Reiko's Trip - DS Question.

[subramanian.k.viswa]

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.

[vinny4nyc]

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.

Hi subramanian
I too agree with you , I just took a CAT test and am stumped by this logic.

Regards
Vinayak

[jlucero]

First off, make sure you are posting in the correct thread Manhattan CAT exam questions have their own forum section.

As for your question, the algebraic solutions for these types of questions are often difficult to understand, so in this case, let's come up with some very straightforward numbers to illustrate why statement (1) is sufficient. The statement talks about rates of 40 and 80 on a roundtrip, so I'm going to say that the distance from A to B is 40 miles and the roundtrip distance would be 80 miles.

If the entire trip is going to be finished at an average speed of 80mph, that means Reiko has less than 1 hour to get from A to B and back. If Reiko could travel back INSTANTLY from B back to A, then he could spend the entire hour traveling from A to B and cover 40 miles in 1 hour for a rate of 40mph. Since instant travel is not an available option on the GMAT, that means that Reiko would have had to spend less than an hour from A to B and traveled at a speed of greater than 40mph. Statement 1 is sufficient.

And as for your contention, remember to watch out for the question being asked. If Reiko did travel from A to B in instant time, he would have been speeding at a lot faster than 40mph.

The takeaway is that you can never double your average speed on the second leg of a roundtrip journey. Mathematically, this is because: r = d/t and to double your rate you would need to double the distance without adding any time: r = 2d/t.

[saurabhbanerjeeiimk]

Does that mean that either of the two directions-A to B OR B to A would be (80 miles)/(LESS than total time for round trip) BECAUSE GMAT does not consider instant travel?

[t.sargeantson]

Hello - I'm not following the logic to the answer provided on this question. All we basically know is that Reiko traveled the same distance twice and at two different constant speeds. The question asks if the speed for the first trip, from a to b, was greater than 40.

Statement (1) says that the average speed for the entire trip was 80 mph. I don't see why this answers the question. The fact that the average speed was 80 mph tells us that at least one leg of the trip was faster than 80 mph, but we know nothing about which leg it was (a to b or b to a). If Reiko traveled from b to a at 300 mph, wouldn't his speed from a to b then be less than 40 mph?? What if he traveled back at 1000 mph?

[tim]

read Joe's post. even if the other leg happened at 1000 mph the first leg still must have been greater than 40 mph. and no the GMAT does not allow instantaneous travel between two points; that would imply an infinite speed..

[mickileepaul]

I just wanted to see if this reasoning was flawed. I've been staring at this problem for too long, so the logic sounds okay, but it might not be. I'm sure by morning I'll realize my mistake (if there is one).

The question says we travel from A to B at one particular constant speed, and the reverse direction at a 'different' constant speed.

(1) says the avg speed is 80mph. So if I was to assume this person travelled BOTH legs at 80mph, the distance is 2d, the avg speed works out to be 80mph also. But since the stem says we travel at different speeds in both directions, then if one speed goes up, the other could go down.

now, I didn't work out these values, I'm just giving examples of one speed going up and the other down.

could be 80mph and 80mph we have avg = 80mph,
or 90mph and 70mph
or 100mph and 60mph

of course these speeds don't average out to 80mph, but they are just an example of one speed increasing for one leg, and the other speed being able to decrease for the other leg.

Since we must have different speeds for each leg (as per the stem) then I know that one speed will always be more than the avg (our 80 mph).

Does this make any sense?

[tim]

if your examples don't average 80mph, they are useless examples to use.. :)

[mickileepaul]

With all do respect, that's why I stated that the examples didn't average out to 80mph. Perhaps I should have said they don't "exactly" average out to 80mph. But I do believe the concept is the same. One goes up, the other goes down.

Perhaps it wasn't clear enough.
So let's say this instead. Same distance 'd' each way.

80mph and 80mph averages 80mph
90mph and 72mph averages 80mph
100mph and 66.66667mph averages 80mph
and as it was stated multiple times before
as one side goes up (let's say 3000mph) then the other speed would be 40.53506mph


so if I was to plug arbitrary variables in for the first leg, and assume that I travelled at 40mph for the second leg, could I find a 'speed' that would allow me travel slower than or equal to 40mph?

Tot D / Tot Time = avg
2d / (d/v1) + (d/40) = 80
2d = 80d/v1 + 80d/40
2d = 80d/v1 + 2d
2d - 2d = 80d/v1
0 = 80d/v1

since d is not 0, then v1 would have to be infinity for 80d/v1 to be equal to 0. In this case, I'd be travelling at an impossibly fast speed for the first leg, and AT 40mph for the second leg. We've already noted that an infinity speed is not possible, so then the other leg would have to be at least a teeny tiny bit greater than 40mph.

I guess I just answered my own question.
I never bothered to work out any of the math, I just thought about this in my head and that as one speed went up, the other went down. I might have still needed to work out the speed at the max threshold, or in our case, the min threshold of 40mph. But it's still the same idea.

As for the actual 'numbers' not working out to an average of 80mph, I would have hoped it was clear enough that they were 'close' enough to the real values to justify the 'concept'.

[jnelson0612]

I love it when people can answer their own questions, because then I know that the understanding is there. :-)

Let us know if we can help you further.

[dhlee922]

Does that mean that either of the two directions-A to B OR B to A would be (80 miles)/(LESS than total time for round trip) BECAUSE GMAT does not consider instant travel?

i agree with this. what if A to B was less than 40 mph and B to A was greater than 40 mph. statement 1 doesnt clarify which leg was faster or slower so it should be E

[tim]

Statement 1 absolutely guarantees that BOTH legs were strictly faster than 40mph. Read the entire thread for the full explanation.

[aruchap]

After I get confused for a while and tired of conceptualisation, I tried to set the expression in RTD. I'm not sure whether I understand is correct. Please rectify what I think.

Let assume Total Distance is 80 mile and total time is 1 hour so that it will fix to the statement 1 - AVG speed is 80 mile/hour.

A to B: Time = X hrs, Distance = 40 miles, and thus Rate = 40/X mile/hour

B to A: Time = 1-X hrs (different time from A to B), Distance = 40 miles, and thus Rate = 40/(1-X) mile/hrs.

Since time of each path can be neither 0 (must have spent time) nor 1 (cant be more than the total time) => 0<X<1.

I then tried plug in number of X and found that all numbers make Rate of either path more than 40 mile/hour. Finally, I think I understand what you guys are discussing. Hope this helps any other average students like me.

[jlucero]

Thanks, aruchap. There's always a lot of great ways to solve questions and the best method is always the one that makes most sense to you.

[hereiamnikhil]

This is how I solved (without using values):

For travel from A to B, we can say
S1 = D/T1 (lets call this equation 1)
Question asks if S1>40?

For travel from B to A
S2= D/T2

Also we know from (A) that for the total travel from A to B and back:
S = (D+D) / (T1+T2)
=> 80 = (2D) / (T1+T2)
=> D = 40 (T1+T2)

Use this to solve equation 1
S1 = D/T1.
S1 = [40 (T1+T2)] / T1
S1 = 40 + 40(T2/T1)

Hence we can say that S1 > 40.