MGMAT

Bhandaripreeti · Posted: Mon Sep 24, 2007 12:07 pm Post subject: MGMAT

Bob and Wendy left home to walk together to a restaurant for dinner. They started out walking at a constant pace of 3 mph. At precisely the halfway point, Bob realized he had forgotten to lock the front door of their home. Wendy continued on to the restaurant at the same constant pace. Meanwhile, Bob, traveling at a new constant speed on the same route, returned home to lock the door and then went to the restaurant to join Wendy. How long did Wendy have to wait for Bob at the restaurant?

(1) Bob’s average speed for the entire journey was 4 mph.

(2) On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

saurabh_1922@yahoo.co.in · Posted: Mon Sep 24, 2007 12:37 pm Post subject:

IMO C
using A we can find the new speed with which he returns midway to hime and then to restraunt.
We need to know the distance between their home and the restaurant.
Using B alone we cannot find anything as there will be two unknowns.
Using both we can find the distance and hence the time difference.
Whats the OA

kajcha · Posted: Mon Sep 24, 2007 1:23 pm Post subject:

Another vote for C

Guest · Posted: Mon Sep 24, 2007 8:33 pm Post subject:

Surprisingly, the OA is B .Can anybody pls explain how?

samirpandeyit62 · Posted: Mon Sep 24, 2007 11:17 pm Post subject:

stmt 1: 2 unknowns so we cannot find the time

stmt 2: On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

suppose distance is x hence windy travelled x whereas bob travelled 2x

now in this 32 mins windy travelled x/2 wheras bob travelled 1.5 x

time taken by windy = x/2 / 3 =x/6

time taken by bob = 32 mins = .5 hrs approx or 1.5x/S(new speed of bob)

so the time windy had to wait was .5 - x/6

but here still we cannot find x coz we dont know the new constant speed of bob.

so IMO ans should be C

mallard906 · Posted: Tue Sep 25, 2007 11:42 am Post subject:

I also think the answer is C. I have no idea why the OA might be B.

saurabh_1922@yahoo.co.in · Posted: Wed Sep 26, 2007 4:47 pm Post subject:

Guys, We all are wrong Smile

. The answer is indeed B
Here is the explanation.Yesterday while taking MGMT CAT I encountered this question.
To see why this statement is sufficient, it is helpful to think of Bob's journey in two legs: the first leg walking together with Wendy (t1), and the second walking alone (t2). Bob's total travel time tb = t1 + t2. Because Wendy traveled halfway to the restaurant with Bob, her total travel time tw = 2t1. Substituting these expressions for tb – tw,
t1 + t2 – 2t1 = t2 – t1

tb – tw = t2 – t1

Statement (2) tells us that Bob spent 32 more minutes traveling alone than with Wendy. In other words, t2 – t1 = 32. Wendy waited at the restaurant for 32 minutes for Bob to arrive.

The correct answer is B.

tw=2t1 is the catch that we all missed Smile

Neo2000 · Posted: Wed Sep 26, 2007 7:47 pm Post subject:

Oops

Worked out the problem. B is correct. The trap is in the wordings Smile

samirpandeyit62 · Posted: Wed Sep 26, 2007 10:38 pm Post subject:

Hi Saurabh,

stmt 2: On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

now you quoted tb= t1 + t2

tw= 2t1

& tb – tw = t2 – t1

now here acooridng to the stmt 2, t2 =32 mins = time Bob walked alone i.e time he walked without windy

so if t2-t1 =32, in that case t1 =0, which is not possible

So I'm not convinced that ans could be B

What are ur thoughts

kajcha · Posted: Thu Sep 27, 2007 5:57 am Post subject:

samir, stmt 2 says Bob spends 32 MORE minutes. so t2-t1=32.

I reworked it and ans is B

samirpandeyit62 · Posted: Thu Sep 27, 2007 7:58 am Post subject:

Thanks kajcha,
I agree with u coz equations give the ans as B, but I wonder if we get such a question on the GMAT, what will we answer?, the stmt 2 here is somewhat misleading.

saurabh_1922@yahoo.co.in · Posted: Thu Sep 27, 2007 1:53 pm Post subject:

on the lighter side :If we get such questions in GMAT we will mark the wrong answer and still come back with a high score Laughing

Stacey Koprince · Posted: Thu Sep 27, 2007 3:22 pm Post subject:

This is a REALLY hard question. If you see something like this on the test, be happy, even if you don't get it right! Smile

Yes, the wording is a bit tricky and, yes, you can sometimes get similar wording on the real test. Most of the time the wording is fairly straightforward but sometimes on the hardest questions, if you miss even a single word, it will change the whole problem.

For example, check out this OG problem (10th edition, PS #284):

On a certain road, 10 percent of the motorists exceed the posted speed limit and receive speeding tickets, but 20 percent of the motorists who exceed the posted speed limit do not receive speeding tickets. What percent of the motorists on that road exceed the posted speed limit?

(A) 10.5 %
(B) 12.5 %
(C) 15 %
(D) 22 %
(E) 30 %

When I give this one in class, the vast majority get it wrong the first time. Once they know how to do it, though, they realize the catch, and they realize they need to REMEMBER that catch so that they won't make the same mistake when they see a problem of this type again. The key is to learn the setup so that you can avoid the trap the next time.

(I'll post the OA down below in a separate post for people who want to try it without looking at the answer.)

Stacey Koprince · Posted: Thu Sep 27, 2007 3:24 pm Post subject:

The answer to the above PS question is B. The answer is NOT E, though most people pick it (even people who learn our double-set matrix method for doing this problem). Why? The wording. (What's the key word?)

I won't put an explanation up yet - play with it.

kajcha · Posted: Thu Sep 27, 2007 6:32 pm Post subject:

B it is.

Suppose there are x total motorists.

Suppose there are y motorists that exceed speed limit

x/10 gets ticket.

y/5 don't get ticket.

x/10 + y/5 = y . Solve to get y/x = 5/40.

% of motorists who exceed speed limit = 5*100/40 = 12.5%