The value of gold of 400gm is inverse to the square root of the fraction Of the impurities. The price of 25gm impuries in gold is 45000 then how many impurity’s in 90000 gold
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Answer:
V1/V2. =√i2/i1
45000/90000. =√i2/i1
i2=6.25
Pure gold in a bar costing Rs.90000 = 400 - 6.25 = 393.75
Your answer is 393.75
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