CHAPTER VI-MIXED PROBLEMS. 1.47 optionally chosen varna (of the originally given gold; and then all these products are to be added). From this sum, the sum of the (various) fractional (exchanged) parts (of the original gold) is to be subtracted. (If now) the observed excess (in the weight of gold due to the exchange) is divided by this resulting remainder. what comes out here happens to be the original wealth of gold. An example in illustration thereof. ● 207-208. A certain small ball of gold of 16 varnas belonging to a merchant is taken; and, and parts thereof are in order exchanged for (different kinds of) gold characterised (respectively) by 12, 10 and 9 vurnas. (The weights of these exchanged varie- ties of gold are) added to what remains (unexchanged) of the original gold. Then 1,000 is observed to be in excess on removing from the account the weight of the original gold What then is (the weight of this) original gold? The rule for arriving at the desired varna with the help of the (mutual) gift of a desired fractional part of the gold (owned by the other), and also for arriving at the (weights of) gold (respectively) corresponding to those optionally gifted parts:- 209 to 212. One divided by (the numerical measure of each of two specifically gifted) iparts is to be noted down in reverse order; and (if each of the quotients so obtained is) multiplied by an 209--212. The rule will be clear from the following working of the problem. in 213-215- Dividing 1 by and, we get respectively 2, 3, altering their position and mnltiplying them by any optionally chosen number, say 1, we get 3, 2. These two numbers iepresent the quantities of gold owned respectively by the two merchants. Choosing 9 as the varna of the gold owned by the first merchant, we can easily arrive, from the exchange proposed by him, at 13 as the varna of the gold owned by the second merchant. These varnas, 9 and 13, give, in the exchange proposed by the second merchant, the average vana of , while the averago varna as given in the sum has to be 12 or 8. Therefore the varnas 9 and 13 have to be altered. If 8 is chosen instead of 9, 13 has to be increased to 16 in the first exchange. Ising these two varnas, 8 and 16, in the second exchange, we obtain as the average varna, instead of
