Question

Jen and Karen are travelling for the weekend. They both leave from Jen's house and meet at their destination 250 miles away. Jen drives 45mph the whole way. Karen drives 60mph but leaves a half hour after Jen. How long does it take for Karen to catch up with Jen?

• A. 1.5 hours
• B. She can't catch up.
• C. 3 hours
• D. 1 hour
• E. 2 hours

Hint:

In catch-up problems, define your time variable carefully. It's usually easiest to let 't' be the travel time of the person who starts later. Then the person who started earlier will have a travel time of 't + (head start time)'.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a "catch-up" problem, which is a type of distance-rate-time problem. To find when Karen catches up with Jen, we need to find the time at which they have both traveled the same distance from the starting point.
Step 2: Key Formula or Approach:
The fundamental formula is Distance = Rate \(\times\) Time.
Let \(t\) be the time in hours that Karen has been traveling.
Since Jen left 0.5 hours earlier, Jen's travel time is \(t + 0.5\) hours.
The distance traveled by Jen is \(d_J = 45(t + 0.5)\).
The distance traveled by Karen is \(d_K = 60t\).
To catch up, their distances must be equal: \(d_J = d_K\).
Step 3: Detailed Explanation:
Set the distance equations equal to each other: \[ 45(t + 0.5) = 60t \] Distribute the 45 on the left side: \[ 45t + 45(0.5) = 60t \] \[ 45t + 22.5 = 60t \] Subtract \(45t\) from both sides to isolate the term with \(t\): \[ 22.5 = 60t - 45t \] \[ 22.5 = 15t \] Solve for \(t\) by dividing both sides by 15: \[ t = \frac{22.5}{15} = \frac{45/2}{15} = \frac{45}{30} = 1.5 \] The time is in hours, as the speeds were in mph.
The 250 miles information is extra, used only to confirm that they catch up before reaching the destination. At 1.5 hours, Karen travels \(60 \times 1.5 = 90\) miles, which is less than 250.
Step 4: Final Answer:
It takes Karen 1.5 hours to catch up with Jen.