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143
CHAPTER VI--MIXED PROBLEMS.

known weight of) the corresponding (variety of gold, gives rise as a matter of course, to the required varņas.

An example in illustration thereof.

188. If, the (component) varņas not being known, the resulting varņa obtained by means of two (different kinds of) gold weighing 16 and 10 (respectively) happens to be 11, what would be the (respective) varņas of those two (different kinds of gold ?

Again, the rule for arriving at the unknown varņas of two (known quantities of gold, when tho resulting varņa of the mixture is known):-

189. Choose an optional varņa 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 varņas 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 varņa is made out to be 10. Think out and say (what) the varņas 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 varņas. Give out the varņas (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 varņa of every stick is to be separately divided by the (given) maximum varņa, 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