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Alex could arrive on time for his appointment if he leaves now and drives 50mph. However, Alex leaves the house 15 minutes earlier, drives 40mph, and arrives on time. How far from Alex's house is the appointment?

User Serbin
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1 Answer

2 votes

Given the following question:

15 = 0.25

Write the question as a equation and then solve to find the distance:


\begin{gathered} 50(x-0.25+0.05)=40x \\ 50\left(x-0.2\right)=40x \\ 50x-10=40x \\ 50x-40x=10 \\ 10x=10 \\ 10x/10=x \\ 10/10=1 \\ x=1 \\ 40(1) \\ 40(1)=40 \\ \text{ 40 miles} \end{gathered}

User Tamas Rev
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