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# Veritas Prep 10 Year Anniversary Promo Question #5

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

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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?

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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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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Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
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Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
Re: Veritas Prep 10 Year Anniversary Promo Question #5  [#permalink]

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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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20 Sep 2012, 08:34
1

We have a winner!

Winner:

Vips0000

Official Explanation:

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

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20 Sep 2012, 09:01
Bunuel wrote:

Winner:

Vips0000

YAY!!!
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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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.

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

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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.

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.

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

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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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