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CHAPTER VI-MIXBD PROBLIMS.
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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 samo component part thereof).
an example in illustration thereof.
170 to 171). There are 1 part (of gold) of 1 varna, 1 part of 2 varras, 1 part of 3 varnis, 2 parts of 4 varnar, parts of 5 varnas, 7 parts of 14 varnas, and 8 parts of 15 varus. Throwing thoso into the fire, make them all into one (mans), and then (say what the varna of the mixoil gold is. This mixed gold is distribuwd among the owners of the foregoing parts. What does caoh of thom get?
The rule for arriving nt the reynired weight of gold (of any desired varna equivalent in value to given quantities of gold) of given wirnar:
172. The given quantities of gold are all (soparatoly) multiplied by their respective varnııs, and the products are addod. The resulting sum is divided by tho total weight of the mixed gold ; the quotiont is to be understood as the resulting avorago virna, This (above-mentioned sum of the products) is separately divided by the desired virnus (to nrrive at the required oquivalent weight of this gold).
Examples in illustration tharenf. 173. Twenty panas (in weight of gold) of 16 varnas bave boon oxchanged for (gold of; 10 rarnas in quality; you give out how many purūnas (in weight) they become now.
1744. One hundred and eight (in weight of) gold of 112 durnas is exchangod for (gold of) 14 varnas. What is the oquivalent quantity of this new) gold?
The rule for finding out the unknown warna :
1759. Froin the product obtained by multiplying the total quantity of gold by tho resulting varna of the mixture, the sum of
