DS from Kaplan. Experts Please Help!

[srini1988]

The Johnsons drove from their home to their cabin in the woods at an average speed of 75 mph. If they returned home later that weekend, what was the their average speed for the entire trip to and from the cabin?

1.The trip back home took 30% longer than the trip to the cabin

2.The distance from the Johnson's home to their cabin is 220 miles

[tanyasethi]

IMO the answer should be C. What is the OA?

[Ozlemg]

my answer is C.

in order to calculate average speed for return, we have to know the distance and time for going or return. Neither is supplied in a statement, so by using the info (1) and (2) we can calculate thw average speed

Thnks.

[top_business_2011]

The Johnsons drove from their home to their cabin in the woods at an average speed of 75 mph. If they returned home later that weekend, what was the their average speed for the entire trip to and from the cabin?

1.The trip back home took 30% longer than the trip to the cabin

2.The distance from the Johnson's home to their cabin is 220 miles

Given: Speed 1 = 75mph. Speed 2 = Unknown
Time 1 = t Time 2 = Unknown
Distance 1 = D Distance 2= D
Required: Overall average speed = Total distance/Total time = 2D/(t + time 2)?

Statement 1: Time 2 = 1.3t
D = Speed 1 * t
So 2D = 2(75 *t)
Hence, 2D/(t + time 2)= [2 *(75*t)]/ t + 1.3t
= (2*75)/2.3 So, Sufficient.
Statement 2: Total Distance = 2D = 220
Here, we can know time 1, but we know nothing about time 2. So insufficient

The answer is A.

[GMATGuruNY]

The Johnsons drove from their home to their cabin in the woods at an average speed of 75 mph. If they returned home later that weekend, what was the their average speed for the entire trip to and from the cabin?

1.The trip back home took 30% longer than the trip to the cabin

2.The distance from the Johnson's home to their cabin is 220 miles

Since statement 2 discusses THE distance, the solution below assumes that the same route is traveled in each direction.

Statement 1: The trip back home took 30% longer than the trip to the cabin.
If the time for the trip home was 30% longer, then the rate for the trip to the cabin was 30% faster.
Since we can determine the two rates, we can determine the average speed for the whole trip.
Sufficient.

To illustrate using easier numbers:
Let d = 260 miles.
Let rate to cabin = 13 miles per hour.
Time to cabin = 260/13 = 20 hours.
Time increased by 30% = 20 + 6 = 26 hours.
Rate home = 260/26 = 10 hours.
Rate to the cabin (13 miles per hour) is 30% faster than the rate home (10 miles per hour).
Average speed for the whole trip = (total distance)/(total time) = 520/46 = 11.3.

(For a discussion about how to quickly determine the average rate when given two speeds, please check here: https://www.beatthegmat.com/cant-figure- ... tml#390213.)

Statement 2: The distance from the Johnson's home to their cabin is 220 miles.
No information about the rate or the time for the trip home.
Insufficient.

The correct answer is A.

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THE ANSWER SHOULD BE E BECAUSE WE ARE NOT SURE ABOUT HIS SPEED WHEN HE RETURN BACK TO HIS HOME. IT MAY BE 1 km/sec OR 1000 km/sec.
I AM 1000 % SURE THAT THE ANSWER IS E

[[email protected]]

FROM A: IT IS WRITTEN THAN THE TRIP IS 30% LONGER. IT IS NOT CLEAR WHETHER THE DISTANCE IS LONGER OR THE TIME TO COVER THE DISTANCE IS LONGER.
IT IT IS TIME TO COVER THE DISTANCE THAN THE ANSWER IS A

[sss2534]

FROM A: IT IS WRITTEN THAN THE TRIP IS 30% LONGER. IT IS NOT CLEAR WHETHER THE DISTANCE IS LONGER OR THE TIME TO COVER THE DISTANCE IS LONGER.
IT IT IS TIME TO COVER THE DISTANCE THAN THE ANSWER IS A

The problem is referring to the time taken to travel from the cabin to the house. This type of phrasing and verbiage is common in word problems. It's not like you need to take a leap of faith to figure out that the problem is referring to the time taken -- and NOT the speed or the distance.

[mirantdon]

+1 FOR A.

Since the time taken is 1.3 times the original time taken . therefore ,the speed is accordingly reduced in that inverse proportion .
Also the average speed is the harmonic mean of the speeds back and forth .
s= 2s1s2/(s1+s2)

[GmatKiss]

IMO:A

[y_roy82]

speed = distance/time
speed inversely proportional to time when d is constant
therefore, speed * time = k (constant)
a) gives
s1= 75 (onward journey)
s2= ? (return journey)
t1= x (onward journey)
t2= 13x/10 (return journey)

so, 75 * x = s2 * 13x/10

get s2...

avg speed
= total distance/total time
= 2d /[(d/75) + (d/s2)] s2 from above... eliminate d to get average speed
= SUFFICIENT
we dont have to do any calculations... i am putting forth the tought process involved the problem can be solved without a pen

b) no info - insufficient

[gmatblood]

IMO:A