पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/२८६

90

GAŅIITASĀRASAŃGRAHA.

(at the rate of) $2{\tfrac {1}{2}}$ aṅgulas in a day and half; the water (thereof) flows out through a pump (at the rate of) $2{\tfrac {1}{2}}$ aṅgulas (of the well in depth) in $1{\tfrac {1}{2}}$ days; $1{\tfrac {1}{5}}$ aṅgulas of water (in depth) are lost in a day by evaporation owing to the (heating) rays of the sun; a tortoise below pulls down $5{\tfrac {1}{4}}$ aṅgula of the stalk of the lotus plant in $3{\tfrac {1}{2}}$ days. By what time will the lotus be on the same level with the water (in the well) ?

31. A powerful unvanquished excellent black snake, which is 32 hastas in length, enters into a hole (at the rate of) $7{\tfrac {1}{2}}$ aṅgulas in ${\tfrac {5}{14}}$ of a day; and in the course of ${\tfrac {1}{4}}$ of a day its tail grows by $2{\tfrac {3}{4}}$ of an aṅgu la. O ornament of arithmeticians, tell me by what time this same (serpent) enters fully into the hole. ^[*]

Thus end the (problems bearing on associated) forward and backward movements.

The rule of operation relating to double, treble and quadruple rule-of-thrē.

^[32]. Transpose the Phala from its own place to the other place (wherein a similar concrete quantity would occur); (then, for the purpose of arriving at the required result), the row consisting of the larger number (of different quantities) should be, (after they are all multiplied together, divided by the row consisting of the

 

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32.^  The transference of the Phala and the other operations herein mentioned will be clear from the following worked out example.

The data in the problem in stanza No.36 are to be first represented thus:-

9Mānīs	1Vāha + 1 Kumbha.
3 Yōjanas.	10 Yōjanas.
60 Paṇas

When the Phala here, viz., 60 pounds, is transferred to the other row we have--

9 Mānīs.	1 Vāha + 1 Kumbha = $1{\tfrac {1}{2}}$ Vāha.
3 Yōjanas.	10 Yōjanas
	60 Paṇas.

Now the right hand row, consisting of a larger number of different quantities, should be, after they are all multiplied together, divided by the smaller left hand row similarly dealt with.

Then we have

${\tfrac {1{\tfrac {1}{4}}\times 10\times 60}{9\times 3}}$

The result here gives the number of paṇas to be found out.