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maggiejdyer: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Sun Sep 21, 2014 11:18 am

by maggiejdyer » Sun Mar 01, 2015 10:17 am

Can someone show how to get to the answer algebraically?

Tim and Glenn are running laps around a circular track. If they start at exactly the same time, in how many seconds will Tim have run exactly one lap further than Glenn?

(1) Tim runs each lap in 48 seconds and Glenn runs each lap in 60 seconds.

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by DavidG@VeritasPrep » Sun Mar 01, 2015 10:32 am

Because this is data sufficiency, you wouldn't actually finish solving the algebra, but the set up would look like this.

Tim's rate is 1 lap every 48 seconds, or 1/48.

Glenn's rate is 1 lap every 60 seconds or 1/60.

If they each run for t seconds, Tim will have gone t/48 laps. And Glenn will have gone t/60 laps. (This is just R*T = D)

If Tim went one more lap than Glenn, and Glenn went t/60 laps, then Tim would have gone t/60 + 1 laps. We already know that Time went t/48 laps, so we can make the following equation: t/48 = t/60 + 1. Clearly, we can solve for t here. Sufficient.

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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

by DavidG@VeritasPrep » Sun Mar 01, 2015 10:46 am

(And generally speaking, when two entities are moving in the same direction, it's fair game to find the net rate by taking the difference. For example, if I'm driving 60 mph, and I'm behind another car (moving in the same direction) driving 50 mph, I'm clearly making up 60 - 50 = 10 miles each hour.)

Using this logic in the above question, the net rate is 1/48 - 1/60. If they are both running for t seconds, and we want them to be 1 lap apart, we can go right to: (1/48 - 1/60)*t = 1.

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by [email protected] » Sun Mar 01, 2015 12:58 pm

Hi maggiejdyer,

This is an example of a rare 'combined rate' question (sometimes called a "chase down" question). The key to these types of questions is to figure out the DIFFERENCE in speeds of the two entities (when they're moving at the same time) and use that number as a basis for comparison.

Here, we know that Tim runs a lap in 48 seconds and Glenn runs a lap in 60 seconds. The difference in their rates is 12 seconds/lap.

Tim will "catch up" 12 seconds on Glenn every lap. Since Glenn needs 60 seconds to finish a lap, Tim needs 60/12 = 5 laps to catch Glenn.

You can see that the results are correct by doing the following math:

Tim: (5 laps)(48 seconds per lap) = 240 seconds

Glenn: (X laps)(60 seconds per lap) = 240 seconds
X = 240/60 = 4 laps

In 240 seconds, Tim runs 5 laps and Glenn runs 4 laps. At this point, Time has "lapped" Glenn.

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by GMATGuruNY » Sun Mar 01, 2015 5:29 pm

maggiejdyer wrote:Can someone show how to get to the answer algebraically?

Tim and Glenn are running laps around a circular track. If they start at exactly the same time, in how many seconds will Tim have run exactly one lap further than Glenn?

(1) Tim runs each lap in 48 seconds and Glenn runs each lap in 60 seconds.

The TIME RATIO for Tim and Glenn = 48:60 = 4:5.
Since time and rate are RECIPROCALS, the RATE RATIO for Tim and Glenn = 5:4.
Implication:
For every 5 laps that Tim runs, Glenn runs only 4 laps.
Thus, after running 5 laps, Tim will have run 1 more lap than Glenn.
At a rate of 48 seconds per lap, the time for Tim to run 5 laps = 48*5 = 240 seconds.

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by DavidG@VeritasPrep » Mon Mar 02, 2015 5:40 am

For those in search of more "catch-up" problems:

Car A is 20 miles behind Car B, which is traveling in the same direction along the same route as Car A. Car A is traveling at a constant speed of 58 miles per hour and Car B is traveling at a constant speed of 50 miles per hour. How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?

(A) 1.5

(B) 2.0

(C) 2.5

(D) 3.0

(E) 3.5

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prachi18oct: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 27, 2014 10:33 pmFollowed by: 5 members; Thanked: 8 times

by prachi18oct » Mon Mar 02, 2015 8:04 am

(E) 3.5

The relative distance to be travelled = 20 +8(miles ahead after overtake)
The speed difference = 58-50 = 8m/h

time taken = 28/8 = 3.5

OR by plugging options:-

Lets put D

If car A travels 3 hours => 58*3 = 174 miles covered
Car b travels in 3 hours => 150 miles
initial distance b/w A & b = 20 miles
so 174-150 = 24 miles distance between A & B i.e only 4 miles ahead but we want 8 miles ahead, so clearly need some more time => hence 3.5 hours.

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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 amFollowed by: 128 members; Location: Boston, MA; Thanked: 1153 times; GMAT Score:770

by DavidG@VeritasPrep » Mon Mar 02, 2015 12:26 pm

Excellent explanation!

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