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3 hard problems

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Re: 3 hard problems [#permalink]

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New post 14 Feb 2009, 23:58
tusharvk wrote:
I am not sure I understood the problem though. If they are asking us to find all the five integers so that the sum can be inferred, we certainly need more equations (5 equns to solve 5 unknowns).
If we are just to find what the sum is equal to then just 1 is sufficient.

How do I distinguish this? If faced with this question (without the subject line saying hard problems), I would definitely have answered E because we need 5 equations.


I read problem slowly and at least twice, monitoring "red flags" and other special words. It is typical for GMAT to trick us using words like "integer, even, distinct nonzero, consecutive". Here, the question begins: "What is the sum of .....?". By the way, I think the answer is E: sum could be 8 or 9.

matematikconsultant wrote:
But sum of 4 the least from 5 in ours case always < 5 and ........?

Sorry, don't get it. By the way, where did you find such nice problems?
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New post 15 Feb 2009, 05:26
Walker!
Sorry, not sum. Product. If f is largest, then m/f <1, n/f<1, etc and m/f +n/f + p/f +K/f +1 = mnpk ----->
mnpk<5.
As for problems, I am a trainer on GMAT and I have thought up them

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Re: 3 hard problems [#permalink]

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New post 15 Feb 2009, 06:55
Cool!

You may add your signature to each post with information that you are a tutor. So, somebody may contact you if he or she is interesting in your services. Of course, if it is appropriate for you. Just a suggestion...
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New post 15 Feb 2009, 07:56
Walker!
What answers ?

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Re: 3 hard problems [#permalink]

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New post 15 Feb 2009, 09:24
I've already posted them: C, B, E
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Re: 3 hard problems [#permalink]

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New post 15 Feb 2009, 09:50
matematikconsultant wrote:
Walker!
Sorry, not sum. Product. If f is largest, then m/f <1, n/f<1, etc and m/f +n/f + p/f +K/f +1 = mnpk ----->
mnpk<5.
As for problems, I am a trainer on GMAT and I have thought up them


ok. Then, let me ask you the question regarding the 3rd problem.
What is the sum could be thought to understand how we can express the sum and not trying to find the actual integers.
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New post 15 Feb 2009, 09:54
Walker!
Unfortunately C,C,E

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New post 15 Feb 2009, 10:08
Very simply,If f is largest, then m/f <1, n/f<1, etc and m/f +n/f + p/f +K/f +1 = mnpk ----->
mnpk<5 or in other words, mnpk = 1,or 2,or 3,or 4 ------>It is only 1,1,2,2 and 2 or 1,1,1,2 and 5

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Re: 3 hard problems [#permalink]

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New post 15 Feb 2009, 10:11
What is my flaw in second problem?
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New post 15 Feb 2009, 10:32
It is the second impotant condition.
2t/3=L/3y +L/2z
L/t=2x -------> 1/3x = 1/3y +1/2z, if y<x then the decision is absent

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Re: 3 hard problems [#permalink]

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New post 15 Feb 2009, 10:55
matematikconsultant wrote:
It is the second impotant condition.
2t/3=L/3y +L/2z
L/t=2x -------> 1/3x = 1/3y +1/2z, if y<x then the decision is absent


The decision cannot be absent because "A car traveled from town A to town B...". In other words, a car did travel, so solution exists. We cannot forget real contest behind formulas.
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Re: 3 hard problems [#permalink]

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New post 15 Feb 2009, 19:59
how can 2) be B or C???

lets say that example1) distance from A to B=45

speed x=12mph, speed y=15mph and speed z=18mph

then the avg speed is 15mph

however if

distance=207 miles

speed x=12mph, speed y=15mph and speed z=180 mph then avg speed is 69mph..

so knowing x or y is insuff..

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Re: 3 hard problems [#permalink]

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New post 16 Feb 2009, 00:22
Hi, FN!

x,y,z must satisfy conditions given in the problem.

For first example, time of first part of the trip is t/3, or 1/3*45/15=1 hour.
But, length of first part 12*1=12 miles is not equal 1/6*45~7.5 miles.

the same for your second condition.
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Re: 3 hard problems [#permalink]

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New post 16 Feb 2009, 10:54
hi walker OK i see your logic but lets say the total distance is 3 miles

speed x=1 mph..speed y=1 mph..and speed z=100 mph..

so the avg speed is still dependednt on the speed of z?

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Re: 3 hard problems [#permalink]

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New post 16 Feb 2009, 11:38
z cannot be arbitrary because of initial information. z influences on time, and time influences on x. In other words, your example again doesn't fit the information. If you want to give an example, check it for fit the initial conditions and I'm pretty sure you will get av=2x.
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Re: 3 hard problems   [#permalink] 16 Feb 2009, 11:38

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