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# If P, Q, R, and S are positive integers, and P/Q = R/S, is R

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If P, Q, R, and S are positive integers, and P/Q = R/S, is R [#permalink]  02 Jul 2008, 21:20
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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

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

Many thanks

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Re: Manhattan workbook wednesday question [#permalink]  02 Jul 2008, 21:33
P/Q = R/S
PS = QR

Statement 1 : P = 140y

140y * S = Q * R

we can have y = 1, S = 1, Q = 14, R = 10, divisible by 5
OR we can have y=1, S=1, Q=10, R=14, not divisible by 5

Not Suff

Statement 2 : Q = 7x, R can take any vaue... Not suff

Combine 140y * S = 7x * R
R = 20y * S / x

if x = 20, y = 1, S = anything ..... we can not determine R's divisibility with 5

IMO E.
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Re: Manhattan workbook wednesday question [#permalink]  02 Jul 2008, 21:41
rohit929 wrote:
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

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

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

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Re: Manhattan workbook wednesday question [#permalink]  02 Jul 2008, 21:53
vdhawan1 wrote:
rohit929 wrote:
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

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

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

so R=140/49 is not possible
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Re: Manhattan workbook wednesday question [#permalink]  02 Jul 2008, 21:57
thanks for pointing out my mistake. 7^x and 7*x.

The answer should be C, because as per question stem, R has to be an interger.
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Re: Manhattan workbook wednesday question [#permalink]  03 Jul 2008, 02:56
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

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

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

so R=140/49 is not possible[/quote]

Ok got it

thanks for pointing this out

wonder where my attention was

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Re: Manhattan workbook wednesday question [#permalink]  03 Jul 2008, 03:28
I get answer C too, following the same reasoning as vdhawan1 did

Regards

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I'm not linked to GMAT questions anymore, so, if you need something, please PM me

I'm already focused on my application package

My experience in my second attempt
http://gmatclub.com/forum/p544312#p544312
My experience in my third attempt
630-q-47-v-28-engineer-non-native-speaker-my-experience-78215.html#p588275

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Re: Manhattan workbook wednesday question [#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).

The units are
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
7^6 = 9
7^7 = 3
7^8 = 1
7^9 = 7
7^10 = 9
7^11 = 3

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

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Re: Manhattan workbook wednesday question [#permalink]  03 Jul 2008, 06:01
P/Q = R/S

R = (P/Q)*S

1) P = 140*k -> obviously insufficient (P=140,Q=70, S=1,R=2), (P=140, Q=70, S = 5, R = 10)
2) Q = 7^x -> insufficient

1&2: R = (P/Q)*S = (7*20*k/7^x)*S = 20*(k/7^(x-1))*S -> divisible by 5, sufficient -> C
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Re: Manhattan workbook wednesday question [#permalink]  03 Jul 2008, 06:01
The answer is up there and the answer is C. It's a good explanation.

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Re: Manhattan workbook wednesday question [#permalink]  03 Jul 2008, 06:40
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

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

Many thanks

S1: Insuff. Try 140/10 = 140/10 or 140/10 = 28/2

S2: Insuff. Tells us nothing.

Together:

Suff.
Re: Manhattan workbook wednesday question   [#permalink] 03 Jul 2008, 06:40
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