PS Question on Time Rate and Distance-Not able to solve

[dhiren8182]

Carlos can run a mile in 6 minutes. Kevin can run a mile in 8 minutes. If Kevin goes out to practice and starts 1 minute before Carlos starts on his practice run, will Carlos catch up with Kevin before he reaches the mile marker? (Assume a straight track and average speed.)
A)3/2 mile
B)2 mile
C)1/2 mile
D)1/4 mile
[spoiler]OA=C[/spoiler]
Can anyone solve it for me with a nice explanation.I normally follow table matrix(algebric way) as mentioned underneath

time rate distance
Carlos
kevin

Regards,
Dhiren[/quote]

[GMATGuruNY]

I believe the following reflects the intent of the problem:

Carlos can run a mile in 6 minutes. Kevin can run a mile in 8 minutes. If Kevin goes out to practice and starts 1 minute before Carlos starts on his practice run, how far will Carlos have run when he catches up to Kevin?

A)3/2 mile
B)2 mile
C)1/2 mile
D)1/4 mile

Since Carlos can run 1 mile in 6 minutes, Carlos's rate = d/t = 1/6 mile per minute.
Since Kevin can run 1 mile in 8 minutes, Kevin's rate = d/t = 1/8 mile per minute.

Distance traveled by Kevin in the first minute = 1/8 mile.

Carlos must CATCH-UP by 1/8 mile.
The catch-up rate is equal to the DIFFERENCE between Carlos's rate and Kevin's rate:
1/6 - 1/8 = 4/24 - 3/24 = 1/24 mile.
Implication:
Since Carlos travels 1/6 mile each minute, and Kevin travels 1/8 mile each minute, every minute Carlos travels 1/24 mile MORE than Kevin, with the result that Carlos CATCHES UP by 1/24 mile each minute.

Time for Carlos to catch up to Kevin = (catch-up distance)/(catch-up rate) = (1/8)/(1/24) = 3 minutes.
At a rate of 1/6 mile per minute, the distance traveled by Carlos in 3 minutes = r*t = (1/6)(3) = 1/2 mile.

The correct answer is C.

An alternate approach is to GRIND OUT how far Kevin and Carlos travel each minute.

To make the math easier, put the rates over a common denominator.
Since Carlos can run 1 mile in 6 minutes, Carlos's rate = d/t = 1/6 = 4/24 mile per minute.
Since Kevin can run 1 mile in 8 minutes, Kevin's rate = d/t = 1/8 = 3/24 mile per minute.

In the first minute, K travels alone.
Every minute thereafter, K travels 3/24 mile more, while C travels 4/24 mile more.

Distances traveled minute by minute:
1 minute --> K = 3/24 mile, C = 0 miles.
2 minutes --> K = 3/24 + 3/24 = 6/24 mile, C = 4/24 mile.
3 minutes --> K = 6/24 + 3/24 = 9/24 mile, C = 4/24 + 4/24 = 8/24 mile.
4 minutes --> K = 9/24 + 3/24 = 12/24 mile, C = 8/24 + 4/24 = 12/24 mile.

Carlos catches up to Kevin after 4 minutes.
Distance traveled by Carlos = 12/24 = 1/2 mile.

[DavidG@VeritasPrep]

And if you wanted to use the matrix, it would look like this:

[Matt@VeritasPrep]

How about this approach:

When Carlos catches Kevin, their distances traveled will be the same, so

Carlos's D = Kevin's D

We know that D = RT, so let's find each runner's rate and time.

Carlos's R = 1/6 of a mile per minute
Carlos's T = t minutes
Kevin's R = 1/8 of a mile per minute
Kevin's T = (t + 1) minutes

Since the distances are equal, we have (1/6)t = (1/8)(t+1), or (1/24)t = 1/8, or t = 3. So when the two men meet, Carlos has been running for 3 minutes. Since he runs a mile in 6 minutes, he must've run 1/2 of a mile in 3 minutes, and the answer is C.