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Time consuming rate problem

by katejohn7 » Sat Feb 27, 2010 9:56 pm
I find the following rate question very time consuming. I would appreciate any tips!

"Ally, Betty and Cathy are taking turns driving in a car from town a to town b, which is 100 miles apart. They all drive at constant rates. Ally covers a certain distance in half the time Cathy does, and Cathy can cover the same distance in twice the time Betty does. If Cathy drives for 20 miles and the total time taken for the journey is 2 hours, what is Ally's speed?

A: 15
B: 30
C: 45
D: 60
E: It can't be determined from the information given.

D
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by ajith » Sat Feb 27, 2010 10:23 pm
katejohn7 wrote:I find the following rate question very time consuming. I would appreciate any tips!

"Ally, Betty and Cathy are taking turns driving in a car from town a to town b, which is 100 miles apart. They all drive at constant rates. Ally covers a certain distance in half the time Cathy does, and Cathy can cover the same distance in twice the time Betty does. If Cathy drives for 20 miles and the total time taken for the journey is 2 hours, what is Ally's speed?

A: 15
B: 30
C: 45
D: 60
E: It can't be determined from the information given.

D
Say Ally's Speed is A, Betty's B and Cathy's C

A = 2C (Ally covers a certain distance in half the time Cathy does, so Ally's speed must be double that of Cathy's )
B = 2C (Cathy can cover the same distance in twice the time Betty does, Cathy's speed is half that of Betty)

Ally's and Betty's speed are the same, so it doesnt matter who drives (both drive at same speed)

Time taken for Cathy to drive 20 miles = Time taken for others to drive 40 miles = say, k
k+2k = 2*60 = 120 [ together they drove 100 miles, 80 miles with twice the speed and 20 miles with Cathy's speed)
k = 40

Ally covers 40 miles in k minutes
40 miles 40 minutes

Her speed = 1mile/minute = 60 mile per hour
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by harshavardhanc » Sat Feb 27, 2010 11:53 pm
katejohn7 wrote:I find the following rate question very time consuming. I would appreciate any tips!

"Ally, Betty and Cathy are taking turns driving in a car from town a to town b, which is 100 miles apart. They all drive at constant rates. Ally covers a certain distance in half the time Cathy does, and Cathy can cover the same distance in twice the time Betty does. If Cathy drives for 20 miles and the total time taken for the journey is 2 hours, what is Ally's speed?

A: 15
B: 30
C: 45
D: 60
E: It can't be determined from the information given.

D
observe that A & B each cover twice the distance as C does in same time. ----------------- (I)

From the question, as C has already covered 20 miles of the total 100 , remaining is 80 which will be split equally between A & B (their rates are equal). Therefore, A & B each cover 40 miles.

and using (I), we now know that A & B will each drive for a time equal to C. ( 40 = 2*20)

Hence, total 2 hours of journey is equally split in 3 parts, each part being 2/3 hr.

therefore, speed of A = 40 / (2/3) = 60 :)
Regards,
Harsha
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by kstv » Sun Feb 28, 2010 6:09 am
Cathy drives 20 miles Starting with this info.
Lets say it takes Cathy 2x mins to drive 20 miles.
Ally covers a certain distance in half the time Cathy does, no harm as taking this distance as 20 miles. So Ally takes x mins to cover 20 miles.
Cathy can cover the same distance in twice the time Betty does. So Betty takes x min to cover 20 miles.
Ally and Betty have same speed and so will cover the remaining 80 miles divided equally among them and take 2x mins each in doing so
6x = 2 hrs, x= 1/3 hrs
Speed of Ally = 20 * 3 = 60 miles per hrs
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by sarthak » Sun Feb 28, 2010 11:38 pm
hi all,
I have a doubt how do you know apart from cathy other two girls drove 40miles each. This is not stated in the problem. I understand it says they took turns in driving, but it never says they drove for equal amount of time each. They could have drove equal miles or might not be equal at all. Can anyone please explain their reasoning behind this assumption.
thanking you
regards
Sarthak
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by sairamGmat » Mon Mar 01, 2010 4:10 am
I agree with Sarthak..All the answers stated above are towards deriving the value 60. There is no statement that Cathy travelled FIRST 20 miles and she stops after 20 miles so that whole REMAINING distance will be shared by remaining two for running. That means will Cathy stop running after 20 miles?

Simple reasoning i can think of is -> Cathy is travelling at 20 miles in 2 hrs...so, her speed in 10 mph.
So, the speeds of Ally and Bett should be 20 mph...

This is because problem ONLY stated, they run at Constant speed.

Going to backsolving, if A runs at 60 mph, how can it be like running in the half time that CAthy runs?

Cathy runs in 2hrs... so, with 60 mph, will Ally run in 1/3 hrs => is that half the time that cathy runs????

IMO, Either the given answer should be wrong...or some info in the question is missing
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by sarthak » Mon Mar 01, 2010 5:45 am
agreeing with sairam IMO is E
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by harshavardhanc » Mon Mar 01, 2010 9:22 am
sairamGmat wrote:I agree with Sarthak..All the answers stated above are towards deriving the value 60. There is no statement that Cathy travelled FIRST 20 miles and she stops after 20 miles so that whole REMAINING distance will be shared by remaining two for running. That means will Cathy stop running after 20 miles?

Simple reasoning i can think of is -> Cathy is travelling at 20 miles in 2 hrs...so, her speed in 10 mph.
So, the speeds of Ally and Bett should be 20 mph...

This is because problem ONLY stated, they run at Constant speed.

Going to backsolving, if A runs at 60 mph, how can it be like running in the half time that CAthy runs?

Cathy runs in 2hrs... so, with 60 mph, will Ally run in 1/3 hrs => is that half the time that cathy runs????

IMO, Either the given answer should be wrong...or some info in the question is missing
I'm really sorry Sairam and Sarthak if you did not understand the solution above and found it as backtracking. I assumed that a person reading my solution would have done first few steps of understanding the problem completely and coming up with some equations.

Anyway, if you see the problem again, there is no indication when did they switch the turn? In other words, for how long did each one drove.

In such a condition, it will be the case that they will drive for a time proportional to the distance they cover. Now, problem boils down : "how much distance did each travel ?"

We already know C has traveled 20 miles of the total. So the remaining is 80.

Also, the question says :

about relation between A & C :

A travels X miles in 1/2 the time taken by C ,

i.e A travels 2X miles in taken by C.

or in same time C will cover half the distance as A-------------------- (i)


similarly, about B & C:

it is given C can cover the same distance in twice the time as B does.

or in same time C will cover half the distance as B. ---- ------ (ii)


Hence, from (i) and (ii), in same time C will cover half the distance as A & B. (observe that rates of A & B are same)


So, time remaining same, the ratio of distances covered by A, B and C is 2:2:1.


or if C covers 20 miles, the rest 80 would have to be split equally among A & B to satisfy the rates constraint.

or when in C's it covers 20 miles, A's turn would last for 40 miles and B's for another forty.

which will be done in a time equal to 2/3hr.


Phew!!! that was something!


Hope this is clear now!
Regards,
Harsha
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by sairamGmat » Mon Mar 01, 2010 10:32 am
Hi Harsha and all,

It was my mistake..I did not read the first sentence of the question very well...

It clearly states that "Ally, Betty and Cathy are taking turns driving in a car from town a to town b, which is 100 miles apart. This implies that they are travelling together 100 miles. All my doubts solved...

Sorry for my exuberant brain...
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by RJ43 » Mon Mar 01, 2010 4:08 pm
Is this an acceptable solution? :

A = Rate Ally
B = Rate Betty
C= Rate Cathy

From the info given, we can see that:

A = 2C
B = 2C

Also, we know the average rate for the trip in total is:

100 miles / 2hr = 50mph therefore:

A + B + C / 3 = 50 mph OR 2C + 2C + C / 3 = 50

5C / 3 = 50
C = 30

Find Ally's speed = A = 2C = 60 MPH D
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by Stuart@KaplanGMAT » Mon Mar 01, 2010 4:27 pm
harshavardhanc wrote: observe that A & B each cover twice the distance as C does in same time. ----------------- (I)

From the question, as C has already covered 20 miles of the total 100 , remaining is 80 which will be split equally between A & B (their rates are equal). Therefore, A & B each cover 40 miles.

and using (I), we now know that A & B will each drive for a time equal to C. ( 40 = 2*20)

Hence, total 2 hours of journey is equally split in 3 parts, each part being 2/3 hr.

therefore, speed of A = 40 / (2/3) = 60 :)
Small correction: we don't know that A, B and C each drive the same distance. However, that's irrelevant.

What we do know, as ajith points out, is that A and B travel at the same rate. Since they travel at the same rate, it doesn't matter how much driving each one does.

If C covers 20 miles, then A and B together cover 80 miles. So,

distance(A&B) = 4*distance(C)

Since A and B travel twice as quickly as does C, they cover the same distance in half the time as C does. So, if they're covering 4 times C's distance, it will take them (1/2)(4) = 2 times as long.

(Reasoning it through: if they traveled at the same speed, 4 times distance would take 4 times as much time; double speed, 4 times distance takes 2 times as much time.)

So, our total time is time(C) + 2(time(C)) = 3(time(C)).

Accordingly, we know that C uses up 1/3 of the total time, = 40 minutes. Consequently, A&B use up 2/3 of the total time, = 80 minutes.

Finally, we have:

Rate(A&B) = Distance(A&B)/Time(A&B)

= 80miles/80minutes = 1mile/1minute = 60miles/hour
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by dear_xavier » Sat Mar 06, 2010 12:12 pm
20/r + 80/2r = 2

2r = 60

"and the total time taken for the journey is 2 hours" Made me think it was the total time of Cathy's journey.
Where is this problem from? I hope it would be better worded if it were official.
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by chintudave » Sat Mar 06, 2010 11:19 pm
Two faster ways to answer this problem without the mumbo jumbo!

Common Realization for both the methods

1. From the problem it is clear that Ally and Betty have the same speed. So in my mind, i would eliminate Betty from the equation. i.e. I would make the problem simpler by assuming that there are only 2 people driving, cathy and ally and cathy's speed is half of ally.

2. Now 100 miles are traveled in 2 hours. So the average speed is 50 miles per hour.


Method I - Leveraging the Options
I know that given the data (a. Relationship between the speeds; b. Total distance and time; c. cathy drove 20 miles) i can surely come up with an answer. So E is definitely out.

Also for the average speed to be 50, Ally has to be above 50 and Cathy below 50. The only option above 50 is option D!

I will mark it and move on.


Method II - Substitution
Based on their speed, I know Ally would have gone 40 miles had she driven instead of Cathy. So the total miles driven in that case would have been 80(Ally's original distance) + 40(miles driven by Ally during the time taken by Cathy to drive 20 miles) = 120. So the average speed is 60 (120/2), which corresponds to Ally's speed!


Let me know your thoughts/questions.
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by dear_xavier » Sun Mar 07, 2010 6:19 am
I think that your strategies are fast but only once you get to think them out.
For a problem like this, opting for a solution other than a straightforward algebraic one would be unwise during a test.
Also, I believe that your knowledge of the answer helped you think out the alternative strategies, something that would not happen during a test.
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by arzanr » Sun Mar 21, 2010 1:15 pm
Wow, this was a really tricky problem! I made the mistake of confusing the rate of A, B and C for the actual time taken - Stuart, thanks for the easy to follow explanation!
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