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:
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:
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:
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 |
 |
| Fig.5.32 |
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 7th 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:
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.
No comments:
Post a Comment