Jane walked for 4 miles. What was her average speed for the first 2 mi

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EX

At 7:00am, Claire leaves the spa, driving her car at 80kph she reaches her house at 7:15am, packs her artwork, and at 7:45am, leaves for the design lab, which is 10km from her house. If she drives her car at 40kph in going to the design lab, what will her average speed be from the spa to the lab ?

O 30 kph
O 40 kph
O 53.3 kph
O 60 kph
O 66.7 kph

The answer is below.








AA

In this case, total time taken = 1 hr
Distance covered = 80 * 1/4 + 10 = 30 km

Avg speed = 30 km/h

Juaz wrote:

Here we have "average speed", so stopover doesn't make any difference. Question is fine IMO.

(1) This gves time taken for whole trip..i.e. 75 min.
INSUFF

(2) x = time taken for first 2 miles
total time take x+x+15
INSUFF

From 1 & 2

2x+15 = 75
=> x = 30 min
Average speed for first 2 miles = 2/(30/60) = 4 miles/hr
SUFF...Ans C.

----------------------------------------------------------------



I agree that 1. gives total time = 75 minutes.

However, 2. can give (x) + (stopover time) + (x+15).

So, combining 1. and 2., you get:
2x + 15 + (stopover time) = 75.
2x + (stopover time) = 60.

This is why I think that the answer should be E. But of course, I may be wrong in my assumptions (should I consider stopovers?)



NOTE
Average speed in traveling from point A to point B is defined to be the total distance from point A to point B, divided by the total time it takes to go from point A to point B, and this means that the stopover time will be important.

I think the question is fine. we should just be relying on the info given. I won't lie, some of these probs get me too.

back to the orig prob. I was toying around w/this a bit and don't understand why this approach won't work (admittedly, Juaz's is much cleaner and quicker). Can anyone tell me why?


1) t = 4/3.2

2)

1st leg:
rt = 2
t = 2/r

2nd leg:
r(t+15) = 2
rt + 15r = 2
t = 2 - 15r/r

2 - 15r/r + 2/r = 4/3.2 =

4-15r/r = 4/3.2

I tried solving for r and kept coming up empty. what gives???

I understand now why stopovers do not count, cuz the stem says "It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles." which means stopovers already excluded, if there were any.

As for the last post, since it is DS, no need for calculations. in fact quite few DS`s have unrealistic answers, if you try to solve them.

In principle, I did the same thing that Juaz did (at least I think I did). but his worked. why didn't the above?

yep, it is just miscalculation. Look at the text in red.

Susan drove at an average speed of 30 miles per hour for the first 30 miles of a trip and then at an average speed of 60 miles per hour for the remaining 30 miles of the trip. If she made no stops during the trip, what was Susan's average speed, in miles per hour, for the entire trip?



There is a reason for them to STATE
"If she made no stops during the trip,", right?

I really find it confusing. Could it be that they made a mistake, a typo, if you will, in writing the answer key to the DS item above (choice C)?

Please vote.

IrinaOK,

please look at my text below your first edit. isn't that the same as what you did? t = 2 - 15r/r

The speed for the first 2 miles?

Statement I

speed =3.2
distance = 4
time=4/3.2
Insufficient, tells us nothing about the first 2 miles or last 2 miles.


Statement II

if time taken for first 2 mile = x
then, time taken for last 2 miles = x+15
we dont know anything about speed, therefore insufficent.

Statement I & II

(4/3.2)*60 = x+x+15

x=12.5

d=2 miles
time = 12.5
speed = 2/12.5 miles/minute
Sufficient.

Hence C.

For what it's worth, although this question is worded a bit more vaguely than most official GMAT questions, I've never seen an official DS question where the 'trick' was that they didn't mention a stopover that happened. That really isn't the kind of trick that the GMAT likes to play. That would make it more like a lateral thinking problem or a riddle, testing whether you suddenly realized what the missing piece was. Instead, the GMAT prefers reasoning problems, where if you follow the right process beginning to end, you'll end up on the right answer.

d = 4 miles, average speed = ??? for the first 2 miles
Let's give x1 as the average speed for the first 2 miles---> x1 =???
x2: the average speed for the second 2 miles-
x: the average speed for the whole journey.

ST1: x = 3.2mph
--->No info about time to travel the whole journey---->NS

ST2:
t1: time to walk in the first 2 miles
t2: time to walk in the second 2 miles
t1 = t2 - (15/60) = t2 - (1/4) (2)
--->No info about speed--->NS

ST1 + ST2: t = (4/3.2) = 5/4 = t1 + t2 (1)
(1) + (2) ------>From here we can find time for each 2 miles and then average speed.
---->The answer is D

Given: Jane walked for 4 miles.
Asked: What was her average speed for the first 2 miles?

(1) Jane's average speed for the 4 miles was 3.2 miles per hour.
Total time taken = 4/3.2 = 1.25 hours = 1 hours 15 minutes
No information is provided for first 2 miles
NOT SUFFICIENT

(2) It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles.
Since total travel time is not provided
NOT SUFFICIENT

Combining (1) & (2)
(1) Jane's average speed for the 4 miles was 3.2 miles per hour.
(2) It took Jane 15 minutes longer to walk the second 2 miles than it took her to walk the first 2 miles.
Total time taken = 4/3.2 = 1.25 hours = 1 hours 15 minutes
Total time = 1.25 hours
x + x + .25 = 1.25
x =.5 hours = 30 minutes
Time taken to cover first 2 miles = x + .25 = .75 hours
Jane's average speed for the first 2 miles = 2 miles / .75 hours = 2.67 miles/hour
SUFFICIENT

IMO C

Shouldn't the first 2 miles be 'x' instead of 'x + .25'? Since statement 2 says Jane took 15 mins longer to walk the second 2 miles.

Hi BrentGMATPrepNow, Could you help with St 2 please especially in time calculation of average speed since it's asking for the first 2miles therefore not quite sure in terms of time here? Thanks

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