knight247 wrote:Ron and Shawn run a race of 2000 m. First, Ron gives Shawn a 200 m start and beats him by 30 seconds. Next, Ron gives Shawn a 3 minute start and is beaten by a 1000 m. Find the time in minutes in which Ron and Shawn run the race separately.

(A) 8,10
(B) 4,5
(C) 5,9
(D) 6,9
(E) 8,12

We can plug in the answers, which represent the two times.
The correct times are likely to be factors of 2000 (the total distance).
Since 9 and 12 do not divide evenly into 2000, the correct answer choice is unlikely to be C, D, or E.

Answer choice B: Ron's time = 4 minutes, Shawn's time = 5 minutes
Shawn's rate = d/t = 2000/5 = 400 meters per minute.

First race: Ron gives Shawn a 200-meter head start and beats him by 30 seconds.
Here, when Ron and Shawn race against each other, Ron travels the entire 2000 meters, while Shawn travels only 1800 meters, since he is given a 200-meter head start.
Since Ron travels the entire 2000 meters, Ron's time = 4 minutes.
Time for Shawn to travel 1800 meters = d/r = 1800/400 = 4.5 minutes.
Success!
Since Ron takes 1/2 minute less than Shawn, he beats Shawn by 30 seconds.

The correct answer is B.

Please note that there is no reason to check what happens in the second race.
Only ONE combination of times will yield the required difference of 30 seconds in the first race.

There was a request (by message) to solve it without using the answers, so let's give it a shot.

For the first race:

Ron's D = 2000
Ron's R = r
Ron's T = t minutes

Shawn's D = 1800
Shawn's R = s
Shawn's T = (t + 1/2) minutes

Now for the second race!

Ron's D = 1000
Ron's R = r
Ron's T = t/2 (since he only gets half as far as he did in the first race)

Shawn's D = 2000
Shawn's R = s
Shawn's T = (t/2 + 3) (since he had three minutes more than Ron had in the second race)

So we have 1000 = r(t/2) and 2000 = s(t/2 + 3).

Let's put our equations together:

2000 = rt = s(t + 1/2) + 200 = s(t/2 + 3)

We want to solve for t, so we don't care about the rt part. Let's start with the last two equations.

s(t + 1/2) + 200 = s(t/2 + 3)
200 = s(t/2 + 3) - s(t + 1/2)
200 = s((t/2 + 3) - (t + 1/2))
200 = s(5/2 - t/2)

So we know 200 = s(5/2 - t/2). From our earlier equation, we know that 2000 = s(t/2 + 3). Let's multiply the first equation by 10, then set the equations equal.

2000 = 10s(5/2 - t/2) and 2000 = s(t/2 + 3)
10s(5/2 - t/2) = s(t/2 + 3)
s isn't 0, so divide both sides by s
10(5/2 - t/2) = (t/2 + 3)
25 - 5t = t/2 + 3
22 = 5.5t
t = 4

Since t is Ron's time, Ron runs in 4 minutes. We know the answer is B (since that's the only one with Ron's time as 4), but to find Shawn's time, we can plug into our first equation:

1800 = s(t + 1/2)
1800 = s(4 + 1/2)
s = 400

2000 = 400 * Shawn's time

Shawn's time = 5 minutes.

knight247 wrote:Ron and Shawn run a race of 2000 m. First, Ron gives Shawn a 200 m start and beats him by 30 seconds. Next, Ron gives Shawn a 3 minute start and is beaten by a 1000 m. Find the time in minutes in which Ron and Shawn run the race separately.

(A) 8,10
(B) 4,5
(C) 5,9
(D) 6,9
(E) 8,12

OA is B

An alternate algebraic approach:

Let t = Ron's time to run the entire 2000 meters.

First race:
Since Shawn gets a 200-meter head start, Shawn's distance = 1800 meters.
Since Shawn is beaten by 30 seconds, he travels for 1/2 minute longer than Ron.
Thus, Shawn's time = t + .5.
Since r = d/t, we get:
Shawn's rate = 1800/(t + .5).

Second race:
Here, Ron travels half the total distance.
Thus, Ron's time = .5t.
Since Shawn is given a 3-minute head start, Shawn's time is 3 minutes longer than Ron's time.
Thus, Shawn's time = .5t + 3.
Since Shawn finishes the race, his distance = 2000 meters.
Since r = d/t, we get:
Shawn's rate = 2000/(.5t + 3).

Since Shawn's rate is the same in each case, we get:
1800/(t + .5) = 2000/(.5t + 3)

900t + 5400 = 2000t + 1000

4400 = 1100t

t = 4.

The correct answer is B.