In the previous section we formulated a rule to calculate y for the case of 'paint and turpentine'. In this section we will try to do it for the other two examples.
Take table 3.2. Take any columns and calculate y/x.
• Column (iii) → y/x = 19/0.5 = 38
• Column (v) → y/x = 190/5 = 38
• Column (vii) → y/x = 456/12 = 38
• Column (viii) → y/x = 532/14 = 38
Here also y/x is a constant. So we can calculate the bill amount just by multiplying the quantity of rice with 38. That is., y = 38x
Take table 3.3. Take any columns and calculate y/x.
• Column (iii) → y/x = 0.5/1 = 1/2
• Column (v) → y/x = 2.5/5 = 1/2
• Column (viii) → y/x = 6/12 = 1/2
• Column (ix) → y/x = 7.5/15 = 1/2
Here also y/x is a constant. So we can calculate the 'no. of spoons of tea dust required', just by multiplying the 'no. of cups of tea' with 1/2. That is., y = x/2
So we have seen three examples which satisfy the relation: y = a constant × x. The constant is usually denoted by 'k'. So we write the relation as: y = kx. This is an equation. Which means both sides of the '=' sign will always be the same. So, if one side increase, the other side will also increase. And if one side decrease, the other side will also decrease. But k is a constant. It cannot increase or decrease. So, to maintain the equality, it is 'x' that will be increasing or decreasing on the right side of the '=' sign.
Let us summarize what we have learned so far in this chapter:
• We have seen three examples of situations where variations occur
• In all those examples, the variation is in such a way that, when one quantity increase, the other quantity also increase
• Also, when one quantity decrease, the other quantity also decrease
• This increase or decrease follows a definite 'rule'
• The rule is: y can always be obtained by multiplying x with a constant k
■ We can see that the quantity y is obtained by multiplying x with a ‘constant’. So what ever the value of x, we multiply it with the same constant. Thus the 'change in y' will be proportional to the 'change in x'.
■ Also we have seen: When there is increase in x, there is increase in y. And when there is decrease in x, there is decrease in y. So the change is direct. The opposite to ‘direct’ is ‘inverse’. In inverse, when there is increase in one quantity, the other quantity decreases. We will learn about it in the next chapter. Here we concentrate on changes which are ‘direct’.
So we have seen two properties:
(1) The changes are proportional
(2) The changes are direct.
If the changes in two quantities satisfy above two properties, then we say that each is directly proportional to the other. And if they are directly proportional to each other, they will satisfy the equation: y = kx.
We will now see some solved examples that will illustrate what we have discussed so far.
Solved example 3.1
In a house, when 12 units of electricity is used, the bill amount is Rs.51.00. When 21 units is used, the bill amount is Rs.89.25. When 18 units is used, the bill amount is Rs.76.50. Check whether the two quantities: (1) Units of electricity used and (2) Bill amount, are in direct proportion.
Solution:
Let us make a table of the given data. It is shown below as Table 3.4:
The next step is to take the y/x ratios in each of the columns from (iii) to (v)
Column (iii) → y/x = 51/12 = 4.25
Column (iv) → y/x = 89.25/21 = 4.25
Column (v) → y/x = 76.5/18 = 4.25
Thus we find that all the y/x ratios give the same result. So y/x = A constant k. Which is same as: y =kx. Also, when x increases, y also increases. The two quantities are in direct proportion.
Solved example 3.2
A car uses 3 litres of petrol to travel 48 km. When it travels 80 km, the quantity of petrol used is 5 litres. And when it travels 128 km, the quantity is 8 litres.
(i) Check whether the two quantities: (1) Quantity of petrol used and (2) Distance travelled, are in direct proportion.
(ii) If they are in direct proportion, find
(a) The distance that can be travelled using 6 litres of petrol.
(b) The quantity of petrol required to travel 20 km.
Solution:
Let us make a table of the given data. It is shown below:
The next step is to take the y/x ratios in each of the columns from (iii) to (v)
Column (iii) → y/x = 48/3 = 16
Column (iv) → y/x = 80/5= 16
Column (v) → y/x = 128/8 = 16
Thus we find that all the y/x ratios give the same result. So y/x = A constant k. Which is same as: y = kx. Also, when x increases, y also increases. The two quantities are in direct proportion.
(ii) Here we have to find
(a) the distance corresponding to a quantity of 6 litres, and
(b) the quantity corresponding to a distance of 20 km.
If there is direct proportion, the above [quantity of 6 litres, and the distance y1 corresponding to it] and the [distance of 20 km and the quantity of petrol x2 corresponding to it] will also have appropriate places in the table 3.5 given above. So the table can be modified as shown below:
6 litres is given in a new column (iv), and 20 km is given in another new column (vii). These new values will also satisfy the relation y/x = 16. Thus we get:
• y1/6 = 16 ⇒ y1 = 16 x 6 =96 km
• 20/x2 = 16 ⇒ x2 = 20/16 = 1.25 litres
It may be noted that, once we understand the basics, it will not be necessary to make a modified table as table 3.5 above. We can apply the equation y =kx directly. Thus:
y = kx ⇒ y1 = 16 x 6 =96 km
y = kx ⇒ 20 = 16 × x2 ⇒ x2 = 20/16 = 1.25 litres
In the next section we will see more solved examples.
Take table 3.2. Take any columns and calculate y/x.
• Column (iii) → y/x = 19/0.5 = 38
• Column (v) → y/x = 190/5 = 38
• Column (vii) → y/x = 456/12 = 38
• Column (viii) → y/x = 532/14 = 38
Here also y/x is a constant. So we can calculate the bill amount just by multiplying the quantity of rice with 38. That is., y = 38x
Take table 3.3. Take any columns and calculate y/x.
• Column (iii) → y/x = 0.5/1 = 1/2
• Column (v) → y/x = 2.5/5 = 1/2
• Column (viii) → y/x = 6/12 = 1/2
• Column (ix) → y/x = 7.5/15 = 1/2
Here also y/x is a constant. So we can calculate the 'no. of spoons of tea dust required', just by multiplying the 'no. of cups of tea' with 1/2. That is., y = x/2
So we have seen three examples which satisfy the relation: y = a constant × x. The constant is usually denoted by 'k'. So we write the relation as: y = kx. This is an equation. Which means both sides of the '=' sign will always be the same. So, if one side increase, the other side will also increase. And if one side decrease, the other side will also decrease. But k is a constant. It cannot increase or decrease. So, to maintain the equality, it is 'x' that will be increasing or decreasing on the right side of the '=' sign.
Let us summarize what we have learned so far in this chapter:
• We have seen three examples of situations where variations occur
• In all those examples, the variation is in such a way that, when one quantity increase, the other quantity also increase
• Also, when one quantity decrease, the other quantity also decrease
• This increase or decrease follows a definite 'rule'
• The rule is: y can always be obtained by multiplying x with a constant k
■ We can see that the quantity y is obtained by multiplying x with a ‘constant’. So what ever the value of x, we multiply it with the same constant. Thus the 'change in y' will be proportional to the 'change in x'.
■ Also we have seen: When there is increase in x, there is increase in y. And when there is decrease in x, there is decrease in y. So the change is direct. The opposite to ‘direct’ is ‘inverse’. In inverse, when there is increase in one quantity, the other quantity decreases. We will learn about it in the next chapter. Here we concentrate on changes which are ‘direct’.
So we have seen two properties:
(1) The changes are proportional
(2) The changes are direct.
If the changes in two quantities satisfy above two properties, then we say that each is directly proportional to the other. And if they are directly proportional to each other, they will satisfy the equation: y = kx.
We will now see some solved examples that will illustrate what we have discussed so far.
Solved example 3.1
In a house, when 12 units of electricity is used, the bill amount is Rs.51.00. When 21 units is used, the bill amount is Rs.89.25. When 18 units is used, the bill amount is Rs.76.50. Check whether the two quantities: (1) Units of electricity used and (2) Bill amount, are in direct proportion.
Solution:
Let us make a table of the given data. It is shown below as Table 3.4:
 |
| Table 3.4 |
Column (iii) → y/x = 51/12 = 4.25
Column (iv) → y/x = 89.25/21 = 4.25
Column (v) → y/x = 76.5/18 = 4.25
Thus we find that all the y/x ratios give the same result. So y/x = A constant k. Which is same as: y =kx. Also, when x increases, y also increases. The two quantities are in direct proportion.
Solved example 3.2
A car uses 3 litres of petrol to travel 48 km. When it travels 80 km, the quantity of petrol used is 5 litres. And when it travels 128 km, the quantity is 8 litres.
(i) Check whether the two quantities: (1) Quantity of petrol used and (2) Distance travelled, are in direct proportion.
(ii) If they are in direct proportion, find
(a) The distance that can be travelled using 6 litres of petrol.
(b) The quantity of petrol required to travel 20 km.
Solution:
Let us make a table of the given data. It is shown below:
 |
| Table 3.5 |
Column (iii) → y/x = 48/3 = 16
Column (iv) → y/x = 80/5= 16
Column (v) → y/x = 128/8 = 16
Thus we find that all the y/x ratios give the same result. So y/x = A constant k. Which is same as: y = kx. Also, when x increases, y also increases. The two quantities are in direct proportion.
(ii) Here we have to find
(a) the distance corresponding to a quantity of 6 litres, and
(b) the quantity corresponding to a distance of 20 km.
If there is direct proportion, the above [quantity of 6 litres, and the distance y1 corresponding to it] and the [distance of 20 km and the quantity of petrol x2 corresponding to it] will also have appropriate places in the table 3.5 given above. So the table can be modified as shown below:
 |
| Table 3.6 |
• y1/6 = 16 ⇒ y1 = 16 x 6 =96 km
• 20/x2 = 16 ⇒ x2 = 20/16 = 1.25 litres
It may be noted that, once we understand the basics, it will not be necessary to make a modified table as table 3.5 above. We can apply the equation y =kx directly. Thus:
y = kx ⇒ y1 = 16 x 6 =96 km
y = kx ⇒ 20 = 16 × x2 ⇒ x2 = 20/16 = 1.25 litres
In the next section we will see more solved examples.
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