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Re: Ratio Tough one [#permalink]
01 Feb 2011, 10:23

1

This post received 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.

Re: Ratio Tough one [#permalink]
01 Feb 2011, 13:22

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). _________________

Re: Ratio Tough one [#permalink]
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 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?

Re: Ratio Tough one [#permalink]
01 Jul 2013, 13:20

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.

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 \(7^{positive \ integer}\).

Re: Ratio Tough one [#permalink]
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 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?

Re: Ratio Tough one [#permalink]
30 Jan 2014, 20:50

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? 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]
20 Nov 2014, 20:02

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. _________________

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