Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 31 May 2016, 11:35

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Director
Joined: 07 Jun 2004
Posts: 613
Location: PA
Followers: 3

Kudos [?]: 523 [1] , given: 22

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

### Show Tags

01 Feb 2011, 11:01
1
KUDOS
2
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

57% (02:21) correct 43% (01:12) wrong based on 130 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

_________________

If the Q jogged your mind do Kudos me : )

Math Expert
Joined: 02 Sep 2009
Posts: 33104
Followers: 5786

Kudos [?]: 71025 [1] , given: 9857

### Show Tags

01 Feb 2011, 11:23
1
KUDOS
Expert's post
rxs0005 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

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.

_________________
Director
Joined: 07 Jun 2004
Posts: 613
Location: PA
Followers: 3

Kudos [?]: 523 [0], given: 22

### Show Tags

01 Feb 2011, 14:10
Bunuel thanks

the way i approached this DS was

i see P / Q = R / S as a proportion

after S1 and S2 let us say p / q is

140 / 49 = 2.87

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
_________________

If the Q jogged your mind do Kudos me : )

Math Expert
Joined: 02 Sep 2009
Posts: 33104
Followers: 5786

Kudos [?]: 71025 [0], given: 9857

### Show Tags

01 Feb 2011, 14:22
Expert's post
rxs0005 wrote:
Bunuel thanks

the way i approached this DS was

i see P / Q = R / S as a proportion

after S1 and S2 let us say p / q is

140 / 49 = 2.87

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).
_________________
Manager
Joined: 27 Oct 2010
Posts: 189
Followers: 1

Kudos [?]: 10 [0], given: 20

### Show Tags

01 Feb 2011, 16:23
(P / Q) * S = R

Stmt 1 says P = 140 * X

(140 * X / Q) * S = R

We do not know anything about Q. So Stmt 1 is insuff.

Stmt 2 says Q can be 7, 49 ... But we do not know anythng abt P. So Stmt 2 is insuff.

Combining both, we know R is +ve integer. So, Q has to be 7 and R is multiple of 5. C
Manager
Joined: 17 Oct 2012
Posts: 58
Followers: 0

Kudos [?]: 7 [0], given: 0

### Show Tags

01 Jul 2013, 14:15
Bunuel wrote:
rxs0005 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

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.

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?
Math Expert
Joined: 02 Sep 2009
Posts: 33104
Followers: 5786

Kudos [?]: 71025 [0], given: 9857

### Show Tags

01 Jul 2013, 14:20
Expert's post
BankerRUS wrote:
Bunuel wrote:
rxs0005 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

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.

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 $$7^{positive \ integer}$$.

Hope it's clear.
_________________
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 43

Kudos [?]: 465 [0], given: 355

### Show Tags

30 Jan 2014, 17:43
Bunuel wrote:
rxs0005 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

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.

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?

Cheers
J
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6585
Location: Pune, India
Followers: 1796

Kudos [?]: 10822 [0], given: 213

### Show Tags

30 Jan 2014, 21:50
Expert's post
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?

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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6585 Location: Pune, India Followers: 1796 Kudos [?]: 10822 [0], given: 213 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R   [#permalink] 20 Nov 2014, 21:02
Similar topics Replies Last post
Similar
Topics:
2 If p, q, and r are integers, is pq + r even? 3 07 Apr 2013, 03:26
8 In the rectangular coordinate system, are the points (p,q) and (r,s) 8 27 Nov 2011, 18:54
If p,q and r are integers, is pq+r even? 1) p+r is even 3 10 Mar 2011, 13:18
4 If P, Q, R, and S are positive integers, and P/Q = R/S, is R divisible 12 02 Jul 2008, 22:20
2 If P, Q, R, and S are positive integers, and P/Q=R/S, is R 6 01 Sep 2006, 06:21
Display posts from previous: Sort by