Chapter 6.5 - Metric weights - Solved examples

In the previous section we learned how the metric weights are expressed as decimals. In this section we will see some solved examples.

Solved example 6.7
The weight of a piece of steel tube was balanced by
• 3 blocks of 1 kg each +
• 5 blocks of 1 hectogram each +
• 7 blocks of 1 decagram each +
• 4 blocks of 1 gram each. 
What is the weight of the tube in kg?
Solution:
• 3 blocks of 1 kg each = 3 kg
• 5 blocks of 1 hectogram each = 5 hectograms.
1 hectogram = ¹⁄₁₀ kg = 0.1 kg
∴ 5 hectograms = 0.5 kg
• 7 blocks of 1 decagram each = 7 decagrams.
1 decagram = ¹⁄₁₀₀ kg = 0.01 kg
∴ 7 decagrams = 0.07 kg
• 4 blocks of 1 gram each = 4 grams.
1 gram = ¹⁄₁₀₀₀ kg =  0.001 kg
∴ 4 grams= 0.004 kg

Thus total weight = 3 + 0.5 + 0.07 + 0.004 = 3.574 kg

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Second method:We know that, the number of kilograms fall before the decimal point, the number of hectograms fall in the tenths place, number of decagrams fall in the hundredths place and the number of grams fall in the thousandths place. (Details here)
We have:
Number of Kilograms = 3; hectograms = 5; decagrams = 7 and grams = 4
So the wt. in kg = 3.574

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Solved example 6.8The weight of a bag of rice was balanced by 5 blocks of 1 kg each plus 3 blocks of 1 hectogram each plus 8 blocks of 1 gram each. What is the weight of the bag of rice in kg? If the wt. of the bag alone is 0.254 kg, what is the net wt. of rice?
Solution:
• 5 blocks of 1 kg each = 5 kg
• 3 blocks of 1 hectogram each = 3 
1 hectogram = 0.1 kg
∴ 3 hectograms = 0.3 kg
• 8 blocks of 1 gram each = 8 grams.
1 gram= 0.001 kg
∴ 8 grams= 0.008 kg

Thus total weight = 5 + 0.3 + 0.008 = 5.308 kg

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Second method:
We know that, the number of kilograms fall before the decimal point, the number of hectograms fall in the tenths place, number of decagrams fall in the hundredths place and the number of grams fall in the thousandths place. 
We have:
Number of Kilograms = 5; hectograms = 3; decagrams = 0 and grams = 8
So the wt. in kg = 5.308

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Wt. of bag alone = 0.254 kg.
There fore wt. of rice = 5.308 - 0.254 = 5.054 kg

Solved example 6.9
The weight of a parcel brought through courier is known to be 2.839 kg. How many blocks each of kg, hectogram, decagram and gram will have to be used to balance this weight ?
Solution:
Weight = 2.839 kg. 
The ‘whole number part’ is 2. So there are full 2 kilograms.
The decimal portion 0.839 kg gives the quantity between 2 kg and 3 kg
This can be split as: 8/10 + 3/100 + 9/1000  - - - (1)
[proof:
• 8/10 = 800/1000
• 3/100 = 30/1000
• 9/1000 = 9/1000
• Total = 800/1000 + 30/1000 + 9/1000 = 839/1000 = 0.839]
From (1), we can say there are 8 ‘one tenths of a kg’ in 0.839. But 1 one tenth of a kg is 1 hectogram. So there are 8 hectograms.
From (1), we can say there are 3 ‘one hundredths of a kg’ in 0.839. But 1 one hundredth of a kg is one decagram. So there are 3 decagrams.
From (1), we can say there are 9 ‘one thousandths of a kg’ in 0.839. But 1 one thousandth of a kg is one gram. So there are 9 grams.

Thus we can write: 2.839 = 2 kg + 8 hectograms + 3 decagrams + 9 grams.

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Second method:
We know that, the number of kilograms fall before the decimal point, the number of hectograms fall in the tenths place, number of decagrams fall in the hundredths place and the number of grams fall in the thousandths place.
We have:
Wt. = 2.839 kg
So, Number of Kilograms = 2; hectograms = 8; decagrams = 3 and grams = 9

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Solved example 6.10
When some Tomatoes were weighed on an electronic balance, the reading was 3.402 kg. Split this weight into kg, hectograms, decagrams and grams.
Solution:
Second method:
We know that, the number of kilograms fall before the decimal point, the number of hectograms fall in the tenths place, number of decagrams fall in the hundredths place and the number of grams fall in the thousandths place.
We have:
Wt. = 3.402 kg
So, Number of Kilograms = 3; hectograms = 4; decagrams = 0 and grams = 2

Solved example 6.11
Weight of a certain object is 3.256 kg. Express this in decagrams
Solution:
Weight = 3.256 kg. That is., 3 kg + 0.2 kg + 0.05 kg + 0.006 kg = 3 kg + 2 hectograms + 5 decagrams + 6 grams.

We have to convert each item into decagrams:
• 3 kg = 30 hectograms (∵ 1 kg = 10 hectogram)
30 hectograms = 300 decagrams (∵ 1 hectogram = 10 decagrams)
• 2 hectograms = 20 decagrams (∵ 1 hectogram = 10 decagrams)
• 5 decagrams = 5 decagrams
• 6 grams = 0.6 decagrams (∵ 1 decagram = 10 grams ⇒ 1 gram = 0.1 decagram)

So we get 3.256 kg = 300 + 20 + 5 + 0.6 = 325.6 decagrams

Solved example 6.12
Weight of a certain object is 5.029 kg. Express this in grams
Solution:
Weight = 5.029 kg. That is., 5 kg + 0.0 kg + 0.02 kg + 0.009 kg = 5 kg + 0 hectograms + 2 decagrams + 9 grams.

Now we have to convert each item into grams:

• 5 kg = 50 hectograms (∵ 1 kg = 10 hectogram)
50 hectograms = 500 decagrams (∵ 1 hectogram = 10 decagrams)
500 decagrams = 5000 grams (∵ 1 decagram = 10 grams)
[Once we understand the basics, we need not write the detailed steps. We need write only this:
5 kg = 5000 grams (∵ 1 kg = 1000 grams)]
• 0 hectograms = 0 grams
• 2 decagrams = 20 grams (∵ 1 decagram = 10 grams)
• 9 grams = 9 grams
So we get 5.029 kg = 5000 + 0 + 20 + 9 = 5029 grams

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So we have learned how to 

• split a given kg weight into smaller quantities like hectograms, decagrams and grams. 

• combine given smaller quantities into kg. 

• express the given wt in any one unit. 

Now we will see how we can use the set of ‘standard weights’ in day to day life, for finding the weights using a balance. 
In day to day life, we do not use hectograms and decagrams. We use only kilograms and grams. Hectograms and decagrams are used only for some special purposes like ‘quantitity of agricultural products’ obtained from a certain area of farm land. Centigrams and decigrams are also not used. For small quantities we use grams and milligrams only.

So we have to learn to find the quantities in terms of kilograms and grams only. For example, suppose an object weighs 2 kg and 6 hectograms. We do not have hectograms in the set of ‘standard weights’. But we do have kilograms and grams. After putting a 2 kg weight on the right side of the balance, we must put 600 grams above it. Because 6 hectograms = 600 grams. Then the two sides will balance.

We must be able to do the reverse also. That is., if we are given a weight, we must be able to express it in terms of kilograms and/or grams. Consider an example: The reading in an electronic balance is 4.283 kg. We can use this reading to convey the idea. People will understand it. For those who want finer details, we can split it into kg and grams. 4.283 kg is 4 kg plus 283 grams.

proof:
0.283 kg = 2 hectograms + 8 decagrams + 3 grams
• 2 hectograms = 20 decagrams  (∵ 1 hectogram = 10 decagrams)
20 decagrams = 200 grams (∵ 1 decagram = 10 grams)
• 8 decagrams = 80 grams (∵ 1 decagram = 10 grams)
• 3 grams = 3 grams

Total = 200 + 80 + 3 = 283 grams.

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From the above proof, we can note the following points:
• We get a wt. in kg
     ♦ The tenths give us hectograms
     ♦ The hundredths give us decagrams
     ♦ The thousandths give us grams
• We want the 'hectograms' and the 'decagrams' to go. For that:
     ♦ Multiply the digit in the tenths place by 100
     ♦ Multiply the digit in the hundredths place by 10
     ♦ keep the digit in the thousandths place as such
• Add the three items. This will give the 'quantity after the decimal point' in grams

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An even easier method is to multiply the decimal part by 1000.

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Some examples:
3.041 kg = 3 kg + [.041 × 1000] grams = 3 kg + 41 grams
5.002 kg = 5 kg + [.002 × 1000] grams = 5 kg + 2 grams
9.305 kg = 9 kg + 305 grams
2.3 kg = 2 kg + 300 grams
2.03 kg = 2 kg + 30 grams

Readers are advised to write the proof for each of the above examples in all the 3 methods.

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So we have seen how the metric weights are expressed as decimals. In the next section we will see the expression of metric volumes as decimals.

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Chapter 6.4 - Metric weights expressed as Decimals

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:

Fig.6.24

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.

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

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

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:

Fig.6.28

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

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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

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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 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

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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

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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 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

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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

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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

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Based on the discussions that we had so far in this section, we get the following Table 6.1:

Table 6.1

This system which uses milligrams, grams, decagrams, kilograms etc., is called the Metric system of weights. 

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In the next section we will see some solved examples.

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