CHAPTER VI-MIXED PROBLEMS. 143 known weight of) the corresponding (variety of) gold, gives rise as a matter of course, to the required varnas. An example in illustration thereof. 188. If, the (component) varnas not being known, the resulting varna obtained by means of two (different kinds of) gold weighing 16 and 10 (respectively) happens to be 11, what would be the (respective) varnas of those two (different kinds of) gold ? Again, the rule for arriving at the unknown tarnas of two (known quantities of gold, when the resulting varna of the mixture is known) :- 189. Choose an optional rarna in relation to one (of the two given quantities of gold); what remains (to be found out) may then be arrived at as before. In relation to (the known quantities of all) the numerous varieties of gold excepting one, the varnas are optional; then (proceed) as before. An example in illustration thereof. 190. On fusing together (two different kinds of) gold which are 12 and 14 (respectively in weight), the resulting varna is made out to be 10. Think out and say (what) the varmas of those two (kinds of gold are). An example to illustrate the latter half of the rule. 191. On fusing together 7, 9, 3, and 10 (in weight respectively of four different kinds) of gold, the resulting mixture turns out to be (gold of) 12 varnas. Give out the rurnas (of the various component kinds of gold) separately. The rule regarding how to arrive at (an estimate of the value of) the test sticks (of gold) :- 192. The varna of every stick is to be separately divided by the (given) maximum varna, and (the quotients so obtained) are (all) to be added together. The resulting sum gives (the measure of) the required quantity of (pure) gold. From the summed up
