Trairāśika (Rule of Three)

Introduction
In the ancient Indian mathematical texts topics like ratio, proportion etc are dealt under the section rule of three. Ratio is used whenever comparison involving numbers.

For example: Cost of a bicycle is Rs. 10,000 and that of a motor bike is Rs. 1,00,000.

when we compare the cost of both the items. $$\frac{100000}{10000} = \frac{10}{1} = 10:1$$

Hence the cost the motorbike is ten times the cost of the bicycle. Ratio is the comparison by division. Ratio is denoted by ":". Ratio expresses the number of times one quantity with the other. The two quantities must be in the same unit.

The two values are said to be in direct proportion when an increase/decrease in one results in an increase/decrease in the other by the same factor.

Direct proportion is seen in the following instances.


 * 1) Cost of fuel increases as quantity of fuel increases
 * 2) Time taken increases with increase in pages to be typed.
 * 3) Cost of vegetable increases as weight of the vegetable increases.
 * 4) Number of units manufactured by a machine increases with the number of hours the machine works.

Trairāśika (Rule of Three)
The Hindu name for the Rule of Three is called "trairāśika" (three terms, hence the rule of three). The term trairāśika occurs in Bakshālī  manuscript, Āryabhaṭīya. Bhāskara I (c. 525) remarked on the origin of this name as "Here three quantities are needed (in the statement and calculation) so the method is called trairāśika (the rule of three terms)". A problem on the rule of three has this form : if p yields f, what will i yield ?. The three terms used are p, f, i. Hindus called the term p (pramāṇa - argument), f(phala - fruit) and i (icchā - requisition). Sometimes they are referred to simply as the first ,second and third respectively.

Āryabhaṭa II gave different names as māna, vinimaya and icchā respectively to the three terms.

Brahmagupta gives the rule as "In the rule of three pramāṇa (argument), phala(fruit) and icchā(requisition) are the (given) terms; the first and the last terms must be similar. The icchā multiplied by the phala and divided by the pramāṇa gives the fruit (of the demand) ".

Bhāskara I in his Āryabhaṭīya-bhaṣya talks about the Trairāśika

त्रयो राशयः समाहृताः त्रिराशिः । त्रिराशिः प्रयोजनमस्य गणितस्येति त्रैराशिकः । त्रैराशिके फलराशिः त्रैराशिकफलराशिः ।  (Āryabhaṭīya-bhaṣya by Bhāskara I on 11.26, p.116) 

"Trairāśi is the three quantities assembled . It is (called) Trairāśika because of this computation with these quantities. Trairāśika -phalarāśi is the desired result in the Rule of Three. "

Trairāśika involves three known quantities and one unknown quantity. The known quantities are pramāṇa (known measure),  pramāṇaphala (result related to known measure) and icchā (desired measure). The term used for the unknown quantity is icchāphala (result related to desired measure).

Example: A car covers 30 kms with 2 litres of petrol. To cover 150 kms, how many litres of petrol are required.?

Solution: For 30 kms, petrol needed = 2 litres

For 150 kms, petrol needed = 'x' litres

Here pramāṇa = 30 ; pramāṇaphala = 2 ; icchā = 150 ; icchāphala = 'x ' litres

pramāṇa -> pramāṇaphala  ( 30 -> 2)

icchā -> (icchā X pramāṇaphala) / pramāṇa  = icchāphala

150 -> ( 150 x 2) / 30 = 300/30 = 10

x= 10 ; 10 litres of petrol are required to cover 150 kms.

Solution on Trairāśika by another mathematician Śrīdhara states: "Of the three quantities, the pramāṇa ("argument") and icchā ("requisition") which are of the same denomination are the first and the last; the phala ("fruit") which is of a different denomination stands in the middle; the product of this and the last is to be divided by the first."

Example from -Līlāvatī vs.74,p.72 : If $$2\frac{1}{2}$$ palas (a weight measure) of saffron costs $$\frac{3}{7}$$ niṣkas (a unit of money), O expert businessman, tell me quickly what quantity of saffron can be bought for $$9$$ niṣkas.

Solution:

pramāṇa and pramāṇaphala - $$\frac{3}{7}$$ niṣkas and  $$2\frac{1}{2}$$ palas

icchā and icchāphala - $$9$$ niṣkas and x

As per Rule of Three - place the quantities indicated by niṣkas in first (pramāṇa ) and third (pramāṇaphala) column. place the remaining quantity in the middle column. Result = $$\frac{Middle\, quantity \ X \ Last\,quantity}{First\,quantity}$$

icchāphalam = $$\frac{\frac{5}{2}\ X\ 9}{\frac{3}{7}}$$= $$\frac{5 \ X \ 9\  X\ 7}{2\ X\ 3}= \frac{105}{2}$$ palas

Hence the quantity of saffron that can be bought for $$9$$ niṣkas is $$52\frac{1}{2}$$ palas.

Inverse Rule of Three
The Hindu name for the Inverse Rule of Three is Vyasta-trairāśika ("inverse rule of three terms").

In Trairāśika when the icchā increases, icchāphala also increases. In Vyasta-trairāśika when the icchā increases, icchāphala decreases.

Two values are said to vary inversely when the increase in one results in a decrease in the other. Example: If 5 men can do a work in 10 days, then 10 men can do the work in lesser number of days. When number of men increases, the number of days decreases. Hence the number of persons and the time taken are said to vary inversely with each other.

Bhāskara II defines Vyasta-trairāśika as "When the desired measure increases, the fruit (result related to desired measure) decreases and when the desired measure decreases, the fruit (result related to desired measure) increases  "

Examples related to inverse proportion are:

Solution on Vyasta-trairāśika by another mathematician Śrīdhara states: "When there is change in the unit of measurement, the middle quantity multiplied by the first quantity and divided by the last quantity gives the result"
 * if the speed of a vehicle is more, the time taken to cover the distance will be less.
 * if more customer support agents are utilized, the time taken to serve a customer will be less.

$$Result =\frac{Middle\, quantity \ X\ First\,quantity }{\ Last\,quantity}$$

In Trairāśika, pramāṇa and pramāṇaphala vary in such a way that $$\frac{pramanaphala}{pramana}$$  is a constant.

Hence in Rule of Three (Trairāśika) $$\frac{icchapalam}{iccha}=\frac {pramanaphala}{pramana}$$

$$\frac{Result \ related \ to \ desired \ measure}{Desired \ measure}=\frac {Result \ related \ to \ known \ measure }{Known \ measure }$$

In Vyasta-trairāśika, pramāṇa and pramāṇaphala vary in such a way that pramāṇaphala X pramāṇa is a constant. Hence in Inverse Rule of Three (Vyasta-trairāśika) icchāphala X icchā = pramāṇaphala X pramāṇa

i.e Result related to desired measure X Desired measure = Result related to known measure X Known measure

Example: With a measure of 7 āḍhakas, a certain quantity of grain measures 100 units. How many units will there be if the measure is 5 āḍhakas?(āḍhakas is a unit of measure of grains.)

Solution: 7 āḍhakas ⇒ 100 units

5 āḍhakas ⇒ x units $$Result =\frac{Middle\, quantity \ X\ First\,quantity }{\ Last\,quantity}$$

$$Number\ of\ Units=\frac{100 \ X \ 7 }{5} = 140$$

Hence the number of units for the measure of 5 āḍhakas is 140.

Pañca-rāśika (Rule of Five)
Trairāśika given by Āryabhaṭa is the basis for Pañca-rāśika (Rule of Five), Sapta-rāśika (Rule of Seven) , Nava-rāśika (Rule of Nine) and

Ekādaśa-rāśika (Rule of eleven).

Pañca-rāśika (Rule of Five) involves finding an unknown quantity with five known quantities.

Sapta-rāśika (Rule of Seven) involves finding an unknown quantity with seven known quantities.

Nava-rāśika (Rule of Nine) involves finding an unknown quantity with nine known quantities.

Ekādaśa-rāśika (Rule of eleven) involves finding an unknown quantity with eleven known quantities.

These problems involves two sets of data. The first set is pramāṇa-pakṣa (known measure side)  where all the quantities are given. The second set is icchā-pakṣa (desired measure side) where one quantity is to be found out.

Trairāśika comes under the Rule of Odd Terms.

Śrīdhara has given the Rule of Odd Terms as "After transposing the fruit from one side to the other, and then having transposed the denominators (in like manner) and having multiplied the numbers (so obtained on either side), divide the side with larger number of quantities (numerators) by the other."

Example: if a rectangular piece of stone with length, breadth, and thickness equal to 9, 5 and 1 cubits (respectively) costs 8, what will two other rectangular pieces of stone of dimensions 10, 7 and 2 cubits cost ?

Solution: This problem belongs to Nava-rāśika (Rule of Nine) involving nine known quantities.

pramāṇa-pakṣa (known measure side) : 1,9,5,1,8

icchā-pakṣa (desired measure side): 2,10,7,2,x Interchange the row containing fruit (cost) as shown below. Divide the numbers in the side with larger number of known quantities by the numbers of the other side. Here 2nd column has large number of known quantities.

$$x = \frac{2 \ X \ 10 \ X  \ 7 \  X  \ 2 \ X  \ 8} {1 \  X  \  9 \  X  \ 5 X \ 1} = \frac {448} {9} =49\frac{7}{9} $$

The Cost of two rectangular pieces of stone of dimensions 10, 7, and 2 cubits is $$49\frac{7}{9} $$

Simple Interest
In ancient Indian Mathematical works miśraka-vyavahāra dealt with the problems related to find interest, principal or time.

Interest - fee paid for a loan received.

Principal - the amount borrowed

Interest will be expressed as a percentage of the principal for a given time duration. In ancient Indian Mathematical works simple interest, not the compound interest was dealt.

Here are the samskrit terms used : Example: If a principal of 1000 rupees gets an interest of R rupees for one month, then what will be the interest received by the principal of P rupees for a period of N months.

This belongs to pañca-rāśika ↓ Formula for Simple Interest is

$$x = \frac{PNR}{100 \ X \ 1}=\frac{PNR}{100} $$

Śrīdhara has stated the formula for Simple interest as "Multiply the argument (Po) by its time (No) and the other time (N) by the fruit (R) ; divide each of those (products) by their sum and multiply by the amount (A) (i.e. capital plus interest). The results give the capital and the interest (respectively)." Po PO P0

Principal $$P = \frac{A \ X \ Po \ X \ No}{(Po \ X \ No) \ + \ (R \ X \ N)} $$

Interest $$I = \frac{A \ X \ R \ X \ N}{(Po \ X \ No) \ + \ (R \ X \ N)} $$

Interest = Amount - Principal

Here If Po = 100 and No= 1 month

$$P = \frac{A \ X \ 100 \ X \ 1}{(100 \ X \ 1) \ + \ (R \ X \ N)} = \frac{100 \ X \ A}{100 \ + \ (R \ X \ N)} $$

Example: If 1½ units is the interest on 100½ units for one third of a month, what will be the interest on 60¼ units for 7½ months?

Solution :

This belongs to pañca-rāśika

pramāṇa-pakṣa (known measure side) : 100½ units, ⅓ months , 1½ interest. converting this mixed fraction to improper fraction.

$$\frac{201}{2}$$ units, $$\frac{1}{3}$$ months , $$\frac{3}{2}$$ interest

icchā-pakṣa (desired measure side): 60¼ units, 7½ months , x interest

$$\frac{241}{4}$$ units, $$\frac{15}{2}$$ months , $$\frac{x}{1}$$ interest ↓ Interchange the row containing Interest (fruit) as shown below. ↓ Interchange the denominators as shown below. this is required for the terms which are fractions. Divide the 2nd column (large number of known quantities) by 1st column (numbers of the other side).

$$x = \frac{ 241 \ X \ 2 \ X \ 15 \ X \ 3 \ X \ 3 \ X \ 1} {201 \ X \ 4 \ X \ 1 \ X \ 2 \ X \ 2 } = \frac{10845}{536}= 20 \frac{125}{536}   $$

Hence Interest on 60¼  units, 7½ months = $$20 \frac{125}{536}   $$

Amount becoming 'n' times the principal :
Śrīdhara has stated the formula to find when the principal will double or triple or quadruple after 'N' months at R% per month.

कालप्रमाणघातः फलभक्तो व्येकगुणहतः कालः । (Pāṭīgaṇita III R.52, p.60)

"The product of the time and the argument divided by the fruit and (then) multiplied by the multiple minus one, gives the required time."

Here time is standard time, argument is standard principal and fruit is rate of interest.

The formula will be

$$Time \ N = \frac{Standard \ principal \ X \ Standard \ time \ X  \ (n -1)}{R} $$

Here Standard Principal = 100 ; Standard time = 1 month ; Rate of Interest = R

Example: If 6 drammas is the interest in 200 (drammas) per month, when will the sum be three times?

Solution:

Given: P = 200 drammas, N = 1 month, I = 6 drammas

$$I = \frac{P \ X \ N \ X \ R}{100}$$

$$6 = \frac{200 \ X \ 1 \ X \ R}{100}$$

R= 3%

To calculate the period in which the sum becomes three times the principal

$$Time \ N = \frac{Standard \ principal \ X \ Standard \ time \ X  \ (n -1)}{R} $$

Here Standard principal = 100 ; Standard tine = 1 month ; n = 3 times

$$Time \ N = \frac{(100 \ X \ 1) \ X \ (3 - 1)}{3}   = \frac{200}{3} = 66 \frac{2}{3} $$  months

Hence the sum becomes three time after $$66\frac{2}{3} $$ months i.e 5 years $$6\frac{2}{3}  $$months