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Chapter 8: Working With Fractions > Fractional Relations

Fractional Relations

Here is a square with some lines drawn inside.

What fraction of area of the whole square does the shaded region occupy?

There are different ways to solve this problem. Here is one of them: Let the area of the whole square be 1 square unit. We can see that the top right square (in figure below), occupies 1 4 of the area of the whole square.

Area of red square = square units.

Let us look at this red square. The area of the triangle inside it (coloured yellow) is half the area of the red square. So,

the area of the yellow triangle = × = square units.

What fraction of this yellow triangle is shaded?

The shaded region occupies of the area of the yellow triangle. Are you able to see why?

The area of shaded part = × = square units.

Thus, the shaded region occupies of the area of the whole square.

In each of the figures given below, find the fraction of the big square that the shaded region occupies.

We will solve more interesting problems of this kind in a later chapter.

A Dramma-tic Donation

The following problem is translated from Bhāskarāchārya’s
(Bhāskara II’s) book, Līlāvatī, written in 1150 CE.

“O wise one! A miser gave to a beggar 15 of 116 of 14 of 12 of 23 of 34 of a dramma. If you know the mathematics of fractions well, tell me O child, how many cowrie shells were given by the miser to the beggar.” a dramma. If you know the mathematics of fractions well, tell me O child, how many cowrie shells were given by the miser to the beggar.”

Dramma refers to a silver coin used in those times. The tale says that 1 dramma was equivalent to 1280 cowrie shells. Let’s see what fraction of a dramma the person gave:

(12 × 23 × 34 × 15 × 116 × 14) th part of a dramma.

Evaluating it gives .

Upon simplifying to its lowest form, we get 67680 = .

So, one cowrie shell was given to the beggar. You can see in the answer Bhāskarāchārya’s humour! The miser had given the beggar only one coin of the least value (cowrie).

Around the 12th century, several types of coins were in use in different kingdoms of the Indian subcontinent. Most commonly used were gold coins (called dinars/gadyanas and hunas), silver coins (called drammas/tankas), copper coins (called kasus/panas and mashakas), and cowrie shells. The exact conversion rates between these coins varied depending on the region, time period, economic conditions, weights of coins and their purity.

Gold coins had high-value and were used in large transactions and to store wealth. Silver coins were more commonly used in everyday transactions. Copper coins had low-value and were used in smaller transactions. Cowrie shells were the lowest denomination and were used in very small transactions and as change.

If we assume 1 gold dinar = 12 silver drammas, 1 silver dramma = 4 copper panas, 1 copper pana = 6 mashakas, and 1 pana = 30 cowrie shells,

1 copper pana = gold dinar (112 × 14)

1 cowrie shell = copper panas

1 cowrie shell = gold dinar

A Pinch of History

As you have seen, fractions are an important type of number, playing a critical role in a variety of everyday problems that involve sharing and dividing quantities equally. The general notion of non-unit fractions as we use them today — equipped with the arithmetic operations of addition, subtraction, multiplication, and division — developed largely in India. The ancient Indian geometry texts called the Śhulbasūtra — which go back as far as 800 BCE, and were concerned with the construction of fire altars for rituals — used general non-unit fractions extensively, including performing division of such fractions as we saw in Example 3.

Fractions even became commonplace in the popular culture of India as far back as 150 BCE, as evidenced by an offhand reference to the reduction of fractions to lowest terms in the philosophical work of the revered Jain scholar Umasvati.

General rules for performing arithmetic operations on fractions — in essentially the modern form in which we carry them out today — were first codified by Brahmagupta in his Brāhmasphuṭasiddhānta in 628 CE. We have already seen his methods for adding and subtracting general fractions. For multiplying general fractions, Brahmagupta wrote:

“Multiplication of two or more fractions is obtained by taking the product of the numerators divided by the product of the denominators.” (Brāhmasphuṭasiddhānta, Verse 12.1.3)

That is, ab × cd = a×cb×d

For division of general fractions, Brahmagupta wrote: “The division of fractions is performed by interchanging the numerator and denominator of the divisor; the numerator of the dividend is then multiplied by the (new) numerator, and the denominator by the (new) denominator.”

Bhāskara II in his book Līlāvatī in 1150 CE clarifies Brahmagupta‘s statement further in terms of the notion of reciprocal: “Division of one fraction by another is equivalent to multiplication of the first fraction by the reciprocal of the second.” (Līlāvatī, Verse 2.3.40) Both of these verses are equivalent to the formula:

ab ÷ cd = ab × dc = a×db×c

Bhāskara I, in his 629 CE commentary Āryabhaṭīyabhāṣhya on Aryabhata’s 499 CE work, described the geometric interpretation of multiplication of fractions (that we saw earlier) in terms of the division of a square into rectangles via equal divisions along the length and breadth.

Many other Indian mathematicians, such as Śhrīdharāchārya (c. 750 CE), Mahāvīrāchārya (c. 850 CE), Caturveda Pṛithūdakasvāmī (c. 860 CE), and Bhāskara II (c. 1150 CE) developed the usage of arithmetic of fractions significantly further.

The Indian theory of fractions and arithmetic operations on them was transmitted to, and its usage developed further, by Arab and African mathematicians such as al-Hassâr (c. 1192 CE) of Morocco. The theory was then transmitted to Europe via the Arabs over the next few centuries, and came into general use in Europe in only around the 17th century, after which it spread worldwide. The theory is indeed indispensable today in modern mathematics.

Figure It Out

1. Evaluate the following:

2. For each of the questions below, choose the expression that describes the solution. Then simplify it.

(a) Maria bought 8 m of lace to decorate the bags she made for school. She used 14 m for each bag and finished the lace. How many bags did she decorate? (i) 8 × 14 (ii) 18 × 14(iii) 8 ÷ 14 (iv) 14 ÷ 8
(iii) 8 ÷ 14. Why? We need to find how many 14 m pieces fit into 8 m. Simplification: 8 ÷ 14 = 8 × = bags
(b) 12 meter of ribbon is used to make 8 badges. What is the length of the ribbon used for each badge? (i) 8 × 12 (ii) 12 ÷ 18 (iii) 8 ÷ 12 (iv) 12 ÷ 8
(iv) 12 ÷ 8. Why? We're dividing 12 m of ribbon equally among 8 badges. Simplification: 12 ÷ 8 = 12 × = meter per badge
(c) A baker needs 16 kg of flour to make one loaf of bread. He has 5 kg of flour. How many loaves of bread can he make? (i) 5 × 16 (ii) 16 ÷ 5 (iii) 5 ÷ 16(iv) 5 × 6
(iii) 5 ÷ 16. Why? We need to find how many 16 kg portions fit into 5 kg. Simplification: 5 ÷ 16 = 5 × = loaves

3. If 14 kg of flour is used to make 12 rotis, how much flour is used to make 6 rotis?

6 rotis is half of 12 rotis, so we need half the flour.
Thus, 14 ÷ 2 = 14 × = kg
Alternate method (finding flour per roti first): Flour per roti = 14 ÷ 12 = 14 × = kg
Flour for 6 rotis = 148 × = = kg

4. Pāṭīgaṇita, a book written by Sridharacharya in the 9th century CE, mentions this problem: “Friend, after thinking, what sum will be obtained by adding together 1 ÷ 16 , 1 ÷ 110, 1 ÷ 113, 1 ÷ 19, and 1 ÷ 12”. What should the friend say?

Find the sum of 1 ÷ 16, 1 ÷ 110, 1 ÷ 113, 1 ÷ 19, and 1 ÷ 12
Dividing by a fraction means multiplying by its .
So 1 ÷ 1n = 1 × =
Calculations: 1 ÷ 16 = 1 × = ; 1 ÷ 110 = 1 × = ; 1 ÷ 113 = 1 × = ; 1 ÷ 19 = 1 × = ; 1 ÷ 12 = 1 × =
Sum: 6 + 10 + 13 + 9 + 2 =
The friend should say: "The sum is 40!"

5. Mira is reading a novel that has 400 pages. She read 15 of the pages yesterday and 310 of the pages today. How many more pages does she need to read to finish the novel?

Total pages:
Pages read yesterday: 15 × = ÷ = pages
Pages read today: 310 × = ( × ) ÷ = ÷ = pages
Total pages read so far: + = pages
Pages remaining: = pages
Mira needs to read 200 more pages to finish the novel.

6. A car runs 16 km using 1 litre of petrol. How far will it go using 2 34 litres of petrol?

Given: Car runs 16 km per litre of petrol
Petrol available: 2 34 litres = litres
Calculation: Distance = Mileage × Petrol = × = ÷ = km
The car will go 44 km using 2 34 litres of petrol.

7. Amritpal decides on a destination for his vacation. If he takes a train, it will take him 5 16 hours to get there. If he takes a plane, it will take him 12 hour. How many hours does the plane save?

Time by train: 5 16 hours while Time by plane: 12 hour
Time saved:
Time saved =
Finding common denominator: = 312 = == hours
The plane saves 4 13 hours (or 4 hours minutes).

8. Mariam’s grandmother baked a cake. Mariam and her cousins finished 45 of the cake. The remaining cake was shared equally by Mariam’s three friends. How much of the cake did each friend get?

Cake eaten by Mariam and cousins:
Cake remaining: =
Shared equally among 3 friends: ÷ = × =
Each friend got 115 of the cake.

9. Choose the option(s) describing the product of (565465 × 707676):

(a) > 565465

(b) < 565465

(c) > 707676

(d) < 707676

(e) > 1

(f) < 1

First, let's analyze each fraction: 565465: Since 565 > 465, this fraction is greater than .
707676: Since 707 > 676, this fraction is greater than .
When you multiply a number by something greater than 1, the product is than the original number.
(a) 565465 × 707676> 565465
Since 707676 1, the product is greater than 565465.
(b) 565465 × 707676< 565465
(c) 565465 × 707676 > 707676
Since 565465 1, the product is greater than 707676.
(d) 565465 × 707676 < 707676
(e) 565465 × 707676 > 1
Product of two numbers each greater than is always greater than 1
(f) 565465 × 707676 < 1
Correct answers: (a), (c), and (e)

10. What fraction of the whole square is shaded?

Divide the whole square. The large square is divided into 4 equal parts (quadrants). Each quadrant = of the whole square.
The shaded region is entirely within the bottom-right quadrant. This quadrant is further divided into smaller parts: small squares (2 × 2 grid)
Each small square = of the quadrant = × = of the whole.
Looking at the bottom-right quadrant: Area 1: The bottom-left small square of this quadrant = ; Area 2: A triangle in the top-left small square = × =
Total shaded area = + = + = =
18 of the whole square is shaded.

11. A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown) and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source?

Starting Point (Bottom)- At the 1st junction: Ants split into equal parts → each path has
Left Branch (toward Mango Tree) where of ants go left.
At the 2nd junction: This splits into equal parts → each path has
At the 3rd junction: The left splits into equal parts → each path has
At the 4th junction: The left splits into equal parts → each path has
Ants reaching Mango Tree: paths arrive i.e. + + + + =
Right Branch (toward Sugarcane Field)- At the 3rd junction: each path has
At the 4th junction: each path has
Ants reaching Sugarcane Field: paths arrive i.e. + =
Thus, ants reaching Mango Tree: 532 and ants reaching Sugarcane Field: 332
12. What is 1 − 12?
(1 − 12) × (1 − 13)?
(1 − 12) × (1 − 13) × (1 − 14) × (1 − 15)?
112×113×114×115×116×117×118×119×1110?
Make a general statement and explain.
Simplify each bracket: 112 = ; 113 = ; 114 = ; 115 = ; 116 = ; 117 = ; 118 = ; 119 = ; 1110 =
Question 1: 112 =
Question 2: 112×113 = × = =
Question 3: 112×113×114×115 = × × × = ÷ =
Question 4: 112×113×114×115×116×117×118×119×1110 = × × × × × × × × =
General Statement: =
When we simplify each bracket: Most terms cancel out! The 2 in the numerator cancels with 2 in the denominator. The 3 in the numerator cancels with 3 in the denominator. And so on...
After all cancellations, only remains in the numerator and remains in the denominator. Result =