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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Looking for hard questions

by kayser » Fri Feb 06, 2009 3:08 pm
Hey team,

I'm a little tired of questions that seem too easy... Wondering if there is a good source of super tough questions out there. 700+ level.

Thanks a lot.
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Re: Looking for hard questions

by sanju09 » Wed Feb 18, 2009 4:11 am
kayser wrote:Hey team,

I'm a little tired of questions that seem too easy... Wondering if there is a good source of super tough questions out there. 700+ level.

Thanks a lot.
:idea: Try this and ask for more of this kind...

Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?


A• 45 minutes

B• 50 minutes

C• 40 minutes

D• 55 minutes

E• 35 minutes
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by ssilver0210 » Sun Feb 22, 2009 7:08 pm
This isn't so hard if you look at it from another perspective.

We know that the wife left at the time that she normally leaves, because she had no idea that the husband had arrived early. We know that the wife and husband arrived home 20 minutes earlier than usual, even though the wife left home at the regular time. For them to arrive home 20 minutes early, the wife would have had to have encountered the husband after he had walked the distance covered by 10 minutes of travel in a car. 20 minutes would be saved because the wife would save the 10 minutes that would have been required to drive from the point she saw her husband to the station, as well as the ten minutes back from the station to the point she saw her husband.

So, let's say the husband usually arrives at 3pm. Today he arrives at 2pm. The wife leaves 2:30pm to pick him up. She will meet up with the husband at 2::50 pm, because of the above analysis (by meeting up with him at 2:50, rather than her normal 3 o'clock meeting, she saves 10 minutes from husband to station, and back from station to husband, and they arrive home 20 minutes early, as stated in the facts.

So, husband begins walking at 2pm, and wife meets up with him at 2:50=50 minutes. It does not matter which numbers you plug in for the time, the result is always the same. If husband arrives usually at 4:30pm, then that means today he arrived at 3:30 pm. Wife needs to meet up with him at 4:20 pm to save the 20 minutes back and forth. 3:30-4:20=50 minutes.

Make sense?
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by rsadana1 » Thu Feb 26, 2009 7:08 am
You can access certain tough questions on Permutation and Combination at:

https://yourperfectscore.articulate-onli ... 9309672040

Let me know if you find these problems tough.
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Probablity question based on Housie/Bingo

by 007.r.mason » Thu Feb 26, 2009 3:35 pm
1. There are three people who are playing a game of Housie/Bingo/Tambola (depending on where you are). What is the probability that out of the first 10 numbers drawn by the administrator, one of the players gets 5 numbers crossed on his/her ticket.

Please note that in Housie each player gets a ticket that has 15 numbers from 1 to 90 and the administrator draws numbers one by one.

This is a question that came to mind while I was playing the game some days ago. Naturally the stakes were high!!
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by rsadana1 » Thu Feb 26, 2009 4:03 pm
Wow that is a question of practical significance.

I will give it a shot.

First the total number of outcomes can be given by 90C10.
This the number of ways in which 10 numbers can be selected by the administrator from the given 90 numbers.
I did the calculation using my calculator and got 5720645481903!!!

Lets assume for the time being that only 1 person is playing:
The event (or desired outcome) is the administrator selecting 5 numbers from the 15 numbers in this person's ticket and the remaining 5 numbers from the remaining 75 numbers (that are not on this person's ticket).

The number of ways in which this event can take place: 15C5 * 75C5, which again calculated using calculator gives 3003*17259390

Thus probability can be calculating by dividing the desired by the total outcomes = 0.009.

Lets define this as Probability of success. or p(success)
Thus P(failure)=1-p(success) = 0.9909

Now coming back to the question on 3 players playing the game.
P(one gets 5 numbers crossed in first 10 calls)=3C1*p(success)*p(failure)*p(failure)

3C1 since we need to select 1 person who succeeds.
We multiple the three probabilities as shown since 1 person has to succeed and remaining two have to fail.

This gives the answer as 0.0266 or 2.7%.

Wow...there is only 2.7% chance that one of the 3 players will get 5 numbers on his tickets crossed in the first 10 calls of the numbers.

Good question 007.r.mason...However, I am pretty sure on GMAT we will not be required to perform such lengthy calculations. But it surely was fun getting to the final answer.

I hope I am on the right track in this solution. Any comments... B-)
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..your questions in level 2 and 2 were gud

by 007.r.mason » Sat Feb 28, 2009 7:08 pm
Thanks rsadana1, your response is long but if one ignores that long numbers, it makes sense.

By the way, I tried some of your lessons and they seem good. The questions in Lesson 3 (or level 3.. I dont remember) were challenging. Keep the good work....

Thanks,

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Re: Looking for hard questions

by farooq » Sun Mar 01, 2009 2:45 am
sanju09 wrote:
kayser wrote:Hey team,

I'm a little tired of questions that seem too easy... Wondering if there is a good source of super tough questions out there. 700+ level.

Thanks a lot.
:idea: Try this and ask for more of this kind...

Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?


A• 45 minutes

B• 50 minutes

C• 40 minutes

D• 55 minutes

E• 35 minutes

I tried and come up with answer C i.e. 40 min.

Correct me, if my approach is wrong.

Lets say man arrive at station at 7am.
and usually couple reached home at 8am.

Now as per the above question.
man reached at station at 6 am.
and couple reached home at 7:40am i.e 20min before.

Usually they take 1 hr.
but today they took 1hr and 40 min.

So man walked 40 min.
Regards,
Farooq Farooqui.
London. UK

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Re: Looking for hard questions

by simba » Mon Mar 16, 2009 12:56 am
farooq wrote:
sanju09 wrote:
kayser wrote:Hey team,

I'm a little tired of questions that seem too easy... Wondering if there is a good source of super tough questions out there. 700+ level.

Thanks a lot.
:idea: Try this and ask for more of this kind...

Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?


A• 45 minutes

B• 50 minutes

C• 40 minutes

D• 55 minutes

E• 35 minutes

I tried and come up with answer C i.e. 40 min.

Correct me, if my approach is wrong.

Lets say man arrive at station at 7am.
and usually couple reached home at 8am.

Now as per the above question.
man reached at station at 6 am.
and couple reached home at 7:40am i.e 20min before.

Usually they take 1 hr.
but today they took 1hr and 40 min.

So man walked 40 min.
No, the initial post is correct.

Lets make some assumptions to make things easier. Imagine the normal journey time one way is 30 mins and normally the wife leaves at 630 to arrive in time for the husbands 7:00 train. Then half an hour travel time back to get back for 7:30

Instead today he arrives at 6:00 and begins walking. The wife leaves at 630 and begins driving. At 650 (she has now driven for 20mins) she spots her husband, picks him up. She now drives back home for 20 mins. The time she gets home is now 710 - 20mins earlier than normal.

You may be thinking that the assumption I made about 30 mins travel time was important, but whatever time we substitute in the answer is the same. eg lets make it 15 mins travel time is normal
So normally leaves at 645, picks him up at 7:00 and drives back for 715. Today, he begins walking at 6, she starts driving at 645, see's him at 650 (hes been walking for 50mins), then drives home for five minutes and is back for 655 - again 20mins early. so whatever time we sub will still mean this dudes been walking for 50mins. hope this helps!
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Re: Looking for hard questions

by sanju09 » Mon Mar 16, 2009 1:21 am
simba wrote:
farooq wrote:
sanju09 wrote:
kayser wrote:Hey team,

I'm a little tired of questions that seem too easy... Wondering if there is a good source of super tough questions out there. 700+ level.

Thanks a lot.
:idea: Try this and ask for more of this kind...

Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?


A• 45 minutes

B• 50 minutes

C• 40 minutes

D• 55 minutes

E• 35 minutes

I tried and come up with answer C i.e. 40 min.

Correct me, if my approach is wrong.

Lets say man arrive at station at 7am.
and usually couple reached home at 8am.

Now as per the above question.
man reached at station at 6 am.
and couple reached home at 7:40am i.e 20min before.

Usually they take 1 hr.
but today they took 1hr and 40 min.

So man walked 40 min.
No, the initial post is correct.

Lets make some assumptions to make things easier. Imagine the normal journey time one way is 30 mins and normally the wife leaves at 630 to arrive in time for the husbands 7:00 train. Then half an hour travel time back to get back for 7:30

Instead today he arrives at 6:00 and begins walking. The wife leaves at 630 and begins driving. At 650 (she has now driven for 20mins) she spots her husband, picks him up. She now drives back home for 20 mins. The time she gets home is now 710 - 20mins earlier than normal.

You may be thinking that the assumption I made about 30 mins travel time was important, but whatever time we substitute in the answer is the same. eg lets make it 15 mins travel time is normal
So normally leaves at 645, picks him up at 7:00 and drives back for 715. Today, he begins walking at 6, she starts driving at 645, see's him at 650 (hes been walking for 50mins), then drives home for five minutes and is back for 655 - again 20mins early. so whatever time we sub will still mean this dudes been walking for 50mins. hope this helps!
B-) that's the OA too, and this was one of a few better ways to explain the situation; good!
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by muna_m » Wed Apr 01, 2009 7:08 am
I have a question...

Somebody mentioned...
Instead today he arrives at 6:00 and begins walking. The wife leaves at 630 and begins driving. At 650 (she has now driven for 20mins) she spots her husband, picks him up. She now drives back home for 20 mins. The time she gets home is now 710 - 20mins earlier than normal.

How do we know the wife drove for 20 mins and not more or less?? I don't see it mentioned anywhere in the question. Am i missing something here?
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by muna_m » Wed Apr 01, 2009 7:21 am
ALSO, how do I access https://yourperfectscore.articulate-onli ... 9309630199

I cant open that link... it asks for some login info..

Thanks!
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by sanju09 » Thu Apr 02, 2009 2:21 am
muna_m wrote:I have a question...

Somebody mentioned...
Instead today he arrives at 6:00 and begins walking. The wife leaves at 630 and begins driving. At 650 (she has now driven for 20mins) she spots her husband, picks him up. She now drives back home for 20 mins. The time she gets home is now 710 - 20mins earlier than normal.

How do we know the wife drove for 20 mins and not more or less?? I don't see it mentioned anywhere in the question. Am i missing something here?
I think your question needs some repair for the damages in parts where you probably tried to supply some extra effects. Please write your question as a plain text only, it will be given the same attention.
The mind is everything. What you think you become. -Lord Buddha



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by muna_m » Thu Apr 02, 2009 6:10 am
Sorry about that!!

My question was.....

Somebody mentioned..

At 6:50 (she has now driven for 20mins) she spots her husband, picks him up. She now drives back home for 20 mins. The time she gets home is now 710 - 20mins earlier than normal.

How do we know the wife drove for 20 mins and not more or less?? I don't see it mentioned anywhere in the question. Am i missing something here?
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by sanju09 » Thu Apr 02, 2009 6:33 am
muna_m wrote:Sorry about that!!

My question was.....

Somebody mentioned..

At 6:50 (she has now driven for 20mins) she spots her husband, picks him up. She now drives back home for 20 mins. The time she gets home is now 710 - 20mins earlier than normal.

How do we know the wife drove for 20 mins and not more or less?? I don't see it mentioned anywhere in the question. Am i missing something here?
I think you are referring to this question:

Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?

Explanation related to your doubt only:

Wherever the wife sees her man, she picks him up and takes him home 20 minutes earlier than usual. Those 20 minutes were saved as the wife avoided to go to the station and come back to the same point where she saw her man. So in a way the one side time from the point of sight to the station was 10 minutes. Assumptions by some links are just to support their point of view. No need to take those assumptions as implications of the question stem.
The mind is everything. What you think you become. -Lord Buddha



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