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Marla starts running around a circular track at the same [#permalink]
07 Oct 2011, 08:09

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00:00

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Difficulty:

25% (medium)

Question Stats:

77% (02:15) correct
23% (01:04) wrong based on 60 sessions

Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes 32 laps around the track per hour and Nick completes 12 laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed 4 more laps around the track than Nick?

Re: Good Time and rate question [#permalink]
08 Oct 2011, 02:14

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gmatcracker24 wrote:

Q.Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes 32 laps around the track per hour and Nick completes 12 laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed 4 more laps around the track than Nick? (A) 5 (B) 8 (C) 12 (D) 15 (E) 20

...

Thanks guys for sharing your approaches.

The way i solved it is :

In one hour Maria covers 20 extra laps than Nick.

So, in 3 minutes -> Maria would cover 1 extra lap.

Hence , for 4 extra lap -> 3*4 = 12 minutes. _________________

Re: Good Time and rate question [#permalink]
14 Oct 2011, 10:07

Howdy, I usually post just in the SC forum--find me there if you have SC questions--but I wanted to add something to the OP's excellent solution.

Whenever a difference in distances is described, you can subtract distances. The paradigm GMAT case is two people/cars/trains/whatever going in the same direction. Whenever a difference in distances is described, and both parties move simultaneously, you can also subtract rates: (difference in rates)*time=(difference in distance).

This question asks for a difference in the distances covered by two people walking simultaneously. The difference in rates is 20 laps/hour, or 1 lap/3 minutes. (1/3)t=4. t-12.

Many good test-takers find this formula, (difference in rates)*time=(difference in distance), so useful that they tweak problems with different times so that they can apply this formula. Let me give you an example, based on this question.

Q.Nick starts walking around a circular track at noon. Marla starts running around the same circular track at 12:20. Nick completes 12 laps per hour and Marla completes 32 laps per hour. At what time will Marla have completed as many laps as Nick? (A) 12:25 (B) 12:28 (C) 12:32 (D) 12:35 (E) 12:40

Here's how to answer this as one hard question: (Marla's time)(Marla's rate)=Marla's distance. (Nick's time)(Nick's rate)=Nick's distance. Nick's time is 20 minutes greater than Marla's. You could say, Let t= Marla's time in hours and t+1/3=Nick's time in hours. But we'll go with, Let t=Marla's time in minutes and t+20=Nick's time in minutes. Marla's distance is Nick's. Let d= Nick's distance and d=Marla's distance. This yields two equations: 32t=d 12(t+20)=d Solve for 12:20+t. I'll trust you to solve algebraically.

Here's the same question rephrased as three easy questions: What did Nick do before Marla started running? Walk 4 laps. Once Marla started running--once they were moving simultaneously--how much faster did she go than Nick? 20 laps/hour, or 1 lap/3 minutes. How long will it take her to gain 4 laps, if it takes her 3 minutes to gain 1 lap? 12 minutes. Add 12 minutes to 12:20, the time from which they moved simultaneously. 12:32.

Notice that this second method would be harder to apply if we complicated the problem a bit. If we asked at what time Marla would have gone twice as far as Nick, the more formal algebraic method works just as neatly (just make Marla's distance 2d instead of d), but the alternative method gets much messier.

Okay, here's one more for you to solve. It's not a real GMAT question, and I've rigged it so that the second method described above will be much easier for most people. Answer below and I'll check in tomorrow with my explanation.

At 9:00, Nick started walking the 9 miles from A to B. At 9:40, Marla started walking the 9 miles from A to B. Until Marla caught up to Nick, Marla walked at a constant rate of 5 miles per hour, and Nick walked at a constant rate of 3 miles per hour. Once Marla caught Nick, they walked together at a constant rate of 4 miles per hour. At what time did they arrive together at B? _________________

Re: Good Time and rate question [#permalink]
14 Oct 2011, 11:44

Thanks catfreak, you're exactly right. I wrote the problem, wrote the answers, wrote the explanation, then changed the problem to make it harder, without making other changes. I'll edit it right now. Good catch. _________________

Re: Good Time and rate question [#permalink]
18 Oct 2011, 19:59

MichaelS wrote:

Howdy, I usually post just in the SC forum--find me there if you have SC questions--but I wanted to add something to the OP's excellent solution.

Whenever a difference in distances is described, you can subtract distances. The paradigm GMAT case is two people/cars/trains/whatever going in the same direction. Whenever a difference in distances is described, and both parties move simultaneously, you can also subtract rates: (difference in rates)*time=(difference in distance).

This question asks for a difference in the distances covered by two people walking simultaneously. The difference in rates is 20 laps/hour, or 1 lap/3 minutes. (1/3)t=4. t-12.

Many good test-takers find this formula, (difference in rates)*time=(difference in distance), so useful that they tweak problems with different times so that they can apply this formula. Let me give you an example, based on this question.

Q.Nick starts walking around a circular track at noon. Marla starts running around the same circular track at 12:20. Nick completes 12 laps per hour and Marla completes 32 laps per hour. At what time will Marla have completed as many laps as Nick? (A) 12:25 (B) 12:28 (C) 12:32 (D) 12:35 (E) 12:40

Here's how to answer this as one hard question: (Marla's time)(Marla's rate)=Marla's distance. (Nick's time)(Nick's rate)=Nick's distance. Nick's time is 20 minutes greater than Marla's. You could say, Let t= Marla's time in hours and t+1/3=Nick's time in hours. But we'll go with, Let t=Marla's time in minutes and t+20=Nick's time in minutes. Marla's distance is Nick's. Let d= Nick's distance and d=Marla's distance. This yields two equations: 32t=d 12(t+20)=d Solve for 12:20+t. I'll trust you to solve algebraically.

Here's the same question rephrased as three easy questions: What did Nick do before Marla started running? Walk 4 laps. Once Marla started running--once they were moving simultaneously--how much faster did she go than Nick? 20 laps/hour, or 1 lap/3 minutes. How long will it take her to gain 4 laps, if it takes her 3 minutes to gain 1 lap? 12 minutes. Add 12 minutes to 12:20, the time from which they moved simultaneously. 12:32.

Notice that this second method would be harder to apply if we complicated the problem a bit. If we asked at what time Marla would have gone twice as far as Nick, the more formal algebraic method works just as neatly (just make Marla's distance 2d instead of d), but the alternative method gets much messier.

Okay, here's one more for you to solve. It's not a real GMAT question, and I've rigged it so that the second method described above will be much easier for most people. Answer below and I'll check in tomorrow with my explanation.

At 9:00, Nick started walking the 9 miles from A to B. At 9:40, Marla started walking the 9 miles from A to B. Until Marla caught up to Nick, Marla walked at a constant rate of 5 miles per hour, and Nick walked at a constant rate of 3 miles per hour. Once Marla caught Nick, they walked together at a constant rate of 4 miles per hour. At what time did they arrive together at B?

Hi:

Is the answer to the highlighted practice problem above 11:40?

Re: Good Time and rate question [#permalink]
22 Oct 2011, 01:11

1

This post was BOOKMARKED

At 9:00, Nick started walking the 9 miles from A to B. At 9:40, Marla started walking the 9 miles from A to B. Until Marla caught up to Nick, Marla walked at a constant rate of 5 miles per hour, and Nick walked at a constant rate of 3 miles per hour. Once Marla caught Nick, they walked together at a constant rate of 4 miles per hour. At what time did they arrive together at B?

Is the answer 11:40 for this ???

Explanation: Suppose they both meet at point C somewhere in the path from A to B. Nick reached point C at time t and Marla reached point C at time t+40/60 (as she started 40 mins late so equating time taken on both sides). Suppose distance between A and C is d. Then both of them covered the distance d in time t (for Nick) and t+40/60 (for Marla).

Using t=d/s, we have the equation:

d/3=d/5+40/60 -> d=5.

so, d=5 distance was covered by Marla in 1 hour.

Remaining distance is 9-5=4, which they cover with constant speed of 4, Another 1 hour to reach C. So total of 2 hours from when Marla started.

Re: Good Time and rate question [#permalink]
31 Oct 2011, 11:26

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Thanks for letting me hijack your thread gmatcracker.

Here's the question I asked:

At 9:00, Nick started walking the 9 miles from A to B. At 9:40, Marla started walking the 9 miles from A to B. Until Marla caught up to Nick, Marla walked at a constant rate of 5 miles per hour, and Nick walked at a constant rate of 3 miles per hour. Once Marla caught Nick, they walked together at a constant rate of 4 miles per hour. At what time did they arrive together at B?

Here's a relatively easy way to solve it:

What happened from 9:00 to 9:40? Nick walked 2 miles (3mph*2/3 of a mile) What was the difference in the rates from that point till Marla caught Nick? 2mph (5mph-3mph) How long did it take for Marla to catch Nick? 1 hour (2mph*1 hour=2 miles)

What time was it then? 10:40

How long had each walked? 5 miles each (Marla had walked for 1 hour at 5mph)

How many miles remained? 4 miles (9 miles-5 miles)

How long did it take them to cover that distance? 1 hour

What time was it then? 11:40

I think that that is substantially the same as ashimood's response. Certainly it's the same result. _________________

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