CHAPTER VI-MIXED PROBLEMS. the mixed gold gives rise to the (resulting) varna. (The original varna of any component part thereof),, when divided by the latter resulting varna (of the mixed up whole), and multiplied by the (given) quantity of gold (in that component part), gives rise to (that) corresponding quantity of (the mixed) gold (which is equal in value to that same component part thereof). 139 An example in illustration thereof. 170 to 171. There are 1 part (of gold) of 1 varna, 1 part of 2 varnas, 1 part of 3 varnas, 2 parts of 4 varnas, parts of 5 varnas, 7 parts of 11 varnas, and 8 parts of 15 varnus. Throwing these into the fire, make them all into one (mass), and then (say) what the varna of the mixed gold is. This mixed gold is distributed among the owners of the foregoing parts. What does each of them get ? · The rule for arriving at the required weight of gold (ot any desired var na equivalent in value to given quantities of gold) of given varnas :- 172 The given quantities of gold are all (separately) multi- plied by their respective varnas, and the products are added. The resulting sum is divided by the total weight of the mixed gold; the quotient is to be understood as the resulting average varna. This (above-mentioned sum of the products) is separately divided by the desired varnas (to arrive at the required equivalent weight of this gold). Examples in illustration thereof. 173. Twenty panas (in weight of gold) of 16 varnas have been exchanged for (gold of) 10 varnas in quality; you give out how many puranas (in weight) they become now. 174. One hundred and eight (in weight of) gold of 11 varnas is exchanged for (gold of) 14 varnas. What is the (equivalent quantity of this new) gold? The rule for finding out the unknown varna :- 175. From the product obtained by multiplying the total quantity of gold by the resulting varna of the mixture, the sum of
