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If P, Q, R, and S are positive integers, and P/Q = R/S, is R [#permalink]
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01 Feb 2011, 10:01
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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
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Re: Ratio Tough one [#permalink]
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01 Feb 2011, 10:23
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.
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Re: Ratio Tough one [#permalink]
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01 Feb 2011, 13:10
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
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Re: Ratio Tough one [#permalink]
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01 Feb 2011, 13:22



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Re: Ratio Tough one [#permalink]
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01 Feb 2011, 15:23
(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



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Re: Ratio Tough one [#permalink]
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01 Jul 2013, 13:15
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?



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Re: Ratio Tough one [#permalink]
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01 Jul 2013, 13:20
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.
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Re: Ratio Tough one [#permalink]
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30 Jan 2014, 16:43
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



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Re: Ratio Tough one [#permalink]
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30 Jan 2014, 20:50
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.
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Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R [#permalink]
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20 Nov 2014, 20:02
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.
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If P, Q, R, and S are positive integers, and P/Q = R/S, is R [#permalink]
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20 Mar 2017, 21:43
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.
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Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R [#permalink]
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04 Nov 2017, 10:56
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
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Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R
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