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25 Nov 2022, 11:09
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D
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Difficulty:
35%
(medium)
Question Stats:
75% (01:47) correct
25%
(01:37)
wrong
based on 72
sessions
History
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Not Attempted Yet
Had Frank driven at an average speed of 40 miles per hour, he would have reached his office 10 minutes earlier than he usually did. At what average speed should Frank drive to reach his office 20 minutes earlier than he usually did?
(1) Frank usually takes 5/6 hours to reach his office
(2) The distance to his office is 80/3 miles
(1) Frank usually takes 5/6 hours to reach his office
(2) The distance to his office is 80/3 miles
29 Nov 2022, 22:07
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Statement 1
Frank usually takes 5/6 hours to reach his office
Frank usually takes 5/6 * 60 = 50 mins to reach his office.
On a particular day he took 40 mins, and the average speed was 40. So the we can obtain the distance to his office.
\(\frac{40}{60} * 40 = \frac{80}{3}\)
We need to find the speed at which Frank should drive to reach his office in 30 mins ( 50 - 20).
As distance and time is known to us, we can calculate the speed.
The information is sufficient.
Statement 2
The distance to his office is 80/3 miles
As the distance is given to us, and we know the speed from the question stem, we can find the time Frank took to reach office on that day. Let that time be t.
His usual time is t + 10.
Required time = usual time - 20 = t - 10
As the time and distance is known, we can calculate the speed.
This statement is also sufficient.
Option D
Frank usually takes 5/6 hours to reach his office
Frank usually takes 5/6 * 60 = 50 mins to reach his office.
On a particular day he took 40 mins, and the average speed was 40. So the we can obtain the distance to his office.
\(\frac{40}{60} * 40 = \frac{80}{3}\)
We need to find the speed at which Frank should drive to reach his office in 30 mins ( 50 - 20).
As distance and time is known to us, we can calculate the speed.
The information is sufficient.
Statement 2
The distance to his office is 80/3 miles
As the distance is given to us, and we know the speed from the question stem, we can find the time Frank took to reach office on that day. Let that time be t.
His usual time is t + 10.
Required time = usual time - 20 = t - 10
As the time and distance is known, we can calculate the speed.
This statement is also sufficient.
Option D
29 Nov 2022, 22:46
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Bunuel
Statement One Alone:
\(\Rightarrow\) Frank usually takes 5/6 hours to reach his office
Since we know Frank's usual time, we can calculate the time of his trip when he reaches the office 10 minutes early. Since we know the time and the average rate, we can calculate the distance between his home and his office. Finally, since we know Frank's usual time, we can calculate the travel time which gets Frank to his office 20 minutes earlier than his usual time. Since we know the distance and the time he should travel that distance, we can calculate the average speed that Frank should drive to reach his office 20 minutes earlier than usual.
Remember that in DS questions, we don't actually need to answer the question, we only need to determine whether we have enough information to answer the question. So, we know at this point that statement one alone is sufficient. However, let's go ahead and calculate the average speed which would take Frank to his office 20 minutes earlier than the usual time.
We are told that it usually takes Frank 5/6 hours = 50 minutes to reach the office. We are told that if he drives at 40 mph, he reaches the office 10 minutes earlier, so the trip would take 50 - 10 = 40 minutes = 2/3 hours. Then, the distance between Frank's home and his office is 40 * 2/3 = 80/3 miles. If he wants to reach his office 20 minutes earlier than the usual time, he should cover the distance of 80/3 miles in 50 - 20 = 30 minutes = 1/2 hours. Thus, his average speed should be (80/3)/(1/2) = 160/3 mph.
Statement one alone is sufficient. Eliminate answer choices B, C, and E.
Statement Two Alone:
\(\Rightarrow\) The distance to his office is 80/3 miles
Knowing the distance between home and the office, we can calculate the amount of time it would take to get to the office at 40 mph. Since this is 10 minutes earlier than the usual time, we can subtract 10 minutes = 1/6 hours from this time to calculate the time that gets Frank to his office 20 minutes earlier than the usual time. Since we know the distance between home and the office, and since we know the required time to get to the office 20 minutes earlier, we can calculate the average speed which would enable Frank to do so.
We know at this point that statement two alone is also sufficient and the answer is D, but let's again calculate the average speed as an exercise.
Since the distance to Frank's office is 80/3 miles, it would take him (80/3)/40 = 2/3 hours = 40 minutes to reach his office driving at 40 mph. We are told that this is 10 minutes earlier than his usual time, so in order to reach his office 20 minutes earlier, he should cover the distance of 80/3 miles in 40 - 10 = 30 minutes = 1/2 hours. Thus, his average speed should be (80/3)/(1/2) = 160/3 mph.
Statement two alone is sufficient.
Answer: D
