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Veritas Prep 10 Year Anniversary Promo Question #5

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Re: Veritas Prep 10 Year Anniversary Promo Question #5  [#permalink]

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New post 20 Sep 2012, 00:56
1
yashii9 wrote:
Vips0000 wrote:
yashii9 wrote:
A.

q(R+S) =(R+S)

Let the value of r and s be anything q will always be equal to 1.


Not correct approach.
what if R=S=0 then?
Q(R+S) = 1*(R+S) => Q=1 ?
as well as Q(R+S)=10*(R+S) =>Q=10 ?
and it can be shown to be true for any other number... So first you can not claim Q=1 this way.
Second, the conclusion u reached
Q(R+S) =(R+S) => Q=1
this is arrived when you've divided both sides by (R+S). Which can not be done if R+S = 0, as it will be illegal operation to divide by 0. Unless you are certain about r+s <>0, you can not reach to conclusion.


Dear vips - if r or s is 0... the equation will be 0 on both side..and u can not get the value of q.

if either r or s is 0 u still get q=1
isnt it?

Also if we are to consider this as an unanswerable case - then even point 2 wont help. r not equal to -s...could well mean r =s =0 again no answer and then reach the conclusion to be E.

What do you say?


Try to follow these steps:
Q(R+S) = 1*(R+S) => Q(R+S) - (R+S) = 0
=>
(Q-1)(R+S)=0
Apply number property here.. This statement (Q-1)(R+S)=0 means
it is possible only if,
case a. Q-1 =0
or
case b. R+S = 0
or
case c. both are 0

We dont have any clue as to which one of these 3 is true.
Now if use statement 2 in question R <> -S which means R+S <> 0, hence if we combine it to the 3 possiblities found in statement 1 : We get choce b is not possible , choice c is not possible. Hence we are left with choice a and Q-1 has to be 0.
So combining the statement gives u ans.. not individual statements.
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Re: Veritas Prep 10 Year Anniversary Promo Question #5  [#permalink]

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New post 20 Sep 2012, 01:25
Good explanation.
However I dont understand how can u solve this as algebra?
point 2 only says that R <> -S but it does not talk abt the value of either r or s.....it can be fraction, positive or zero
unless the values of r and s are specified how do we find value of Q?

Hence E.

I might be thinking a bit too much but its better to get this clarified here than screwing up in the exam. :)
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Re: Veritas Prep 10 Year Anniversary Promo Question #5  [#permalink]

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New post 20 Sep 2012, 01:57
1
yashii9 wrote:
Good explanation.
However I dont understand how can u solve this as algebra?
point 2 only says that R <> -S but it does not talk abt the value of either r or s.....it can be fraction, positive or zero
unless the values of r and s are specified how do we find value of Q?

Hence E.

I might be thinking a bit too much but its better to get this clarified here than screwing up in the exam. :)


yes, It can be fraction, positive or zero.. but whatever it is, one thing we know is that R <> -S which means R+S <> =0. Which means for (Q-1)(R+S) to be equal to 0, Q has to be 1.
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Re: Veritas Prep 10 Year Anniversary Promo Question #5  [#permalink]

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New post 20 Sep 2012, 08:34
1

We have a winner!



Winner:

Vips0000

Official Explanation:

Answer is C

Statement 1 at first appears to be sufficient, as one can factor out the common \(q\) to get: \(q(r + s) = r + s\). This would suggest that \(q\) equals one. But you must ask about statement 2, “why are you here?”. Statement 2 is clearly not sufficient, but it sheds a bit of light on something you may not have considered with statement 1. If \(r\) were to equal \(-{s}\), then \(r + s\) would be 0. And in that case, our revised equation for statement 1, \(q(r + s) = r + s\), is true for any value of \(q\). So statement 2 is critical - it shows us that statement 1 is not sufficient alone, but that along with statement 2 we can rest assured that \(q\) is 1. The correct answer is C.
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Re: Veritas Prep 10 Year Anniversary Promo Question #5  [#permalink]

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New post 20 Sep 2012, 09:01
Bunuel wrote:

Winner:

Vips0000



YAY!!! :twisted:
Thanks.. Hopefully I'd be able to prepare much better with this package!! ;)
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Re: Veritas Prep 10 Year Anniversary Promo Question #5  [#permalink]

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New post 20 Sep 2012, 09:18
thevenus wrote:
why not (A)?
(i)q(r+s)=(r+s)
q=? ...(doesn't matter what r or s is , we have to find the q ;not to check whether is positive negative or zero);the value cant be found out insufficient so (A) is
sufficient
(ii) lacks info-insufficient


Exactly, even i dont understand why everybody is trying to solve for r & s, when its clear from the 1st equation that q(r+s) = r+s => q =1.

And answer would be A. But i'm worried,looking at other answers!!!
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Re: Veritas Prep 10 Year Anniversary Promo Question #5  [#permalink]

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New post 20 Sep 2012, 10:12
1
shekbhi wrote:
thevenus wrote:
why not (A)?
(i)q(r+s)=(r+s)
q=? ...(doesn't matter what r or s is , we have to find the q ;not to check whether is positive negative or zero);the value cant be found out insufficient so (A) is
sufficient
(ii) lacks info-insufficient


Exactly, even i dont understand why everybody is trying to solve for r & s, when its clear from the 1st equation that q(r+s) = r+s => q =1.

And answer would be A. But i'm worried,looking at other answers!!!


What is the value of integer \(q\)?

(1) \(qr + qs = r + s\) --> \(q(r+s)=r+s\) --> \(q(r+s)-(r+s)=0\) --> \((r+s)(q-1)=0\) --> EITHER \(r+s=0\) (\(r=-s\)) OR \(q-1=0\) (\(q=1\)). Notice that if \(r+s=0\), then \(q\) can take ANY value, not necessarily 1. Not sufficient.

(2) \(r\neq{-s}\). Clearly insufficient.

(1)+(2) Since from (2) \(r\neq{-s}\), then from (1) it must be true that \(q=1\). Sufficient.

Answer: C.

Hope it's clear.
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Re: Veritas Prep 10 Year Anniversary Promo Question #5  [#permalink]

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New post 18 Jan 2019, 18:03
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Re: Veritas Prep 10 Year Anniversary Promo Question #5   [#permalink] 18 Jan 2019, 18:03

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