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:

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

Not convinced by the OA

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.

R / S can be any values that can lead to 2.87 so R by itself need not be a multiple of 5

what is the error in my reasoning so i went for E

First of all 140/49 does not equal to 2.87 it equals to 20/7 (it'll be a recurring decimal). So R/S=20/7 --> R is a multiple of 20 so it's a multiple of 5 too (note that we are told that all variables are positive integers). _________________

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

Not convinced by the OA

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.

Answer: C.

Q can also be 35..so then the 5 would also be reduced. So how do you know that R is then still a multiple of 5? what am I missing?

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

Not convinced by the OA

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.

Answer: C.

Q can also be 35..so then the 5 would also be reduced. So how do you know that R is then still a multiple of 5? what am I missing?

Q cannot be 35 or any other multiple of 5, since it equals to \(7^{positive \ integer}\).

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

Not convinced by the OA

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.

Answer: C.

Bunuel sorry, I don't quite get it

So we have P,Q,R,S are positive integers and we're trying to figure out whether R = PS/Q is divisible by 5 or if PS / 5Q is an integer right?

So Statement 1

P is divisible by 130, but I don't know nothing about the other two only that they are integers

Not sufficients

Statement 2

Q= 7^X

Clearly Insuff

Both together

I have the quesiton: is 140S / 5Q an integer where Q = 7^X

Well x could be anything and hence not sufficient

E

But I'm pretty sure there's something I'm missing, would anybody please clarify what's wrong with this line of reasoning?

I have the quesiton: is 140S / 5Q an integer where Q = 7^X

Well x could be anything and hence not sufficient

E

But I'm pretty sure there's something I'm missing, would anybody please clarify what's wrong with this line of reasoning?

Cheers J

Yeah, the concept here is quite basic but we often overlook it.

Think about it: Is \(3^5 * 7^6 * 11^3\) divisible by 13? The answer is simply 'No'.

For the numerator to be divisible by the denominator, the denominator MUST BE a factor of the numerator. 3^5 is only five 3s. 7^6 is only six 7s. 11^3 is only three 11s. In the entire numerator, there is no 13 so the numerator is not divisible by 13.

On the other hand, is \(3^5 * 7^6 * 11^3 * 13\) divisible by 13? Yes, it is. 13 gets cancelled and the quotient will be \(3^5 * 7^6 * 11^3\).

Is 2^X divisible by 3? No. No matter what X is, you will only have X number of 2s in the numerator and will never have a 3. So this will not be divisible by 3.

On the same lines, in this question,

Given that \(R = \frac{(140a)*S}{7^X}\) where R is an integer. \(R = \frac{2^2 * 5*7*a*S}{7^X}\)

So whatever X is, 7^X will get cancelled out by the numerator and we will be left with something. That something will include 5 since only 7s will be cancelled out from the numerator. Hence R is divisible by 5. _________________

Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R [#permalink]

Show Tags

20 Nov 2014, 21:02

Expert's post

VeritasPrepKarishma wrote:

jlgdr wrote:

Both together

I have the quesiton: is 140S / 5Q an integer where Q = 7^X

Well x could be anything and hence not sufficient

E

But I'm pretty sure there's something I'm missing, would anybody please clarify what's wrong with this line of reasoning?

Cheers J

Yeah, the concept here is quite basic but we often overlook it.

Think about it: Is \(3^5 * 7^6 * 11^3\) divisible by 13? The answer is simply 'No'.

For the numerator to be divisible by the denominator, the denominator MUST BE a factor of the numerator. 3^5 is only five 3s. 7^6 is only six 7s. 11^3 is only three 11s. In the entire numerator, there is no 13 so the numerator is not divisible by 13.

On the other hand, is \(3^5 * 7^6 * 11^3 * 13\) divisible by 13? Yes, it is. 13 gets cancelled and the quotient will be \(3^5 * 7^6 * 11^3\).

Is 2^X divisible by 3? No. No matter what X is, you will only have X number of 2s in the numerator and will never have a 3. So this will not be divisible by 3.

On the same lines, in this question,

Given that \(R = \frac{(140a)*S}{7^X}\) where R is an integer. \(R = \frac{2^2 * 5*7*a*S}{7^X}\)

So whatever X is, 7^X will get cancelled out by the numerator and we will be left with something. That something will include 5 since only 7s will be cancelled out from the numerator. Hence R is divisible by 5.

Responding to a pm:

"The second statement says that R is multiple of 5 for any x . So why are we combining the two statements ? Could you please help."

From the second statement alone, all we know is that Q is a power of 7. We have no idea about what R will be. Statement 1 tells us that P = 140 i.e. a multiple of 5. Hence we know that P must have 5 as a factor. Hence R will have a factor of 5 too. _________________

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...