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Raul bikes 6 mi 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?

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znk

Answer:

1.5 h

Explanation:

We know that distance travelled (d) equals rate of travel (r) times time (t) spent in travel. In symbols,

d = rt or rt =d

When Leah starts, Raul is already 30 min (0.5 h) ahead.

So, let t time for Leah and t + 0.5 be the time for Raul.

For Raul: 3(t+0.5) = d

For Leah:         4t = d

When Leah catches up with Raul, both will have travelled the same distance. So,

3(t+0.5) = 4t     Remove parentheses

3t + 1.5 = 4t      Subtract 3t from each side

t = 1.5 h

It takes Leah 1.5 h to catch up with Raul.

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Check:

Distance travelled by Raul  = 3 × (1.5 + 0.5) = 3 × 2 = 6 mi

Distance travelled by Leah =                        4 × 1.5 = 6 mi

Lanuel

The number of hours it will take Leah to catch up with Raul is equal to 1.5 hour.

Give the following data:

  • Distance covered by Raul = 6 miles
  • Raul's rate = 3 mi/h
  • Leah's rate = 4 mi/h

To calculate the number of hours it will take Leah to catch up with Raul;

First of all, we would convert the value of time in minutes to hour:

Conversion:

60 minutes = 1 hour

30 minutes = 0.5 hour

Mathematically, distance with respect to rate and time is given by the formula:

[tex]Distance = rate \times time[/tex]

Translating the word problem into an algebraic expression, we have;

For Raul:

[tex]3(t+0.5) = 6[/tex]

For Leah:

[tex]4t = 6\\\\t=\frac{6}{4}[/tex]

Time, t = 1.5 hour

Read more on distance here: https://brainly.com/question/8898885