In the previous section we completed the discussion on decimals. In this chapter, we will discuss about ratios.
Ratios are used to compare quantities. Consider an example: There are two steel rods. Rod A and Rod B, as shown in fig.7.1 below:
One is longer than the other. This, we may be able to say just by looking at them. Especially if the longer one has a far greater length than the other. When we say that one is longer than the other, we are making a comparison. But in mathematics, just saying 'one is longer than the other' is not sufficient. We must be more specific. Let us see how we can give a specific comparison:
• Measure the length of A. Let it be 12 cm
• Measure the length of B. Let it be 24 cm
We know that 24 = 2 × 12. So we can write a specific comparison:
The length of Rod B is 'two times' the length of Rod A.
The mathematical way of writing this is:
The ratio of the length of Rod A to the length of Rod B is 12:24. This ratio 12:24 can be reduced to it's simplest form. Dividing both sides by 12 we get 1:2. [Just as we reduce a fraction to it's simplest form by dividing both the numerator and the denominator by the same number, here also, we must divide both sides of the ':' sign by the same number.]
So we can write:
■ The ratio of the length of Rod A to the length of Rod B is 1:2
Another way of writing this is:
■ Length of Rod A : Length of Rod B = 1:2
Yet another way is:
■ Lengths of Rod A and Rod B are in the ratio 1:2
It is very important to preserve the order of mentioning the items. In all the above three methods, Rod A was mentioned first, and B second. So in the final ratio 1:2, '1' must correspond to Rod A and '2' must correspond to Rod B.
Mentioning Rod B first and Rod A second is the inverse of the above ratio. Let us see how it is written:
We know that 12 = 24 × 1⁄2. So we can write: The length of the smaller rod is 'half' the length of the longer rod. The mathematical ways of writing this is:
■ The ratio of the length of Rod B to the length of Rod A is 24 : 12 = 2:1
■ Length of Rod B : Length of Rod A = 2:1
■ Lengths of Rod B and Rod A are in the ratio 2:1
Another example: A student is 150 cm tall. Her younger brother is 100 cm tall. Compare the heights.• Height of student = 150 cm
• Height of younger brother = 100 cm
We know that 100 = 150 × 2⁄3
So we write: Height of younger brother is 2⁄3 times the height of the student
The mathematical ways are:
■ The ratio of the height of the student to the height of her younger brother is 150:100 = 3:2 (dividing both sides by 50)
■ Height of student : Height of her younger brother = 3:2
■ Heights of the student and her younger brother are in the ratio 3:2
We can write the inverse also:
■ The ratio of the height of the younger brother to the height of the student is 100:150 = 2:3
■ Height of the younger brother: Height of the student = 2:3
■ Heights of the younger brother and the student are in the ratio 2:3
Yet another example: Consider two vessels A and B. A has a capacity of 27 litres. B has a capacity of 45 litres. Compare the capacities.
• Capacity of vessel A = 27 litres
• Capacity of vessel B = 45 litres
The ratio can be written as:
■ The ratio of the capacity of vessel A to the capacity of vessel B is 27 : 45 = 3:5 (dividing both sides by 9)
■ Capacity of vessel A: Capacity of vessel B = 3:5
■ Capacities of vessel A and vessel B are in the ratio 3:5
The inverse ratio:
■ The ratio of the capacity of vessel B to the capacity of vessel A is 45 : 27 = 5:3
■ Capacity of vessel B: Capacity of vessel A = 5:3
■ Capacities of vessel B and vessel A are in the ratio 5:3
It is very important to ensure that the quantities on both sides of the ':' symbol have the same unit. For example, if the length or height on one side is in cm, the other side must also be in cm. If the capacity on one side is in litres, the other side must also be in litres.
Now we will see some solved examples:
Solved example 7.1
Weight of a parcel A is 4 kg. Weight of another parcel B is 12 kg. What is the ratio of the weight of A to weight of B
Solution:
Weight of A : Weight of B = 4 : 12 = 1:3 (dividing both sides by 4)
Solved example 7.2
Length of a bridge A is 1.2 km. Length of another bridge B is 800 m. What is the ratio of their lengths?
Solution:
To write the ratio, both quantities must be in the same unit. So we will convert km to m.
We have: 1.2 km = 1.2 × 1000 = 1200 m
So we can write:
Length of bridge A : Length of bridge B = 1200 : 800 = 3 : 2 (dividing both sides by 400)
We can take a same size print out of the picture. It will be small enough to be pasted in a small pocket album. But if we want to paste it in an album, we can take a larger picture. Fig.7.3 below shows some enlarged pictures.
Fig.7.3(a) shows the original picture. Fig.(b) shows an enlarged picture. But some thing just does not seem to be right. The flower does not appear the same. Some thing went wrong while the enlargement was done. What could it be? Let us analyse:
The enlarged picture has a height of 10 cm and a width of 12 cm. It is evident that while enlarging the picture, it’s original width 6 was doubled to 12 cm. But then, the height also should have doubled. The original height 8 should have become 16 cm. But here it is only 10. This is a 'disproportionate enlargement'. Such an enlargement will cause distortion to the picture. To avoid this, when width increases, height should increase proportionately.
Let us take ratios. Ratios of height to width.
• In the original picture, this ratio is 8:6 which is same as 4:3.
• In the enlarged picture this ratio is 10:12 which is same as 5:3.
In the two ratios, the right side is the same 3. But the left sides are different: In the original, it is 4, and in the enlarged one, it is 5.
Now consider fig.(c). It is also an enlarged picture of the original. It has a height of 16 cm and a width of 7.5 cm. It is evident that while enlarging the picture, it’s original height 8 was doubled to 16 cm. But then, the width also should have doubled. The original width 6 should have become 12 cm. But here it is only 7.5. This is also 'disproportionate enlargement'. Such an enlargement has caused distortion to the picture. To avoid this, when height increases, width should increase proportionately.
Let us take ratios. Ratios of height to width.
• In the original picture, this ratio is 8:6 which is same as 4:3.
• In the enlarged picture in (c), this ratio is 16:7.5 which is same as 4:1.875.
In the two ratios, the left side is the same 4. But the right sides are different. In the original, it is 3, and in the enlarged one, it is 1.875.
Now consider fig.7.4 below. Fig.7.4(a) shows the original, and (b) shows the enlarged picture. Here, both the height and width are doubled. The original height 8 cm has become 16, and the original width 6 has become 12. The enlarged picture shows no distortion.
Let us take ratios as before.
• In the original picture, this ratio is 8:6 which is same as 4:3.
• In the enlarged picture in (b), this ratio is 16:12 which is same as 4:3
We see that the two ratios are the same.
So, distortion can be avoided by keeping the ratios the same. When the ratios are kept the same, there will be a 'proportionate increase' for both the height and width.
This is applicable when reducing the sizes also. If we want to make a picture smaller, we must keep the ratio the same. Then, there will be a proportionate decrease in both the height and width.
For example, the original picture which is 8 x 6 can be reduced to 5 x 3.75. Let us take the ratio of length to width for the new reduced size:
• 5:3.75. Dividing both sides by 1.25 we get 4:3, which is the same original ratio. So in this case, there will be proportionate decrease in height and width, and so distortion can be avoided.
Let us consider the different sizes of the picture in which there is no distortion:
■ Original size: 8 x 6. In ratio form it is 8:6
■ Enlarged size: 16 x 12. In ratio form it is 16:12
■ Reduced size: 5 x 3.75. In ratio form it is 5:3.75
All the above 3 ratios are related. Any one of them can be derived from any other by dividing or multiplying both sides by the same number. Example: 5:3.75 can be derived from 16:12 by dividing both sides by 3.2. Such ratios are called Equivalent ratios. This is similar to equivalent fractions which are obtained by dividing or multiplying both numerator and the denominator by the same number.
Equivalent ratios have many applications. Consider the national flag. National flags of many countries are rectangular in shape. They can be made in different sizes. But the ratio must be maintained. In other words, the ratio of length to width in every size must be equivalent.
Another example: In a computer lab, there are 2 computers for every 4 students. How many computers will be required for 30 students?
Solution:
Let us analyse the problem: there are 2 computers for every 4 students. That means 4 students can use 2 computers. This is same as 2 students using 1 computer. When the number of students increase, the number of computers must also increase proportionately. Other wise more than 2 students will have to gather around each computer.
Let us take ratios:
ratio of number of computers to number of students. This ratio is 2:4 which is same as 1:2 (dividing both sides by 2)
This ratio must be always maintained. For that, the ‘new ratio’ that we take, with the new number of students must be an equivalent ratio of 1:2. Our task is to find this equivalent ratio.
Let us put ‘x’ in the place of the ‘new number of computers’ which is the unknown quantity. So the new ratio is x:30. This must be an equivalent ratio of 1:2. That means., we can multiply or divide both sides of 1:2 to get x:30.
If we multiply the right side 2 with 15, we will get 30. So we must multiply the left side 1 also by 15 So we can write:
• 1:2 is equivalent to (1 × 15) : (2 × 15)
• 1:2 is equivalent to 15 :30
Therefore x = 15.
So we can say: When the number of students increase to 30, the number of computers must be 15.
Scales of maps and technical drawings use the same concept of equivalent fractions. Details can be seen here. More about proportions can be seen here.
In the next section we will see the relation between ratios and fractions.
Ratios are used to compare quantities. Consider an example: There are two steel rods. Rod A and Rod B, as shown in fig.7.1 below:
 |
| Fig.7.1 |
• Measure the length of A. Let it be 12 cm
• Measure the length of B. Let it be 24 cm
We know that 24 = 2 × 12. So we can write a specific comparison:
The length of Rod B is 'two times' the length of Rod A.
The mathematical way of writing this is:
The ratio of the length of Rod A to the length of Rod B is 12:24. This ratio 12:24 can be reduced to it's simplest form. Dividing both sides by 12 we get 1:2. [Just as we reduce a fraction to it's simplest form by dividing both the numerator and the denominator by the same number, here also, we must divide both sides of the ':' sign by the same number.]
So we can write:
■ The ratio of the length of Rod A to the length of Rod B is 1:2
Another way of writing this is:
■ Length of Rod A : Length of Rod B = 1:2
Yet another way is:
■ Lengths of Rod A and Rod B are in the ratio 1:2
It is very important to preserve the order of mentioning the items. In all the above three methods, Rod A was mentioned first, and B second. So in the final ratio 1:2, '1' must correspond to Rod A and '2' must correspond to Rod B.
Mentioning Rod B first and Rod A second is the inverse of the above ratio. Let us see how it is written:
We know that 12 = 24 × 1⁄2. So we can write: The length of the smaller rod is 'half' the length of the longer rod. The mathematical ways of writing this is:
■ The ratio of the length of Rod B to the length of Rod A is 24 : 12 = 2:1
■ Length of Rod B : Length of Rod A = 2:1
■ Lengths of Rod B and Rod A are in the ratio 2:1
Another example: A student is 150 cm tall. Her younger brother is 100 cm tall. Compare the heights.• Height of student = 150 cm
• Height of younger brother = 100 cm
We know that 100 = 150 × 2⁄3
So we write: Height of younger brother is 2⁄3 times the height of the student
The mathematical ways are:
■ The ratio of the height of the student to the height of her younger brother is 150:100 = 3:2 (dividing both sides by 50)
■ Height of student : Height of her younger brother = 3:2
■ Heights of the student and her younger brother are in the ratio 3:2
We can write the inverse also:
■ The ratio of the height of the younger brother to the height of the student is 100:150 = 2:3
■ Height of the younger brother: Height of the student = 2:3
■ Heights of the younger brother and the student are in the ratio 2:3
Yet another example: Consider two vessels A and B. A has a capacity of 27 litres. B has a capacity of 45 litres. Compare the capacities.
• Capacity of vessel A = 27 litres
• Capacity of vessel B = 45 litres
The ratio can be written as:
■ The ratio of the capacity of vessel A to the capacity of vessel B is 27 : 45 = 3:5 (dividing both sides by 9)
■ Capacity of vessel A: Capacity of vessel B = 3:5
■ Capacities of vessel A and vessel B are in the ratio 3:5
The inverse ratio:
■ The ratio of the capacity of vessel B to the capacity of vessel A is 45 : 27 = 5:3
■ Capacity of vessel B: Capacity of vessel A = 5:3
■ Capacities of vessel B and vessel A are in the ratio 5:3
It is very important to ensure that the quantities on both sides of the ':' symbol have the same unit. For example, if the length or height on one side is in cm, the other side must also be in cm. If the capacity on one side is in litres, the other side must also be in litres.
Now we will see some solved examples:
Solved example 7.1
Weight of a parcel A is 4 kg. Weight of another parcel B is 12 kg. What is the ratio of the weight of A to weight of B
Solution:
Weight of A : Weight of B = 4 : 12 = 1:3 (dividing both sides by 4)
Solved example 7.2
Length of a bridge A is 1.2 km. Length of another bridge B is 800 m. What is the ratio of their lengths?
Solution:
To write the ratio, both quantities must be in the same unit. So we will convert km to m.
We have: 1.2 km = 1.2 × 1000 = 1200 m
So we can write:
Length of bridge A : Length of bridge B = 1200 : 800 = 3 : 2 (dividing both sides by 400)
Equivalent ratios
Consider the picture of a flower shown in fig.7.2 below. It is a small rectangular picture. The rectangle has a height of 8 cm and a width of 6 cm. |
| Fig.7.2 |
 |
| Fig.7.3 |
The enlarged picture has a height of 10 cm and a width of 12 cm. It is evident that while enlarging the picture, it’s original width 6 was doubled to 12 cm. But then, the height also should have doubled. The original height 8 should have become 16 cm. But here it is only 10. This is a 'disproportionate enlargement'. Such an enlargement will cause distortion to the picture. To avoid this, when width increases, height should increase proportionately.
Let us take ratios. Ratios of height to width.
• In the original picture, this ratio is 8:6 which is same as 4:3.
• In the enlarged picture this ratio is 10:12 which is same as 5:3.
In the two ratios, the right side is the same 3. But the left sides are different: In the original, it is 4, and in the enlarged one, it is 5.
Now consider fig.(c). It is also an enlarged picture of the original. It has a height of 16 cm and a width of 7.5 cm. It is evident that while enlarging the picture, it’s original height 8 was doubled to 16 cm. But then, the width also should have doubled. The original width 6 should have become 12 cm. But here it is only 7.5. This is also 'disproportionate enlargement'. Such an enlargement has caused distortion to the picture. To avoid this, when height increases, width should increase proportionately.
Let us take ratios. Ratios of height to width.
• In the original picture, this ratio is 8:6 which is same as 4:3.
• In the enlarged picture in (c), this ratio is 16:7.5 which is same as 4:1.875.
In the two ratios, the left side is the same 4. But the right sides are different. In the original, it is 3, and in the enlarged one, it is 1.875.
Now consider fig.7.4 below. Fig.7.4(a) shows the original, and (b) shows the enlarged picture. Here, both the height and width are doubled. The original height 8 cm has become 16, and the original width 6 has become 12. The enlarged picture shows no distortion.
 |
| Fig.7.4 |
• In the original picture, this ratio is 8:6 which is same as 4:3.
• In the enlarged picture in (b), this ratio is 16:12 which is same as 4:3
We see that the two ratios are the same.
So, distortion can be avoided by keeping the ratios the same. When the ratios are kept the same, there will be a 'proportionate increase' for both the height and width.
This is applicable when reducing the sizes also. If we want to make a picture smaller, we must keep the ratio the same. Then, there will be a proportionate decrease in both the height and width.
For example, the original picture which is 8 x 6 can be reduced to 5 x 3.75. Let us take the ratio of length to width for the new reduced size:
• 5:3.75. Dividing both sides by 1.25 we get 4:3, which is the same original ratio. So in this case, there will be proportionate decrease in height and width, and so distortion can be avoided.
Let us consider the different sizes of the picture in which there is no distortion:
■ Original size: 8 x 6. In ratio form it is 8:6
■ Enlarged size: 16 x 12. In ratio form it is 16:12
■ Reduced size: 5 x 3.75. In ratio form it is 5:3.75
All the above 3 ratios are related. Any one of them can be derived from any other by dividing or multiplying both sides by the same number. Example: 5:3.75 can be derived from 16:12 by dividing both sides by 3.2. Such ratios are called Equivalent ratios. This is similar to equivalent fractions which are obtained by dividing or multiplying both numerator and the denominator by the same number.
Equivalent ratios have many applications. Consider the national flag. National flags of many countries are rectangular in shape. They can be made in different sizes. But the ratio must be maintained. In other words, the ratio of length to width in every size must be equivalent.
Another example: In a computer lab, there are 2 computers for every 4 students. How many computers will be required for 30 students?
Solution:
Let us analyse the problem: there are 2 computers for every 4 students. That means 4 students can use 2 computers. This is same as 2 students using 1 computer. When the number of students increase, the number of computers must also increase proportionately. Other wise more than 2 students will have to gather around each computer.
Let us take ratios:
ratio of number of computers to number of students. This ratio is 2:4 which is same as 1:2 (dividing both sides by 2)
This ratio must be always maintained. For that, the ‘new ratio’ that we take, with the new number of students must be an equivalent ratio of 1:2. Our task is to find this equivalent ratio.
Let us put ‘x’ in the place of the ‘new number of computers’ which is the unknown quantity. So the new ratio is x:30. This must be an equivalent ratio of 1:2. That means., we can multiply or divide both sides of 1:2 to get x:30.
If we multiply the right side 2 with 15, we will get 30. So we must multiply the left side 1 also by 15 So we can write:
• 1:2 is equivalent to (1 × 15) : (2 × 15)
• 1:2 is equivalent to 15 :30
Therefore x = 15.
So we can say: When the number of students increase to 30, the number of computers must be 15.
Scales of maps and technical drawings use the same concept of equivalent fractions. Details can be seen here. More about proportions can be seen here.
In the next section we will see the relation between ratios and fractions.
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