Chapter 3.11 - Inverse proportions in Angular measurements

In the previous sections we saw a number of cases where two quantities are in inverse proportion. In this section we will see inverse proportions where angular measurements are involved. 

We know that the central angle of a circle is 360^o. We can divide the area of a circle into sectors by drawing radial lines. We saw this when we learned about pie charts. In the case of pie charts, the size of each sector will depend upon the value of the quantity that it represents. But on many occasions, we will need to divide a circle into a number of equal sectors. Some examples are: Design of emblems and logos, design of toys, machine parts, etc.,

When two radial lines are drawn, a circle will be divided into 2 sectors. Fig.3.21(a) below shows a circle divided into 2 sectors in this way. In the fig.(a), one sector is small and the other is large.

Fig.3.21 Dividing circle into equal sectors

But we want the 2 sectors to be of the same size. For that, we have to draw the two radial lines exactly opposite to each other. This is shown in (b). Now the two sectors are of the same size. As the number of sectors is 2, the central angle of both the sectors is 360/2 = 180^o deg.

• When we draw 3 radial lines, we get 3 sectors as shown in the fig.(c). If all the three are of the same size, central angle of each will be equal to 360/3 = 120^o.  
• When we draw 4 radial lines, we get 4 sectors as shown in the fig.(d). If all the four are of the same size, central angle of each will be equal to 360/4 = 90^o .
• When we draw 5 radial lines, we get 5 sectors as shown in the fig.(e). If all the five are of the same size, central angle of each will be equal to 360/5 = 72^o .
• When we draw 6 radial lines, we get 6 sectors as shown in the fig.(f). If all the six are of the same size, central angle of each will be equal to 360/6 = 60^o .

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We can continue like this to draw any number of radial lines, and we can calculate the central angle in each case. We will now form a table of the above results:

From the table we can see that when the number of sectors increase, the central angle decrease and vice versa. Also the product of the two quantities is always equal to 360. Thus it is a case of inverse proportion. 

Using this information, we can divide a circle into any number of equal sectors. For example, if we want a circle to be divided into 9 equal sectors, the table can be formed as:

The number of sectors 9 and the corresponding angle y1 should be given appropriate places in the table. Also 9 and y1 should give the same constant 360. So we can write:
9 × y1 = 360 ⇒ y1 = 360 / 9 = 40^o

Another application:
A certain  central angle (say 30^o) is given. We have to divide the circle into a number of equal sectors, each having the same 30^o as central angle. How many sectors will be there?

The table can be formed as follows:

The angle 30 deg and the corresponding number of sectors x1 should be given appropriate places in the table. Also 30 and x1 should give the same constant 360. So we can write:
30 × x1 = 360 ⇒ x1 = 360 / 30 = 12 sectors.

Graphs of Inverse proportions

We have earlier seen that the graphs of Direct proportions are straight lines. Let us see if we can get such a definite graphical form for inverse proportions. We will plot the values in the table of the second example because it has more coordinates. The table is given here again for easy verification.

The plot will be as follows:

Fig.3.22 Graph of Inverse proportion

We can see that all the points fall in a curved shape. So the graph for an inverse proportion is a curve. We have plotted only a small portion (between x = 1.0 and x = 2.66) of the curve. The full shape of the curve is as shown in fig.3.23 below:

Fig.3.23 Graph of Inverse proportion

From this graph we can note the following points:
• When we move along the x axis towards the right, the x value increases. The corresponding y values decreases because the graph is falling as we move towards the right.
• When we move along the x axis towards the left, the x value decreases. The corresponding y values increases because the graph is rising as we move towards the left.

This is what is expected from an inverse proportion: When x increases, y decreases and vice versa. We will learn about the applications of such graphs in later chapters.

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So we have completed the discussion on inverse proportions. In the next chapter we will discuss about integers.

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Chapter 3.10 - Inverse proportions - Solved examples

In the previous sections we saw some solved examples on inverse proportions. In this section, we will see a few more.
Solved example 3.21 
A student cycles to school at an average speed of 8 km/hr. He reaches the school in 15 minutes. At what speed should he ride, if he has to reach the school in 10 minutes?
Solution: We will form the table directly from the data. It is shown below:

The distance from home to school remains the same. So if speed increases, the time decreases, and vice versa. Thus it is a case of inverse proportion. 
• The new time 10 minutes and the corresponding new speed x1 should be given appropriate places in the table. 
• Also, x1 and 10 will give the same constant 120. 
So we can write:
10 × x1 = 120 ⇒ x1 = 120 / 10 = 12 km/hr
Solved example 3.22
A work force of 32 men can complete a job in 90 days. 
(i) How many days will it take if the work force is reduced to 24. 
(ii) How many more men should join the present work force, if the job is to be completed in 75 days?
Solution:
The table is given below:

(i) We know that it is a case of inverse proportion because, when the work force increases, time decreases and vice versa. 
• So the new work force 25 and the corresponding new time y1 is given appropriate places in the table as shown above. 
• Also 25 and y1 should give the same constant 2880. 
So we can write:
24 × y1 = 2880 ⇒ y1 = 2880 / 24 = 120 days


(ii) This is an indirect question. They are asking how many 'more'?. But it should not cause any difficulty. We can find the total new number required. This we do by the usual method. Then subtracting the 'old total number of workers' from the 'new total number of workers' will give us the 'more' number of workers.
• The new time 48 days and the corresponding new work force x1 is given appropriate places in the table as shown above. 
• Also 48 and x1 should give the same constant 2880. 
So we can write:
48 × x1 = 2880 ⇒ x1 = 2880 / 48 = 60 workers
So the number of workers needed more = 60 -32 = 28
Solved example 3.23
In a hostel, there are 75 people. The available provisions would last for 35 days. Some people left the hostel on vacation. Now it was found that the provisions would last for 105 days. How many people leave on vacation?
Solution:
This is an indirect question just as in part (ii) of the previous question. We will do it in the usual way to find the final number and then do the subtraction. The table is given below:

• The new time 105 days and the corresponding new no. of people x1 is given appropriate places in the table as shown above. 
• Also 105 and x1 should give the same constant 2625. 
So we can write:
105 × x1 = 2625 ⇒ x1 = 2625 / 105 = 25
So the number of people who left = 75 -25 = 50
Solved example 3.24

Two quantities x and y are inversely proportional to each other. Fill up the missing values in the table given below:

Solution:
• In the table, some columns have x value. They miss the corresponding y value. We have to calculate this missing y value in such columns.
• Some columns have y value. They miss the corresponding x value. We have to calculate this missing x value in such columns.
• We know the equation xy =k. If we have the value of k, and any one of x and y, the other missing value can be calculated (∵ x = k/y and y = k/x). For this method, we have to know the value of k. 
• But k is not given. Or is it?
• k is hidden in the table. If we look at column (vi), we will find that, in it, neither x nor y is missing. We can multiply those values to get k. Thus we get k = 45 × 8 = 360
• Once we have k, we can calculate any missing value. So we proceed as follows:

♦ column (ii):    y = k/x ⇒ y = 360/12 = 30
♦ column (iii):   x = k/y ⇒ x = 360/90 = 4
♦ column (iv):   y = k/x ⇒ y = 360/3 = 120
♦ column (v):    x = k/y ⇒ x = 360/45 = 8
♦ column (vii):  x = k/y ⇒ x = 360/9 = 40

In the next section, we will see inverse variation of angular measurements.

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Chapter 3.9 - Inverse Proportions - Solved examples

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:

Type	A	B	C	D
Cost (₹)	300	250	375	224

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.

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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.

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• 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.

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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.

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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

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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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