1.

The pollution in a normal atmosphere is less than `0.01%`. Due to leakage of a gas from a factory, the pollution is increased to 20%. If every day 80% of pollution is neutralised, in how many days the atmosphere will be normal?

Answer» Let the pollution on 1st day =20
The pollution on 2nd day `=20xx20%=20(0.20)`
The pollution on 3rd day `=20(0.20)^(2)`
`" " vdots " "vdots " "vdots " "vdots `
Let in n days the atmosphere will be normal
` therefore " " 20(0*20)^(n-1)lt0.01`
` implies " " ((2)/(10))^(n-1)lt(1)/(2000)`
Taking logarithm on base 10, we get
`(n-1)(log2 -log10)ltlog1- log2000`
`(n-1)(0*3010-1)lt0-(0*3010+3)`
` implies n-1gt(3*3010)/(0*6990)`
` implies ngt5*722`
Hence, the atmosphere will be normal in 6 days.


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