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It is a 150 mile drive. If John traveled first 100 miles at 50 mph and he also traveled last 100 mile at 50 mph. That means he travelled the whole 150 mile at 50 mph !

Answer should be C


U dont have the speed (ie speed may vary for those 100km) breakup

avg. speed = total distance/ total time
suppose for 100 km he may covered 75mph @ 75mph and another 25miles @ 25mph then his speed is 50mph
or,
100miles covered in 15 miles @ 10mph and another 85 miles @ 170mph then again his speed is 50mph.

also notice such variance is speed may be for either of the 100miles covered in both options.
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==> In the original condition, he travels the total 150 miles by dividing it to two trips of 50 miles each. Hence, since there are 6 variables, E is most likely to be the answer. In order for C to be the answer, there must be a word “constant rate” mentioned.

Therefore, E is the answer.
Answer: E
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Combining both (1) and (2), it can be inferred that the avg speed is 50 mph. I think the answer is C.
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==> In the original condition, he travels the total 150 miles by dividing it to two trips of 50 miles each. Hence, since there are 6 variables, E is most likely to be the answer. In order for C to be the answer, there must be a word “constant rate” mentioned.

Therefore, E is the answer.
Answer: E

1) John traveled the first 100 miles at the rate of 50 miles per hour
2) John traveled the last 100 miles at the rate of 50 miles per hour

Dear MathRevolution,

I do not understand why the correct answer it E.

Booth statements gives us the actual rate and not any average. So we know for sure that in the first 100miles his speed was always 50mph. The same goes for the last 100 miles. Therefore we know for sure that he was driving at 50mph the whole 150 miles trip. Or where did I go wrong?
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the question mentions traveling at rate of 50 mile/hr. It does not say average speed of first 100 miles is 50 mile/hr.
The answer is C. Since first 100 miles he travels at 50 mile/hr and last 100 also at 50 mile/hr, which means that during any time of the trip, he is travelling at 50 mile/hr.

I was about to choose E but then re-read answer and it said rate, not average.
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Combining both (1) and (2), it can be inferred that the avg speed is 50 mph. I think the answer is C.

Let a, b and c be numbers of hours for the first 50 miles, the second 50 miles and the last 50 miles.

If a = 1, b = 1 and c = 1, then the average speed of the trip is 150 / 3 = 50 miles/hr.
If a = 0.5, b = 1.5 and c = 0.5, then the average speed of the trip is 150 / ( 0.5 + 1.5 + 0.5 ) = 150/2.5 = 60 miles/hr.

Since we don't have a unique solution, E is the answer.
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==> In the original condition, he travels the total 150 miles by dividing it to two trips of 50 miles each. Hence, since there are 6 variables, E is most likely to be the answer. In order for C to be the answer, there must be a word “constant rate” mentioned.

Therefore, E is the answer.
Answer: E

1) John traveled the first 100 miles at the rate of 50 miles per hour
2) John traveled the last 100 miles at the rate of 50 miles per hour

Dear MathRevolution,

I do not understand why the correct answer it E.

Booth statements gives us the actual rate and not any average. So we know for sure that in the first 100 miles his speed was always 50 mph. The same goes for the last 100 miles. Therefore we know for sure that he was driving at 50 mph the whole 150 miles trip. Or where did I go wrong?

Since the second 50 miles are overlapped for both conditions. The following is the explanation.

Let a, b and c be numbers of hours for the first 50 miles, the second 50 miles and the last 50 miles.

If a = 1, b = 1 and c = 1, then the average speed of the trip is 150 / 3 = 50 miles/hr.
If a = 0.5, b = 1.5 and c = 0.5, then the average speed of the trip is 150 / ( 0.5 + 1.5 + 0.5 ) = 150/2.5 = 60 miles/hr.

Since we don't have a unique solution, E is the answer.
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Poor quality question;gmat always gives enough clarity on question.

Posted from my mobile device
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==> In the original condition, he travels the total 150 miles by dividing it to two trips of 50 miles each. Hence, since there are 6 variables, E is most likely to be the answer. In order for C to be the answer, there must be a word “constant rate” mentioned.

Therefore, E is the answer.
Answer: E

1) John traveled the first 100 miles at the rate of 50 miles per hour
2) John traveled the last 100 miles at the rate of 50 miles per hour

Dear MathRevolution,

I do not understand why the correct answer it E.

Booth statements gives us the actual rate and not any average. So we know for sure that in the first 100 miles his speed was always 50 mph. The same goes for the last 100 miles. Therefore we know for sure that he was driving at 50 mph the whole 150 miles trip. Or where did I go wrong?

Since the second 50 miles are overlapped for both conditions. The following is the explanation.

Let a, b and c be numbers of hours for the first 50 miles, the second 50 miles and the last 50 miles.

If a = 1, b = 1 and c = 1, then the average speed of the trip is 150 / 3 = 50 miles/hr.
If a = 0.5, b = 1.5 and c = 0.5, then the average speed of the trip is 150 / ( 0.5 + 1.5 + 0.5 ) = 150/2.5 = 60 miles/hr.

Since we don't have a unique solution, E is the answer.

Hello MathRevolution,
Could you please explain in detail? I still did not get why have you chosen different hours in 2 different cases? Also, you had mentioned that if it were a constant speed then maybe we would have gotten the answer so could please explain what's the difference and how exactly would constant term affect this question in detail?
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