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CHAPTER VI-XIXED PROBLEMS.
147
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 bolonging to a merchant is taken; and 4, and parts thereof are in order exchanged for (different kinds of) gold characterised (respectively) by 12, 10 and 9 varnas. (The weights of these exchanged varieties of gold are) added to what remains (unexchanged) of tho original gold. Then 1,000 is observed to be in excess on removing from the account the weight of the original gold. What thon is (the weight of this) original gold?
The rule for arriving at the desired earna 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) parts is to he noted down in reverse order; aud (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 213-815:
Dividing 1 by and, we get respectively 2,; altering their position and multiplying them by any optionally chosen number, say 1, we get 2, 2. These two numbers represent 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 vary of the gold owned by the second merchant. These varnas, 9 and 13, give, in the exchange proposed by the second merchant, the average varna of, while the average varpa as given in the sum has to be 12 or .
If 8 is chosen instead of
Therefore the varnas 9 and 13 have to be altered. 9, 13 has to he increased to 18 in the first exchange. Using these two varnas,
8 and 16, in the second exchange, we obtain as the average varna, instead of 14.
