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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]

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01 Feb 2011, 11:01
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57% (01:17) correct 43% (01:22) wrong based on 327 sessions

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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, 11:23
3
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.

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Re: Ratio Tough one  [#permalink]

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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
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Re: Ratio Tough one  [#permalink]

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01 Feb 2011, 14:22
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).
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Re: Ratio Tough one  [#permalink]

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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
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Re: Ratio Tough one  [#permalink]

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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?
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Re: Ratio Tough one  [#permalink]

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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 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.
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Re: Ratio Tough one  [#permalink]

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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
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Re: Ratio Tough one  [#permalink]

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30 Jan 2014, 21:50
1
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, 21:02
1
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, 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 cross-multiplying 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, 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

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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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05 Oct 2018, 16:53
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

if we rewrite the equation to be P * S = R * Q

1) P is divisible by 140. P is a multiple of 140

140 * S = R * Q. Let S=1, let R = 140, and let Q = 1 then yes divisible. Let S = 1, let R = 1 and let Q = 140. No not divisible.

Insufficient

2) it provides information on Q only which is not enough to decide.

Combine both now

140 * S = R * $$7^x$$

20 * 7 * S = R * $$7^x$$

P can be 140,280,420 etc.. the 7 will always have a power of 1.

R is divisible by 5 sufficient.

C
Re: If P, Q, R, and S are positive integers, and P/Q = R/S, is R &nbs [#permalink] 05 Oct 2018, 16:53
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# If P, Q, R, and S are positive integers, and P/Q = R/S, is R

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