Question

In a pollution study of 1500 Indian rivers the following data were reported.520 were polluted by sulphur compounds, 335 were polluted by phosphates, 425 were polluted by crude oil, 100 were polluted by both crude oil and sulphur compounds, 180 were polluted by both sulphur compound and phosphates, 150 were polluted by both phosphates and crude oil and 28 were polluted by sulphur compounds, phosphates and crude oil. Let "n" be the number of the rivers were polluted by at least one of the three impurities.Find sum of digits of "n" ?

Answer 1

Given, $n\left(S\right)=520,n\left(P\right)=335,n\left(C\right)=425,n(S\cap P)=180,n(C\cap S)=100,n(P\cap C)=150,n(C\cap P\cap S)=28$
Now, number of rivers polluted by at least one of the impurities is
$n(P\cup S\cup C)=n\left(P\right)+n\left(S\right)+n\left(C\right)-n(P\cap S)-n(S\cap C)-n(P\cap C)+n(P\cap S\cap C)$
$=520+335+425-100-180-150+28=878$
Now, number of rivers polluted by exactly sulphur compounds
$n(S\cap P^{\prime }\cap C^{\prime })=n(S\cap (P\cup C{)}^{\prime })$
$=n\left(S\right)-n(S\cap (P\cup C\left)\right)$
$=n\left(S\right)-\left[n\right(S\cap P)\cup n(S\cap C\left)\right]$
$=n\left(S\right)-\left[n\right(S\cap P)+n(S\cap C)-n(S\cap P\cap C)$
$=520-180-100+28=268$
Now, number of rivers polluted by exactly phosphorous compounds
$n(P\cap S^{\prime }\cap C^{\prime })=n(P\cap (S\cup C{)}^{\prime })$
$=n\left(P\right)-\left[n\right(P\cap (S\cup C)\left)\right]$
$=n\left(P\right)-\left[n\right(P\cap S)\cup n(P\cap C\left)\right]$
$=n\left(P\right)-n(P\cap S)-n(P\cap C)+n(P\cap C\cap S)$
$=335-180-150+28=33$
Now, number of rivers polluted by exactly crude oil
$n(C\cap S^{\prime }\cap P^{\prime })=n(C\cap (S\cup P{)}^{\prime })$
$=n\left(C\right)-\left[n\right(C\cap (S\cup P)\left)\right]$
$=n\left(C\right)-\left[n\right(C\cap S)\cup n(C\cap P\left)\right]$
$=n\left(C\right)-n(C\cap S)-n(C\cap P)+n(P\cap C\cap S)$
$=425-150-100+28=203$
So, the number of rivers polluted by exactly one of the three impurities
$=268+33+203=504$