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Manager
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Are you ready for GMAT? then try this [#permalink]
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24 Apr 2010, 16:32
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This attached document has really nice questions for practice. If you think you are ready for GMAT, try this



Manager
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26 Apr 2010, 12:09
Nice thanks for sharing I will try to resolve them



Manager
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27 Apr 2010, 23:23
Thanks



Intern
Joined: 26 Mar 2010
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02 May 2010, 14:51
My God, I got stuck with the following seemingly simple problem:
"If 7 workers can build 7 cars in 7 days, then how many days would it take 5 workers to build 5 cars?"
I solved this as follows:
49 worker/days  7 cars x  5 cars
x = 35 worker/days are necessary to build 5 cars.
then let's divide 35 worker/days by 5 workers to receive 7 days.
But my approach may seem overcomplicated, so I will be grateful if you share other approaches to solving this problem. Thanks.



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02 May 2010, 19:08
joluwarrior wrote: Thanks I am not sure if you are aware of the RTD method but Rate * Time = Distance Workers R T D 7 7 7 => rate for 7workers is 1car per day => rate for 1 worker is : 7workers  rate 1 1worker  rate x x= (1*1)/7=1/7 Workers R T D 5 * 1/7 * X = 5 ===> X=(5*7)/5 = 7days



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02 May 2010, 19:29
3. If p and q are prime numbers, how many divisors does the product (p^3)(q^6) have?
A) 9 B) 12 C) 18 D) 28 E) 36
OA: D
help ! cannt think of a way to figure this out...should be combinatorics.......



Intern
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03 May 2010, 00:16
Thanks intern. Of course i know RTD formula, but somehow I did not know how to apply to in this particular problem. thanks.



Intern
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03 May 2010, 00:39
Regarding the above problem with prime numbers, I do not know the answer as well, but I would like to share my guess:
a prime number is a number which can be divided by 1 or itself only. So the prime number has only two possible divisors.
The number p3 can seen as a product of 3 prime numbers, while each prime number has only 2 divisiors (1 and itself), so the maximum possible number of divisiors for p3 is 3x2=6.
Similarly with the number q raised to 6th power. It consists of 6 prime numbers with each prime number having only 2 possible divisors, so the maximum number of divisiors for q6 = 6x2=12.
Finally, the product of p3 and q6 will be all those divisors together (meaning 6+12). So my answer is 18. Please someone with more expertise confirm or correct my approach.



Intern
Joined: 23 Jan 2010
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03 May 2010, 02:18
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I think the answer should be 28 as :
1. +1 for '1' which is a a factor for every number. 2. As they are prime , then factors of p=3(p,p2,p3) and for q6 ( similarly) 6 factors , total=9 3. The combonation of pq , say p(1) and for q 6 values total combinations=6 , similarly for p2 and p3 also 6 , which is total of 18.
All combined =1+9+18=28.
Please let me know if this is not correct..



Intern
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03 May 2010, 02:55
dairymilk,
Your approach seems rather reasonable, especially in that you use "1" only once.
However, I think that in the number 18 (the number of combinations of p and q) there must be some repetitive combinations, because the three "p" are the same numbers, and the three "q" are the same numbers, and the order does not matter here.
Once again, I appreciate your contribution, dairymilk, and agree that it might be correct, but it would be great to finally hear the officially correct answer.
Thanks.



Intern
Joined: 14 Feb 2010
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03 May 2010, 05:05
I solved the question on prime numbers using a formula that I had learnt a couple of years back >
if N  p^2 x q^3, where p and q are prime numbers, the total number of divisors = (p+1)(q+1).
I had learnt this formula for CAT exam, probably from Arun Sharma's book on Quant or IMS material....I will be checking the book today to reassure that I remember the formula correctly.....will post my findings



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03 May 2010, 10:47
zisis wrote: 3. If p and q are prime numbers, how many divisors does the product (p^3)(q^6) have?
A) 9 B) 12 C) 18 D) 28 E) 36
OA: D
help ! cannt think of a way to figure this out...should be combinatorics....... For me the easy way is: 3+1 = 4 and 6+1 = 7 4x7 = 28 just add 1 to the exponents (if they are prime numbers) and then multiply I hope it help



Manager
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03 May 2010, 10:47
Awesome! Thanks for sharing.



Manager
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03 May 2010, 10:58
Actually p and q must be different prime numbers because if p = q then you have a different result



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03 May 2010, 10:59
krishna3891 wrote: Awesome! Thanks for sharing. the thanks are for whom?



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03 May 2010, 11:07
netrix wrote: krishna3891 wrote: Awesome! Thanks for sharing. the thanks are for whom? for the pdf contributor.



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03 May 2010, 13:05
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joluwarrior wrote: I solved the question on prime numbers using a formula that I had learnt a couple of years back >
if N  p^2 x q^3, where p and q are prime numbers, the total number of divisors = (p+1)(q+1).
I had learnt this formula for CAT exam, probably from Arun Sharma's book on Quant or IMS material....I will be checking the book today to reassure that I remember the formula correctly.....will post my findings will specify 2 rules here  1 directed at above question and the other at a similar question involving primes and divisors. These might help for quickly answering such questions. 1. The number of divisors of a composite number:If D = (a^p)(b^q)(c^r), where a, b and c are primes, then the number of divisors of D, represented by 'n' is given by n = (p+1)(q+1)(r+1) 2. The sum of divisors of a composite number:For the same expression above D, sum of divisors S, is given by S = [{a^(p+1)  1}{b^(q+1)  1}{c^(r+1)  1}]/[(a1)(b1)(c1)]



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03 May 2010, 13:38
kudos netrix & joluwarrior
i was aware of the formula but totally forgot to use it in this scenario
I am working on a pdf document for explanations for all of the questions so feel free to contribute the shortest way of solving each question of the pdf on this topic.



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03 May 2010, 13:41
zisis wrote: kudos netrix & joluwarrior
i was aware of the formula but totally forgot to use it in this scenario
I am working on a pdf document for explanations for all of the questions so feel free to contribute the shortest way of solving each question of the pdf on this topic. Hi zisis, thanks! great work there please let me know if I can help you with another problem. There are always different approach to a problem and we can learn them from the contribution of others



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03 May 2010, 13:47
Very helpful. Thanks. I'm much more enlightened now




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