Team time percentage

All three team members took turns fulfilling the competition task without delay. The first member used up 30% of the total competition time. The second needed 10 minutes more than the first, and only 10 minutes remained for the third. What % of the total time did the other member need?

Final Answer:

p₂ =  50 %

Step-by-step explanation:

$$\begin{gathered} p_1=30\%=\frac{30}{100}=\frac{3}{10}=0.3 \\ b=10+a \\ c=10\text{min} \\ a=p_1\cdot (a+b+c) \\ a=p_1\cdot (a+(10+a)+c) \\ a=0.3\cdot (a+(10+a)+10) \\ 0.4a=6 \\ a=\frac{6}{0.4}=15 \\ a=15 \\ a=15\text{min} \\ b=10+a=10+15=25\text{min} \\ s=a+b+c=15+25+10=50\text{min} \\ {p}_2=100\cdot \frac{b}{s}=100\cdot \frac{25}{50}=\frac{100\cdot 25}{50}=\frac{2500}{50}=50\% \end{gathered}$$

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