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Time , speed and distance problem

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Time , speed and distance problem

by hemant_rajput » Wed Jan 16, 2013 9:03 am
Q1 . Three runner A, B and C run a race, with runner A finishing 24 meters ahead of runner B and 36 meters ahead of runner C, while runner B finishes 16 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race?

a. 72
b. 96
c. 120
d. 144
[spoiler]oa: b[/spoiler]
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by adthedaddy » Wed Jan 16, 2013 10:47 am
Let the total length of race be x meters.

When A finishes the race, it is 24mts ahead of B & 36 mts ahead of C.
When B finishes the race, the distance between B & C is 16mts.

Thus, when B covers 24mts, C covers 20mts. So ratio of speeds of B & C = 24:20 = 6:5

Length of Race : (x-24)/(x-36)=6/5
Thus x=96

OA=(b)
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by GMATGuruNY » Wed Jan 16, 2013 6:50 pm
hemant_rajput wrote:Q1 . Three runner A, B and C run a race, with runner A finishing 24 meters ahead of runner B and 36 meters ahead of runner C, while runner B finishes 16 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race?

a. 72
b. 96
c. 120
d. 144
[spoiler]oa: b[/spoiler]
When A finishes:
B is 24 meters behind A.
C is 36 meters behind A.
Thus, the distance between B and C = 36-24 = 12 meters.

When B finishes:
B has traveled 24 more meters.
The distance between B and C increases by 4 meters to 16 meters.

Thus:
For every 24 meters traveled by B, B pulls 4 more meters ahead of C.
Since B finishes 16 meters ahead of C, we can set up the following proportion to determine the total distance traveled by B:

(24 meters traveled by B)/(4 meters ahead of C) = (x total meters traveled by B)/(16 meters ahead of C)
6/1 = x/16
x = 96.

The correct answer is B.
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by hemant_rajput » Thu Jan 17, 2013 5:49 am
GMATGuruNY wrote:
hemant_rajput wrote:Q1 . Three runner A, B and C run a race, with runner A finishing 24 meters ahead of runner B and 36 meters ahead of runner C, while runner B finishes 16 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race?

a. 72
b. 96
c. 120
d. 144
[spoiler]oa: b[/spoiler]
When A finishes:
B is 24 meters behind A.
C is 36 meters behind A.
Thus, the distance between B and C = 36-24 = 12 meters.

When B finishes:
B has traveled 24 more meters.
The distance between B and C increases by 4 meters to 16 meters.

Thus:
For every 24 meters traveled by B, B pulls 4 more meters ahead of C.
Since B finishes 16 meters ahead of C, we can set up the following proportion to determine the total distance traveled by B:

(24 meters traveled by B)/(4 meters ahead of C) = (x total meters traveled by B)/(16 meters ahead of C)
6/1 = x/16
x = 96.

The correct answer is B.
Hi GMATGuruNY,

Thanks for the response. I wasn't able to understand how to you come up the magic equation. Can you please elaborate that?
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by GMATGuruNY » Thu Jan 17, 2013 6:18 am
hemant_rajput wrote: Thanks for the response. I wasn't able to understand how to you come up the magic equation. Can you please elaborate that?
When B runs the final leg of the race:
The additional distance traveled by B = 24 meters.
The additional distance between B and C = 4 METERS.

The values above imply the following RATIO:
(distance traveled by B) : (increase in the distance between B and C) = 24:4 = 6:1.
In other words, FOR EVERY 6 METERS traveled by B, the distance between B and C INCREASES BY 1 METER.

Let x = the total distance of the race.
When B finishes the race -- having traveled a total of x meters -- the distance between B and C = 16 meters.
Since EVERY 6 METERS traveled by B yields a 1-METER DIFFERENCE between B and C, we get the following proportion:
6/1 = x/16
x = 96.
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