SOLUTION: Carlton goes for a jog on a 5 mile loop. He has completed two miles when David begins the same loop. Carlton will jog at a constant rate of 0.1 miles/minute for the remainder of

Question 1006347: Carlton goes for a jog on a 5 mile loop. He has completed two miles when David begins the same loop.
Carlton will jog at a constant rate of 0.1 miles/minute for the remainder of his jog. David will jog at a constant rate of 0.2 miles/minute for the duration of his jog.
How many miles will Carlton have jogged at the point when David catches up with him?
My attempt:
I can see David is jogging twice as fast as Carlton, but i don't see how to calculate the number of miles when he'll catch up to Carlton. The answer is supposed to be 4 miles.
Thank you for any help.

Found 2 solutions by Theo, Alan3354:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
rate * time = distance.

when david catches up with carlton, they both will have jogged the same amount of time.

the distance that carlton jogs will be 2 less than the distance that david jogs because carlton has already jogged 2 miles and david needs to run that same 2 miles plus whatever carlton continues to jog in order to catch up.

let R1 * T1 = D1 be the rate * time = distance formula for carlton.
let R2 * T2 = D2 be the rate * time = distance formula for david.

since the time each will jog once david starts jogging is the same, then let T = T1 = T2 and the formulas become:

R1 * T = D1
R2 * T = D2

since carlton needs to jog 2 miles less than david needs to jog after david starts jogging, then D1 = D2 - 2

the formulas become:

R1 * T = D2 - 2
R2 * T = D2

you are told that carlton's rate is .1 miles per minute and that david's rate is .2 miles per minute.

the formulas become:

.1 * T = D2 - 2
.2 * T = D2

solve each of these formulas for T and you get:

T = (D2 - 2) / .1
T = D2 / .2

replace T in the first equation with the equivalent value of D2 / .2 from the second equation to get:

D2 / .2 = (D2 - 2) / .1

cross multiply to get:

.1 * D2 = .2 * (D2 - 2)

divide both sides of this equation by .1 to get:

D2 = 2 * (D2 - 2)

simplify to get:

D2 = 2 * D2 - 4

subtract D2 from both sides of this equation and add 4 to both sides of this equation to get:

4 = 2 * D2 - D2

simplify to get:

4 = D2

this can also be shown as:

D2 = 4

D2 is how far david needs to run.

D1 = how far carlton has run after david has run 4 miles.

D1 = D2 - 2 = 4 - 2 = 2

this means that carlton will have run 2 miles when david catches up to him.

carlton will have jogged 2 more miles after david starts.
david will have jogged 4 miles when he catches up with carlton.

you can solve the the time as well.

you get for carlton, rate * time = distance becomes .1 * T = 2.
this gets you T = 2/.1 = 20 minutes.

you get for david, rate * time = distance becomes .2 * T = 4.
this gets you T = 4/.2 = 20 minutes.

they both run for 20 minutes when david catches up to carlton.

carlton has run .1 * 20 = 2 miles.
david has run .2 * 20 = 4 miles.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Carlton goes for a jog on a 5 mile loop. He has completed two miles when David begins the same loop.
Carlton will jog at a constant rate of 0.1 miles/minute for the remainder of his jog. David will jog at a constant rate of 0.2 miles/minute for the duration of his jog.
How many miles will Carlton have jogged at the point when David catches up with him?
---------------
C is 2 miles away when D starts.
D "gains on" C at 0.1 mi/minute (0.2 - 0.1)
2 mi/(0.1 mi/min) = 20 minutes to catch up
C jogs 2 miles + 20 min*(0.1 mi/minute)
= 2 + 2
= 4 miles
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