पृष्ठम्:Ganita Sara Sangraha - Sanskrit.djvu/३३९

CHAPTER VI-MIXED PROBLEMS. 141 An example in illustration thereof. 181. Three pieces of gold, of 3 each in weight, aud of 2, 3, and 4 varnus (respectively), are added to (an unknown weight of) gold of 13 vurnas. The resulting varna comes to be 10. Tell me, O friend, the measure (of the unknown weight) of gold. The rule for arriving at (the weights of) gold (corresponding to two given varnas) from (the known weight and varna of) the mixture of two (given specimens of) gold of (given) varnas :- 182. Obtain the differences between the resulting varna (of the mixture on the one hand) and the known higher and lower varnas (of the unknown component quantities of gold on the other hand); divide one by these differences (in order); then carry out as before the operation of prah sepuku (or proportionate distribution with the aid of these various quotients). In this manner it is possible to arrive even at the value of many component quantities of gold also Again, the rule for arriving at (the weights of) gold (corre- sponding to two given varnas) from (the known weight and varna of) the mixture of two (given specimens of) gold of (given) varnas: 183. Write down in inverse order the difference between the resulting varna and the higher (of the two given varnas of the two component quantities of gold), and also the difference between the resulting varna and the lower (of the two given rarnas). The result arrived at by means of the operation of proportionate distri- bution (carried out with the aid of these inversely arranged differences),--that (result) gives the required (weights of the component quantities of) gold. An example in illustration thereof. 184. If gold of 10 varnas, on being combined with gold of 16 varnas, produces as result 100 in weight of gold of 12 varnas, give out separately (the measures in weight of) the two different varieties of gold.