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Chapter 8: Working With Fractions

    Multiplication of Fractions

    Division of Fractions

    Some Problems Involving Fractions

    Fractional Relations

    Summary

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Chapter 8: Working With Fractions > Some Problems Involving Fractions

Some Problems Involving Fractions

Example 3: Leena made 5 cups of tea. She used 14 litre of milk for this. How much milk is there in each cup of tea?

Leena used 14 litres of milk in 5 cups of tea. So, in 1 cup of tea the volume of milk should be:
Writing this as multiplication, we have: 5 × (milk per cup) = 14.
We perform the division as follows as per Brahmagupta’s method: The reciprocal of 5 (the divisor) is .
Multiplying this reciprocal by the dividend (14), we get: 15 × 14 =
So, each cup of tea has 120 litre of milk.

Example 4: Some of the oldest examples of working with non-unit fractions occur in humanity’s oldest geometry texts, the Śhulbasūtra.

Here is an example from Baudhāyana’s Śhulbasūtra (c. 800 BCE).

Cover an area of 7 12 square units with square bricks each of whose sides is 15 units.

How many such square bricks are needed? Each square brick has an area of 15 × 15 = 125 square units.
The total area to be covered is 7 12 sq. units = sq. units
As (Number of bricks) × (Area of a brick) = Total Area, Number of bricks = ÷ . The reciprocal of the divisor is .
Multiplying the reciprocal by the dividend, we get: 25 × 152 = 25×152 =

Example 5: This problem was posed by Chaturveda Pṛithūdakasvāmī (c. 860 CE) in his commentary on Brahmagupta’s book Brāhmasphuṭasiddhānta. Four fountains fill a cistern. The first fountain can fill the cistern in a day. The second can fill it in half a day. The third can fill it in a quarter of a day. The fourth can fill the cistern in one fifth of a day. If they all flow together, in how much time will they fill the cistern?

Let us solve this problem step by step.

In a day, the number of times the first fountain will fill the cistern is 1 ÷ 1 =
In a day, the number of times the second fountain will fill the cistern is ÷ =
In a day, the number of times the third fountain will fill the cistern is ÷ =
In a day, the number of times the fourth fountain will fill the cistern is ÷ =
The number of times the four fountains together will fill the cistern in a day is 1 + 2 + 4 + 5 =
Thus, the total time needed by the four fountains to fill the cistern together is days.