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एतत् पृष्ठम् परिष्कृतम् अस्ति
141
CHAPTER VI--MIXED PROBLEMS.

An example in distribution thereof.

181. Three pieces of gold, of 3 each in weight, and of 2, 3, and 4 varņas (respectivel ), are added to (an unknown weight of) gold of 18 varņas. The, resulting varņa comes to be 10. Tell me, O friend, the measure (of the unknown weight) of gold.

Tho rule for arriving at (the weights of gold (corresponding to two given varņas) from (the known weight and varņa of) the mixture of two (given specimens of) gold of (given) varņas:-

182. Obtain the differences between the resulting varņa (of the mixture on the one hand) and the known higher and lower varņas (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 prakșēpaka (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 (corresponding to two given varņas) from (the known weight and varņa of) the mixture of two (given specimons of gold of (given) varņas:-

183. Write down in inverse order the difference between the resulting varņa and the higher (of the two given varņas of the two component quantities of gold), and also the difference between the resulting varņa and the lower (of the two given varņas). The result arrived at by means of the operation of proportionate distribution (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 varņas, on being combined with gold of 16 varņas produces as result 100 in weight of gold of 12 varņas, give out separately (the measures in weight of) the two different varieties of gold.