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CHAPTER VI--MIXED PROBLEMS.

gold obtained in exchange as multiplied by the second of the specified varņas out of the exchanged gold-these two differences) have to be written down. If then, they are altered in position and divided by the difference between the (two specified) varņas (of the two varieties) of the exchanged gold, the result happens to be the (two required) quantities (of the two kinds) of gold (obtained in exchange).

An example in illustration, thereof.

198. Seven hundred in weight of gold characterised by 16 varņas produces, on being exchanged, 1,008 (in weight) of two kinds of gold characterised (respectively) by 12 and 10 varņas. Now, what is the weight (of each of these two varieties) of gold?

The rule for finding out the (various weights of) gold obtained as the result of many (specified) kinds of exchange :-

199. If the (given) weight of gold (to be exchanged) as multiplied by the varņa (thereof) is divided by (the quantity of) the desired gold (obtained in exchange), there arises the uniform average varņa. On carrying out (further) operations as mentioned before, the result arrived at gives the required weights of the various kinds of gold obtained in exchange.

An example in illustration thereof.

200-201. In the case of a man exchanging 800 in weight of gold characterised by 14 varņas; the gold (obtained in exchange) is seen to be altogether 500 in weight, (the various parts whereof are respectively) characterised by 12, 10, 8 and 7 varņas. What is the weight of gold separately corresponding to each of these (different) varņas?

The rule for arriving at (the various weights of) gold obtained in exchange which are characterised by known varņas and are (definite) multiples in proportion:-

202-203. The sum of the (given) proportional multiple numbers is to be divided by the sum of the products (obtained) by

 

 

199. The operation which is stated here as having been mentioned before is what it given in stanza 185 above.