Question

raul bikes 6 mi home from school at a rate of 3 mi/h. leah leaves school 30 min after raul and bikes the same route at a rate of 4 mi/h. how many hours does it take leah to catch up with raul?
a. 1.5 h
b. 3 h
c. 6 h
d. \\(\frac{3}{3}h\\) (note: it may be a typo, probably \\(\frac{3}{2}h\\) or other correct form, but keep the original as much as possible)

Explanation:

Step1: Convert time to hours

30 minutes is $\frac{30}{60} = 0.5$ hours. Let $t$ be the time Leah bikes until she catches up. Then Raul has been biking for $t + 0.5$ hours.

Step2: Set up distance equations

Distance = rate × time. Raul's distance: $d = 3(t + 0.5)$. Leah's distance: $d = 4t$. Since they meet at the same distance, set equal:
$$3(t + 0.5) = 4t$$

Step3: Solve for t

Expand left side: $3t + 1.5 = 4t$. Subtract $3t$: $1.5 = t$. Wait, but let's check again. Wait, maybe I misread the problem. Wait, Raul is biking 6 mi home? Wait, no, maybe the problem is about catching up before Raul reaches home? Wait, no, let's re - examine. Wait, Raul's rate is 3 mph, Leah's is 4 mph. Leah starts 0.5 h later. Let's find when their distances are equal.

Wait, maybe the initial approach is wrong. Let's do it again. Let $t$ be Leah's time. Raul's time is $t + 0.5$. Distance for Raul: $3(t + 0.5)$, distance for Leah: $4t$. Set equal: $3t + 1.5 = 4t$ → $t = 1.5$? But wait, let's check Raul's distance: $3(1.5 + 0.5)=3\times2 = 6$ miles, Leah's distance: $4\times1.5 = 6$ miles. Oh, so Raul was biking the 6 - mile route, and Leah catches up when they both have biked 6 miles? Wait, but Raul's time to bike 6 miles is $6\div3 = 2$ hours. Leah starts 0.5 hours later, so she bikes for $2 - 0.5 = 1.5$ hours, which matches. So the answer should be 1.5 h. But wait, the options have A. 1.5 h. But wait, let's check the options again. Wait, maybe I made a mistake. Wait, let's re - solve the equation:

$3(t + 0.5)=4t$

$3t+1.5 = 4t$

$4t - 3t=1.5$

$t = 1.5$ hours. So the answer is A.

Wait, but let's check the problem statement again. "Raul bikes 6 mi home from school at a rate of 3 mph. Leah leaves school 30 min after Raul and bikes the same route at a rate of 4 mph. How many hours does it take Leah to catch up with Raul?"

So Raul is going home, 6 miles. His time to complete the trip is 2 hours. Leah starts 0.5 hours later. We need to find when Leah's distance equals Raul's distance. As we saw, when $t = 1.5$ hours, Leah has biked $4\times1.5 = 6$ miles, and Raul has biked $3\times(1.5 + 0.5)=6$ miles. So they meet at the end of the route? But that's when Raul finishes. So that makes sense. So the answer is 1.5 h, which is option A.

Answer:

A. 1.5 h