SOLUTION: Sheila leaves on a long trip driving at a steady rate of 25 miles per hour. Her sister Allison leaves from the same location traveling to the same destination 2 hours later. She dr

Question 767533: Sheila leaves on a long trip driving at a steady rate of 25 miles per hour. Her sister Allison leaves from the same location traveling to the same destination 2 hours later. She drives at a steady rate of 50 miles per hour. How long after Sheila leaves home will Allison catch up?

I have tried this problem for approximately 3 hours. I'm in 9th grade and I'll be honest with you and say, this is a graded assignment. Thank you in advance for any help you can provide.
-Faith

Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
Distance = rate x time

Sheila's rate = 25 mph
Sheila's time = t
Sheila's distance = 25t {distance = rate x time}

Allison's rate = 50mph
Allison's time = t - 2 {she started 2 hours later or lost two hours off of the starting time}
Allison's distance = 50(t - 2) = 50t - 100 {distance = rate x time}

25t = 50(t - 2) {when Allison catches up with Sheila the distances are equal}
25t = 50t - 100 {used distributive property}
-25t = -100 {subtracted 50t from each side}
t = 4 {divided each side by -25}
{t corresponds with Sheila's time, which is what the question is asking for}

In 4 hours Allison will catch up with Sheila
{Your mistake was probably on Allison's time. You probably had t + 2.}

For more help from me, visit:
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