Ganitasara Sangraha

SR No.: 020800
Original Sutra Author: N/A
Author: Mahaviracharya, M Rangacharya
Publisher: Government of Madras
Publication Year: 1912
Total Pages: 523
Language: Hindi, Sanskrit
Classification: Book_Devnagari
File Size: 22 MB

Page 350

CHAPTER VI-MIXED PROBLEMS.
153
The rule for arriving at the value of) the wages corresponding to the various stages (over which varying numbers of persons carry a given weight) :
230. The distance (travelled over by the various numbers of men) are (respectively) to be divided by the numbers of the men that are (doing the work of carrying) there. The quotients (80 obtained) have to be combined so that the first of them is taken at first separately and then) has (1, 2, and 3, etc., of) the following (quotients) added to it. (These quantities so resulting are to be respectively) multiplied by the numbers of the men that turn away (from the journey at the various stages. Then) by adopting (in relation to these resulting products the process of proportionate distribution (praksēpaka), the wages (due to the men leaving at the different stages) may be found out.
An example in illustration thereof.
231-232. Twenty men have to carry a palanquin over (a distance of) 2 yojanas, and 720 dināras form their wages. Two men stop away after going over two krošas; after going over two (more) krāsas, three others (stop away); after going over half of the remaining distance, five men stop away. What wages do they (the various bearers) obtain ?
The rule for arriving at (the value of the money contents of) a purse which (when added to what is on hand with each of certain persons) becomes a specified multiple (of the sum of what is on hand with the others): -
233-235. The quantities obtained by adding one to each of the specified) multiple numbers in the problem, and then)
233-235. In the problem given in 236--237, let a, y, z represent the money on hand with the three merchants, and u the money in the purse. Then + s = a (y + 2)
where a, b, c represent the multiples
where u+y=b (2 + x)}
1 given in the problem. u +2 = (x + y)) Now u + 2+ y +2= (a + 1) (y + x)
= (6+1) (+ 2) = (c+1) (x + y).