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Mr. Alex usually starts at 9:00am and reaches his office

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Mr. Alex usually starts at 9:00am and reaches his office

by AAPL » Tue Jul 31, 2018 6:25 am

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Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual 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 reached 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.

The OA is A.

from 1:

Let x be the speed and t be the time.
as distance traveled is equal
x*t=.8x(t+1/3)
on solving gives
t=4/3 or 80 minutes

Now for the question
as again distance traveled is equal
1.25x*t1=4/3*x

t1=64 minutes. Therefore if he started at 9:30 he did not reach on time. As he had to reach by 10:20

2) does not give us an equation.

Therefore, the correct answer is A.

Has anyone another strategic approach to solving this DS question? Regards!
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Source: — Data Sufficiency |
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mr. alex

by GMATGuruNY » Tue Jul 31, 2018 8:16 am

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AAPL wrote:Mr. Alex usually starts at 9:00am and reaches his office just in time, driving at his regular speed. Last Wednesday, he started at 9:30am and drove 25% faster than his usual 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 reached 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.

Since 20% slower = 1/5 slower and 25% faster = 1/4 faster, let the usual rate = the LCM of the two denominators = 5*4 = 20 feet per minute.
Let the usual time = t.
Thus, the distance to work = rt = 20t feet.

Statement 1:
Since Mr. Alex leaves 20 minutes early last Monday -- giving himself an additional 20 minutes -- the time last Monday = t + 20 minutes.
Since Mr. Alex drives 20% slower, the rate last Monday = 20 - (20% of 20) = 16 feet per minute.
Thus, the distance traveled last Monday = 16(t + 20) = 16t + 320 feet.
Since the two blue distances must be THE SAME, we get:
20t = 16t + 320
4t = 320
t = 80 minutes.
Thus, the distance to work = (usual rate)(usual time) = 20*80= 1600 feet.

Last Wednesday, he started at 9:30am and drove 25% faster than his usual speed. Did he reach the office in time?
9:30am is 30 minutes later than the normal departure time of 9am.
Since Mr. Alex leaves 30 minutes late, the time last Wednesday = (usual time) - (30 minutes) = 80-30 = 50 minutes.
To arrive at his office in time, Mr. Alex must travel 1600 feet in 50 minutes
Since he drives 25% faster, the rate last Wednesday= 20 + (25% of 20) = 25 feet per minute.
Distance traveled last Wednesday = rt = 25*50 = 1250 feet.
Since the distance traveled is too small, Mr. Alex does not arrive at work in time.
SUFFICIENT.

Statement 2:
Since he started 10 minutes early and reached his office 10 minutes early, Mr. Alex traveled at his usual speed last Tuesday.
No new information about last Wednesday.
INSUFFICIENT.

The correct answer is A.
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