SOLUTION: Good day. I'd be really thankful if anyone will be able to help me with this conundrum of mine. It is as follows: A boy starts walking to school at 7:30 a.m. at the rate of 3 mile

Question 282750: Good day. I'd be really thankful if anyone will be able to help me with this conundrum of mine. It is as follows:
A boy starts walking to school at 7:30 a.m. at the rate of 3 miles per hour. At 7:40 a.m., his brother starts for school on his bicycle at 6 miles per hour. When will the boy on the bicycle overtake his brother?
Thank you very much for the help!
Found 2 solutions by Mathematicians, Theo:
Answer by Mathematicians(84) (Show Source): You can put this solution on YOUR website!
You want to overuse this formula:

We are looking for time, the trick in this problem is the boy on the bike begins to overtake his brother, they will have equal distances.
Guy walking:
because he started 10 minutes early, that is 1/6 an hour.

Guy Bike:

Since the distances are equal:

distributive property
subtract 3t on both sides:

It will take 1/6th an hour (10 minutes) to catch up!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
Rate * Time = Distance

Walking Brother travels at 3 miles per hour.
Biking Brother travels at 6 miles per hour.

Walking Brother starts at 7:30 am.
Biking Brother starts at 8:30 am.

At some point in time, Biking Brother catches up to Walking Brother.
At that point in time, they will have both traveled the same distance.
At that point in time, Walking Brother will have traveled for 10 more minutes than Biking Brother.

10 more minutes is the same as 1/6 hours = .166667 hours.

Everything needs to be converted to hours so the rate and the time are consistent.

Basic Formula to use is Rate * Time = Distance.

Let D = the distance they both traveled.
Let T+.166667 = the time it took Walking Brother to travel that distance.
Let T = the time it took Biking Brother to travel that same distance.
Rate of Travel for Walking Brother is 3 miles per hour.
Rate of Travel for Biking Brother is 6 miles per hour.

The formulas to use are:

Walking Brother:

3 * (T+.166667) = D

Biking Brother:

6 * T = D

Since they both = D, then both equations are equal to each other, so you get:

3 * (T+.166667) = 6*T

You have to solve for T.

Simplify the left side of the equation to get:

3*T + .5 = 6*T

Subtract 3*T from both sides of the equation to get:

.5 = 3*T

Divide both sides of the equation by 3 to get:

T = .166667

Biking Brother traveled for .166667 hours.

Walking Brother traveled for .33333 hours.

Plug those times into both equation to get:

3*.33333 = 1 mile.

6*.166667 = 1 miles.

Biking Brother catches up to Walking Brother in 1 mile.

You could also have translated everything to minutes and you would have gotten the same answer.

Here's how that would have been done.

Walking Brother travels at 3 miles per hour which is equivalent to .05 miles per minute.

Biking Brother travels at 6 miles per hour which is equivalent to .1 miles per minute.

Walking Brother travels for 10 more minutes than Biking Brother when they have both traveled the same distance.

Formulas are:

.05 * (T+10) = D for Walking Brother.

.1 * T = D for Biking Brother.

Make the equations equal to each other to get:

.05 * (T+10) = .1 * T

Simplify to get

.05 * T + .05 * 10 = .1 * T

Subtract .05 * T from both sides of the equation to get

.05 * 10 = .05 * T

Divide both sides of the equation by .05 to get:

10 = T

Biking Brother travels for 10 minutes.

Walking Brother travels for 20 minutes.

10 * .1 = 1 mile traveled for Biking Brother.

20 * .05 = 1 mile traveled for Walking Brother.

same answer.

Biking Brother catches up to Walking Brother in 1 mile.

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