Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
we will have to wait for the OA for this for one week
let us discuss it
here is my attempt at the problem
IMO A is the answer to this one
my line of reasoning is as follows
statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time
now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5
statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent
Accordingly, A is the answer for this
I hope i have taken the correct approach to solve the question and arrived at the correct answer
we will have to wait for the OA for this for one week
let us discuss it
here is my attempt at the problem
IMO A is the answer to this one
my line of reasoning is as follows
statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time
now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5
statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent
Accordingly, A is the answer for this
I hope i have taken the correct approach to solve the question and arrived at the correct answer P/Q = R/S PS=RQ (1) P is divisible by 140 140 k*s=RQ
now if k=2 and Q=140 and s=1 then r is not divisible by 5 Insuff
(2)Q = 7^x , where x is a positive integer
Insuff
combine 140 k*s=R 7^x
R=(140 k*s/(7^x))
since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5
Comments please
Many thanks
Ok i see i mistake in statement 1 (thanks for pointing this out) but i just have one question
u say that
Quote:
140 k*s=R 7^x
R=(140 k*s/(7^x))
since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5
agreed that (7^x) can never be a multiple of 10, but let us try some numbers then let us say x =2 and k and s = 1 what we get is R=140/49 which then clearly is not divisible by 5
we will have to wait for the OA for this for one week
let us discuss it
here is my attempt at the problem
IMO A is the answer to this one
my line of reasoning is as follows
statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time
now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5
statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent
Accordingly, A is the answer for this
I hope i have taken the correct approach to solve the question and arrived at the correct answer P/Q = R/S PS=RQ (1) P is divisible by 140 140 k*s=RQ
now if k=2 and Q=140 and s=1 then r is not divisible by 5 Insuff
(2)Q = 7^x , where x is a positive integer
Insuff
combine 140 k*s=R 7^x
R=(140 k*s/(7^x))
since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5
Comments please
Many thanks
Ok i see i mistake in statement 1 (thanks for pointing this out) but i just have one question
u say that
Quote:
140 k*s=R 7^x
R=(140 k*s/(7^x))
since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5
agreed that (7^x) can never be a multiple of 10, but let us try some numbers then let us say x =2 and k and s = 1 what we get is R=140/49 which then clearly is not divisible by 5
so then the answer to this should be E
what do u think
why are you people forgetting If P, Q, R, and S are positive integers
we will have to wait for the OA for this for one week
let us discuss it
here is my attempt at the problem
IMO A is the answer to this one
my line of reasoning is as follows
statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time
now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5
statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent
Accordingly, A is the answer for this
I hope i have taken the correct approach to solve the question and arrived at the correct answer P/Q = R/S PS=RQ (1) P is divisible by 140 140 k*s=RQ
now if k=2 and Q=140 and s=1 then r is not divisible by 5 Insuff
(2)Q = 7^x , where x is a positive integer
Insuff
combine 140 k*s=R 7^x
R=(140 k*s/(7^x))
since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5
Comments please
Many thanks[/quote][/quote]
Ok i see i mistake in statement 1 (thanks for pointing this out) but i just have one question
u say that
Quote:
140 k*s=R 7^x
R=(140 k*s/(7^x))
since R is an integer and (7^x) can never be a multiple of 10 ,R is divisible by 5
agreed that (7^x) can never be a multiple of 10, but let us try some numbers then let us say x =2 and k and s = 1 what we get is R=140/49 which then clearly is not divisible by 5
so then the answer to this should be E
what do u think[/quote]
why are you people forgetting If P, Q, R, and S are positive integers
Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R divisible [#permalink]
03 Jul 2008, 05:26
I think the answer should be E.
vdhawan1 wrote:
If P, Q, R, and S are positive integers, and P/Q = R/S, is R divisible by 5 ?
(1) P is divisible by 140
(2)Q = 7^x , where x is a positive integer
Statement 1) First, all we know is P is divisible by 140, we don't know at all what Q is. The value of Q will help determine what value R can be.
If p = 140 (which is divisible by 140), we must pick a number for Q. We could pick 280 for Q, and so 140 / 280 = 1/2. R could be 1 and that is not divisible by 5. The question asked of us is essentially "Is R always or never divisible by 5?" If we can come to a conclusion of "always" or "never" - the statement is sufficient. Becuase we can't do that for #1, it is insufficient.
Statement 2) This fails for a similar reason to Statement 1. We know possible values of Q, but we don't know the value of P. If x = 2, then Q = 49 (7^2=49). We are not limited at all as to what q can be. Insufficient.
Together) This should prove to be sufficient. If P is a multiple of 140, and Q is 7^x, we need to see what values are available for R/S.
Factors of 140 will help us determine if (when reduced) the value will be divisible by 5.
1-2-4-5-10-14-28-35-70-140.
Some available numbers for Q: 7-49-343-2401...(I stopped doing the math, because there is a useful principle here).
The only factors of (7^x) will be the number itself and the numbers ahead of it in the progression of 7^x (i.e., factors of 7^5 = 7^1, 7^2, 7^3, 7^4,7^5) There is a pattern that emerges here with the units column (the column that tells us if it is divisible by 5).
You see the pattern. So no matter what P/Q is, it will never be reduced down to something that makes P divisible by 5. Any multiple of 140 divided by 7^x will not leave a 5 or 0 in the units column because any multiple of 140 will have a 0 in the units column and in order to divide anything with a 0 in the units column and result in a 0 or 5 in the units column you mus divide by multiples of 5 or 10 (and mulltiples of 2 [that are also not multiples of 6]).
After all of this analysis and I was ready to write the answer is C, the thought struck me.
If P/Q = R/S, and P=140 and Q = 7^x, P/Q could be 140/49. and if that = RS, there is nothing that would keep R/S from being 140/49 and R is then divisible by 5. The reasoning above shows that we can find plenty of situations that R is not divisible by 5, but because we found one where R is divisible by 5
Answer has to be E. _________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
we will have to wait for the OA for this for one week
let us discuss it
here is my attempt at the problem
IMO A is the answer to this one
my line of reasoning is as follows
statement 1 tells us that P is divisible by 140 and the stem tell us that P,Q,R,S are positive numbers so it means that P will be a multiple of 140 at each time
now let us try picking and plugging some numbers in the equation as per the stem let us say p = 280 Q = 3. now since P/Q equals R/S therefore we know that R/S will always represent 280/3 . which means that R will always be a multiple of 5
statement 2, we only know that Q=7^x and x is any positive number, but we do not know about what are the other numbers viz, P,R,S. therefore this statement alone is insufficent
Accordingly, A is the answer for this
I hope i have taken the correct approach to solve the question and arrived at the correct answer
Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R divisible [#permalink]
02 Mar 2015, 05:07
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R divisible [#permalink]
02 Mar 2015, 07:03
Expert's post
1
This post was BOOKMARKED
If P, Q, R, and S are positive integers, and P/Q=R/S, is R divisible by 5?
Given: \(R=\frac{PS}{Q}\)
(1) P is divisible by 140 --> P is multiple of 5 --> now, if all 5-s from P and S (if there are any) are reduced by 5-s in Q then the answer will be No (for example P=140=5*28, S=1 and Q=5) but if the sum of powers of 5 in P and S is higher than the power of 5 in Q then not all 5-s will be reduced and R will be a multiple of 5 (for example P=140=5*28, S=1 and Q=1 or P=140*5=5^2*28, S=1 and Q=5). Not sufficient.
(2) Q= 7^x, where x is a positive integer. Clearly insufficient.
(1)+(2) Q is not a multiple of 5 at all thus 5-s in P won't be reduced so as \(R=\frac{PS}{Q}=5*integer\) then R is indeed a multiple of 5. Sufficient.
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...