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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In planning for a trip, Joan estimated both the distance of the trip,

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Math Expert V
Joined: 02 Sep 2009
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In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 65% (02:05) correct 35% (02:03) wrong based on 2383 sessions

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In planning for a trip, Joan estimated both the distance of the trip, in miles, and her average speed, in miles per hour. She accurately divided her estimated distance by her estimated average speed to obtain an estimate for the time, in hours, that the trip would take. Was her estimate within 0.5 hour of the actual time that the trip took?

(1) Joan’s estimate for the distance was within 5 miles of the actual distance.

(2) Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.

Kudos for a correct solution.

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Re: In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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Bunuel wrote:
In planning for a trip, Joan estimated both the distance of the trip, in miles, and her average speed, in miles per hour. She accurately divided her estimated distance by her estimated average speed to obtain an estimate for the time, in hours, that the trip would take. Was her estimate within 0.5 hour of the actual time that the trip took?

(1) Joan’s estimate for the distance was within 5 miles of the actual distance.

(2) Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.

Kudos for a correct solution.

Here, all you need to do is test one case to convince yourself that the answer is (E).

We know that if her actual speed and actual distance are dot on estimated speed and distance, her time taken would be equal to estimated time.
All we have to do is prove that her actual time can vary by more than 0.5 hrs.

Try to increase her time:
Say, estimated distance 10 miles. Actual distance = 15 miles
Estimated speed = 20 mph. Actual speed = 10 mph
Estimated time = 10/20 = 0.5 hrs
Actual time = 15/10 = 1.5 hrs
Her actual time varies by 1 hr.

So both statements together are not sufficient. Answer (E)
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Re: In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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Bunuel wrote:
In planning for a trip, Joan estimated both the distance of the trip, in miles, and her average speed, in miles per hour. She accurately divided her estimated distance by her estimated average speed to obtain an estimate for the time, in hours, that the trip would take. Was her estimate within 0.5 hour of the actual time that the trip took?

(1) Joan’s estimate for the distance was within 5 miles of the actual distance.

(2) Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.

Kudos for a correct solution.

Statement 1:
The distance was original distance + 5 or -5 miles . This statement doesn't provide any information about the average speed.
Even if we take some random values of distance and speed, the estimate of time can be both more or less than 0.5 hours.
INSUFFICIENT

Statement 2:
The average speed was +10 or -10 miles/hr. This statement does not provide any information about distance.
The estimate of time can again be more or less than 0.5 hours on taking random values of distance and speed.
INSUFFICIENT

On combining 1 & 2 , we do not get any additional information or concrete values. And the estimate would again be more or less than 0.5 hours on taking random distance and speed values.

##### General Discussion
Manager  Joined: 10 Aug 2015
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Re: In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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Solution :
Lets take her estimate for distance,d =100 for all calc.

Statement1: This is not sufficient because nothing is given about the relation between her estimation of speed and actual speed.
If you are not satisfied, then lets try by considering her estimate of speed as exactly correct.
Let s = 5. Then t = 100/5 = 20. If actual distance of 95, then t = 95/5 =19. So, no.
Let s =50 . Then t =100/50 = 2. If actual distance of 95, then t = 95/50 =1.9. So, Yes.
Clearly its giving different answers. So, insufficient.

Statement2: This is not sufficient because nothing is given about the relation between her estimation of distance and actual distance.We can prove it in the same way as in statement1.

Combined: If s = 50, then t =100/50=2 and if actual distance is 95 and actual speed is 60, then t = 95/60=1.58.So,yes.
But if s=25,then t=100/25=4 and if actual distance is 95 and actual speed is 35, then t = 95/35=2.7.So,No.
Therefore, clearly not sufficient.

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Re: In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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Was estimate within 0.5 hour of the actual time that the trip took?

(1) Joan’s estimate for the distance was within 5 miles of the actual distance.
only distance is given, no speed

(2) Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.
only speed given, no distance

Combined, we have distance and speed where we can find a rate. However, thinking about it conceptually, we can see that the range of +/- 5 for the distance and +/- 10 for the speed can give us an estimate of within .5hrs or not depending on the extremes.
Insufficient

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Re: In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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1
St1 and St2 individually clearly INSUF

1+2)
10/1= 10hrs 5/11~.5 Hrs -->No (use values at the opposite extremes to get the highest time diff)
10/10=1Hr 15/20=.75Hrs -->Yes (use values at the same extremes to get the lowest time diff)

Ans E
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Re: In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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2
In planning for a trip, Joan estimated both the distance of the trip, in miles, and her average speed, in miles per hour. She accurately divided her estimated distance by her estimated average speed to obtain an estimate for the time, in hours, that the trip would take. Was her estimate within 0.5 hour of the actual time that the trip took?

(1) Joan’s estimate for the distance was within 5 miles of the actual distance.

(2) Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.

In the original condition and the question, it is estimated:v1*t1=d1, actual:v2*t2=d2. There are 2 equations(v1*t1=d1, v2*t2=d2) and 6 variables(v1,v2,t1,t2,d1,d2), which should match with the number of equations. So, you need 4 more equations. For 1) 1 equation, for 2) 1 equation and overall you need 2 more equations, which is likely to make E the answer. In fact, E is the answer.

-> For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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Bunuel wrote:
In planning for a trip, Joan estimated both the distance of the trip, in miles, and her average speed, in miles per hour. She accurately divided her estimated distance by her estimated average speed to obtain an estimate for the time, in hours, that the trip would take. Was her estimate within 0.5 hour of the actual time that the trip took?

(1) Joan’s estimate for the distance was within 5 miles of the actual distance.

(2) Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.

Target question: Was Joan's ESTIMATE within 0.5 hour of the ACTUAL TIME that the trip took?

Statement 1: Joan’s ESTIMATE for the distance was within 5 miles of the ACTUAL distance.
Travel time = distance/speed

Statement 1 provides information regarding the accuracy of Joan's estimation of the travel distance, BUT it does not provide any information regarding her accuracy in estimating her speed.
As such, statement 1 is NOT SUFFICIENT

Statement 2: Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.
Statement 2 provides information regarding the accuracy of Joan's estimation of her average speed, BUT it does not provide any information regarding her accuracy in estimating the travel distance.
As such, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Let's test some numbers.
There are several possible scenarios that satisfy BOTH statements. Here are two:
Case a: Joan's estimates were PERFECTLY accurate. In this case, her ACTUAL travel time was definitely WITHIN 0.5 hours of her ESTIMATED travel.

Case b: Joan's ESTIMATED distance and average speed were 8 miles and 8 miles per hour respectively, and the ACTUAL distance and average speed were 5 miles and 1 mile per hour respectively. So, Joan's ESTIMATED travel time = 8/8 = 1 hour, and her ACTUAL travel time = 5/1 = 5 hours. In this case, Joan's ACTUAL travel time was NOT WITHIN 0.5 hours of her ESTIMATED travel.

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Cheers,
Brent
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If you enjoy my solutions, you'll love my GMAT prep course. Originally posted by BrentGMATPrepNow on 15 Nov 2017, 16:34.
Last edited by BrentGMATPrepNow on 20 Dec 2019, 07:37, edited 1 time in total.
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Re: In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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Bunuel wrote:
In planning for a trip, Joan estimated both the distance of the trip, in miles, and her average speed, in miles per hour. She accurately divided her estimated distance by her estimated average speed to obtain an estimate for the time, in hours, that the trip would take. Was her estimate within 0.5 hour of the actual time that the trip took?

(1) Joan’s estimate for the distance was within 5 miles of the actual distance.

(2) Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.

Kudos for a correct solution.

Joan estimated both the distance of the trip, in miles, and her average speed, in miles per hour. She accurately divided her estimated distance by her estimated average speed to obtain an estimate for the time, in hours, that the trip would take.

so basically d/s = Time

Target question : Was her estimate within 0.5 hour of the actual time that the trip took?

(1) Joan’s estimate for the distance was within 5 miles of the actual distance.
No info on Speed so Clearly not sufficient.

(2) Joan’s estimate for her average speed was within 10 miles per hour of her actual average speed.
N info on time. Not sufficient

Together

We have a range for the distance ie within 5 miles of the actual distance. and range for the speed : her average speed was within 10 miles per hour of her actual average speed.

Case a: Joan's estimates were PERFECTLY accurate. In this case, her ACTUAL travel time was definitely WITHIN 0.5 hours of her ESTIMATED travel.

Case b: Joan's ESTIMATED distance and average speed were 8 miles and 8 miles per hour respectively, and the ACTUAL distance and average speed were 5 miles and 1 mile per hour respectively. So, Joan's ESTIMATED travel time = 8/8 = 1 hour, and her ACTUAL travel time = 5/1 = 5 hours. In this case, Joan's ACTUAL travel time was NOT WITHIN 0.5 hours of her ESTIMATED travel.

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

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Re: In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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(1) Clearly insufficient.

(2) Clearly insufficient similar to statement1.

(1) + (2) All we have to show is combining we can get 2 different answers.

For e.g. if real speed is 10m/hr, and estimated speed 20m/hr, and if real distance is 10m and estimated distance is 15m
Therefore, difference between estimated and real time, delta T = (10/10) - (15/20) = 1 - 0.75 = 0.25. In this case answer is YES.

Now we just reverse the distances, so that, if real speed is 10m/hr and estimated speed is 20m/hr, and if real distance is 15m and estimated distance is 10m.
Therefore, difference between estimated and real time, delta T = (15/10) - (10/20) = 1.5 - 0.5 = 1 hours. In this case answer is NO.

So we have both possibilities even if we combine the two statements, hence insufficient.

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Re: In planning for a trip, Joan estimated both the distance of the trip,  [#permalink]

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After choosing C, I revisited the question and solved it as following:

Assume values for d=10 s=2
Estimated time =D/S=10/2=5
Actual Time (1): (10+5)/(2+10)=15/12 ---not within 0.5 hr of estimated time
Assume values for d=15 s=3.1
Actual time (2): (10+5)/(2+1.1) T=15/3.1=4.83 ---within 0.5 hr of estimated time

I fell for the trap that the difference in distance and speed would give difference in time which is not the case since we dont have actual numbers which can significantly vary the difference in estimated time.

Happy to hear any criticism on my approach - rate problems are a real boon for me! Bunuel VeritasKarishma
GMATPrepNow Re: In planning for a trip, Joan estimated both the distance of the trip,   [#permalink] 20 Dec 2019, 07:12

# In planning for a trip, Joan estimated both the distance of the trip,  