## Relative Rates problem

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### Relative Rates problem

by gmattesttaker2 » Sun Jul 29, 2012 10:05 am
Hello,

This question is from P.35 MGMAT Guide 3. I was not clear with the explanation given in the book. I was wondering if you can please assist:

Car X is 40 miles west of Car Y. Both cars are travelling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y doing?

Ans: [spoiler]If r = 30, then Car Y was going 30 miles per hour[/spoiler]

Thanks a lot for your help.

Best Regards,
Sri

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by AbhiJ » Sun Jul 29, 2012 11:35 am
If you draw a diagram with N , S , E, W conventions you would find that X needs to travel 40 miles east to reach Y.
Let speed of Y = V ,
speed of X = 1.5 V,
relative speed of X wrt Y (catch up speed of X) = 0.5V.
Now catch up speed = (distance between X and Y)/time.

0.5 V = 40 miles / (8/3 hour),
so V = 30 miles/hr.

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by GMATGuruNY » Sun Jul 29, 2012 4:18 pm
gmattesttaker2 wrote:Hello,

This question is from P.35 MGMAT Guide 3. I was not clear with the explanation given in the book. I was wondering if you can please assist:

Car X is 40 miles west of Car Y. Both cars are travelling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y doing?

Ans: [spoiler]If r = 30, then Car Y was going 30 miles per hour[/spoiler]

20 30 40 50 60

Best Regards,
Sri
X is BEHIND by 40 miles.
To CATCH UP to Y in 8/3 hours, X must travel 40 more miles than Y in 8/3 hours:
40/(8/3) = 15.
In other words, X must travel 15 miles per hour FASTER than Y.
X's rate - Y's rate = 15.

From here, we could plug in the answers, which an actual GMAT question would include.
20 30 40 50 60
Since X travels 50% faster, X's rate must be one of the following:
20 + .5(20) = 30.
30 + .5(30) = 45.
40 + .5(40) = 60.
50 + .5(50) = 75.
60 + .5(60) = 90.

Only answer choice B yields the required difference of 15 miles per hour:
X's rate - Y's rate = 45-30 = 15.

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by Brent@GMATPrepNow » Mon Jul 30, 2012 6:54 am
gmattesttaker2 wrote:Car X is 40 miles west of Car Y. Both cars are travelling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y doing?
Let's let Car X's original position be the initial starting point. So, when Car X is at the initial starting point, Car Y has already traveled 40 miles.

My "word" equation involves the conditions when Car X catches up to Car Y. At that point, we can say:
Car X's total distance = Car Y's total distance

Car Y's total distance
Let V = Car Y's speed (our goal is to find the value of V)
From the time that Car X begins moving, Car Y drives for 2 2/3 hours (2 hours, 40 minutes).
So Car Y's total distance = (time)(rate) = (2 2/3)(V) + 40 miles

Car X's total distance
We know that Car X is going 50% faster than Car Y. If Car Y's rate is V, then Car X's rate must be 1.5V
We also know that Car X travels for 2 2/3 hours.
So Car X's total distance = (time)(rate) = (2 2/3)(1.5V) miles
Simplify: (2 2/3)(1.5V) = (8/3)(3/2) = 4V

We're now ready to write our algebraic equation.
Car X's total distance = Car Y's total distance
4V = (2 2/3)(V) + 40 miles
4/3V = 40
V = 40(3/4)
V=30

Cheers,
Brent[spoiler][/spoiler]
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by gmattesttaker2 » Fri Aug 03, 2012 9:49 pm
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:Car X is 40 miles west of Car Y. Both cars are travelling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y doing?
Let's let Car X's original position be the initial starting point. So, when Car X is at the initial starting point, Car Y has already traveled 40 miles.

My "word" equation involves the conditions when Car X catches up to Car Y. At that point, we can say:
Car X's total distance = Car Y's total distance

Car Y's total distance
Let V = Car Y's speed (our goal is to find the value of V)
From the time that Car X begins moving, Car Y drives for 2 2/3 hours (2 hours, 40 minutes).
So Car Y's total distance = (time)(rate) = (2 2/3)(V) + 40 miles

Car X's total distance
We know that Car X is going 50% faster than Car Y. If Car Y's rate is V, then Car X's rate must be 1.5V
We also know that Car X travels for 2 2/3 hours.
So Car X's total distance = (time)(rate) = (2 2/3)(1.5V) miles
Simplify: (2 2/3)(1.5V) = (8/3)(3/2) = 4V

We're now ready to write our algebraic equation.
Car X's total distance = Car Y's total distance
4V = (2 2/3)(V) + 40 miles
4/3V = 40
V = 40(3/4)
V=30

Cheers,
Brent[spoiler][/spoiler]

Hello Brent,

Thank you very much for your detailed and thorough explanation (as always!). It is very clear now. Thank you so much again.

Best Regards,
Sri

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by gmattesttaker2 » Fri Sep 13, 2013 6:58 pm
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:Car X is 40 miles west of Car Y. Both cars are travelling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y doing?
Let's let Car X's original position be the initial starting point. So, when Car X is at the initial starting point, Car Y has already traveled 40 miles.

My "word" equation involves the conditions when Car X catches up to Car Y. At that point, we can say:
Car X's total distance = Car Y's total distance

Car Y's total distance
Let V = Car Y's speed (our goal is to find the value of V)
From the time that Car X begins moving, Car Y drives for 2 2/3 hours (2 hours, 40 minutes).
So Car Y's total distance = (time)(rate) = (2 2/3)(V) + 40 miles

Car X's total distance
We know that Car X is going 50% faster than Car Y. If Car Y's rate is V, then Car X's rate must be 1.5V
We also know that Car X travels for 2 2/3 hours.
So Car X's total distance = (time)(rate) = (2 2/3)(1.5V) miles
Simplify: (2 2/3)(1.5V) = (8/3)(3/2) = 4V

We're now ready to write our algebraic equation.
Car X's total distance = Car Y's total distance
4V = (2 2/3)(V) + 40 miles
4/3V = 40
V = 40(3/4)
V=30

Cheers,
Brent[spoiler][/spoiler]

Hello Brent,

I was trying to solve the following problem with a similar approach:

Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.
(2) Abernathy is driving 25 meters per minute faster than Berdoff.

OA: B

My approach was:

At time t, distance covered by Abernathy = Distance covered by Berdoff

1) In-sufficient

2) Let, Abernathy's rate be x met/min
So Berdoff's rate is x + 25 met/min

Time t is when the distances covered will be equal.

So at time t, xt = (x+25)t + 200 (Since Berdoff was already 200 met. ahead of Abernathy)

I think I am making a mistake here since the above has a negative answer.

Best Regards,
Sri

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by Brent@GMATPrepNow » Sat Sep 14, 2013 6:01 am
gmattesttaker2 wrote: Hello Brent,

I was trying to solve the following problem with a similar approach:

Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.
(2) Abernathy is driving 25 meters per minute faster than Berdoff.

OA: B

My approach was:

At time t, distance covered by Abernathy = Distance covered by Berdoff

1) In-sufficient

2) Let, Abernathy's rate be x met/min
So Berdoff's rate is x + 25 met/min

Time t is when the distances covered will be equal.

So at time t, xt = (x+25)t + 200 (Since Berdoff was already 200 met. ahead of Abernathy)

I think I am making a mistake here since the above has a negative answer.

Best Regards,
Sri
Hey Sri,

First let's examine a quick way to tackle statement 2.
If Abernathy is driving 25 meters per minute faster than Berdoff, then the GAP between Abernathy and Berdoff will decrease by 25 meters EVERY MINUTE.
There is currently as 200-meter gap
So, it will take 8 seconds to decrease the gap from 200 meters to 0 meters.

Now let's take a look at your solution.
You're equation is almost perfect.
To see the small error, let's begin with a "word equation"
By the time Abernathy catches up to Berdoff, we know that Abernathy will have traveled 200 meters So, we can write: Abernathy's distance = Berdoff's distance + 200 You had added the 200 to the wrong side of the equation. Tip: before adding or subtracting anything, take a look at the two distances. We know that, in order for Abernathy to reach Berdoff, Abernathy must travel 200 meters more. So, Abernathy's distance is GREATER than Berdoff's distance. So, how do we make those distances equal? We can either subtract 200 from Abernathy's distance or we can add 200 to Berdoff's distance
So, (x+25)t = xt + 200
When you solve for t, you get 8 seconds

Cheers,
Brent
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by gmattesttaker2 » Sat Sep 14, 2013 4:08 pm
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote: Hello Brent,

I was trying to solve the following problem with a similar approach:

Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.
(2) Abernathy is driving 25 meters per minute faster than Berdoff.

OA: B

My approach was:

At time t, distance covered by Abernathy = Distance covered by Berdoff

1) In-sufficient

2) Let, Abernathy's rate be x met/min
So Berdoff's rate is x + 25 met/min

Time t is when the distances covered will be equal.

So at time t, xt = (x+25)t + 200 (Since Berdoff was already 200 met. ahead of Abernathy)

I think I am making a mistake here since the above has a negative answer.

Best Regards,
Sri
Hey Sri,

First let's examine a quick way to tackle statement 2.
If Abernathy is driving 25 meters per minute faster than Berdoff, then the GAP between Abernathy and Berdoff will decrease by 25 meters EVERY MINUTE.
There is currently as 200-meter gap
So, it will take 8 seconds to decrease the gap from 200 meters to 0 meters.

Now let's take a look at your solution.
You're equation is almost perfect.
To see the small error, let's begin with a "word equation"
By the time Abernathy catches up to Berdoff, we know that Abernathy will have traveled 200 meters So, we can write: Abernathy's distance = Berdoff's distance + 200 You had added the 200 to the wrong side of the equation. Tip: before adding or subtracting anything, take a look at the two distances. We know that, in order for Abernathy to reach Berdoff, Abernathy must travel 200 meters more. So, Abernathy's distance is GREATER than Berdoff's distance. So, how do we make those distances equal? We can either subtract 200 from Abernathy's distance or we can add 200 to Berdoff's distance
So, (x+25)t = xt + 200
When you solve for t, you get 8 seconds

Cheers,
Brent

Hello Brent,

Thank you very much for your excellent and detailed explanation.

Best Regards,
Sri

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