GANITASĀRASANGRAHA. The rule for arriving at the (weights of) many (component quantities of) gold (of known varnas in a mixture of known varna and weight):- 185. (In relation to all the known component varnas) 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. 142 An example in illustration thereof. 186. The (given) varnas (of the component quantities of gold) are 5, 6, 7, 8, 11, and 13 (respectively); and the resulting van na 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 varnas of two (known quantities of gold when the resulting varna of the mixture is known) :- 187. Divide one (separately) by the two (given weights of) gold; multiply (separately cach of the quotients thus obtained) by (the weight of) the (sorresponding quantity of) gold and (also) by the (resulting) varna; write down (both the products so obtained) in two (different) places; (cach of these in each of the two sets,) if diminished and increased alternately by one as divided by (the 185. The rude referred to here is found in stanza 180 above. 187. The rule will become clear by the following working of the problem in stanza 188.- - 1 16 x 16 x 13 and thus. 1 16 Then sets thus. and 11 11 11 + 1 10 x 10 x 11 are written down in two places. 1 10 are added and subtracted alternately in each of the two 11 - 1 16 11 11 1 10. and (1-4) 11 16 11 + 10 . These give the two sets of answers.
