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

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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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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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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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Bunuel
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

Originally posted by BrentGMATPrepNow on 15 Nov 2017, 17:34.
Last edited by BrentGMATPrepNow on 20 Dec 2019, 08:37, edited 1 time in total.
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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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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.

(1) No information is about speed; Insufficient,

(2) No information is about distance; Insufficient.

Using both options;
Distance within 5 means the estimate can be 0--------5-----------10
The speed with 10 means, the estimate can be 0--------10---------20

Thus if the distance is 5 and speed is 10 then the time will be 5/10 = 0.5 Yes,
But if the distance is 10 and speed is 1 then the time will be 10/1 = 10 No

Originally posted by MHIKER on 05 Oct 2020, 13:07.
Last edited by MHIKER on 16 Dec 2020, 13:47, edited 1 time in total.
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Re: In planning for a trip, Joan estimated both the distance of the trip, [#permalink]
Video solution from Quant Reasoning:
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Re: In planning for a trip, Joan estimated both the distance of the trip, [#permalink]
Bunuel
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)

VeritasKarishma The first time I tried to solve this, I plugged in numbers similar to yours (see below), but got an answer that shows a difference of only .25 hrs... is there a way to solve this with variables? Or is the only way to solve this is by plugging in numbers? What if it takes to long to keep plugging in/guessing numbers to find an answer that proves a difference of .5 hrs?

Estimated distance = 10 miles. Actual distance = 15 miles
Estimated speed = 10 mph. Actual speed = 20 mph
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Re: In planning for a trip, Joan estimated both the distance of the trip, [#permalink]
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dc2880
Bunuel
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)

VeritasKarishma The first time I tried to solve this, I plugged in numbers similar to yours (see below), but got an answer that shows a difference of only .25 hrs... is there a way to solve this with variables? Or is the only way to solve this is by plugging in numbers? What if it takes to long to keep plugging in/guessing numbers to find an answer that proves a difference of .5 hrs?

Estimated distance = 10 miles. Actual distance = 15 miles
Estimated speed = 10 mph. Actual speed = 20 mph

There are too many unknown variables here and what you need is whether their value must lie within a range. Even if you take variables and try to form equations, you will not get any exact values. Hence, all you need to focus on are the extremes.

You know that her estimate of everything could be absolutely on target and then her estimated time = actual time (so yes, within 0.5 hrs of the actual time)

Now we just need to find a case in which her estimated time was way off. This would happen when the estimates are within the range given but actuals are small too such that even with these ranges, the estimates are way off.

What I mean is this:

Say actual distance is 10 miles. Her estimated distance could be 14.999999 miles i.e. almost 15 miles.
Say her actual speed was 20 mph. Her estimated speed could be 10.00000001 miles i.e. almost 10 mph.

Actual time taken = 10/20 = 0.5 hrs
Estimated time = 15/10 = 1.5 hrs

We increased the estimated distance and decreased the estimated speed so that estimated time increases by a whole lot.
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Re: In planning for a trip, Joan estimated both the distance of the trip, [#permalink]
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Video solution from Quant Reasoning:
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Re: In planning for a trip, Joan estimated both the distance of the trip, [#permalink]
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Bunuel
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.
Solution:

Question Stem Analysis:

We need to determine whether Joan’s estimated time is within 0.5 hour of the actual time the trip took.

Statement One Alone:

Since we don’t know Joan’s estimated average speed, we can’t determine whether her estimated time is within 0.5 hour of the actual time the trip took. Statement one alone is not sufficient.

Statement Two Alone:

Since we don’t know Joan’s estimated distance, we can’t determine whether her estimated time is within 0.5 hour of the actual time the trip took. Statement two alone is not sufficient.

Statements One and Two Together:

Both Statements together are still not sufficient. For example, if the actual distance is 100 miles and the actual average speed is 50 mph, then the actual time the trip took is 100/50 = 2 hours. However, the estimated distance can be any value between 95 and 105, inclusive, and the estimated average speed can be any value between 40 and 60, inclusive. If she estimated the distance to be 98 miles and the average speed to be 49 mph, then her estimated time of 98/49 = 2 hours is exactly the actual time of 2 hours (hence it’s within 0.5 hour of the actual time). However, if she estimated the distance to be 104 miles and the average speed to be 40 mph, then her estimated time of 104/40 = 2.6 hours, which is not within 0.5 hour the actual time of 2 hours.

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Re: In planning for a trip, Joan estimated both the distance of the trip, [#permalink]
Speed = 100, Distance = 100, Time becomes 1

estimated speed = 110, estimated distance = 105, estimated time becomes within 0,5 hours

OR

Speed = 1, Distance = 1, Time becomes 1

estimated speed = 10, estimated distance = 1, estimated time is not within 0,5 hours

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

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

If Joan were as slow as a snail: Joan's distance traveled 0.0001 miles and her average speed was 0.0001 miles per hour.

Clearly, her estimated values can be significantly off her actual distance/speed given the ranges in statements 1 and 2.

On the other hand, her estimated values can be exactly the same as her actual values.

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