examples.
105 × y1 = 1200 ⇒ y1 = 1200 / 105 = 11.43 days (same as before).
Note that for direct proportion, the constant was y⁄x. Here, for inverse proportion, the constant is xy
Solved example 3.20
14 workers can do a work in 42 days. How many days will it take if the number of workers is decreased to 12 ?
Solution:
First we will see the table with basic details:
The table is self explanatory.
• ‘u’ is the average work done by a single worker in a single day.
• So 14 workers will do a work of 14u on a single day.
• In 42 days, these 14 workers will do 42 x 14 u = 588u. This is the total work to be completed.
• If the number of workers increase, the number of days will decrease and vice versa. So this is a case of inverse proportion
• If there are 12 workers, the work done 12u by them on a single day, and the corresponding no. of days y1 should also be given appropriate places in the table.
• 12u and y1 will also give the same total work 588u. So we can write:
12u × y1 = 588u ⇒ y1 = 588u / 12u = 49 days
Once we understand the basics, we can write the table directly from the given data:
12 and y1 should give the same constant 588. So we can write:
12 × y1 = 588 ⇒ y1 = 588 / 12 = 49 days (same as before).
In the next section, we will see a few more solved examples.
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