Reiko's Trip - DS Question.

[RonPurewal]

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

in statement 1 we know that both legs were faster than 40 mph, so it doesn't matter which one you choose. (it's a symmetric situation, because it's a round trip and no difference between the two parts is stated -- so, by definition, anything that must be true for the first half of the trip must also be true for the second half of the trip.)

remember that, in statement 1, the average speed has to be 80 mi/h in this problem.
basically, we discovered (above) that, if one leg is only 40 mi/h, then the other leg would have to be "infinity" mi/h to make this average speed.
that's not possible, so, basically, here are the three extreme cases:
* first leg = barely over 40 mi/h; second leg = some gigantic speed
* first leg = second leg = 80 mi/h
* first leg = some gigantic speed; second leg = barely over 40 mi/h

--

[RonPurewal]

if this is all too abstract, then plug in a concrete value for the length of the trip, say 400 miles each way = 800 miles total (which are convenient multiples of both 40 and 80). then it should be clear what's going on.

in that case, the question becomes ... "Did Reiko take less than 10 hours to go the first 400 miles?"
(since 400/40 = 10, d/r = t, and dividing by "greater than" makes "less than" with positive numbers)

but 800/80 = 10, so statement 1 says... "Reiko took 10 hours to go the whole 800 miles."
if this is the case, then the answer to the question formulated above is a definite "yes".

[mcmebk]

I didn't make it right in the exam, but now I see how it can be resolved:

T=T, + T,,

since T = 2D/80 = D/40

Thus T, must be less than D/40

S, = D/T, > D/(D/40) = 40

S, >40

as Ron said, it is a symmetrical situation, so both S, and S,, are greater than 40. The reason why we made it wrong because it feels a little against our intuitive sense.

[RonPurewal]

I didn't make it right in the exam, but now I see how it can be resolved:

T=T, + T,,

since T = 2D/80 = D/40

Thus T, must be less than D/40

S, = D/T, > D/(D/40) = 40

S, >40

what are the commas? their meaning isn't immediately obvious to me. it seems they have something to do with the first and second parts of the trip, but i'm not 100% sure what is going on here.

if you are going to introduce notation, please explain the notation before you start using it. thanks!

[nishankerd]

Another way to look at this problem.

This approach is useful for those if us who prefer equations over picking numbers.

Let's define t1=time for leg AB, t2=time for leg BA, d=distance for one leg.

Then: v1=d/t1 but the question stems is asking if v1>40. In other words, "is (d/t1)>40?" This can rewritten (since all values are positive) as "is t1<(d/40)?"

From Statement 1 you get: v(avg) =2d/(t1+t2) = 80
Therefore, (t1+t2)=2d/80=d/40. Statement 1 is telling you that t1=d/40 - t2 --> which answers our rewritten version of the stem question: t1 will alwasy be less than d/40 because you will have to substract t2, which is a number greater than zero.

[tim]

cool! thanks for sharing.

[steve.tan.chee.wei]

so i guess the best way to answer this question in the short given time would be to use the equation method?

because trying to think it about it logically has a big trap that we cannot imagine to make one part of the journey to < 40mph would require the other part to require impossible speed.

[RonPurewal]

so i guess the best way to answer this question in the short given time would be to use the equation method?

because trying to think it about it logically has a big trap that we cannot imagine to make one part of the journey to < 40mph would require the other part to require impossible speed.

This "trap" is not a trap at all, if you just test extremes.

If you can get a case in which the speed from A to B was greater than 40 mi/h AND a case in which it was not greater than (i.e., ≤) 40 mi/h, then "not sufficient".

For statement 1, it's easy to find the former case (you can just do the entire trip at a constant speed of 80 mi/h). So, you can devote all your resources to finding the latter case.
Just test extremes.
There's no point in trying any values less than 40 mi/h from A to B, since you're shooting for an overall average of 80.
So, just try 40 from A to B.
Then work out the numbers.
The numbers don't work out (you'd need to travel the rest of the trip in zero time).

No "trap" unless you are trying to do the whole thing in your head, which is pretty much always a bad choice.

[bruno]

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.

Thanks hereiamnikhil, that took me out of a massive headhache on this Statement 1!

[tim]

Glad to hear it!

[camerons957]

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.

Man, I was following you beautifully until the last line. How does s1= [40 (t1+t2)]/t1 go to 40+40(t2/t1) and then that ends in s1 > 40?

Thanks!

[tim]

Distribute the numerator and divide each piece by T1. Then S1 = 40 + some positive number, which means S1 > 40.

[RonPurewal]

Man, I was following you beautifully until the last line. How does s1= [40 (t1+t2)]/t1 go to 40+40(t2/t1) and then that ends in s1 > 40?

Thanks!

If you understand why (6x + 9y)/3 is equal to 2x + 3y, it's the same thing.

[camerons957]

Man, I was following you beautifully until the last line. How does s1= [40 (t1+t2)]/t1 go to 40+40(t2/t1) and then that ends in s1 > 40?

Thanks!

If you understand why (6x + 9y)/3 is equal to 2x + 3y, it's the same thing.

Yeahhhh, I see it now. Feel a bit foolish. Thanks

Small world by the way, I remember fondly taking your geometry class in high school :)

[tim]

Ron, I didn't realize you taught high school! Fun times.