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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Chapter 6.1 - Place values in Decimals

In the previous section we saw the relation between:
• the 'position' of a digit after the decimal point and
• the 'power' of 10 in the denominator of it's fraction form
In this section we will see a pictorial representation of that relation:

We have seen that 0.36 = ³⁶⁄₁₀₀ = ³⁄₁₀  + ⁶⁄₁₀₀
 ³⁄₁₀ is 3 parts taken out of 10 equal parts. This can be shown pictorially as in fig.6.6(a):

Fig.6.6

In fig.(a), the whole is divided into 10 equal parts by the horizontal lines. From those 10, 3 parts (shown in green) are taken out.  But we need 6 more. This 6 cannot be taken from the divisions in fig.(a) because:
• The smallest division here is 10. 
• Our requirement 6 is smaller than the smallest division 10. 
So to take out 6, we have to divide the whole into still smaller parts. So we divide the whole into 100 equal parts. This is shown in fig.(b). The horizontal and vertical lines together divide the whole into 100 equal parts. Among those 100, 6 nos. are marked in magenta color. Those 6 form 6/100 of the whole.

So to get 0.36, we have to add 3 from the 10 equal parts and 6 from the 100 equal parts. That is.,
0.36 = ³⁄₁₀  + ⁶⁄₁₀₀ 
• 3 is having the 1^st power of 10 in the denominator and
• 6 is having the 2^nd power of 10 in the denominator

Thus we have a pictorial representation of the relation. Based on this relation, we can give accurate 'place value names' for each of the digits coming after the decimal point:

■ 3, which is the 1^st digit coming just after the decimal point is obtained from dividing the whole into 10 equal parts. So the place value of the 1^st digit is called tenths.
■ 6, which is the 2^nd digit coming after the decimal point is obtained from dividing the whole into 100 equal parts. So the place value of the 2^nd digit is called hundredths


This is shown in the fig.6.7 below:

Fig.6.7

In the same way, the 3^rd position after the decimal point is called thousandths, and so on.

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We will now see some solved examples based on the above discussion:
Solved example 6.1
Write the following fractions in decimal form: (i) 2/5  (ii) 6/25  (iii) 1/8  (iv) 5/8  (v) 3/20
Solution:
(i) • We have to convert ²⁄₅ into an equivalent fraction which has a denominator 10, or 100, or 1000 ... so on, which ever is suitable. Let us try 10:
• ²⁄₅ = ^x⁄₁₀. It is clear that if we multiply the denominator 5 by 2, we will get 10. So the numerator must also be multiplied by 2. We will get x = 2 × 2  = 4
• So the equivalent fraction is calculated as:  ²⁄₅  =   ^{(2 × 2)}⁄_{(5 × 2)}  =  ⁴⁄₁₀   
• Thus the decimal form is 0.4

(ii) • We have to convert ⁶⁄₂₅ into an equivalent fraction which has a denominator 10, or 100, or 1000 ... so on, which ever is suitable. Let us try 10:
• ⁶⁄₂₅ = ^x⁄₁₀. There is no whole number which when multiplied with 25 will give 10. So we will try 100:
• ⁶⁄₂₅ = ^x⁄₁₀₀. It is clear that if we multiply the denominator 25 by 4, we will get 100. So the numerator must also be multiplied by 4. We will get x = 6 × 4  = 24
• So the equivalent fraction is calculated as:  ⁶⁄₂₅  =   ^{(6 × 4)}⁄_{(25 × 4)}  =  ²⁴⁄₁₀₀   
• Thus the decimal form is 0.24

(iii) • We have to convert ¹⁄₈ into an equivalent fraction which has a denominator 10, or 100, or 1000 ... so on, which ever is suitable. Let us try 10:
• ¹⁄₈ = ^x⁄₁₀. There is no whole number which when multiplied with 8 will give 10. So we will try 100:
• ¹⁄₈ = ^x⁄₁₀₀. There is no whole number which when multiplied with 8 will give 100. So we will try 1000:
• ¹⁄₈ = ^x⁄₁₀₀₀. It is clear that if we multiply the denominator 8 by 125, we will get 1000. So the numerator must also be multiplied by 125. We will get x = 1 × 125  = 125
• So the equivalent fraction is calculated as:  ¹⁄₈  =   ^{(1 × 125)}⁄_{(8 × 125)}  =  ¹²⁵⁄₁₀₀₀   
• Thus the decimal form is 0.125

(iv) • We have to convert ⁵⁄₈ into an equivalent fraction which has a denominator 10, or 100, or 1000 ... so on, which ever is suitable. From previous example, we know that, for the denominator 8, 10 and 100 are not possible. We have to use 1000. We also know that 125 is the factor that has to be used for multiplication.
• So the equivalent fraction is calculated as:  ⁵⁄₈  =   ^{(5 × 125)}⁄_{(8 × 125)}  =  ⁶²⁵⁄₁₀₀₀   
• Thus the decimal form is 0.625

(v) • We have to convert ³⁄₂₀ into an equivalent fraction which has a denominator 10, or 100, or 1000 ... so on, which ever is suitable. Let us try 10:
• ³⁄₂₀ = ^x⁄₁₀. There is no whole number which when multiplied with 20 will give 10. So we will try 100:
• ³⁄₂₀ = ^x⁄₁₀₀. It is clear that if we multiply the denominator 20 by 5, we will get 100. So the numerator must also be multiplied by 5. We will get x = 3 × 5  = 15
• So the equivalent fraction is calculated as:  ³⁄₂₀  =   ^{(3 × 5)}⁄_{(20 × 5)}  =  ¹⁵⁄₁₀₀   
• Thus the decimal form is 0.15

Solved example 6.2

Write the place value of each of the digits in the following numbers:

(i) 0.85  (ii) 0.639  (iii) 0.079 (iv) 0.02  (v) 0.0208

Solution:

The place values are shown in the following table:

Solved example 6.3

Write the decimal represented by each of the following figs

Fig.6.8

Solution:

(a) • 8 parts are taken out of 10 equal parts + 4 parts are taken out of 100 equal parts

• So in fractional form it is ⁸⁄₁₀ + ⁴⁄₁₀₀

• From this we get the decimal form as 0.84 (8 in the tenths place and 4 in the hundredths place)

(b) • 5 parts are taken out of 10 equal parts + 9 parts are taken out of 100 equal parts

• So in fractional form it is ⁵⁄₁₀ + ⁹⁄₁₀₀

• From this we get the decimal form as 0.59 (5 in the tenths place and 9 in the hundredths place)

(c) • 3 parts are taken out of 10 equal parts + 8 parts are taken out of 100 equal parts

• So in fractional form it is ³⁄₁₀ + ⁸⁄₁₀₀

• From this we get the decimal form as 0.38 (3 in the tenths place and 8 in the hundredths place)

(d) • 7 parts are taken out of 10 equal parts + 2 parts are taken out of 100 equal parts

• So in fractional form it is ⁷⁄₁₀ + ²⁄₁₀₀

• From this we get the decimal form as 0.72 (7 in the tenths place and 2 in the hundredths place)

Solved example 6.4

An artist is trying to make a particular shade of yellow colour for his painting. He has several blocks (all of the same size) of 'perfect yellow' in his shelf. But he does not want perfect yellow. He wants a particular shade of yellow. For that, he must take 0.27 of a block. How would he obtain 0.27 of a block?

solution:

• The artist must take exact 0.27. Any other quantity would not give the required shade.

• We know that 0.27 = ²⁄₁₀ + ⁷⁄₁₀₀

• Given that all blocks are of equal size.

• So first he must take one block, divide it into 10 equal parts and take 2 parts from it

• Then he must take another block, divide it into 100 equal parts and take 7 parts from it

• The '2 out of 10', and the '7 out of 100' together will give 0.27 of a block. This is shown in the fig. below:

Fig.6.9

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In the next section we will discuss about comparison and addition of decimals

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