Chapter 6.9 - Basics of Recurring decimals

In the previous section we saw how money is expressed as decimals. In this section, we will see some more advanced topics related to decimals.
We have seen the basics about decimals here. We know how to convert fractions like ¹⁄₂ and ³⁄₄ into decimal form. 
We know that  ¹⁄₂ = 0.5 and ³⁄₄  = 0.75
To convert fractions like ¹⁄₈, a little more work is involved. We saw such problems here. Let us analyse them again:
• Take ¹⁄₈.  To convert it into decimal for, we must first convert it into an equivalent fraction
• The denominator of this equivalent fraction should be any one of 10¹, 10², 10³ . . . etc., which ever is suitable
• For obtaining such an equivalent fraction, we must multiply both the numerator and denominator of ¹⁄₈ by a 'suitable number'
• There is a clear procedure to obtain this 'suitable number'. Let us see what it is:
1. We have '8' in the denominator. We must factorise it first
2. We have: 8 = 2×2×2. There are 3 'twos' 
3. We must convert each of these 'twos' into a '10'
4. For that, we give each two, a '5' like this: (2×5)×(2×5)×(2×5). Now it becomes 10×10×10 = 1000
5. So 3 external 'fives' are used to get a power of 10
6. 3 external 'fives' give 5 ×5 ×5 = 125.   So the 'suitable number' is 125. If we multiply the denominator by 125, we will get the 'required power of 10'. But the numerator must also be multiplied by 125. So we can write:
7. ¹⁄₈ = ^(1×125) ⁄_(8×125) = ¹²⁵⁄₁₀₀₀ = 0.125

---

The above result will find application in another situation also:
If we have a fraction with denominator 125, we can multiply both the numerator and denominator by '8'. We will get the 'required power of 10' in the denominator.

---

Another example: Convert ³⁄₁₆₀  into decimal form
1. We have '160' in the denominator. We must factorise it first
2. We have: 160 = 2×2×2×2×2×5. There are 5 'twos', and a five. We will separate 1 two and the five
3. So we can write: 160 = (2×2×2×2)×(2×5) =  (2×2×2×2)×(10). So we have 4 twos remaining
4. We must convert each of these 'twos' into a '10'
5. For that, we give each two, a '5' like this: (2×5)×(2×5)×(2×5)×(2×5)×(10). Now it becomes 10×10×10×10×10 = 100000 =10⁵
6. So 4 external 'fives' are used to get a power of 10. 
7. 4 external 'fives' give 5 ×5 ×5×5 = 625. So the 'suitable number' is 625. If we multiply the denominator by 625, we will get the 'required power of 10'. But the numerator must also be multiplied by 625. So we can write:
8. ³⁄₁₆₀ = ^(3×625) ⁄_(160×625) = ¹⁸⁷⁵⁄₁₀₀₀₀₀ = 0.01875

---

We will see some solved examples:
Solved example 6.27
Write each of the fractions below in the decimal form
(i) ¹⁄₅₀,  (ii) ³⁄₄₀,  (ii) ⁵⁄₁₆,   (iv) ¹²⁄₆₂₅
Solution:
(i) ¹⁄₅₀ : We know that when 50 is multiplied by 2, we will get 100. But we will do the steps to get more acquainted with the process:
1. We have '50' in the denominator. We must factorise it first
2. We have: 50 = 2×5×5. There are 2 'fives', and a 'two'. We will separate 1 five and the two
3. So we can write: 50 = (5)×(2×5) = (5)×(10) . So we have one 'five' remaining
4. We must convert this 'five' into a '10'
5. For that, we give this 'five', a '2' like this: (5×2)×(10). Now it becomes 10×10 = 100 =10²
6. So 1 external 'two' is used to get a power of 10. 
7. So the 'suitable number' is 2. If we multiply the denominator by 2, we will get the 'required power of 10'. But the numerator must also be multiplied by 2. So we can write:
8. ¹⁄₅₀ = ^(1×2) ⁄_(50×2) = ²⁄₁₀₀ = 0.02

(ii) ³⁄₄₀ : 1. We have '40' in the denominator. We must factorise it first
2. We have: 40 = 2×2×2×5. There are 3 'twos', and a five. We will separate 1 two and the five
3. So we can write: 40 = (2×2)×(2×5) =  (2×2)×(10). So we have 2 twos remaining
4. We must convert each of these 'twos' into a '10'
5. For that, we give each two, a '5' like this: (2×5)×(2×5)×(10). Now it becomes 10×10×10 = 1000 =10³
6. So 2 external 'fives' are used to get a power of 10. 
7. 2 external 'fives' give 5 ×5  = 25. So the 'suitable number' is 25. If we multiply the denominator by 25, we will get the 'required power of 10'. But the numerator must also be multiplied by 25. So we can write:
8. ³⁄₄₀ = ^(3×25) ⁄_(40×25) = ⁷⁵⁄₁₀₀₀ = 0.075

(iii) ⁵⁄₁₆ : 1. We have '16' in the denominator. We must factorise it first
2. We have: 16 = 2×2×2×2. There are 4 'twos'.
3. We must convert each of these 'twos' into a '10'
4. For that, we give each two, a '5' like this: (2×5)×(2×5)×(2×5)×(2×5). Now it becomes 10×10×10×10 = 10000 =10⁴
5. So 4 external 'fives' are used to get a power of 10. 
6. 4 external 'fives' give 5 ×5 ×5×5 = 625. So the 'suitable number' is 625. If we multiply the denominator by 625, we will get the 'required power of 10'. But the numerator must also be multiplied by 625. So we can write:
7. ⁵⁄₁₆ = ^(5×625) ⁄_(16×625) = ³¹²⁵⁄₁₀₀₀₀ = 0.3125

(iv) ¹²⁄₆₂₅ : 1. We have '625' in the denominator. We must factorise it first
2. We have: 625 = 5×5×5×5. There are 4 'fives'
3. We must convert each of these 'fives' into a '10'
4. For that, we give each five, a '2' like this: (5×2)×(5×2)×(5×2)×(5×2). Now it becomes 10×10×10×10 = 10000 =10⁴
5. So 4 external 'twos' are used to get a power of 10. 
6. 4 external 'twos' give 2 ×2 ×2×2 = 16. So the 'suitable number' is 16. If we multiply the denominator by 16, we will get the 'required power of 10'. But the numerator must also be multiplied by 16. So we can write:

7. ¹²⁄₆₂₅ = ^(12×16) ⁄_(625×16) = ¹⁹²⁄₁₀₀₀₀ = 0.0192

---

Now let us try to write ¹⁄₃ in decimal form. The above method will not work because, there is no natural number, which when multiplied with 3, will give any 'power of 10'. So we will use another method:
1. We know that ¹⁄₃ = ^(1×10) ⁄_(3×10). 
2. Let us rearrange the right side: ¹⁄₃ = ¹⁄₁₀ × ¹⁰⁄₃ 
3. In the above result, we can write ¹⁰⁄₃ as (3 + ¹⁄₃ )
4. So (2) becomes ¹⁄₃ = ¹⁄₁₀ × (3 + ¹⁄₃ ). So we get:
5. ¹⁄₃ = ³⁄₁₀ + ¹⁄₃₀
6. Look at the above result carefully. We have two fractions on the right side: ³⁄₁₀ and ¹⁄₃₀ 
    ♦ Out of these two, ³⁄₁₀ is in an 'ideal form'. Because it has a 'power of 10' in the denominator. So it can be readily converted into a decimal form
    ♦ But the other fraction ¹⁄₃₀ is causing a problem. It cannot be readily converted into a decimal form
    ♦ If this ¹⁄₃₀ is very very small, then we can ignore it. In that case, (5) will become ¹⁄₃ = ³⁄₁₀
    ♦ But unfortunately, ¹⁄₃₀ is not very small, and we cannot ignore it. 
• After reaching (5), if we write ¹⁄₃ = 0.3, we are ignoring ¹⁄₃₀
• That is not a good thing to do because, ¹⁄₃₀ is not a small quantity, that can be 'just ignored'
7. We arrived at (5) by writing ¹⁄₃ as ^(1×10) ⁄_(3×10) in (1).  Now let us write it in a modified form: 

8. We know that ¹⁄₃ = ^(1×100) ⁄_(3×100). 
9. Let us rearrange the right side: ¹⁄₃ = ¹⁄₁₀₀ × ¹⁰⁰⁄₃ 
10. In the above result, we can write ¹⁰⁰⁄₃ as (33 + ¹⁄₃ )
11. So (9) becomes ¹⁄₃ = ¹⁄₁₀₀ × (33 + ¹⁄₃ ). So we get:
12. ¹⁄₃ = ³³⁄₁₀₀ + ¹⁄₃₀₀
13. Look at the above result carefully. We have two fractions on the right side: ³³⁄₁₀₀ and ¹⁄₃₀₀
    ♦ Out of these two, ³³⁄₁₀₀ is in an 'ideal form'. Because it has a 'power of 10' in the denominator. So it can be readily converted into a decimal form
    ♦ But the other fraction ¹⁄₃₀₀ is causing a problem. It cannot be readily converted into a decimal form
    ♦ If this ¹⁄₃₀₀ is very very small, then we can ignore it. In that case, (12) will become ¹⁄₃ = ³³⁄₁₀₀
    ♦ But unfortunately, ¹⁄₃₀₀ is not very small, and we cannot ignore it
• After reaching (12), if we write ¹⁄₃ = 0.33, we are ignoring ¹⁄₃₀₀
• That is not a good thing to do because, ¹⁄₃₀₀ is not a small quantity, that can be 'just ignored'
• It may be noted that ¹⁄₃₀₀ is ten times smaller than ¹⁄₃₀ , which is causing the problem in (5) 
14. We arrived at (12) by writing ¹⁄₃ as ^(1×100) ⁄_(3×100) in (8).  Now let us write it in a modified form:

15. We know that ¹⁄₃ = ^(1×1000) ⁄_(3×1000). 
16. Let us rearrange the right side: ¹⁄₃ = ¹⁄₁₀₀₀ × ¹⁰⁰⁰⁄₃ 
17. In the above result, we can write ¹⁰⁰⁰⁄₃ as (333 + ¹⁄₃ )
18. So (16) becomes ¹⁄₃ = ¹⁄₁₀₀₀ × (333 + ¹⁄₃ ). So we get:
19. ¹⁄₃ = ³³³⁄₁₀₀₀ + ¹⁄₃₀₀₀
20. Look at the above result carefully. We have two fractions on the right side: ³³³⁄₁₀₀₀ and ¹⁄₃₀₀₀
    ♦ Out of these two, ³³³⁄₁₀₀₀ is in an 'ideal form'. Because it has a 'power of 10' in the denominator. So it can be readily converted into a decimal form
    ♦ But the other fraction ¹⁄₃₀₀₀ is causing a problem. It cannot be readily converted into a decimal form
    ♦ If this ¹⁄₃₀₀₀ is very very small, then we can ignore it. In that case, (19) will become ¹⁄₃ = ³³³⁄₁₀₀₀
    ♦ But unfortunately, ¹⁄₃₀₀₀ is not very small, and we cannot ignore it
• After reaching (19), if we write ¹⁄₃ = 0.333, we are ignoring ¹⁄₃₀₀₀
• That is not a good thing to do because, ¹⁄₃₀₀₀ is not a small quantity, that can be 'just ignored'
• It may be noted that ¹⁄₃₀₀₀ is ten times smaller than ¹⁄₃₀₀ , which is causing the problem in (12)
• Also it is 100 times smaller than ¹⁄₃₀ , which is causing the problem in (5)
• So the fractional part is obviously decreasing with each step. It will keep on decreasing with each step and reach very low values. How low can it reach? 
The lowest value possible is 'zero'. So, with each step, the fractional part gets closer and closer to zero
21. We arrived at (19) by writing ¹⁄₃ as ^(1×1000) ⁄_(3×1000) in (15)

First we used 10, then 100, and we used 1000 just above. We can proceed using 10000, 100000, etc.,
But we do not have to write the steps. A pattern has already emerged. Based on that pattern, we can write:
■ ¹⁄₃    =  ³⁄₁₀   +  ¹⁄₃₀       =   0.3 + ¹⁄₃₀
■ ¹⁄₃    =  ³³⁄₁₀₀   +  ¹⁄₃₀₀      =   0.33 + ¹⁄₃₀₀
■ ¹⁄₃    =  ³³³⁄₁₀₀₀   +  ¹⁄₃₀₀₀       =   0.333 + ¹⁄₃₀₀₀
■ ¹⁄₃    =  ³³³³⁄₁₀₀₀₀   +  ¹⁄₃₀₀₀₀       =   0.3333 + ¹⁄₃₀₀₀₀
■ ¹⁄₃     =  ³³³³³⁄₁₀₀₀₀₀   +  ¹⁄₃₀₀₀₀₀        =   0.33333 + ¹⁄₃₀₀₀₀₀

All the above results are true. They are exact values of ¹⁄₃ . We can proceed further as long as we wish. But this much is sufficient for us to understand an important property:

When the number of digits on the 'right side of the decimal point' increases, the remaining fractional portion decreases. 
• For example, if we take 4 places on the right side of the decimal point, ¹⁄₃ = 0.3333, the fractional part then is  ¹⁄₃₀₀₀₀
• If we take 5 places on the right side of the decimal point, ¹⁄₃ = 0.33333, the fractional part then is ¹⁄₃₀₀₀₀₀, which is smaller than ¹⁄₃₀₀₀₀

As the fractional part becomes smaller and smaller, it can be ignored if we take sufficient number of places after the decimal point. In various fields of science and engineering, there are strict rules that tell us the 'number of places' that we have to take after the decimal point.

Another important point can also be noted from the above discussion:
• We have written ¹⁄₃ as the sum of a decimal value and a fractional value 
• The left side is always a constant, which is equal to ¹⁄₃
• So the right side must also be a constant. That is., the sum of the decimal value and the fractional value must also be a constant equal to ¹⁄₃
• But we saw that, when the number of places after the decimal point increases, the fractional value decreases
• When the fractional value decreases, the decimal portion must increase in value. Then only will the sum remain a constant. Thus we can say: 
■ As the number of decimal places increases, the value of the decimal portion gets closer and closer to ¹⁄₃ . This is shown below:

---

So now we know that we cannot convert ¹⁄₃ into an exact decimal form. There will always be a small fraction remaining. We use a special method to represent such decimals.

In the final pattern that we derived above, we saw 0.3, 0.33, 0.333, and so on. The digit '3' will repeat for ever. Such decimals are called recurring decimals. There are three different ways to represent recurring decimals. We will see the details of those methods by taking ¹⁄₃ as an example:

• In Method 1, three dots are placed after the decimal. It indicates that it is a recurring decimal
• In Method 2, a dot is placed above the digit which repeats forever. In our case, 3 repeats for ever. So, the dot is placed over it
• In Method 3, a line is placed above the digit which repeats forever. In our case, 3 repeats for ever. So, the line is placed over it

---

In the next section we will see another example.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 8 - Basics about Percentage

In the previous section we completed the discussion on ratios. In this chapter, we will discuss about percentage. We have seen fractions and decimals. Decimals are the next stage of development from fractions. The stage after decimals is percent. Let us see the details:

We know that, comparison of fractions become easy if they have a common denominator. We have also seen that decimals are fractions with denominator 10, 100, 1000 etc.,. Now, Percent is a special decimal in which the denominator is always 100.  In fact, percent is two words combined: 'per' and 'cent'. Per means 'for every'. And cent means 'hundred'. So percent means: 'for every hundred'. It is indeed true because, the denominator is always 100.

Consider the fraction ³⁵⁄₁₀₀. It means 35 parts taken from 100 equal parts. That is 35 'per 100'. In decimal form it is 0.35. But when the denominator is 100, we can write it as 35%. It is read as 'Thirty five Percent'.

Similarly, ²²⁄₁₀₀  is 22%, ⁷⁴⁄₁₀₀ is 74%. To write a fraction or decimal as a percent, the denominator must always be 100. Let us see some examples. Fig.8.1(a) below shows a square area divided into 25 equal parts.

Fig.8.1

In (b), 4 out of 25 is shaded. In fraction form, it is ⁴⁄₂₅. To write it in decimal form, we must convert it into an equivalent fraction with denominator 10, or 100, or 1000 and so on. We can choose any one out of these 10, 100, 1000 etc., which ever is convenient. But to write it in percent form, there is no choice. The denominator must be exactly 100. So we will select 100.

⁴⁄₂₅ =  ^x⁄₁₀₀. By cross products method, we get x = 16. 
Thus ⁴⁄₂₅ = ¹⁶⁄₁₀₀. But ¹⁶⁄₁₀₀ = 16%. So we can write:
In fig.(b), 16% of the square area is shaded.  

Similarly in fig.(c), ⁶⁄₂₅ is shaded. ⁶⁄₂₅ = ²⁴⁄₁₀₀ = 24%. So we can write:
In fig.(c), 24% of the square area is shaded.

---

Now we will see some solved examples
Solved example 8.1
In a class of 20 students, 12 are boys. What percent of the whole class is boys?
Solution:
• Total number of students = 20
• Number of boys = 12
• Number of boys expressed as a fraction of the total number of students = ¹²⁄₂₀ 
• Equivalent fraction of ¹²⁄₂₀ with denominator 100 = ⁶⁰⁄₁₀₀
Thus, 60% of the whole class are boys.
Solved example 8.2
In a library, there are 1200 books. Out of them, 240 are on the subject Maths. What percent of the whole books are Maths books?
Solution:
• Total number of books = 1200
• Number of Maths books = 240
• Number of maths books expressed as a fraction of the total number of books = ²⁴⁰⁄₁₂₀₀ 
• Equivalent fraction of  ²⁴⁰⁄₁₂₀₀ with denominator 100 = ²⁰⁄₁₀₀
Thus, 20% of the total of books in the library are Maths books.
Solved example 8.3
A cake is divided into equal pieces. Student A took 4 pieces. Student B took 5 pieces. Student C took the remaining 3 pieces. What percentage of the whole cake did each student take?
Solution:
Number of pieces that A took = 4 
Number of pieces that B took = 5
Number of pieces that C took = 3
∴ Total number of pieces = 12
• Fraction taken by A = ⁴⁄₁₂ = ^33.33⁄₁₀₀
Proof:
Let ⁴⁄₁₂ = ^x⁄₁₀₀
Cross multiplying we get 400 = 12 x
∴ x = 400/12 = 100/3 = 33.33
• So A took 33.33% of the whole cake
Fraction taken by B = ⁵⁄₁₂ = ^41.67⁄₁₀₀ 
• So B took 41.67% of the whole cake
Fraction taken by C = ³⁄₁₂ = ²⁵⁄₁₀₀
• So C took 25% of the whole cake

This completes the solution to the problem. But we will do an additional calculation:
■ Let us add the number of pieces:
Number of pieces taken by A = 4
Number of pieces taken by B = 5
Number of pieces taken by C = 3
Sum = 4 +5 + 3 = 12 pieces = One full cake

■ Let us add the fractions:
Fraction taken by A = ⁴⁄₁₂
Fraction taken by B = ⁵⁄₁₂
Fraction taken by C = ³⁄₁₂
Sum = ⁴⁄₁₂ + ⁵⁄₁₂ + ³⁄₁₂ = ¹²⁄₁₂ = 1 One full cake

■ Let us add all the 3 percentages:
Percentage taken by A = 33.33
Percentage taken by B = 41.67
Percentage taken by C = 25
Sum = 33.33 + 41.67 + 25 = 100% = 1 full cake

This is an interesting result. We add the 'percentage of the whole' taken by each student, and the sum that we get is 100 %, which is 1 full cake.

Solved example 8.4
Some bricks were unloaded from a truck. Three workers A, B and C did the unloading. A unloaded 37 bricks. B unloaded 42 bricks. C unloaded the remaining 21 bricks. What percentage of the whole did each worker unload?
Solution:
Number of bricks that A unloaded = 37
Number of bricks that B unloaded = 42
Number of bricks that C unloaded = 21
• Total number of bricks = 37 + 42 + 21 = 100
Fraction unloaded by A = ³⁷⁄₁₀₀
• Here the denominator is already 100. This is because, 'the whole' (the total number of bricks) which gives the denominator is 100. So we do not need to find an equivalent fraction.
So A unloaded 37% of the whole
• Fraction unloaded by B = ⁴²⁄₁₀₀
So B unloaded 42% of the whole
• Fraction unloaded by C = ²¹⁄₁₀₀ 
So C unloaded 21 % of the whole
Solved example 8.5
Convert each of the following fractions into percentage:
(i) ⁵⁄₁₆  (ii) ¹⁄₃  (iii) ⁴⁄₉  (iv) ⁵⁄₈  (v) ⁷⁄₇ 
Solution:
(i) ⁵⁄₁₆ = ^x⁄₁₀₀ ⇒ x = ⁵⁄₁₆ × 100 = ⁵⁰⁰⁄₁₆ = 31.25
∴ ⁵⁄₁₆ = ^31.25⁄₁₀₀  = 31.25%
• The above steps can be written in just one line:
Multiply the given fraction by 100. Put a '%' sign at the end of the product. That is the required percentage.

(ii) ¹⁄₃ = ^x⁄₁₀₀ ⇒ x = ¹⁄₃ × 100 = ¹⁰⁰⁄₃ = 33.33
∴ ¹⁄₃ = ^33.33⁄₁₀₀  = 33.33%
• The above steps can be written in just one line:
Multiply the given fraction by 100. Put a '%' sign at the end of the product. That is the required percentage.

(iii) ⁴⁄₉ = ^x⁄₁₀₀ ⇒ x = ⁴⁄₉ × 100 = ⁴⁰⁰⁄₉ = 44.44
∴ ⁴⁄₉ = ^44.44⁄₁₀₀  = 44.44%
• The above steps can be written in just one line:
Multiply the given fraction by 100. Put a '%' sign at the end of the product. That is the required percentage.
■ From the above examples, we can conclude that, to convert a fraction into a percentage, all we need to do is to multiply the fraction by 100. And then, put a '%' sign at the end of the product.
(iv) ⁵⁄₈ × 100 = ⁵⁰⁰⁄₈ = 62.5
∴ ⁵⁄₈ = 62.5%

(v) ⁷⁄₇ × 100 = ⁷⁰⁰⁄₇ = 100
∴ ⁷⁄₇ = 100%
• It may be noted that this problem does not require any steps to find the solution. This is because ⁷⁄₇ (which is equal to 1) is actually not a fraction. It is a 'whole 1'. And a 'whole 1' is 100%
Solved example 8.6
Convert each of the following decimals into percentage:
(i) 0.25  (ii) 0.74 (iii) 0.3 (iv) 0.03  (v) 0.003  (vi) 0.0003 (vii) 2.32
Solution:
(i) 0.25 = ²⁵⁄₁₀₀ = 25%
(ii) 0.74 = ⁷⁴⁄₁₀₀ = 74%
(iii) 0.3 = ³⁄₁₀ = ³⁰⁄₁₀₀ = 30%
(iv) 0.03 = ³⁄₁₀₀ = 3%
(v) 0.003 = ^0.3⁄₁₀₀  = 0.3%
(vi) 0.0003 = ^0.03⁄₁₀₀  = 0.03%
(vii) 2.32  =  ²³²⁄₁₀₀  = 232%

■ From the above examples we can conclude that, to convert a decimal into a percentage, all that we need to do is write the given decimal as an appropriate fraction with 100 as denominator. Then take out the numerator and give a '%' sign.

Solved example 8.7
Convert each of the following percentages into fraction and decimal forms:
(i) 12%  (ii) 20% (iii) 84%   (iv) 7%
Solution:
■ Whenever we get a fraction with a denominator 100, we can straight away write it into percentage form, just by using the numerator. In the present problem, we are doing the reverse: We are given the percentage. So this given percentage will be the numerator and the denominator will be none other than 100. Using this fact, we can straight away write the fraction form. Also, once we derive the fraction form, we can straight away write the decimal form because the denominator is 100. [The fraction form with denominator 100 may be reduced to the simplest form if possible]

(i) 12%. In the fraction form, this 12 will be the numerator and 100 will be the denominator. So we can write 12% = ¹²⁄₁₀₀ = ³⁄₂₅  (dividing both numerator and denominator by 4)
When the fraction form with the denominator 10, or 100, or 1000 etc., is given, we can straight away write the decimal form. Thus ¹²⁄₁₀₀ = 0.12

So we can write: 12% = ³⁄₂₅  = 0.12
(ii) 20% = ²⁰⁄₁₀₀ = 0.20
(iii) 84% = ⁸⁴⁄₁₀₀ = ²¹⁄₂₅ = 0.84
(iv) 7% = ⁷⁄₁₀₀ = 0.07

---

In the next section we will see more details about percentage.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 6.8 - Money as Decimals

In the previous section we learned how the metric lengths are expressed as decimals. In this section we will see how different amounts of money are expressed as decimals.

We know that the basic unit of our currency is the Rupee. It is denoted by the symbol ₹. Some examples where we use this currency system are when we say:
• Cost of this book is ₹50   • Cost of this pen is ₹ 22   • He sold his bicycle for ₹1780  
In all of the above examples whole numbers are used. On many occasions, the amount of money may not be whole numbers. For example, the cost of the book may be greater than 50, and at the same time, less than 51. The cost of the pen may be greater than 22 but less than 23. In such cases we will require fractions of ₹1. We will need those fractions also when the amount is less than ₹1.

Just as we divided, 1 kilogram, 1 litre and 1 metre, we have to divide ₹1 also into smaller divisions. But the method of division here is different. ₹1 is divided up to a maximum of hundredths only. There is no further division into thousandths. The tenths have no particular name. Each of the hundredths have a particular name. It is 'paise'. We can use the following fig.6.35 to illustrate the divisions.

Fig.6.35

As there are no thousandths, there will only be two decimal places after the decimal point. We will now see an example:
₹ 2.49 =  2 + ⁴⁄₁₀ + ⁹⁄₁₀₀ = 2 + ⁴⁰⁄₁₀₀ + ⁹⁄₁₀₀ = 2 + ⁴⁹⁄₁₀₀
This indicates that, in addition to full Two Rupees, there are '49 parts out of 100 equal parts' of a rupee. These 49 parts is 49 paise, because each part is a paise. So we can say that 2.49 is Rupees 2 and 49 paise. We write it as Rs. 2 Ps. 49.

More examples:
• ₹ 23.82 = Rs. 23 Ps. 82
• ₹ 97.03 = Rs. 97 Ps. 3
• ₹ 97.3 = Rs. 97 Ps. 30

---

Solved example 6.24

Express as ₹ using decimals: (i) Rs. 0 Ps. 8  (ii) Rs. 0 Ps. 70  (iii) Rs. 25  Ps. 85   (iv) Rs. 0 Ps. 875
Solution:
(i)  • Rs. 0 Ps. 8 = 0 + ⁸⁄₁₀₀ (∵ 1 paise is one out of 100 equal parts, which gives 8 paise = 8 out of hundred equal parts)
• 0 + ⁸⁄₁₀₀ = 0 + 0.08 = ₹ 0.08 
(ii) • Rs. 0 Ps. 70 = 0 + ⁷⁰⁄₁₀₀ (∵ 1 paise is one out of 100 equal parts, which gives 70 paise = 70 out of hundred equal parts)
• 0 + ⁷⁰⁄₁₀₀ = 0 + 0.70 = ₹ 0.70
(iii) • Rs. 25 Ps. 85 = 25 + ⁸⁵⁄₁₀₀ (∵ 1 paise is one out of 100 equal parts, which gives 85 paise = 85 out of hundred equal parts)
• 25 + ⁸⁵⁄₁₀₀ = 25 + 0.85 = ₹ 25.85
(iv) • Rs. 0 Ps. 875 
• 875 paise = 800 + 75. But 800 paise = Eight 100 parts = Eight full rupees
• Thus 800 +75 = Rs. 8 +  Ps. 75 = ₹ 8.75
Solved example 6.25
Mr. A spent ₹ 253.6 for buying fruits and ₹ 125.48 for buying vegetables. What is the total amount that he spent?
Solution:
• Amount spent for fruits = ₹ 253.6
• Amount spent for vegetables = ₹ 125.48
∴ Total amount = 253.6 + 125.48 = ₹ 379.08
Solved example 6.26
A student brought ₹ 300 to school. ₹ 262.50 was spent for buying books. How much money is left?
Solution:
• Initial amount of money = ₹ 300
• Amount spent for buying books = ₹ 262.50
∴ Balance amount = 300 - 262.50 = ₹ 37.50 
Steps for the above two examples are shown below:

---

So we have seen how to express money as decimals and to do calculations on them. In the next section we will discuss about recurring decimals.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved