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Distance problem - experts please help

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Distance problem - experts please help

by rahulvsd » Wed Aug 10, 2011 7:38 am
Hi Please help me out with this problem, it will be nice if I can get some general tips for solving advanced distance-rate problems like the one below:

Mr David usually starts at 9:00 AM and reaches his office just in time, driving at regular speed. Last wednesday, he started at 9:30 AM and drove 25% faster than his regular speed. Did he reach the office in time?

1) Last monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reaches his office just in time.
2) Last tuesday, he started to his office 10 minutes early and reached the office 10 minutes early driving at his regular speed.

OA: A
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by gmatboost » Wed Aug 10, 2011 8:24 am
The key to every rate question is (in my opinion) to keep using the RT=D equation very carefully.

Another tip for harder questions is to remember that time is usually in HOURS, so when you are told things in MINUTES, you must convert that info to HOURS.

Let's say R is the normal rate of David, and let's say T is the amount of time it normally takes.

Then, the distance to work is D = RT.

Last Wednesday, he started at 9:30. That means he had 1/2 hour less than usual if he wanted to make it in time. So his time in this case should be (T - 1/2).

He drove 25% faster than usual. I prefer to use fractions, so his rate in this case should be (5/4)R. This is because it is 1/4R faster than usual.

So, the distance he covered on Wednesday in the allotted time was:
((5/4)R)(T - 1/2)
We are asked to figure out whether he got to work in this amount of time. Remember that the distance to work is RT.
So, the question is asking us: ((5/4)R)(T - 1/2) >= RT?

We can simplify a lot:
((5/4)R)(T - 1/2) >= RT ?
(5/4)RT - (5/8)R >= RT ?
(5/4)T - (5/8) >= T ? (Divided by R)
(1/4)T - (5/8) >= 0 ? (Subtracted T)
(1/4)T >= 5/8 ? (Subtracted T)
T >= 20/8 ? (Multiplied by 4)
T >= 5/2 ?

So, that is the question we need to answer:
Is his usual time to work greater than or equal to 2.5 hours?

Statement 1:
He left 20 minutes early, which is 1/3 of an hour -> His time is (T + 1/3)
He drove 20% slower -> His rate is (4/5)R, since 1/5R is 20% of his speed

We are told that he reached his office right on time, which means he covered the exact distance RT on Monday:
((4/5)R)(T + 1/3) = RT
(4/5)RT + (4/15)R = RT
(4/5)T + (4/15) = T
(4/15) = (1/5)T
20/15 = T = 4/3

So, since T = 4/3, we know that the answer to the question "Is his usual time to work greater than or equal to 2.5 hours?" is a DEFINITIVE NO. Sufficient.

Statement 2 just says he left 10 minutes early and got there 10 minutes early. No info, insufficient.
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by GmatKiss » Wed Aug 10, 2011 8:46 am
IMO:A
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by rahulvsd » Wed Aug 10, 2011 8:52 am
Thanks a lot Greg
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by GMATGuruNY » Sat Aug 13, 2011 3:22 am
rahulvsd wrote:Hi Please help me out with this problem, it will be nice if I can get some general tips for solving advanced distance-rate problems like the one below:

Mr David usually starts at 9:00 AM and reaches his office just in time, driving at regular speed. Last wednesday, he started at 9:30 AM and drove 25% faster than his regular speed. Did he reach the office in time?

1) Last monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reaches his office just in time.
2) Last tuesday, he started to his office 10 minutes early and reached the office 10 minutes early driving at his regular speed.

OA: A
Another approach.
Plug in easy numbers to determine how the changes in speed affect the time.

Let distance = 400 miles.
Let r = 20 miles per hour.
t = 400/20 = 20 hours.

25% faster:
r = 25 miles per hour.
t = d/r = 400/25 = 16 hours.
When the speed increases by 25%, the time decreases by 20% (from 20 hours to 16 hours).

Applying this logic to the DS question above:
Traveling 25% faster, David needs 20% less time.
He leaves 30 minutes late.
If 30 minutes is not more than 20% of his typical driving time, David will arrive at work on time.
Thus, to arrive on time:
30 ≤ .2t
t ≥ 150.

Question rephrased: Is t ≥ 150 minutes?

Statement 1: Last monday, he started to his office 20 minutes early, drove 20% slower than his regular speed, and reached his office just in time.

Using our example of 400 miles at a typical speed of 20 miles per hour and a typical time of 20 hours:
20% slower = 16 miles per hour.
t = 400/16 = 25 hours.
When the speed decreases by 20%, the time increases by 25% (from 20 to 25 hours).

Thus, statement 1 implies that 20 minutes is 25% of David's typical driving time:
20 = .25t
t = 80 minutes.
Sufficient.

Statement 2: Last Tuesday, he started to his office 10 minutes early and reached the office 10 minutes early driving at his regular speed.
No way to determine the typical driving time.
Insufficient.

The correct answer is A.
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