Math, asked by Dhita5210, 9 months ago

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

Answers

Answered by malininandankar
1

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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