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.
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
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.
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.
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| Fig.6.35 |
₹ 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:
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.
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