Data Sufficiency

HI, I was stuck with this question in MGMAT CAT. I am not very happy with the explanation provided. Can some please clarify??

Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour.

(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip.
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.

MGMAT Explanation: Correct Answer (A) .. But i selected (E)
(1) SUFFICIENT: Say the trip is d miles long in each direction, so that the round-trip distance is 2d miles. According to this statement, Reiko took (2d miles)/(80 miles/hour) = d/40 hours to drive the entire round trip.

Reiko could not have driven from B to A in zero time, so it must have taken her less than d/40 hours to drive from A to B. Therefore, her speed on the trip from A to B must have been (d miles)/(LESS than d/40 hours) = 1/(LESS than 1/40 hours) = GREATER than 40 miles per hour.

(Note: when dividing by a "less than" number, flip the sign to "greater than." If we had been dividing by a "greater than" number, we would have flipped the sign to "less than.")

MY CONTENTION: This explanation could be valid vice-versa as well right?? For instance:
Reiko could not have driven from A TO B in zero time, so it must have taken her less than d/40 hours to drive from B TO A. Therefore, her speed on the trip from B TO A must have been (d miles)/(LESS than d/40 hours) = 1/(LESS than 1/40 hours) = GREATER than 40 miles per hour.

Which means either of the two trips can be greater than 40 miles per hour, right??

Would appreciate if someone can help clarify this.

Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?

(1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour.

(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip.

Plug in the THRESHOLD.
Since we want to know whether the speed from A to B is greater than 40mph, the threshold here is 40mph.

Statement 1: Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour.
Let the distance in each direction = 40 miles.
Time to travel the 80 miles there and back at a speed of 80 miles per hour = 1 hour.
If he travels from A to B at the threshold speed -- 40 miles per hour -- then the time from A to B = d/r = 40/40 = 1 hour.
Not possible -- since the TOTAL time is 1 hour, the time FROM A TO B must be LESS than 1 hour.
Thus, the speed from A to B must be GREATER than 40 miles per hour.

Try an extreme case:
Let the distance in each direction = 400 miles.
Time to travel the 800 miles there and back at a speed of 80 miles per hour = 800/80 = 10 hours.
If he travels from A to B at the threshold speed -- 40 miles per hour -- then the time from A to B = 400/40 = 10 hours.
Not possible -- since the TOTAL time is 10 hours, the time FROM A TO B must be LESS than 10 hours.
Thus, the speed from A to B must be GREATER than 40 miles per hour.

The two cases above illustrate the following:
If the speed from A to B is 40 miles per hour, then the TIME FROM A TO B will be equal to the TIME FOR THE ENTIRE TRIP.
Clearly not possible.
Thus, the speed from A to B must be GREATER than 40 miles per hour.
SUFFICIENT.

Statement 2: It took Reiko 20 more minutes to drive from A to B than to make the return trip.
No way to determine the speed from A to B.
INSUFFICIENT.

The correct answer is A.

Here's the take-away:
When the same distance is traveled at two different speeds, the average speed for the entire trip must be LESS THAN TWICE the slower speed.
As the cases in statement 1 illustrate, if the average speed for the entire trip is EQUAL to twice the slower speed, then the TIME TRAVELED AT THE SLOWER SPEED will be equal to the TOTAL TIME FOR THE ENTIRE TRIP.
Clearly not possible.
Since statement 1 indicates that the average speed for the entire trip is 80 miles per hour, if s = the slower speed:
80 < 2s
s > 40.
Since the slower speed is greater than 40 miles per hour, the speed from A to B must be greater than 40 miles per hour.

Hi GMATGuruNY,

Thanks much for the reply. I agree with your explanation but unfortunately my contention still remains. I agree with you that one way journey (A to B) or (B to A) has to be greater than 40. As you explained "When the same distance is traveled at two different speeds, the average speed for the entire trip must be LESS THAN TWICE the slower speed. "

But in this particular case, without any additional information for e.g. (Reiko traveled from A to B faster than B to A), how can we conclude that reiko actually faster from A to B than B to A. What if it was the other way round?

Sorry to bother you again. But it would be really helpful if you can throw some more light and help me correct my thought process.

Cheers
Subbu

Oh i think got this. Does this mean that since average speed is 80. Both A to B and B to A will be greater than 40?

satish.coolin wrote:Oh i think got this. Does this mean that since average speed is 80. Both A to B and B to A will be greater than 40?

Hi Satish,
if u understand the expression:
80 < 2s1("s1" is the slower of the two speeds)
u'll agree that 40 < s1.

If the slower speed, s1, is > 40, definitely s2 (the second speed,
will be > 40.

Hence, both s1 and s2 are greater than 40.

2nd Approach:

let V and v represent speeds in the two different directions.
generally, Avg Speed (S) = (2Vv)/(V + v)
If S = 2Vv/(V+v)
Assume v is the speed from A to B as 40; S = 80:
==> 80 = 2*40*V/ (40+V)
==> 80 = 80V/(40+v)
==> 3200 + 80V = 80V....no solution
Again try v = 39 and u get a negative value for V- not correct.

But take v =41, only then u get a +ve value for V.

Hope my explanation helps.

Gmatdriller:
i didnt understand ur 2nd approach.

S = 2Vv/(V+v)

Where is the distance taken into account in this equation?????

gmatdriller wrote:

satish.coolin wrote:Oh i think got this. Does this mean that since average speed is 80. Both A to B and B to A will be greater than 40?

Hi Satish,
if u understand the expression:
80 < 2s1("s1" is the slower of the two speeds)
u'll agree that 40 < s1.

If the slower speed, s1, is > 40, definitely s2 (the second speed,
will be > 40.

Hence, both s1 and s2 are greater than 40.

2nd Approach:

let V and v represent speeds in the two different directions.
generally, Avg Speed (S) = (2Vv)/(V + v)
If S = 2Vv/(V+v)
Assume v is the speed from A to B as 40; S = 80:
==> 80 = 2*40*V/ (40+V)
==> 80 = 80V/(40+v)
==> 3200 + 80V = 80V....no solution
Again try v = 39 and u get a negative value for V- not correct.

But take v =41, only then u get a +ve value for V.

Hope my explanation helps.

karthikpandian19 wrote:Gmatdriller:
i didnt understand ur 2nd approach.

S = 2Vv/(V+v)

Where is the distance taken into account in this equation?????

gmatdriller wrote:

satish.coolin wrote:Oh i think got this. Does this mean that since average speed is 80. Both A to B and B to A will be greater than 40?

Hi Satish,
if u understand the expression:
80 < 2s1("s1" is the slower of the two speeds)
u'll agree that 40 < s1.

If the slower speed, s1, is > 40, definitely s2 (the second speed,
will be > 40.

Hence, both s1 and s2 are greater than 40.

2nd Approach:

let V and v represent speeds in the two different directions.
generally, Avg Speed (S) = (2Vv)/(V + v)
If S = 2Vv/(V+v)
Assume v is the speed from A to B as 40; S = 80:
==> 80 = 2*40*V/ (40+V)
==> 80 = 80V/(40+v)
==> 3200 + 80V = 80V....no solution
Again try v = 39 and u get a negative value for V- not correct.

But take v =41, only then u get a +ve value for V.

Hope my explanation helps.

Karthik,

You don't need the distance (as the distance is equal in both directions). Example - assume speed in one direction is a and speed in the other is b.

Speed = Distance/Time = 2d/((d/a)+(d/b)) = 2ab/(a+b)

Regards
Anup