Kevin got $$80\textbackslash\%$$ of the problems correct on a $$25-$$problem test, $$ 90\textbackslash\%$$ on a $$40-$$problem test, and $$70\textbackslash\%$$ on a $$10-$$problem test. What percent of all the problems did Kevin answer correctly? (2010 AMC8 Problem, Question \#9) 
Overseas Competition->Knowledge Point->Word Problem Modules->Solving Problems Involving Fractions and Percentages->Finding the Rate
Ryan answered $$(0.8)(25)=20$$ problems correct on the first test, $$(0.9)(40)=36$$ on the second, and $$(0.7)(10)=7$$ on the third. This amounts to a total of $$ 20+36+7=63$$ problems correct. The, total number of problems is $$25+40+10=75$$.Therefore, the percendage is $$\frac {63}{75}\to(\rm D)84$$. 