Author 
Message 
TAGS:

Hide Tags

Director
Joined: 07 Jun 2004
Posts: 605
Location: PA

If P, Q, R, and S are positive integers, and P/Q = R/S, is R [#permalink]
Show Tags
01 Feb 2011, 11:01
Question Stats:
56% (01:14) correct 44% (01:18) wrong based on 278 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
If the Q jogged your mind do Kudos me : )



Math Expert
Joined: 02 Sep 2009
Posts: 46297

Re: Ratio Tough one [#permalink]
Show Tags
01 Feb 2011, 11:23
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 5s from P and S (if there are any) are reduced by 5s 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 5s 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 5s 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 07 Jun 2004
Posts: 605
Location: PA

Re: Ratio Tough one [#permalink]
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: 46297

Re: Ratio Tough one [#permalink]
Show Tags
01 Feb 2011, 14:22



Manager
Joined: 27 Oct 2010
Posts: 135

Re: Ratio Tough one [#permalink]
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: 52

Re: Ratio Tough one [#permalink]
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 5s from P and S (if there are any) are reduced by 5s 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 5s 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 5s 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?



Math Expert
Joined: 02 Sep 2009
Posts: 46297

Re: Ratio Tough one [#permalink]
Show Tags
01 Jul 2013, 14:20
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 5s from P and S (if there are any) are reduced by 5s 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 5s 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 5s 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 \(7^{positive \ integer}\). Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



SVP
Joined: 06 Sep 2013
Posts: 1881
Concentration: Finance

Re: Ratio Tough one [#permalink]
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 5s from P and S (if there are any) are reduced by 5s 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 5s 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 5s 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? Cheers J



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8102
Location: Pune, India

Re: Ratio Tough one [#permalink]
Show Tags
30 Jan 2014, 21:50
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.
_________________
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: 8102
Location: Pune, India

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
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



Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1314
Location: Malaysia

If P, Q, R, and S are positive integers, and P/Q = R/S, is R [#permalink]
Show Tags
20 Mar 2017, 22:43
rxs0005 wrote: If P, Q, R, and S are positive integers, and \(\frac{P}{Q} = \frac{R}{S}\), is R divisible by 5?
(1) P is divisible by 140
(2)\(Q = 7^x\), where x is a positive integer OFFICIAL SOLUTION Let's begin by analyzing the information given to us in the question: If P, Q, R, and S are positive integers, and \(\frac{P}{Q} = \frac{R}{S}\), is R divisible by 5 ? It is often helpful on the GMAT to rephrase equations so that there are no denominators. We can do this my crossmultiplying as follows: \(\frac{P}{Q} = \frac{R}{S}\) → \(PS=RQ\) Now let's analyze Statement (1) alone: P is divisible by 140. Most GMAT divisibility problems can be solved by breaking numbers down to their prime factors (this is called a "prime factorization"). The prime factorization of 140 is: \(140=2*2*5*7\). Thus, if P is divisible by 140, it is also divisible by all the prime factors of 140. We know that P is divisible by 2 twice, by 5, and by 7. However, this gives us no information about R so Statement (1) is not sufficient to answer the question. Next, let's analyze Statement (2) alone: \(Q = 7^x\), where x is a positive integer. From this, we can see that the prime factorization of Q looks something like this: \(Q=7*7*7......\) Therefore, we know that 7 is the only prime factor of Q. However, this gives us no information about R so Statement (2) is not sufficient to answer the question. Finally, let's analyze both statements taken together: From Statement (1), we know that P has 5 as one of its prime factors. Since 5 is a factor of P and since P is a factor of PS, then by definition, 5 is a factor of PS. Recall that the question told us that \(PS=RQ\). From this, we can deduce that PS must have the same factors as QR. Since 5 is a factor of PS, 5 must also be a factor of QR. From Statement (2), we know that 7 is the only prime factor of Q. Therefore, we know that 5 is NOT a factor of Q. However, we know that 5 must be a factor of QR. The only way this can be the case is if 5 is a factor of R. Thus, by combining both statements we can answer the question: Is R divisible by 5? Yes, it must be divisible by 5. Since BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient, the correct answer is C.
_________________
"Be challenged at EVERY MOMENT."
“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”
"Each stage of the journey is crucial to attaining new heights of knowledge."
Rules for posting in verbal forum  Please DO NOT post short answer in your post!



Senior Manager
Joined: 15 Jan 2017
Posts: 361

Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R [#permalink]
Show Tags
04 Nov 2017, 11:56
Did this by plugging values:
1) P/140 = no remainder. So P/Q = no remainder And then we get PS = RQ > so PS = R (140) (Lets say that Q is 140). Either way, we don't know anything about R. NS
2) Q =7^x > cn be 7 to power 1, 2,3..least would be 7 to power 1. P/7 = R/ S > PS = RQ, again R is what? NS
1) + 2) > 140 / 7 (coz P's least value is likely to be 140 and Q's least value is 7).
140/7 = R/S S(140) = R(7) S (4x5) = R (cancelling 7 on both sides)Therefore R is divisible by 5. Ans C
Kudos please, if you found this useful!




Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R
[#permalink]
04 Nov 2017, 11:56






