In the previous section we discussed about the place values in decimals. In this section we will learn how to compare different decimal quantities. And later in this section, we will discuss about addition of decimals.
Consider two decimals: 0.3 and 0.5. Can we say which one is larger? Of course 0.5 is larger. We can prove it in this way:
• 0.3 = 3⁄10. So it is 3 parts taken from 10 equal parts
• 0.5 = 5⁄10 . So it is 5 parts taken from 10 equal parts.
5 > 3. Thus we can say 0.5 is larger
Another example: Compare 0.69 and 0.84
Solution: • 0.69 = 69⁄100 . So it is 69 parts taken from 100 equal parts
• 0.84 = 84⁄100. So it is 84 parts taken from 100 equal parts.
84 > 69. Thus we can say 0.84 is larger
Another example: Compare 0.15 and 0.078
Solution: Here the digits in the second decimal are higher than those in the first. So at first appearance it may seem to be the larger one. We will do a detailed analysis as above:
• 0.15 = 15⁄100. So it is 15 parts taken from 100 equal parts
• 0.078 = 78⁄1000. So it is 78 parts taken from 1000 equal parts.
The denominators are different. So comparison is not easy. We will make the denominators same by using equivalent fractions method.
• 15⁄100 = 150⁄1000. So it is 150 parts taken from 1000 equal parts.
150 > 78. So we can say 0.15 is larger than 0.078.
Another example: Compare 0.08 and 0.084
Solution:
• 0.08 = 8⁄100 = 80⁄1000
• 0.084 = 84⁄1000. 84 > 80. So 0.084 is larger than 0.08
Another example: Compare 0.099 and 0.19
Solution:
• 0.099 = 99⁄1000
• 0.19 = 19⁄100 = 190⁄1000 . 190 > 99. So 0.19 is larger than 0.099
Consider two decimals: 0.3 and 0.5. Can we say which one is larger? Of course 0.5 is larger. We can prove it in this way:
• 0.3 = 3⁄10. So it is 3 parts taken from 10 equal parts
• 0.5 = 5⁄10 . So it is 5 parts taken from 10 equal parts.
5 > 3. Thus we can say 0.5 is larger
Another example: Compare 0.69 and 0.84
Solution: • 0.69 = 69⁄100 . So it is 69 parts taken from 100 equal parts
• 0.84 = 84⁄100. So it is 84 parts taken from 100 equal parts.
84 > 69. Thus we can say 0.84 is larger
Another example: Compare 0.15 and 0.078
Solution: Here the digits in the second decimal are higher than those in the first. So at first appearance it may seem to be the larger one. We will do a detailed analysis as above:
• 0.15 = 15⁄100. So it is 15 parts taken from 100 equal parts
• 0.078 = 78⁄1000. So it is 78 parts taken from 1000 equal parts.
The denominators are different. So comparison is not easy. We will make the denominators same by using equivalent fractions method.
• 15⁄100 = 150⁄1000. So it is 150 parts taken from 1000 equal parts.
150 > 78. So we can say 0.15 is larger than 0.078.
Another example: Compare 0.08 and 0.084
Solution:
• 0.08 = 8⁄100 = 80⁄1000
• 0.084 = 84⁄1000. 84 > 80. So 0.084 is larger than 0.08
Another example: Compare 0.099 and 0.19
Solution:
• 0.099 = 99⁄1000
• 0.19 = 19⁄100 = 190⁄1000 . 190 > 99. So 0.19 is larger than 0.099
Addition of decimals
So we have learned how to compare any given decimals. We will now discuss addition of decimals. Consider the addition: 0.41 + 0.38
Let us write it in fraction form:
=41⁄100 + 38⁄100 . So we are adding '41 parts out of 100' with '38 parts out of 100'. That will be 41 + 38 = 79 parts out of 100. That is., 79⁄100. In decimal form it is 0.79.
So we can write: 0.41 + 0.38 = 0.79. This is usual addition procedure as shown below in fig.6.10
Consider another addition: 0.75 + 0.82
So we can write: 0.41 + 0.38 = 0.79. This is usual addition procedure as shown below in fig.6.10
 |
| Fig.6.10 |
Let us write it in fraction form:
75⁄100 + 82⁄100 . So we are adding '75 parts out of 100' with '82 parts out of 100'. That will be 75 + 82 = 157 parts out of 100. That is., 157⁄100. In decimal form it is 1.57.
We must understand the difference of this result '1.57' from the previous result '0.79'. In the previous example, we added 41 parts and 38 parts. Together they give 79 parts. This is less than 100. That means the sum is less than 'one whole'. Even after adding two quantities, it is still less than one. In other words it is still a 'fraction'.
But in the present case, the sum of 75 and 82 parts is 157, which greater than 100. The number of parts greater than 'one whole' is 57. So we have 'one whole' and an additional 57 parts. This is indicated by the 1 before the decimal point and 'Five Seven' after the decimal point.
Also note that '0.75 + 0.82 = 1.57' can be obtained by the usual addition procedure, as shown below in fig.6.11
Such decimals with a whole number part before the decimal point corresponds to mixed fractions or improper fractions that we learned in the previous chapter.
Let us see one more example: 0.2 + 0.005. This is the addition of '2 parts out of 10 equal parts' with '5 parts out of 1000 equal parts'. That is., 2⁄10 + 5⁄1000. We have to convert 2⁄10 into an equivalent fraction with denominator 1000.
We have 2⁄10 = 200⁄1000.
Thus 2⁄10 + 5⁄1000 = 200⁄1000 + 5⁄1000 = 205⁄1000 = 0.205
We will get the same result by usual addition procedure as shown below:
We will see some solved examples based on the above discussion
Solved example 6.5
A large number of solid blocks are placed in a hall. All of them are of the same size as shown in fig.6.12 below.
• Some of them are divided into 10 equal pieces. They are coloured in Green
• Some are divided into 100 equal pieces. They are coloured in Red
• Some are divided into 1000 equal pieces. They are coloured in Blue
• Some are not divided at all. They remain as 'whole'. They are coloured in Yellow.
Each student can pick any number of pieces from any block. They can also pick any number of 'whole' blocks. They must bring all that they pick to the Teacher, and present the quantity in decimal form. The table below shows the number of each piece that the students picked.
What is the quantity that each student picked?
Solution:
Student A: (i) 3 Green pieces. Each is 1⁄10 of a full block. So 3 pieces give 3⁄10 of a full block = 0.3
(ii) 5 Red pieces. Each is 1⁄100 of a full block. So 5 pieces give 5⁄100 of a full block = 0.05
(iii) 2 Blue pieces. Each is 1⁄1000 of a full block. So 2 pieces give 2⁄1000 of a full block = 0.002
Adding the 3 items (fig.6.13) we get :
So, if the Student A stacks all the pieces that he picked in proper order, he will get 0.352 of a full block.
Student B: (i) 1 full block. So we have '1' before the decimal place
(ii) 6 Green pieces. Each is 1⁄10 of a full block. So 6 pieces give 6⁄10 of a full block = 0.6
(iii) 4 Red pieces. Each is 1⁄100 of a full block. So 4 pieces give 4⁄100 of a full block = 0.04
(iv) 7 Blue pieces. Each is 1⁄1000 of a full block. So 7 pieces give 7⁄1000 of a full block = 0.007
Adding the 4 items (fig.6.14) we get:
So, if the Student B stacks all the pieces that he picked in proper order, he will get one full block, and, in addition to that, 0.647 of a full block.
Student C: (i) 3 full blocks. So we have '3' before the decimal place
(ii) 12 Green pieces. (In the fig.6.12, we can see only one green block. It contains only 10 pieces. But here the student has taken 12 green pieces. This is possible because, as given in the question, there are a large number of blocks of each colour arranged in the hall) Each is 1⁄10 of a full block. So 12 pieces give 12⁄10 of a full block = 1.2
This means that, when 12 green pieces are taken, one full block is formed, and in addition, 0.2 of a full block is obtained.
(iii) 0 Red pieces. Each is 1⁄100 of a full block. So 0 pieces give 0⁄100 of a full block = 0
(iv) 5 Blue pieces. Each is 1⁄1000 of a full block. So 5 pieces give 5⁄1000 of a full block = 0.005
Adding the 4 items we get (fig.6.15):
So, if the Student C stacks all the pieces that he picked in proper order, he will get 4 full blocks, and, in addition to that, 0.205 of a full block.
Student D: (i) 5 Green pieces. Each is 1⁄10 of a full block. So 5 pieces give 5⁄10 of a full block = 0.5
(ii) 23 Red pieces. Each is 1⁄100 of a full block. So 23 pieces give 23⁄100 of a full block = 0.23
(iii) 8 Blue pieces. Each is 1⁄1000 of a full block. So 8 pieces give 8⁄1000 of a full block = 0.008
Adding the 3 items we get (fig.6.16):
So, if the Student D stacks all the pieces that he picked in proper order, he will get 0.738 of a full block
Student E: (i) 5 full blocks. So we have '5' before the decimal place
(ii) 32 Green pieces. Each is 1⁄10 of a full block. So 32 pieces give 32⁄10 of a full block = 3.2
This means that, when 32 green pieces are taken, three full blocks are formed, and in addition, 0.2 of a full block is obtained.
(iii) 7 Red pieces. Each is 1⁄100 of a full block. So 0 pieces give 7⁄100 of a full block = 0.07
(iv) 123 Blue pieces. Each is 1⁄1000 of a full block. So 5 pieces give 123⁄1000 of a full block = 0.123
Adding the 4 items we get (fig.6.17):
In the same way, the rest of the problems can be solved:
Student F: 2 + 0.9 + 0.03 + 0.004 = 2.934
Student G: 14 + 0.6 + 0.32 + 0.075 = 14.995
Student H: 21 + 2.9 + 0.15 + 0.006 = 24.056
Student I: 0 + 0.6 + 0.04 + 0.075 = 0.715
We can also calculate the total quantity taken by more than one student.
Example: The total quantity taken by Students A and D are: 0.352 + 0.738 = 1.090
Another example: The total quantity taken by Students B, C and H are: 1.647 + 4.205 + 24.056 = 29.908
Solved example 6.6
Solve: (i) 0.28 + 0.36 (ii) 0.5 + 0.07 (iii) 1.41 + 1.70 (iv) 2.68 +3.84
Solution:
(i) 0.28 + 0.36 = 0.64 (ii) 0.5 + 0.07 = 0.57 (iii) 1.41 + 1.70 = 3.11 (iv) 2.68 +3.84 = 6.52
The steps are shown in the fig.6.18 below:
In the next section we will discuss about subtraction of decimals.
We must understand the difference of this result '1.57' from the previous result '0.79'. In the previous example, we added 41 parts and 38 parts. Together they give 79 parts. This is less than 100. That means the sum is less than 'one whole'. Even after adding two quantities, it is still less than one. In other words it is still a 'fraction'.
But in the present case, the sum of 75 and 82 parts is 157, which greater than 100. The number of parts greater than 'one whole' is 57. So we have 'one whole' and an additional 57 parts. This is indicated by the 1 before the decimal point and 'Five Seven' after the decimal point.
Also note that '0.75 + 0.82 = 1.57' can be obtained by the usual addition procedure, as shown below in fig.6.11
 |
| Fig.6.11 |
Let us see one more example: 0.2 + 0.005. This is the addition of '2 parts out of 10 equal parts' with '5 parts out of 1000 equal parts'. That is., 2⁄10 + 5⁄1000. We have to convert 2⁄10 into an equivalent fraction with denominator 1000.
We have 2⁄10 = 200⁄1000.
Thus 2⁄10 + 5⁄1000 = 200⁄1000 + 5⁄1000 = 205⁄1000 = 0.205
We will get the same result by usual addition procedure as shown below:
We will see some solved examples based on the above discussion
Solved example 6.5
A large number of solid blocks are placed in a hall. All of them are of the same size as shown in fig.6.12 below.
• Some of them are divided into 10 equal pieces. They are coloured in Green
• Some are divided into 100 equal pieces. They are coloured in Red
• Some are divided into 1000 equal pieces. They are coloured in Blue
• Some are not divided at all. They remain as 'whole'. They are coloured in Yellow.
 |
| Fig.6.12 |
What is the quantity that each student picked?
Solution:
Student A: (i) 3 Green pieces. Each is 1⁄10 of a full block. So 3 pieces give 3⁄10 of a full block = 0.3
(ii) 5 Red pieces. Each is 1⁄100 of a full block. So 5 pieces give 5⁄100 of a full block = 0.05
(iii) 2 Blue pieces. Each is 1⁄1000 of a full block. So 2 pieces give 2⁄1000 of a full block = 0.002
Adding the 3 items (fig.6.13) we get :
 |
| Fig.6.13 |
So, if the Student A stacks all the pieces that he picked in proper order, he will get 0.352 of a full block.
Student B: (i) 1 full block. So we have '1' before the decimal place
(ii) 6 Green pieces. Each is 1⁄10 of a full block. So 6 pieces give 6⁄10 of a full block = 0.6
(iii) 4 Red pieces. Each is 1⁄100 of a full block. So 4 pieces give 4⁄100 of a full block = 0.04
(iv) 7 Blue pieces. Each is 1⁄1000 of a full block. So 7 pieces give 7⁄1000 of a full block = 0.007
Adding the 4 items (fig.6.14) we get:
 |
| Fig.6.14 |
Student C: (i) 3 full blocks. So we have '3' before the decimal place
(ii) 12 Green pieces. (In the fig.6.12, we can see only one green block. It contains only 10 pieces. But here the student has taken 12 green pieces. This is possible because, as given in the question, there are a large number of blocks of each colour arranged in the hall) Each is 1⁄10 of a full block. So 12 pieces give 12⁄10 of a full block = 1.2
This means that, when 12 green pieces are taken, one full block is formed, and in addition, 0.2 of a full block is obtained.
(iii) 0 Red pieces. Each is 1⁄100 of a full block. So 0 pieces give 0⁄100 of a full block = 0
(iv) 5 Blue pieces. Each is 1⁄1000 of a full block. So 5 pieces give 5⁄1000 of a full block = 0.005
Adding the 4 items we get (fig.6.15):
 |
| Fig.6.15 |
So, if the Student C stacks all the pieces that he picked in proper order, he will get 4 full blocks, and, in addition to that, 0.205 of a full block.
Student D: (i) 5 Green pieces. Each is 1⁄10 of a full block. So 5 pieces give 5⁄10 of a full block = 0.5
(ii) 23 Red pieces. Each is 1⁄100 of a full block. So 23 pieces give 23⁄100 of a full block = 0.23
(iii) 8 Blue pieces. Each is 1⁄1000 of a full block. So 8 pieces give 8⁄1000 of a full block = 0.008
Adding the 3 items we get (fig.6.16):
 |
| Fig.6.16 |
So, if the Student D stacks all the pieces that he picked in proper order, he will get 0.738 of a full block
Student E: (i) 5 full blocks. So we have '5' before the decimal place
(ii) 32 Green pieces. Each is 1⁄10 of a full block. So 32 pieces give 32⁄10 of a full block = 3.2
This means that, when 32 green pieces are taken, three full blocks are formed, and in addition, 0.2 of a full block is obtained.
(iii) 7 Red pieces. Each is 1⁄100 of a full block. So 0 pieces give 7⁄100 of a full block = 0.07
(iv) 123 Blue pieces. Each is 1⁄1000 of a full block. So 5 pieces give 123⁄1000 of a full block = 0.123
Adding the 4 items we get (fig.6.17):
 |
| Fig.6.17 |
Student F: 2 + 0.9 + 0.03 + 0.004 = 2.934
Student G: 14 + 0.6 + 0.32 + 0.075 = 14.995
Student H: 21 + 2.9 + 0.15 + 0.006 = 24.056
Student I: 0 + 0.6 + 0.04 + 0.075 = 0.715
We can also calculate the total quantity taken by more than one student.
Example: The total quantity taken by Students A and D are: 0.352 + 0.738 = 1.090
Another example: The total quantity taken by Students B, C and H are: 1.647 + 4.205 + 24.056 = 29.908
Solved example 6.6
Solve: (i) 0.28 + 0.36 (ii) 0.5 + 0.07 (iii) 1.41 + 1.70 (iv) 2.68 +3.84
Solution:
(i) 0.28 + 0.36 = 0.64 (ii) 0.5 + 0.07 = 0.57 (iii) 1.41 + 1.70 = 3.11 (iv) 2.68 +3.84 = 6.52
The steps are shown in the fig.6.18 below:
 |
| Fig.6.18 |
In the next section we will discuss about subtraction of decimals.
No comments:
Post a Comment