पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३३८

142

GAŅITASĀRASAṄGRAHA.

The rule for arriving at the (weights of) many (component quantities of) gold (of known varņa in a mixture of known your and weight):-

185. (In relation to all the known component varņas) excepting one of them, optionally chosen weights may be adopted. Then what remains should be worked out as in relation to the previously given cases by means of the rule bearing upon the (determination of an) unknown weight of gold.

An example an illustration thereof.

186. The (given) varņas (of the component quantities of gold) are 5, 6, 7, 8, 11, and 13 (respectively); and the resulting zero is in fact 9; and if (the total) weight (of all the component quantities) of gold be 60, what may be the several measures (in weight of the various component quantities) of gold?

The rule for arriving at the unknown varņas of two (known quantities of gold when the resulting part of the mixture is known):-

187. Divide one (separately) by the two (given weights of) gold; multiply (separately each of the quotients thus obtained) by (the weight of) the (corresponding quantity of) gold and (also) by the (resulting) varņa; write down (both the products so obtained) in two (different) places; (each of these in each of the two sets,) if diminished and increased alternately by one as divided by (the

 

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185. The rule preferred to here is found in stanza 180 above. 187. The rule will become clear by the following working of the problem in stanza 188:-

${\tfrac {1}{16}}\times 16\times 10{\text{ and }}{\tfrac {1}{10}}\times 11$ are written down in two places

thus: $\quad {\begin{aligned}&11\quad &11\\&11\qquad &11\end{aligned}}$

Then ${\tfrac {1}{16}}$ and ${\tfrac {1}{10}}$ are added and subtracted alternately in each of the two sets thus:

${\begin{aligned}&11+{\tfrac {1}{16}}\\&11-{\tfrac {1}{10}}\end{aligned}}{\Bigg \}}{\text{ and }}{\begin{cases}11-{\tfrac {1}{16}}\\11+{\tfrac {1}{10}}\end{cases}}$ These give the two sets of answers.