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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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.

Answer:- E

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

Answer: E

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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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

Answer: E

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
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