In the previous section we discussed about the subtraction of decimals. In this section we will see some day to day situations where we use decimals. First we consider weights of quantities. On many occasions we will want to know the weights of various quantities. For example:
• weight of rice that we buy
• weight of tomatoes that we buy
• weight of a parcel that we send by post or courier
• weight of a bag of cement.
For finding the weight of small quantities like rice, vegetables, etc., we use a simple balance as shown in the fig.6.24 below:
Initially, the two sides of the balance are at the same level. On the left side, we place rice, or sugar, or vegetables or which ever item, the weight of which we need to find. When we place it, that side lowers down. And the other side moves up. The needle at the middle sways to the left. To raise the left side, and thus to bring it back to the initial level, we put some ‘standard weights’ on the right side of the balance. When the standard weights that we put on the right side become equal to the weight on the left side, the two sides will come to the same initial level. Thus the weight of rice or vegetables that we took will be given by the ‘standard weights’ on the right side. In order to use those standard weights properly, we need to learn some of their basic details:
The standard weights are available as a set. A merchant should have atleast one complete set. One such set is shown in the fig.6.25.
Some other types of sets can be seen here. Each member of a set will have a particular weight. And this weight will be clearly marked on it. For example,
• if it is marked as ‘200 g’ on a weight, it means, it’s weight is 200 grams.
• if it is marked as ‘1 kg’ on a weight, it means it’s weight is 1 kilogram.
Now we have to learn how to use these standard weights. We will learn it by discussing an example:
Suppose we want to know the weight of a wooden block. We place the block on the left side of the balance. The left side goes down and the right side goes up as shown in the fig.6.26(a).
Next, we put a standard wt. of 2 kg on the right side. But the left side is showing no sign of moving up. It is still down as shown in the fig.6.26(b). This means that the weight of 2 kg that we put on the right side is less than the weight of the block. In other words:
■ The wt of the block is more than 2kg
So we put more weight. We put an additional 1 kg. on the right side. So the total wt. on the right side is now 3 kg. This time the left side do rise. But it over shoots. It has gone high up. Higher than the right side. This is shown in the fig.6.26(c) This means:
■ The wt of the block is less than 3 kg.
So we can conclude that the wt. of the block is in between 2 kg and 3 kg. We do have to put some additional weights above the 2 kg on the right side. But this additional weights must be less than 1 kg. In other words, the additional weights must be 'suitable fractions’ of 1 kg. Fig.6.27 below shows how these suitable fractions can be obtained.
Fig.6.27(a) shows one full kg. That is., 1 kg. It is divided into 10 equal parts in fig.(b). So each part in (b) is one tenth of a kg. There is a special name given to 'one tenth of a kg'. It is Hectogram. So each part in fig.(b) is one hectogram. We can also say the reverse: 10 hectograms make 1 kg.
So we have successfully obtained the fractions of 1 kg. Let us put these fractions on to the right side of the balance above the 2 kg weight. This is shown in fig.6.28 below:
When 3 hectograms are placed, the two sides are at the same level, and the needle is at the center. So we can say: The weight of the block is 2 kg and 3 hectograms.
We have to write this weight in decimal form. We know from fig.6.27(b) that, one hectogram is 1⁄10 of a kg. That is., 1 hectogram = 0.1 kg. So 3 hectograms = 3⁄10 kg = 0.3 kg. Thus we can write: The wt. of the block = 2.3 kg
We can note a special relation:
■ 3 Hectograms were taken. That means, 3 'one tenths' were taken
■ In the decimal form, this 3 falls in the tenths place value
Another situation that can arise:
The wt. of the block is more than ‘2 kilograms and 3 hectograms’. At the same time, it is less than ‘2 kilograms and 4 hectograms’.
In this situation, we cannot put 4 hectograms on the right side. Neither can we stop at 3. Here arises the need to get 'fractions of a hectogram'. So one hectogram is divided into 10 equal parts. This is same as dividing 1 kg into 100 equal parts as shown in the fig.6.27(c). Each one of the 100 parts in fig.(c) is called a Decagram. So
• 10 decagrams make one hectogram.
• Also 100 decagrams make one kg.
Let us put some decagrams on to the right side of the balance. When 7 decagrams are placed, the balance becomes level. This is shown in the fig.6.29 below. So we can say: The weight of the block is 2 kg + 3 hectograms + 7 decagrams.
We have to write this weight in decimal form. We have seen that 3 hectograms = 0.3 kg. We know from fig.6.27(c) that, one decagram is 1⁄100 of a kg. That is., 1 decagram = 0.01 kg. So 7 decagrams = 7⁄100 kg = 0.07 kg. Thus we can write: The wt of the block = 2 kg + 0.3 kg + 0.07 kg = 2.37 kg
We can note a special relation:
■ 7 Decagrams were taken. That means, 7 'one hundredths' were taken
■ In the decimal form, this 7 falls in the hundredths place value
Yet another situation that can arise:
The wt. of the block is more than ‘2 kg + 3 hectograms + 7 decagrams’. At the same time it is less than ‘2 kg + 3 hectograms + 8 decagrams’. In this situation, we cannot put 8 decagrams on the right side. Neither can we stop at 7. So one decagram is further divided into 10 equal parts. This is same as dividing 1 kg into 1000 equal parts as shown in the fig.6.27(d). Each one of the 1000 parts is called a gram. So
• 10 grams make one decagram.
• Also 1000 grams make one kg.
Let us put some grams on to the right side of the balance. When 4 grams are placed, the balance becomes level. This is shown in the fig.6.30 below. So we can say: The weight of the block is 2 kg + 3 hectograms + 7 decagrams + 4 grams.
We have to write this weight in decimal form. We have seen that 3 hectograms = 0.3 kg. And also 7 decagrams = 0.07 kg. We know from fig.6.27(d) that, one gram is 1⁄1000 of a kg. That is., 1 gram = 0.001 kg. So 4 grams = 4⁄1000 kg = 0.004 kg. Thus we can write: The wt. of the block = 2 kg + 0.3 kg + 0.07 kg + 0.004 kg = 2.374 kg
We can note a special relation:
■ 4 Grams were taken. That means, 4 'one thousandths' were taken
■ In the decimal form, this 4 falls in the thousandths place value
Combining all such relations that we saw above, we can write:
■ The digit in the tenths place indicate how many hectograms are present
■ The digit in the hundredths place indicate how many decagrams are present
■ The digit in the thousandths place indicate how many grams are present
Based on the discussions that we had so far in this section, we get the following Table 6.1:
This system which uses milligrams, grams, decagrams, kilograms etc., is called the Metric system of weights.
In the next section we will see some solved examples.
• weight of rice that we buy
• weight of tomatoes that we buy
• weight of a parcel that we send by post or courier
• weight of a bag of cement.
For finding the weight of small quantities like rice, vegetables, etc., we use a simple balance as shown in the fig.6.24 below:
 |
| Fig.6.24 |
The standard weights are available as a set. A merchant should have atleast one complete set. One such set is shown in the fig.6.25.
 |
| Fig.6.25 |
Some other types of sets can be seen here. Each member of a set will have a particular weight. And this weight will be clearly marked on it. For example,
• if it is marked as ‘200 g’ on a weight, it means, it’s weight is 200 grams.
• if it is marked as ‘1 kg’ on a weight, it means it’s weight is 1 kilogram.
Now we have to learn how to use these standard weights. We will learn it by discussing an example:
Suppose we want to know the weight of a wooden block. We place the block on the left side of the balance. The left side goes down and the right side goes up as shown in the fig.6.26(a).
 |
| Fig.6.26 |
■ The wt of the block is more than 2kg
So we put more weight. We put an additional 1 kg. on the right side. So the total wt. on the right side is now 3 kg. This time the left side do rise. But it over shoots. It has gone high up. Higher than the right side. This is shown in the fig.6.26(c) This means:
■ The wt of the block is less than 3 kg.
So we can conclude that the wt. of the block is in between 2 kg and 3 kg. We do have to put some additional weights above the 2 kg on the right side. But this additional weights must be less than 1 kg. In other words, the additional weights must be 'suitable fractions’ of 1 kg. Fig.6.27 below shows how these suitable fractions can be obtained.
 |
| Fig.6.27 |
So we have successfully obtained the fractions of 1 kg. Let us put these fractions on to the right side of the balance above the 2 kg weight. This is shown in fig.6.28 below:
 |
| Fig.6.28 |
We have to write this weight in decimal form. We know from fig.6.27(b) that, one hectogram is 1⁄10 of a kg. That is., 1 hectogram = 0.1 kg. So 3 hectograms = 3⁄10 kg = 0.3 kg. Thus we can write: The wt. of the block = 2.3 kg
We can note a special relation:
■ 3 Hectograms were taken. That means, 3 'one tenths' were taken
■ In the decimal form, this 3 falls in the tenths place value
Another situation that can arise:
The wt. of the block is more than ‘2 kilograms and 3 hectograms’. At the same time, it is less than ‘2 kilograms and 4 hectograms’.
In this situation, we cannot put 4 hectograms on the right side. Neither can we stop at 3. Here arises the need to get 'fractions of a hectogram'. So one hectogram is divided into 10 equal parts. This is same as dividing 1 kg into 100 equal parts as shown in the fig.6.27(c). Each one of the 100 parts in fig.(c) is called a Decagram. So
• 10 decagrams make one hectogram.
• Also 100 decagrams make one kg.
Let us put some decagrams on to the right side of the balance. When 7 decagrams are placed, the balance becomes level. This is shown in the fig.6.29 below. So we can say: The weight of the block is 2 kg + 3 hectograms + 7 decagrams.
 |
| Fig.6.29 |
We can note a special relation:
■ 7 Decagrams were taken. That means, 7 'one hundredths' were taken
■ In the decimal form, this 7 falls in the hundredths place value
Yet another situation that can arise:
The wt. of the block is more than ‘2 kg + 3 hectograms + 7 decagrams’. At the same time it is less than ‘2 kg + 3 hectograms + 8 decagrams’. In this situation, we cannot put 8 decagrams on the right side. Neither can we stop at 7. So one decagram is further divided into 10 equal parts. This is same as dividing 1 kg into 1000 equal parts as shown in the fig.6.27(d). Each one of the 1000 parts is called a gram. So
• 10 grams make one decagram.
• Also 1000 grams make one kg.
Let us put some grams on to the right side of the balance. When 4 grams are placed, the balance becomes level. This is shown in the fig.6.30 below. So we can say: The weight of the block is 2 kg + 3 hectograms + 7 decagrams + 4 grams.
 |
| Fig.6.30 |
We can note a special relation:
■ 4 Grams were taken. That means, 4 'one thousandths' were taken
■ In the decimal form, this 4 falls in the thousandths place value
Combining all such relations that we saw above, we can write:
■ The digit in the tenths place indicate how many hectograms are present
■ The digit in the hundredths place indicate how many decagrams are present
■ The digit in the thousandths place indicate how many grams are present
Based on the discussions that we had so far in this section, we get the following Table 6.1:
 |
| Table 6.1 |
In the next section we will see some solved examples.
No comments:
Post a Comment