In the previous section we completed the discussion on the 'simplest form' of a fraction. In this section we will discuss about 'like fractions'.
The word 'like fractions' is in the plural form. Because we are saying 'fractions', not 'fraction'. So we are talking about more than one fraction here. If these fractions are to fall under the category of like fractions, their denominators must be the same. Consider the following fig.5.21:
Each of the three bars represents a fraction. All the three bars are divided into the same number of equal parts. So all the three fractions will have the same denominator. Thus they are 'like fractions'.
If the denominators are not the same, they are called unlike fractions. Example: 5/11 and 5/10 are unlike fractions.
• When the cake is divided into 3 equal parts, each piece will be larger than when it is divided into 5 equal parts.
• But uncle is offering '2 out the 5 smaller pieces'. Is it larger than '1 out of the bigger 3'? or smaller?
Let us draw a fig. and find out:
From the fig.5.22 it is clear that 2/5 is larger than 1/3. So the nephew should ask for the 2/5 part. The problem is solved. But let us analyse it further:
What caused the nephew confusion? It is the 'methods of division'.
• In the first option, the cake is to be divided into 5 equal parts.
• In the second option, the cake is to be divided into 3 equal parts.
The size of the equal parts will be different in each option. In addition to this, the numerators are different. So any body would be confused.
The problem was solved by drawing a fig. But we will find a mathematical solution to such problems so that, we will not have to draw figures each time we find ourselves in such situations. We know that the 'unequal division' is the main reason for confusion. Can't we assume that they are divided equally? We can. For this we use the 'Equivalent fractions'.
• We have the first option 1/3. We must change it into a 'suitable' equivalent fraction. Let us call it EF1. How do we decide whether EF1 is suitable? We will see that soon.
• We have the second option 2/5. We must change it also into a 'suitable' equivalent fraction. Let us call it EF2.
The suitability of EF1 and EF2 is decided by one condition:
EF1 and EF2 should have the same denominator. So by this condition we are ensuring that the cake is divided into equal parts in both the two options. Let us now try to find these suitable EF1 and EF2:
We will write the possible equivalent fractions of 1/3 and 2/5 side by side in a tabular form as shown below:
From the table we can see that the 'equivalent fraction 5/15 for 1/3' and the 'equivalent fraction 6/15 for 2/5' have the same denominator 15. So we can make the selection of suitable equivalent fractions:
• 'EF1 for 1/3' is 5/15 and • 'EF2 for 2/5' is 6/15. So:
■ When uncle offers 1/3, he is offering 5 pieces out of 15
■ When he offers 2/5, he is offering 6 pieces out of 15.
6 is greater than 5. So the nephew should select the 2/5 option. Thus we can conclude that, whenever we want to compare unlike fractions, we must convert them into like fractions, by using the Equivalent fractions method.
Solved example 5.16
Pick out the larger fraction from among the following pairs
(a) 3/5, 2/3 (b) 2/5, 1/4
Solution:
(i) We will write a few equivalent fractions for 3/5 and 2/3:
From the above table, the suitable Equivalent fraction for 3/5 is 9/15, and that for 2/3 is 10/15. Because both have the same denominator 15, and so are like fractions. So
• 3/5 is 9 parts out of 15, and
• 2/3 is 10 parts out of 15.
Thus 2/3 is larger than 3/5
(ii) We will write a few equivalent fractions for 2/5 and 1/4:
From the above table, the suitable Equivalent fraction for 2/5 is 8/20, and that for 1/4 is 5/20. Because both have the same denominator 20, and so are like fractions. So
• 2/5 is 8 parts out of 20, and
• 1/4 is 5 parts out of 20.
Thus 2/5 is larger than 1/4
So now we can compare any given fractions and decide which one is the greatest among them. But the above method is lengthy:
The word 'like fractions' is in the plural form. Because we are saying 'fractions', not 'fraction'. So we are talking about more than one fraction here. If these fractions are to fall under the category of like fractions, their denominators must be the same. Consider the following fig.5.21:
 |
| Fig.5.21 |
If the denominators are not the same, they are called unlike fractions. Example: 5/11 and 5/10 are unlike fractions.
Comparing fractions
Consider the situation: Mr. A wants to share a cake with his nephew. He decides to play a trick. He asks his nephew: “Do you want 1/3 of the cake or 2/5 ?” The nephew obviously wants the bigger part. But he cannot decide which is bigger. 1/3 or 2/5? He knows that:• When the cake is divided into 3 equal parts, each piece will be larger than when it is divided into 5 equal parts.
• But uncle is offering '2 out the 5 smaller pieces'. Is it larger than '1 out of the bigger 3'? or smaller?
Let us draw a fig. and find out:
 |
| Fig.5.22 |
From the fig.5.22 it is clear that 2/5 is larger than 1/3. So the nephew should ask for the 2/5 part. The problem is solved. But let us analyse it further:
What caused the nephew confusion? It is the 'methods of division'.
• In the first option, the cake is to be divided into 5 equal parts.
• In the second option, the cake is to be divided into 3 equal parts.
The size of the equal parts will be different in each option. In addition to this, the numerators are different. So any body would be confused.
The problem was solved by drawing a fig. But we will find a mathematical solution to such problems so that, we will not have to draw figures each time we find ourselves in such situations. We know that the 'unequal division' is the main reason for confusion. Can't we assume that they are divided equally? We can. For this we use the 'Equivalent fractions'.
• We have the first option 1/3. We must change it into a 'suitable' equivalent fraction. Let us call it EF1. How do we decide whether EF1 is suitable? We will see that soon.
• We have the second option 2/5. We must change it also into a 'suitable' equivalent fraction. Let us call it EF2.
The suitability of EF1 and EF2 is decided by one condition:
EF1 and EF2 should have the same denominator. So by this condition we are ensuring that the cake is divided into equal parts in both the two options. Let us now try to find these suitable EF1 and EF2:
We will write the possible equivalent fractions of 1/3 and 2/5 side by side in a tabular form as shown below:
• 'EF1 for 1/3' is 5/15 and • 'EF2 for 2/5' is 6/15. So:
■ When uncle offers 1/3, he is offering 5 pieces out of 15
■ When he offers 2/5, he is offering 6 pieces out of 15.
6 is greater than 5. So the nephew should select the 2/5 option. Thus we can conclude that, whenever we want to compare unlike fractions, we must convert them into like fractions, by using the Equivalent fractions method.
Solved example 5.16
Pick out the larger fraction from among the following pairs
(a) 3/5, 2/3 (b) 2/5, 1/4
Solution:
(i) We will write a few equivalent fractions for 3/5 and 2/3:
From the above table, the suitable Equivalent fraction for 3/5 is 9/15, and that for 2/3 is 10/15. Because both have the same denominator 15, and so are like fractions. So
• 3/5 is 9 parts out of 15, and
• 2/3 is 10 parts out of 15.
Thus 2/3 is larger than 3/5
(ii) We will write a few equivalent fractions for 2/5 and 1/4:
From the above table, the suitable Equivalent fraction for 2/5 is 8/20, and that for 1/4 is 5/20. Because both have the same denominator 20, and so are like fractions. So
• 2/5 is 8 parts out of 20, and
• 1/4 is 5 parts out of 20.
Thus 2/5 is larger than 1/4
So now we can compare any given fractions and decide which one is the greatest among them. But the above method is lengthy:
• We wrote several equivalent fractions for each given fraction.
• From among them we selected the suitable fractions.
There is a direct method by which we can calculate the 'suitable equivalent fractions':
In all the examples that we saw above, the suitability was decided by one condition:
• The denominators should be the same.
If we examine those ‘same denominators’ we will see that each is a common multiple of the given denominators’. Let us see them again:
• In the solved example 5.16 (a) 15 is the common multiple of 5 and 3, (where 5 and 3 are the given denominators). That is., 15 is a multiple of both 5 and 3.
• In (b), 20 is the common multiple of 5 and 4.
This gives us a method to calculate the denominator of the ‘suitable equivalent fraction’:
• Just find a common multiple of the given denominators. That will be our required denominator.
Once we find that denominator, we can get the ‘suitable equivalent fraction’. The following solved examples will demonstrate the method.
Solved example 5.17
Pick out the larger fraction from among the following pairs:
(a) 7/15, 3/13 (b) 23/39, 14/34
Solution:
(a) 7/15, 3/13
■ Step 1: Find the suitable equivalent fraction for each given fractions
• Common multiple of 15 and 13.
There will be several common multiples for 15 and 13. Any one of them will serve our purpose. The easiest way to find one is to multiply them. Because the product of two numbers will obviously be a ‘common multiple’ of both.
So we have: common multiple of 15 and 13 = 15 ×13 = 195. This will be the denominator of both our ‘suitable equivalent fractions’
• Suitable equivalent fraction for 7/15:
♦ The denominator of the given fraction is 15.
♦ The denominator of the new equivalent fraction should be 195.
♦ We have seen above that, this 195 is obtained by multiplying 15 with 13
♦ So we have to multiply the numerator also by 13. The calculation steps and the result are given below:
♦ So the required equivalent fraction is 91 ⁄ 195
• Suitable equivalent fraction for 3/13:
♦ The denominator of the given fraction is 13.
♦ The denominator of the new equivalent fraction should be 195.
♦ We have seen above that, this 195 is obtained by multiplying 13 with 15
♦ So we have to multiply the numerator also by 15. The calculation steps and the result are given below:
It may be noted that, in this step, we are multiplying both numerator and denominator by the 'other denominator'. Based on this, we will soon see the easy steps to solve the problem.
■ Step 2: Write down the equivalent fractions
7 ⁄ 15 = 91 ⁄ 195 and 3 ⁄ 13 = 45 ⁄ 195
■ Step 3: Compare the equivalent fractions
91 > 45. So 7/15 is greater than 3/13
(b) 23/39, 14/34
■ Step 1: Find the suitable equivalent fraction for each given fractions
• Common multiple of 39 and 34.
There will be several common multiples for 39 and 34. Any one of them will serve our purpose. The easiest way to find one is to multiply them. Because the product of two numbers will obviously be a ‘common multiple’ of both.
So we have: common multiple of 39 and 34 = 39 ×34 = 1326. This will be the denominator of both our ‘suitable equivalent fractions’
• Suitable equivalent fraction for 23/39:
♦ The denominator of the given fraction is 39.
♦ The denominator of the new equivalent fraction should be 1326.
♦ We have seen above that, this 1326 is obtained by multiplying 39 with 34
♦ So we have to multiply the numerator also by 34. The calculation steps and the result are given below:
♦ So the required equivalent fraction is 782 ⁄ 1326
• Suitable equivalent fraction for 14/34:
♦ The denominator of the given fraction is 34.
♦ The denominator of the new equivalent fraction should be 1326.
♦ We have seen above that, this 1326 is obtained by multiplying 34 with 39
♦ So we have to multiply the numerator also by 39. The calculation steps and the result are given below:
♦ So the required equivalent fraction is 546 ⁄ 1326
It may be noted that, in this step, we are multiplying both numerator and denominator by the 'other denominator'. Based on this, we will soon see the easy steps to solve the problem.
■ Step 2: Write down the equivalent fractions
• 23 ⁄ 39 = 782 ⁄ 1326 and 14 ⁄ 34 = 546 ⁄ 1326
■ Step 3: Compare the equivalent fractions
782 > 546. So 23/39 is greater than 14/34
Solved example 5.18
Compare the fractions in the following pairs:
(a) 5/11, 7/13 (b) 10/17, 8/15 (c) 6/23, 2/7
Solution:
(a) The steps are same as above. So this time we will write only those steps which are absolutely necessary, as shown below:
(b)
(c)
In the above examples we found out the common multiple just by multiplying the two numbers. Some times it will be convenient to calculate the LCM. That is., the least common multiple. Out of the several available ‘common multiples’, The LCM will be the ‘least’ one, or the smallest one. In this way, we will be able to keep the ‘size of the numbers’ down.
Now we are in a position to compare any given fractions. If we are given a group of more than two fractions, write the 'suitable equivalent fraction' for each of them. And then compare. We will even be able to write the given fractions in ascending or descending order.
In some special cases, there will not be any need for 'a paper and pencil' to do the above calculations, to find which is the biggest among the given fractions. We will be able to solve them 'mentally'. We will discuss about them in the next section.
In all the examples that we saw above, the suitability was decided by one condition:
• The denominators should be the same.
If we examine those ‘same denominators’ we will see that each is a common multiple of the given denominators’. Let us see them again:
• In the solved example 5.16 (a) 15 is the common multiple of 5 and 3, (where 5 and 3 are the given denominators). That is., 15 is a multiple of both 5 and 3.
• In (b), 20 is the common multiple of 5 and 4.
This gives us a method to calculate the denominator of the ‘suitable equivalent fraction’:
• Just find a common multiple of the given denominators. That will be our required denominator.
Once we find that denominator, we can get the ‘suitable equivalent fraction’. The following solved examples will demonstrate the method.
Solved example 5.17
Pick out the larger fraction from among the following pairs:
(a) 7/15, 3/13 (b) 23/39, 14/34
Solution:
(a) 7/15, 3/13
■ Step 1: Find the suitable equivalent fraction for each given fractions
• Common multiple of 15 and 13.
There will be several common multiples for 15 and 13. Any one of them will serve our purpose. The easiest way to find one is to multiply them. Because the product of two numbers will obviously be a ‘common multiple’ of both.
So we have: common multiple of 15 and 13 = 15 ×13 = 195. This will be the denominator of both our ‘suitable equivalent fractions’
• Suitable equivalent fraction for 7/15:
♦ The denominator of the given fraction is 15.
♦ The denominator of the new equivalent fraction should be 195.
♦ We have seen above that, this 195 is obtained by multiplying 15 with 13
♦ So we have to multiply the numerator also by 13. The calculation steps and the result are given below:
7 ⁄ 15 = (7 × 13) ⁄ (15 × 13) = 91 ⁄ 195
♦ So the required equivalent fraction is 91 ⁄ 195
• Suitable equivalent fraction for 3/13:
♦ The denominator of the given fraction is 13.
♦ The denominator of the new equivalent fraction should be 195.
♦ We have seen above that, this 195 is obtained by multiplying 13 with 15
♦ So we have to multiply the numerator also by 15. The calculation steps and the result are given below:
3 ⁄ 13 = (3 × 15) ⁄ (13 × 15) = 45 ⁄ 195
♦ So the required equivalent fraction is 45 ⁄ 195It may be noted that, in this step, we are multiplying both numerator and denominator by the 'other denominator'. Based on this, we will soon see the easy steps to solve the problem.
■ Step 2: Write down the equivalent fractions
7 ⁄ 15 = 91 ⁄ 195 and 3 ⁄ 13 = 45 ⁄ 195
■ Step 3: Compare the equivalent fractions
91 > 45. So 7/15 is greater than 3/13
(b) 23/39, 14/34
■ Step 1: Find the suitable equivalent fraction for each given fractions
• Common multiple of 39 and 34.
There will be several common multiples for 39 and 34. Any one of them will serve our purpose. The easiest way to find one is to multiply them. Because the product of two numbers will obviously be a ‘common multiple’ of both.
So we have: common multiple of 39 and 34 = 39 ×34 = 1326. This will be the denominator of both our ‘suitable equivalent fractions’
• Suitable equivalent fraction for 23/39:
♦ The denominator of the given fraction is 39.
♦ The denominator of the new equivalent fraction should be 1326.
♦ We have seen above that, this 1326 is obtained by multiplying 39 with 34
♦ So we have to multiply the numerator also by 34. The calculation steps and the result are given below:
23 ⁄ 39 = (23 × 34) ⁄ (39 × 34) = 782 ⁄ 1326
♦ So the required equivalent fraction is 782 ⁄ 1326
• Suitable equivalent fraction for 14/34:
♦ The denominator of the given fraction is 34.
♦ The denominator of the new equivalent fraction should be 1326.
♦ We have seen above that, this 1326 is obtained by multiplying 34 with 39
♦ So we have to multiply the numerator also by 39. The calculation steps and the result are given below:
14 ⁄ 34 = (14 × 39) ⁄ (34 × 39) = 546 ⁄ 1326
♦ So the required equivalent fraction is 546 ⁄ 1326
It may be noted that, in this step, we are multiplying both numerator and denominator by the 'other denominator'. Based on this, we will soon see the easy steps to solve the problem.
■ Step 2: Write down the equivalent fractions
• 23 ⁄ 39 = 782 ⁄ 1326 and 14 ⁄ 34 = 546 ⁄ 1326
■ Step 3: Compare the equivalent fractions
782 > 546. So 23/39 is greater than 14/34
Solved example 5.18
Compare the fractions in the following pairs:
(a) 5/11, 7/13 (b) 10/17, 8/15 (c) 6/23, 2/7
Solution:
(a) The steps are same as above. So this time we will write only those steps which are absolutely necessary, as shown below:
(c)
In the above examples we found out the common multiple just by multiplying the two numbers. Some times it will be convenient to calculate the LCM. That is., the least common multiple. Out of the several available ‘common multiples’, The LCM will be the ‘least’ one, or the smallest one. In this way, we will be able to keep the ‘size of the numbers’ down.
Now we are in a position to compare any given fractions. If we are given a group of more than two fractions, write the 'suitable equivalent fraction' for each of them. And then compare. We will even be able to write the given fractions in ascending or descending order.
In some special cases, there will not be any need for 'a paper and pencil' to do the above calculations, to find which is the biggest among the given fractions. We will be able to solve them 'mentally'. We will discuss about them in the next section.
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