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OHAPTER VI-MIIRD PROBLEYS.
141
An example in illustration thereof.
181. Threo pioccs of gold, of 3 onoh in weight, and of 2, 3, and 4 varrus (rospectivoly), are aldod to an unkuown woight of) gold of 13 varnas. 'The rosulting varna comes to be 10. Tell mo, O friend, the moasuro (of the unknown weight) of gold.
Tho rule for arriving at the weights of) gold (corresponding to two given varnas) from (the known weight and rarnu of) the mixture of two (given spocimens of) gold of (givon) rarnas :
182. Obtain the differences between the resulting varna (of the mixture on the oue hand) and the known higher and lower eyrnas (of the unknown component quantities of gold on tho other band); divide one by these differences (in order); the carry out as boforo the operation of praksipukul (or proportionate distribution with the aid of these various quotionts). In this mannor it is possiblo to arrive oveu at the value of many coin poucnt quantities of gold also.
Again, the rule for arriving at (the weights of) gold (corrosponding to two given rarnax) from (the known weight and varna of) the mixture of two (givon specimens ot) gold of (given) varnas :
183. Write down in inverse order the difference lictwoon thio resulting varna and the higher (of the two given varnar of the two oomponont quantities of gold), and also the difforonco butween the rosulting varna and the lower of the two given rarnas). The result arrived at by means of the operation of proportionato distri. bution carried out with the aid of these inversely arranged differencos),--that (result) gives the required (weights of the component quantities of) gold.
An example in illustration thereof.
184. If gold of 10 rarnas, on being combined with gold of 16 varnas, produces as result 100 in weight of gold of 12 varnas, give out separately (tbe measuros in weight of the two different varieties of gold.
