A+B+C=one hundred

A+B+C=100. If A=27.5 and the value of B are between 10 and 15 and which has two decimal places, then what can be the greatest value of C?

Correct answer:

C =  62.49

Step-by-step explanation:

$$\begin{gathered} A=27.5 \\ A+B+C=100 \\ B+C=100-27.5=72.5 \\ 10<B<15 \\ B=XX.01 \\ B=10.01 \\ C=72.5-B=72.5-10.01=\frac{6249}{100}=62\frac{49}{100}=62.49 \\ \text{ Verifying Solution: } \\ s=A+B+C=27.5+10.01+62.49=100 \end{gathered}$$

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