Chapter 5.17 - Addition and subtraction of Fractions

In the previous section we completed the discussion on the various properties of fractions. In this section, we will see addition and subtraction of fractions.

We know that,when the denominators are equal, the addition of fractions is easy. The denominator of the result will be the denominator in the given problem. The numerator will be the sum of the numerators.


When adding fractions with different denominators, more work is involved.  We have already seen the process here. Let us see an example:

In the above process, 
• We multiplied the numerator and denominator of first fraction ¹⁄₂ with the 'other denominator', which is '3'
• We multiplied the numerator and denominator of the second fraction ¹⁄₃ also with the 'other denominator', which is '2'

• Thus we get two fractions with the same denominators

We can write the process in algebraic form:

Based on this, we can write the general form:

For subtraction, the general form will be:

Example for addition:

Example for subtraction:

The above operations will become very easy if the numerators are all '1'. Let us check such cases algebraically:

So we can write the general form for addition of fractions with numerator '1' as:

And the general form for the subtraction of fractions with numerator '1' can be written as:

Addition example:

Subtraction example:

---

Fractions with Numerator '1' are called unit fractions. We will now see some properties of unit fractions:

---

Consider the pattern:

We can continue the steps any number of times. But for our present discussion, the few steps written above are sufficient. 

1. Consider the sum: [¹⁄₁ - ¹⁄₂] + [¹⁄₂ - ¹⁄₃ ]. What is the result?
2. -¹⁄₂ and ¹⁄₂ will cancel out each other. So the sum is ¹⁄₁ - ¹⁄₃
3. But from the above pattern, we have: [¹⁄₁ - ¹⁄₂] = ¹⁄_(1×2)  and  [¹⁄₂ - ¹⁄₃ ] = ¹⁄_(2×3) 
4. So comparing (2) and (3), we get: ¹⁄_(1×2) + ¹⁄_(2×3) = ¹⁄₁ - ¹⁄₃ 
Next, let us take three terms:
1. we get [¹⁄₁ - ¹⁄₂] + [¹⁄₂ - ¹⁄₃ ] + [¹⁄₃ - ¹⁄₄ ]. What is the result?
2. -¹⁄₂ and ¹⁄₂ will cancel out each other. Also -¹⁄₃ and ¹⁄₃ will cancel out each other. So the sum is ¹⁄₁ - ¹⁄₄ 
3. But from the pattern, we have: 
[¹⁄₁ - ¹⁄₂] = ¹⁄_(1×2)  ,
[¹⁄₂ - ¹⁄₃ ] = ¹⁄_(2×3)  and 
[¹⁄₂ - ¹⁄₃ ] = ¹⁄_(3×4) 
4. So comparing (2) and (3), we get: ¹⁄_(1×2) + ¹⁄_(2×3) + ¹⁄_(3×4)= ¹⁄₁ - ¹⁄₄ 
Next, let us take four terms:
1. we get [¹⁄₁ - ¹⁄₂] + [¹⁄₂ - ¹⁄₃ ] + [¹⁄₃ - ¹⁄₄ ] + [¹⁄₄ - ¹⁄₅ ]. What is the result?
2. -¹⁄₂ and ¹⁄₂ will cancel out each other. 
-¹⁄₃ and ¹⁄₃ will cancel out each other. 
-¹⁄₄ and ¹⁄₄ will cancel out each other 
So the sum is ¹⁄₁ - ¹⁄₅ 
3. But from the pattern, we have: 
[¹⁄₁ - ¹⁄₂] = ¹⁄_(1×2)  ,
[¹⁄₂ - ¹⁄₃ ] = ¹⁄_(2×3)  ,
[¹⁄₂ - ¹⁄₃ ] = ¹⁄_(3×4) and 
[¹⁄₂ - ¹⁄₃ ] = ¹⁄_(4×5) 

4. So comparing (2) and (3), we get: ¹⁄_(1×2) + ¹⁄_(2×3)  + ¹⁄_(3×4) + ¹⁄_(4×5)= ¹⁄₁ - ¹⁄₅ 

We considered three cases:
Case 1: We took two terms, and got the result:
■  ¹⁄_(1×2) + ¹⁄_(2×3) = ¹⁄₁ - ¹⁄₃  
Case 2: We took three terms, and got the result:
■  ¹⁄_(1×2) + ¹⁄_(2×3) + ¹⁄_(3×4)= ¹⁄₁ - ¹⁄₄ 
Case 3: We took four terms, and got the result:
■  ¹⁄_(1×2) + ¹⁄_(2×3)  + ¹⁄_(3×4) + ¹⁄_(4×5) = ¹⁄₁ - ¹⁄₅ 

We can go on like this and take any number of cases as we like. But that will not be necessary. A pattern has already emerged from the three cases. When we understand the pattern, we will be able to write the answer for any given case. Here is an example:

Case 98:
■  ¹⁄_(1×2) + ¹⁄_(2×3)  + ¹⁄_(3×4) + . . . . + ¹⁄_(99×100) = ¹⁄₁ - ¹⁄₁₀₀  = 1 - 0.01 = 0.99

---

In the next section we will see multiplication and division with fractions.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 5.10 - Subtraction of Fractions

In the previous section we learned about mixed fractions and their addition. In this section we will see how mixed fractions help us to measure distances. Later in this section, we will discuss subtraction of fractions.

In chapter 5.1 we saw how fractions help us to measure distances which are less than 1 metre. We discussed it based on fig.5.10. For convenience, that fig. is shown again below:

Now let us see an 'improved situation': Some modifications were made to the robot. It now walked more than 2 m. But it did not reach the 3 m mark. This is shown in the fig.5.31 below:

Fig.5.31

This time also a question will arise: How far did it walk after the 2 m mark? To answer this question, we have to divide the portion between 2m and 3m into equal parts. As before, let us divide it into 5 equal parts. This is shown in the fig.5.32 below:

Fig.5.32

The distance between the 2 m mark and the 3 m mark is 1 metre. This 1 metre distance is divided into 5 equal parts. So each part is ¹⁄₅ metre. After the 2 m mark, the robot walked 2 units. That 2 units is ²⁄₅ m. 
We can say: After the 2 m mark, the robot walked ²⁄₅ m. Thus the total distance walked is 2 m plus ²⁄₅ m. This is written in the mixed fraction form as 2²⁄₅

---

Thus we have seen how mixed fractions help us to specify distances. We will see some solved examples based on this discussion:

Solved example 5.29

Mr. A bought 4²⁄₅ m of a rope. Mr B bought 5⁷⁄₈ m of the same quality rope from another shop. What is the total length of the ropes they bought?

Solution:

The ropes are of the same quality. But they are of different lengths, and were bought from different shops. The first merchant prefer to divide each metre of rope into 5 equal parts. So after measuring 4 'full metres', he took two 'one fifths'. Thus the length becomes 4²⁄₅

The second merchant prefer to divide each metre of rope into 8 equal parts. So after measuring 5 'full metres', he took seven 'one eighths'. Thus the length becomes 5⁷⁄₈

What ever be the methods of measurements, we need not worry. Because we have learned to add any type of fractions. Thus:

Total length = 4²⁄₅  +  5⁷⁄₈

• 4²⁄₅ = ²²⁄₅       • 5⁷⁄₈  = ⁴⁷⁄₈ 

The remaining steps and the result are shown below:

So we get the total length as 10¹¹⁄₄₀ . It is greater than 10 m but less than 11 m. The fraction after 10 m is 11 out of '40 equal parts of a metre'.

It may be noted that such different divisions by different merchants is not in practice these days. Now, one metre is always divided into 100 equal parts. We discussed about it here. But we must be able to solve any type of problems.

---

Next we will discuss about subtraction of fractions. We will need to use subtraction on many occasions.

Consider the same sliced bread loafs at the camp. One camper took 9 slices out of 14 slices of a loaf. Just when he completed eating the 7^th slice, his stomach was full. So he wrapped the remaining two slices in food grade aluminium foil to eat later. What fraction of the whole loaf did he wrap in the aluminium foil? Let us analyse:
• Initially he took 9 slices. That is ⁹⁄₁₄ of the whole loaf.
• He consumed 7 slices. That is ⁷⁄₁₄ of the whole loaf.
• So the fraction which is left is ⁹⁄₁₄ – ⁷⁄₁₄ = ²⁄₁₄
• He has ²⁄₁₄ of a whole loaf wrapped in the foil.
This is shown in the fig.5.33 below:

Thus we can say that subtraction of fractions can be done using the same procedure as for addition. We can find the difference between even unlike mixed fractions. We will see some solved examples below:
Solved example 5.30
Solve: (a) ⁸⁄₁₉ – ⁶⁄₁₉    (b) ³⁄₄ – ¹⁄₂    (c) 5²⁄₇ – 3⁵⁄₆
Solution:

(a) The given fractions are like fractions. They have the same denominators. So we just need to subtract the second numerator from the first:
⁸⁄₁₉ – ⁶⁄₁₉ = ²⁄₁₉ 
(b) Here we have to convert each fraction into suitable equivalent fraction. The steps and result are shown below:

(c) Here we have to convert each fraction into an improper fraction first. The steps and result are shown below:
• 5²⁄₇ = ³⁷⁄₇       • 3⁵⁄₆  = ²³⁄₆ 

Solved example 5.31
Compare the following two fractions. Then subtract the smaller from the larger. ¹⁴⁄₂₉, ²⁄₇
Solution:
The comparison result is shown below:

So we have to subtract ²⁄₇ from ¹⁴⁄₂₉. When we do the comparison, we get the suitable equivalent fractions also. Thus:
¹⁴⁄₂₉ - ²⁄₇  = ⁹⁸⁄₂₀₃  -  ⁵⁸⁄₂₀₃ = ⁴⁰⁄₂₀₃ 
Solved example 5.32
The original length of a rope was 5²⁄₃ m. From it, a piece of 3¹⁄₄ was cut. What is the length of the remaining piece?
Solution:
• 5²⁄₃ = ¹⁷⁄₃       • 3¹⁄₄  = ¹³⁄₄ 

---

So we have completed the discussion on the basics of fractions. Until now we were doing the following steps:
• We divided a 'whole' into a definite number of equal parts
• Out of those 'equal parts', we took out a definite number
• We expressed this 'number of  equal parts that were taken out' as a fraction

Often in day to day life, we may have to do the above steps in a reverse order. We will now discuss such situations:

A farmer has 120 apples. He wants to give ²⁄₁₀ of it to his friend. How many apples would he give to the friend?

We must clearly understand the difference in this type of problem. In our earlier discussion, we would take out a definite number of apples, and express those apples as a fraction of the whole. But here, the fraction is already given. We want the number of apples to be taken out. So it is a reverse situation. Let us try to solve it:

The given fraction is 2/10. So 2 parts is to be taken out of 10 equal parts. So we must first divide the whole into 10 equal parts. When 120 is divided into 10 equal parts, each part will have 120/10 = 12 apples. 
• Each part has 12 apples. 
• And each part represents 1/10 of the 'whole 120'. 
• The farmer has to take out two such equal parts. 
• So the farmer would give 24 apples to his friend.

Let us analyse the above steps: The first step was to divide the whole 120 into 10 equal parts. For that, we divided 120 by 10. Here 10 is the denominator of the given fraction. So we can say: The first step is to divide the whole by the 'given denominator'. If 'W' is the 'whole', and 'D' is the denominator, then the first step is to calculate ^W⁄_D.  The result ^W⁄_D, is the quantity in each of the 'equal parts'. For our present problem, ^W⁄_D = ¹²⁰⁄₁₀  =12  

The next step is to take out the number of equal parts. We took out 2 parts. Each part has 12 apples. So 2 parts give 2 × 12 = 24. Here 2 is the numerator of the given fraction. So we can say: The second step is to multiply ^W⁄_D with the 'given numerator'. If 'N' is the numerator, we can write the second step as (^W⁄_D) × N. We get the required answer from this step, and so, this second step is the final step. 


We can rearrange the terms:
(^W⁄_D) × N ⇒  W × ^N⁄_D   But ^N⁄_D is the given fraction. So we can find the answer in just one step:
Multiply the given W by the given fraction. This will be same as dividing the whole into equal parts and then taking out the required number of equal parts. So, The required quantity = ^(W×N)⁄_D

Let us see some solved examples based on the above discussion:
Solved example 5.33
A man wants to keep aside ²⁄₅ of his salary to pay bills. If his salary is ₹ 12000/-, how much money would he keep aside?
Solution:
Total salary = ₹ 12000. So W = 12000
Fraction to be kept aside = ²⁄₅. So N = 2, D = 5
Amount to be kept aside = ^(W ×N)⁄_D  = ^{(12000 × 2)}⁄₅ = 24000/5 = ₹ 4800/-
Solved example 5.34
Solve: (a) ³⁄₈ of 72  (b) ⁵⁄₆ of 5400
Solution:
(a) W = 72, N= 3, D = 8
We have ^(W×N)⁄_D = ^(72×3)⁄₈  = ⁽²¹⁶⁾⁄₈ = 27
(b) W = 5400, N= 5, D = 6
We have ^(W×N)⁄_D = ^{(5400 × 5)}⁄₆  = ^(27000)⁄₆  = 4500

---

In the next section we will discuss different forms of fractions.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 5.8 - Addition of Fractions

In the previous section we completed the topic of 'comparison of fractions'. In this section we will discuss about addition of fractions. Consider the following situation:

At a camp, bread loafs such as shown in the fig.5.27 below, are served. All loafs are of the same size. Each loaf is sliced into a number of pieces and kept on tables. Campers can walk around the tables and take as many slices as they wish.

Fig.5.27

One particular loaf was divided into 14 slices as shown in fig.5.28(a) below. Mr.A takes 3 slices (marked in green colour) from that loaf as shown in the fig.(b). So the portion taken by him is 3/14 of the ‘whole loaf’.  Mr.B takes 4 slices (marked in yellow colour) from the same loaf. So the portion taken by him is 4/14 of the ‘whole loaf’. The total number of slices taken by A and B is 7.

Fig.5.28

So the portion taken by A and B together is 7/14 of the ‘whole loaf’. Mathematically we can write this as: 3/14 + 4/14 = 7/14. This is simple addition of numerators. We were able to simply add the numerators because, they are like fractions (denominators are the same). We will see some solved examples on this type of addition:
Solved example 5.22
Calculate the following:
(a) ²⁄₇  +  ⁴⁄₇   (b) ⁸⁄₁₃  +  ⁴⁄₁₃    (c) ¹⁴⁄₂₅  +  ⁷⁄₂₅
Solution:
(a) ²⁄₇  +  ⁴⁄₇  =  ⁶⁄₇    (b) ⁸⁄₁₃  +  ⁴⁄₁₃  =  ¹²⁄₁₃     (c) ¹⁴⁄₂₅  +  ⁷⁄₂₅  =  ²¹⁄₂₅
Solved example 5.23
Fill up the boxes in each of the following

Solution:
The given fractions are all like fractions. We can simply add the numerators. Thus we have:
(a) 2 + x = 7  ∴ x = 7 -2 = 5      (b) x + 8 = 14  ∴ x = 14 -8 = 6

---

Now let us see another type of addition:
We saw that Mr.A took 3 slices out of 14 from a loaf. After consuming the 3 slices, he went to the table again and took 2 slices from another loaf. This is shown in fig.5.29(b) below:

Fig.5.29

But this loaf was divided into 10 equal parts. (We must remember that, though different loafs are divided into different number of slices, all the loafs are of the same size). So the portion taken out by A this time is 2/10 of the ‘whole loaf’. What is the total portion consumed by A? For this we have to add the two portions. That is., 3/14 + 2/10. Here we cannot just add the numerators because the slices are of different sizes as shown in fig.(c).

We have to use another method. We have to find the equivalent fractions of both 3/14 and 2/10, which has a same denominator:
So our task is to add 3/14 and 2/10
■ Step 1: Find the suitable equivalent fraction for each given fractions
• Common multiple of 14 and 10.
There will be several common multiples for 14 and 10. 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 14 and 10 = 14 × 10 = 140. This will be the denominator of both our ‘suitable equivalent fractions’
• Suitable equivalent fraction for 3/14:
    ♦ The denominator of the given fraction is 14.
   ♦ The denominator of the new equivalent fraction should be 140.
   ♦ We have seen above that, this 140 is obtained by multiplying 14 with 10
   ♦ So we have to multiply the numerator also by 10. The calculation steps and the result are given            below:

            ³⁄₁₄  =   ^{(3 × 10)}⁄_{(14 × 10)}  =  ³⁰⁄₁₄₀  

   ♦ So the required equivalent fraction is  ³⁰⁄₁₄₀ 
• Suitable equivalent fraction for 2/10:
    ♦ The denominator of the given fraction is 10.
   ♦ The denominator of the new equivalent fraction should be 140.
   ♦ We have seen above that, this 140 is obtained by multiplying 10 with 14
   ♦ So we have to multiply the numerator also by 14. The calculation steps and the result are given            below:

            ²⁄₁₀  =   ^{(2 × 14)} ⁄ _{(10 × 14)}  =  ²⁸⁄₁₄₀ 

   ♦ So the required equivalent fraction is  ²⁸⁄₁₄₀
■ Step 2: Write down the equivalent fractions
  ³⁄₁₄  =  ³⁰⁄₁₄₀  and ²⁄₁₀  =  ²⁸⁄₁₄₀ 
■ Step 3: Analyse the above result
• Taking 3 slices from a total of 14 equal slices is same as taking 30 slices from a total of 140 equal slices.
• Taking 2 slices from a total of 10 equal slices is same as taking 28 slices from a total of 140 equal slices.
So the slices have become equal in size. We can add them directly. Mr.A took a total of 58 slices out of 140 equal slices. Mathematically this can be written as:
³⁰⁄₁₄₀  +  ²⁸⁄₁₄₀  =  ⁵⁸⁄₁₄₀
58/140 can be reduced to the simplest form as 29/70.  So Mr.A consumed 29/70 of one ‘whole loaf’.

---

Thus we have learned how to add unlike fractions. Let us see some solved examples in this category:
Solved example 5.24
Calculate the following:
(a) 2/5 + 1/3    (b) 3/7 + 5/9
Solution:
(a) Our task is to add 2/5 and 1/3
■ Step 1: Find the suitable equivalent fraction for each given fractions
• Common multiple of 5 and 3
There will be several common multiples for 5 and 3. 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 5 and 3 = 5 × 3 = 15. This will be the denominator of both our ‘suitable equivalent fractions’
• Suitable equivalent fraction for 2/5:
    ♦ The denominator of the given fraction is 5.
   ♦ The denominator of the new equivalent fraction should be 15.
   ♦ We have seen above that, this 15 is obtained by multiplying 5 with 3
   ♦ So we have to multiply the numerator also by 3. The calculation steps and the result are given            below:

            ²⁄₅  =   ^{(2 × 3)}⁄_{(5 × 3)}  =  ⁶⁄₁₅  

   ♦ So the required equivalent fraction is  ⁶⁄₁₅
   ♦ This is same as multiplying both the numerator and the denominator by the other denominator 
• Suitable equivalent fraction for 1/3:
    ♦ The denominator of the given fraction is 3.
   ♦ The denominator of the new equivalent fraction should be 15.
   ♦ We have seen above that, this 15 is obtained by multiplying 3 with 5
   ♦ So we have to multiply the numerator also by 5. The calculation steps and the result are given            below:

            ¹⁄₃  =   ^{(1 × 5)} ⁄ _{(3 × 5)}  =  ⁵⁄₁₅ 

  
   ♦ So the required equivalent fraction is  ⁵⁄_{15
   ♦ This is same as multiplying both the numerator and the denominator by the other denominator}
■ Step 2: Write down the equivalent fractions
  ²⁄₅  =  ⁶⁄₁₅  and ¹⁄₃  =  ⁵⁄₁₅ 
■ Step 3: Analyse the above result
• Taking 2 parts from a total of 5 equal parts is same as taking 6 parts from a total of 15 equal parts
• Taking 1 part from a total of 3 equal parts is same as taking 5 parts from a total of 15 equal parts
So the parts have become equal in size. We can add them directly.
⁶⁄₁₅  +  ⁵⁄₁₅  =  ¹¹⁄₁₅

(b) Our task is to add 3/7 and 5/9. This can be done using the same procedure as above. So we will not write detailed steps. The required steps are shown below:

Solved example 5.25
Solve: (a) 3/10 + 7/15  (b) 4/21 + 3/19   (c) 8/30  + 26/45
Solution:

   

---

In the next section we will discuss about Mixed fractions.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved