# Problem Solving: Find the time for the second bus to catch up

by Katie

(USA)

**Question**

A bus leaves a station at 8:00 a.m. and averages 30 mi/h. Another bus leaves the same station following the same route two hours after the first and averages 50 mi/h. When will the second bus catch up to the first?

**Answer**

**STEP 1: **We are given that the first bus leaves a station at a rate of 30 mi/h at 8.00 am and that the second bus leaves the same station two hours later at a rate of 50mi/h. Assume a variable for the time taken by the first bus to travel, say,

t.

**STEP 2: **Recall the formula:

In this case, the distances are the same, so we just need to set the

rate × time equal for both the cases.

**STEP 3: **We can see that 50 is distributed over the difference

t – 2. To

remove the parentheses in this expression, first multiply

50 by

t – 2. To remove the parentheses in this expression, first multiply

t. We get

50t. Multiply

50 by

-2 next. The result that you get is

30t = 50t – 100.

**STEP 4: **Now, we have to isolate the variable terms on one side and the constant terms on the other side.

Subtract

50t from both sides of the obtained expression and

simplify.

**STEP 5: **Isolate the variable

t, by dividing both sides of the obtained expression by

-20.

This is the time of travel for the first bus till the second bus catches up with it.

**STEP 6: **Now you have to find the time at which the second bus catch up to the first bus. We know that the first bus leaves the station at 8 00 am. So we have to add 5 hrs to the starting time of the first bus:

8.00 am + 5 hours = 1.00 pm

The second bus catches up to the first by 1.00 pm.