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# On a recent trip, Mary drove 50 miles. What was the average

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On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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20 Aug 2012, 02:18
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On a recent trip, Mary drove 50 miles. What was the average speed at which she drove the 50 miles?

(1) She drove 30 miles at an average speed of 60 miles per hour and then drove the remaining 20 miles at an average speed of 50 miles per hour.
(2) She drove a total of 54 minutes.

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Difficulty: 550
[Reveal] Spoiler: OA

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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20 Aug 2012, 02:18
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On a recent trip, Mary drove 50 miles. What was the average speed at which she drove the 50 miles?

$$average \ speed=\frac{total \ distance}{total \ time}$$ --> $$average \ speed=\frac{50}{total \ time}$$. So, as we can see all we need to find is the total time spent on this trip.

(1) She drove 30 miles at an average speed of 60 miles per hour and then drove the remaining 20 miles at an average speed of 50 miles per hour. $$total \ time=\frac{30}{60}+\frac{20}{50}$$. Sufficient.

(2) She drove a total of 54 minutes. Directly gives us the total time spent. Sufficient.

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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21 Aug 2012, 10:19
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D

Avg Speed = Total Distance / Total Time

1) Total Distance = 50

Total time = (30/60) + (20/50)

2) Can be directly calculated

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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22 Aug 2012, 04:54
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Both statements are individually sufficient.

To answer the question, we need to know the total distance and time taken. Then it becomes simple average question.

Statement 2 directly gives the time taken thus average is distance traveled/ average time taken= 50/54= 55 something.

Statement 1, we need to either logically solve it or do a rtd chart;

Rate Time Distance
First part 60 1/2 hr 30km
Second part 50 0.4 hr 20 km

Thus first part is covered in 30 minutes and second in 24 minutes. Total 54 minutes. From here on, we can calculate as has been done for statement 2.

Bandesha1 wrote:
D

Avg Speed = Total Distance / Total Time

1) Total Distance = 50

Total time = (30/60) + (20/50)

2) Can be directly calculated

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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22 Aug 2012, 05:00
Another way to look at it is to use "weighted average"

Average speed= [(Speed 1x distance 1)+ (Speed 2x distance 2)]/(distance 1+distance 2)

60x30+50x20/30+20= 2800/50= 56min

manjeet1972 wrote:
Level 600

Both statements are individually sufficient.

To answer the question, we need to know the total distance and time taken. Then it becomes simple average question.

Statement 2 directly gives the time taken thus average is distance traveled/ average time taken= 50/54= 55 something.

Statement 1, we need to either logically solve it or do a rtd chart;

Rate Time Distance
First part 60 1/2 hr 30km
Second part 50 0.4 hr 20 km

Thus first part is covered in 30 minutes and second in 24 minutes. Total 54 minutes. From here on, we can calculate as has been done for statement 2.

Bandesha1 wrote:
D

Avg Speed = Total Distance / Total Time

1) Total Distance = 50

Total time = (30/60) + (20/50)

2) Can be directly calculated

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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01 Aug 2013, 13:24
On a recent trip, Mary drove 50 miles. What was the average speed at which she drove the 50 miles?

(1) She drove 30 miles at an average speed of 60 miles per hour and then drove the remaining 20 miles at an average speed of 50 miles per hour.

Total time taken = (distance1/rate1) + (distance2/rate2)
Total time taken = (30/60) + (20/50)
Total time taken = (1/2) + (2/5)
Total time taken = 9/10ths of an hour.
SUFFICIENT

(2) She drove a total of 54 minutes.
Rate = distance/time
Rate = 50/54
Rate = .92 miles/minute
Average speed = distance/time
Average speed = 50/.92 hours
Average speed = 54 miles/hour
SUFFICIENT

(D)

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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18 Sep 2013, 01:58
Hello Bunuel,

My initial thought was to choose D since I had the same solution process which you described.
But then I thought to myself that no where in the question or the statements was it mentioned that Mary drove at a constant rate.

I guess I got confused between the types of questions where I can use the total distance/ total time to get the average speed, and others where I have to compute each trip on its own to get the total average.

I'm not sure if you get my point on this? because I faced several problems where if I simply use total distance and total time, it would yield a wrong answer. Please help me to clarify this misunderstanding

Thanks
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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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18 Sep 2013, 02:37
SaraLotfy wrote:
Hello Bunuel,

My initial thought was to choose D since I had the same solution process which you described.
But then I thought to myself that no where in the question or the statements was it mentioned that Mary drove at a constant rate.

I guess I got confused between the types of questions where I can use the total distance/ total time to get the average speed, and others where I have to compute each trip on its own to get the total average.

I'm not sure if you get my point on this? because I faced several problems where if I simply use total distance and total time, it would yield a wrong answer. Please help me to clarify this misunderstanding

Thanks

The question asks: what was the average speed at which she drove the 50 miles?

Now, the average speed equals to (total distance)/(total time) irrespective whether the distance was covered at a constant speed or not.
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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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10 Nov 2013, 07:11
I'm still confused about 2nd option.
Case A:-
Travelled 30 miles in 30 mints. Speed =dist/time = 30/0.5= 60 mph
Travelled 20 miles in 24 mints. Speed = dist/time = 20/(2/5) = 50mph
Avg = 55mph

Case b:-
Travelled 30 miles in 20 mints. Speed = dist/time= 30/(1/3) = 90mph
Travelled 20 miles in 34mints. Speed = dist/time = 20/(34/60) = 35.3mph
Avg = (90+35.3)/2= 62.64mph

Now in both cases we are getting different avg speed. Please comment

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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10 Nov 2013, 10:58
Nilabh_s wrote:
I'm still confused about 2nd option.
Case A:-
Travelled 30 miles in 30 mints. Speed =dist/time = 30/0.5= 60 mph
Travelled 20 miles in 24 mints. Speed = dist/time = 20/(2/5) = 50mph
Avg = 55mph

Case b:-
Travelled 30 miles in 20 mints. Speed = dist/time= 30/(1/3) = 90mph
Travelled 20 miles in 34mints. Speed = dist/time = 20/(34/60) = 35.3mph
Avg = (90+35.3)/2= 62.64mph

Now in both cases we are getting different avg speed. Please comment

You are using info from (1) for (2).

In fact the second statement is very easy. We know that Mary drove 50 miles, and we need to find the average speed at which she drove those 50 miles. (2) says that she drove a total of 54 minutes (0.9 hours), thus $$average \ speed=\frac{total \ distance}{total \ time}=\frac{50}{0.9}$$. Sufficient.

Hope it helps.
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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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10 Nov 2013, 15:27
Hi,
Sorry the numbers I took to prove my doubt for #2 may have co-incidently matched info in #1. My doubt is also simple if we just know the total distance travelled & total time taken how can we calculate the avg. speed? The person may have driven different speed at different time and hence avg speed can be different as I have shown above.
In GMAT we get many questions where we need to calculate the avg. speed by taking weighted avg of distance,time AND/OR speed.
Asuuming that speed was "constant" when it is not mentioned so- is it not "assuming" in a DS question?

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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11 Nov 2013, 00:11
Nilabh_s wrote:
Hi,
Sorry the numbers I took to prove my doubt for #2 may have co-incidently matched info in #1. My doubt is also simple if we just know the total distance travelled & total time taken how can we calculate the avg. speed? The person may have driven different speed at different time and hence avg speed can be different as I have shown above.
In GMAT we get many questions where we need to calculate the avg. speed by taking weighted avg of distance,time AND/OR speed.
Asuuming that speed was "constant" when it is not mentioned so- is it not "assuming" in a DS question?

In your previous post yo calculated the average speed incorrectly. Again (average speed)=(total distance)/(total time).
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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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11 Nov 2013, 00:21
Nilabh_s wrote:
Hi,
Sorry the numbers I took to prove my doubt for #2 may have co-incidently matched info in #1. My doubt is also simple if we just know the total distance travelled & total time taken how can we calculate the avg. speed? The person may have driven different speed at different time and hence avg speed can be different as I have shown above.
In GMAT we get many questions where we need to calculate the avg. speed by taking weighted avg of distance,time AND/OR speed.
Asuuming that speed was "constant" when it is not mentioned so- is it not "assuming" in a DS question?

Hi Nilabh,

Since we know that average speed (is always)= total distance/ total time. Now if we have these two information, it is sufficient to arrive at the average speed. It does not matter whether the speed was constant or not if we know these two variables.
The scenario you presented in which we need to take into account the varying speed happens for two reasons-
1)for calculating the total time or 2) total distance, as the case may be.

Here in option(2) we are already provided with total distance and total time. Hence, the variation in speed will not impact the calculation for average speed. We already know the total distance and total time.

For case 1) see we have the total distance but not total time. hence we use weigthage average to calculate the total time and hence average speed.

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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11 Nov 2013, 00:39
hi All,
I get my mistake. There is a calculation mistake in my post.
Thanks I understand it now.

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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30 Jul 2016, 07:48
Bunuel wrote:
On a recent trip, Mary drove 50 miles. What was the average speed at which she drove the 50 miles?

(1) She drove 30 miles at an average speed of 60 miles per hour and then drove the remaining 20 miles at an average speed of 50 miles per hour.
(2) She drove a total of 54 minutes.

We are being tested on average speed. The formula for average speed is:

average speed = total distance/total time

We are given that the total distance is 50 miles, so we can substitute 50 miles into our formula.

average speed = 50/total time

Thus, if we can determine the total time, we can determine the average speed.

Statement One Alone:

She drove 30 miles at an average speed of 60 miles per hour and then drove the remaining 20 miles at an average speed of 50 miles per hour.

We are given that Mary first drove 30 miles at an average speed of 60 mph.

Since time = distance/rate, we can say:

time for the first 30 miles = 30/60 = ½ hour

We are also given that Mary drove the remaining 20 miles at an average speed of 50 miles per hour. Thus, we can say:

time for the remaining 20 miles = 20/50 = 2/5 hours

So we know that total time = ½ + 2/5 = 5/10 + 4/10 = 9/10 hours. Since we have the total time, we can determine average speed. Statement one is sufficient to answer question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

She drove a total of 54 minutes.

We are given the total time, so statement two is also sufficient to answer the question.

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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19 Mar 2017, 13:16
Hi!

How do you know that this is a 650 level question?

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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19 Mar 2017, 23:39
EliMan wrote:
Hi!

How do you know that this is a 650 level question?

Difficulty level (seen in tags) is calculated automatically based on the timer stats from the users which attempted the question. The difficulty level of this question is thus sub-600. As you can see 77% of the users attempting the question answered it correctly spending at average 01:39 minutes.
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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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20 Mar 2017, 04:26
Bunuel wrote:
On a recent trip, Mary drove 50 miles. What was the average speed at which she drove the 50 miles?

(1) She drove 30 miles at an average speed of 60 miles per hour and then drove the remaining 20 miles at an average speed of 50 miles per hour.
(2) She drove a total of 54 minutes.

Practice Questions
Question: 22
Page: 276
Difficulty: 650

To find the answer, we need the total duration for the journey
ST 1: time taken for 30 miles = 30/60 = 0.5 hrs and for 20 miles = 20/50 = 0.4 hrs. total time taken 0.9 hrs. ANSWER
ST 2: total time is 54 minutes. ANSWER

Option D

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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28 Jun 2017, 09:41
Strange occurrence:

I did weighted average method for 1):

1) (0.6*60)+(0.4*50)=56 Suf.
2) if we plug in the data we get: 50=S*9/10 so 500/9=55.55555 Suf.

But the two answers don't exactly match. Is this normal? Should we expect this on the test?

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Re: On a recent trip, Mary drove 50 miles. What was the average [#permalink]

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28 Jun 2017, 10:35
On a recent trip, Mary drove 50 miles. What was the average speed at which she drove the 50 miles?

$$Speed = \frac{Distance}{Time}$$

$$Avg. Speed = \frac{Total Distance}{Total Time}$$

As the question is asking for Average Speed and we are already aware of distance, which means if we get some information on - how much time Mary drove we will be able to calculate the average speed.

Note, only knowing time would be sufficient for us to determine if the sentence is sufficient or not. We need calculate the exact average speed as this is DS question.

(1) She drove 30 miles at an average speed of 60 miles per hour and then drove the remaining 20 miles at an average speed of 50 miles per hour.

As we are given two different speeds for two split of distances, we should be able to calculate the time taken for these two speeds which will be our total time taken.

As we will be able to calculate time, this statement should be sufficient.

Hence, (1) =====> is SUFFICIENT

(2) She drove a total of 54 minutes.

This directly gives us the total time taken.

Once again, this statement should be sufficient.

Hence, (2) =====> is SUFFICIENT

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Re: On a recent trip, Mary drove 50 miles. What was the average   [#permalink] 28 Jun 2017, 10:35

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