Three runners A, B and C run a race with A finishing 20 m ahead of B and 34 m of C, while B finishes 21 m ahead of C. Each runner travels the entire distance at a constant speed. What was the length of the race?
A) 40M
B) 50M
C) 60M
D) 80M
E) 90M
The OA is C.
I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
Three runners A, B and C run a race with A finishing...
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Let's start with a technical approach:
Distance = Rate * Time
We'll call t A's time to finish the race, and a, b, c the rates of the three runners. We know that
D = a * t
D - 20 = b * t
D - 34 = c * t
Now let's look at what happens after A finishes. Let's say that the time between A's finishing the race and B's finishing the race = v. After A finishes, B goes from being 14 meters ahead of C to finishing 21 meters ahead of C. B also had 20 meters left in the race, so
20 = b * v
and
7 = (b - c) * v
or
c = (13/20)b
Plugging c = (13/20)b into our two equations from the beginning (D - 20 = b * t and D - 34 = c * t), we get
D - 20 = b*t
D - 34 = (13/20)b*t
D = b*t + 20
D = (13/20)b*t + 34
b*t + 20 = (13/20)b*t + 34
(7/20)b*t = 14
b*t = 40
We want to solve for D, so plugging b*t = 40 back into D - 20 = b*t, we get D - 20 = 40, or D = 60, and we're done.
Distance = Rate * Time
We'll call t A's time to finish the race, and a, b, c the rates of the three runners. We know that
D = a * t
D - 20 = b * t
D - 34 = c * t
Now let's look at what happens after A finishes. Let's say that the time between A's finishing the race and B's finishing the race = v. After A finishes, B goes from being 14 meters ahead of C to finishing 21 meters ahead of C. B also had 20 meters left in the race, so
20 = b * v
and
7 = (b - c) * v
or
c = (13/20)b
Plugging c = (13/20)b into our two equations from the beginning (D - 20 = b * t and D - 34 = c * t), we get
D - 20 = b*t
D - 34 = (13/20)b*t
D = b*t + 20
D = (13/20)b*t + 34
b*t + 20 = (13/20)b*t + 34
(7/20)b*t = 14
b*t = 40
We want to solve for D, so plugging b*t = 40 back into D - 20 = b*t, we get D - 20 = 40, or D = 60, and we're done.
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An approach with less algebra would go something like this:
When A finishes the race, we know
A has run D meters
B has run D - 20 meters
C has run D - 34 meters
At this point, B has 20 meters left to go and C has 34 meters left to go.
When B finishes the race, we know C still has 21 meters left to go.
So in the time that passes after A finishes, B runs 20 meters and C runs (34 - 21) meters, or 13 meters. (C was 34 meters from the finish, then was 21 meters from the finish, so (s)he must've run 13 meters.) That means that C's rate = 13/20 of B's.
Now let's return to the first part of the race. We know Rate = Distance / Time, so we can say that
B's rate = (D - 20) / (time it takes A to finish)
C's rate = (D - 34) / (time it takes A to finish)
So B's rate / C's rate = (D - 20) / (D - 34).
But we also just learned that B's rate / C's rate = 20/13! So we can set these equal to each other:
(D - 20)/(D - 34) = 20/13
13 * (D - 20) = 20 * (D - 34)
13D - 260 = 20D - 680
420 = 7D
60 = D
When A finishes the race, we know
A has run D meters
B has run D - 20 meters
C has run D - 34 meters
At this point, B has 20 meters left to go and C has 34 meters left to go.
When B finishes the race, we know C still has 21 meters left to go.
So in the time that passes after A finishes, B runs 20 meters and C runs (34 - 21) meters, or 13 meters. (C was 34 meters from the finish, then was 21 meters from the finish, so (s)he must've run 13 meters.) That means that C's rate = 13/20 of B's.
Now let's return to the first part of the race. We know Rate = Distance / Time, so we can say that
B's rate = (D - 20) / (time it takes A to finish)
C's rate = (D - 34) / (time it takes A to finish)
So B's rate / C's rate = (D - 20) / (D - 34).
But we also just learned that B's rate / C's rate = 20/13! So we can set these equal to each other:
(D - 20)/(D - 34) = 20/13
13 * (D - 20) = 20 * (D - 34)
13D - 260 = 20D - 680
420 = 7D
60 = D
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One last approach that's slow but almost foolproof is backsolving. I'll use two answers to show what the right answer should look like and what a wrong answer could look like.
Suppose I try answer A, and make the race an hour long. If the race is 40 meters long, then A runs 40 meters, B runs 20, and C runs 6. That means B's rate = 20 meters/hr and C's rate = 6 meters/hr. By those numbers, B will finish the race in another hour, but after another hour C will only run 6 more meters, meaning that C has 28 meters to go. We need C to have 21 meters to go, so A is not the right answer.
Now consider answer C, and again make the race an hour long. If the race is 60 meters long, then A runs 60 meters, B runs 40, and C runs 26. Since B's rate = 40 meters per hour, B will take half an hour to finish. In that time, C will run 26 * 0.5, or 13 meters, meaning that C has 21 meters to go. (C has run 26 + 13, and the race is 60 in all: 60 - (26 + 13) = 21.) This is what we want, given the prompt, so C is the right answer.
Suppose I try answer A, and make the race an hour long. If the race is 40 meters long, then A runs 40 meters, B runs 20, and C runs 6. That means B's rate = 20 meters/hr and C's rate = 6 meters/hr. By those numbers, B will finish the race in another hour, but after another hour C will only run 6 more meters, meaning that C has 28 meters to go. We need C to have 21 meters to go, so A is not the right answer.
Now consider answer C, and again make the race an hour long. If the race is 60 meters long, then A runs 60 meters, B runs 40, and C runs 26. Since B's rate = 40 meters per hour, B will take half an hour to finish. In that time, C will run 26 * 0.5, or 13 meters, meaning that C has 21 meters to go. (C has run 26 + 13, and the race is 60 in all: 60 - (26 + 13) = 21.) This is what we want, given the prompt, so C is the right answer.
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We can PLUG IN THE ANSWERS, which represent the length of the race.LUANDATO wrote:Three runners A, B and C run a race with A finishing 20 m ahead of B and 34 m of C, while B finishes 21 m ahead of C. Each runner travels the entire distance at a constant speed. What was the length of the race?
A) 40M
B) 50M
C) 60M
D) 80M
E) 90M
Note the following:
A/C = A/B * B/C.
In the equation above, the values in green CANCEL OUT.
B: 50 meters
Since A finishes 34 meters ahead of C, in the time it takes A to travel the entire 50 meters, the distance traveled by C = 50-34 = 16 meters.
Thus:
A/C = 50/16 = 25/8.
Since A finishes 20 meters ahead of B, in the time it takes A to travel the entire 50 meters, the distance traveled by B = 50-20 = 30 meters.
Thus:
A/B = 50/30 = 5/3.
Since B finishes 21 meters ahead of C, in the time it takes B to travel the entire 50 meters, the distance traveled by C = 50-21 = 29 meters.
Thus:
B/C = 50/29.
Multiplying the ratios for A/B and B/C, we get:
A/C = A/B * B/C = (5/3)(50/29) = 250/87.
Doesn't work:
The red ratio for A/C is LESS than the blue ratio for A/C, since the blue ratio = 25/8 = 250/80.
Eliminate B.
D: 80 meters
Since A finishes 34 meters ahead of C, in the time it takes A to travel the entire 80 meters, the distance traveled by C = 80-34 = 46 meters.
Thus:
A/C = 80/46 = 40/23.
Since A finishes 20 meters ahead of B, in the time it takes A to travel the entire 80 meters, the distance traveled by B = 80-20 = 60 meters.
Thus:
A/B = 80/60 = 4/3.
Since B finishes 21 meters ahead of C, in the time it takes B to travel the entire 80 meters, the distance traveled by C = 80-21 = 59 meters.
Thus:
B/C = 80/59.
Multiplying the ratios for A/B and B/C, we get:
A/C = A/B * B/C = (4/3)(80/59) = 320/177.
Doesn't work:
The red ratio for A/C is GREATER than the blue ratio for A/C, since the blue ratio = 40/23 = 320/184.
Eliminate D.
Notice the PATTERN:
B --> The red ratio for A/C is LESS than the blue ratio for A/C.
D --> The red ratio for A/C is GREATER than the blue ratio for A/C.
Implication:
For the red ratio to be EQUAL to the blue ratio, the correct distance must be BETWEEN B AND D.
The correct answer is C.
Answer choice C: 60
Since A must finish 34 meters ahead of C, in the time it takes A to travel the entire 60 meters, the distance traveled by C = 60-34 = 26 meters.
Thus:
A/C = 60/26= 30/13.
Since A finishes 20 meters ahead of B, in the time it takes A to travel the entire 60 meters, the distance traveled by B = 60-20 = 40 meters.
Thus:
A/B = 60/40 = 3/2.
Since B finishes 21 meters ahead of C, in the time it takes B to travel the entire 60 meters, the distance traveled by C = 60-21 = 39 meters.
B/C = 60/39 = 20/13.
Multiplying the ratios for A/B and B/C, we get:
A/C = A/B * B/C = (3/2)(20/13) = 30/13.
Success!
The red ratio for A/C is EQUAL to the blue ratio for A/C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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