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
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Carlton will jog at a constant rate of 0.1 miles/minute for the remainder of
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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.
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.
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?
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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
