CHAPTER VI-MIXED PROBLEMS. 145 gold obtained in exchange as multiplied by the second of the specified vurnus 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) varnas (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 varnas produces, on being exchanged, 1,008 (in weight) of two kinds of gold characterised (respectively) by 12 and 10 varnas. Now, what is the weight (of cach of these two varicties) of gold? A The rule for finding out the (various weights of) gold obtained as the result of many (specified) kinds of exchange - 1992. If the (given) weight of gold (to be exchanged) as multiplied by the varna (thercof) is divided by (the quantity of) the desired gold (obtained in exchange), there arises the uniform average varna 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 un illustration thereof. 2003-201. In the case of a man exchanging 300 in weight of gold characterised by 14 varnas, 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 varnas. What is the weight of gold separately corresponding to each of these (different) varnas? The rule for arriving at (the various weights of) gold obtained in exchange which are characterised by known varnas and are (definite) multiples in proportion :-- 202-203 The sum of the (given) proportional multiple num- bers is to be divided by the sum of the products (obtained) by 199. The operation which is stated here as having been mentioned before 18 what is given in stanza 185 above.
