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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? |
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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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