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

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19 Sep 2012, 10:00

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A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

61% (01:42) correct
39% (00:29) wrong based on 105 sessions

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Veritas Prep 10 Year Anniversary Promo Question #5

One quant and one verbal question will be posted each day starting on Monday Sept 17th at 10 AM PST/1 PM EST and the first person to correctly answer the question and show how they arrived at the answer will win a free Veritas Prep GMAT course ($1,650 value). Winners will be selected and notified by a GMAT Club moderator. For more questions and details please check here: veritas-prep-10-year-anniversary-giveaway-138806.html

To participate, please make sure you provide the correct answer (A,B,C,D,E) and explanation that clearly shows how you arrived at it. Winners will be announced the following day at 10 AM Pacific/1 PM Eastern Time.

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

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19 Sep 2012, 10:00

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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. _________________

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

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19 Sep 2012, 10:09

(1) qr + qs = r + s <=> q (r+s) = (r+s) <=> r + s = 0 or q = 1 Can't determine q -> not sufficient (2) r is different to -s => r+s <> 0 . No information about q -> not sufficient

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

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19 Sep 2012, 10:09

The answer is: ( C) Solution: What is the value of integer ?

We have: (1) qr+qs = r+s Then q(r+s) = r+s => (q-1)(r+s)=0 => q=1 or r=-s. So it’s not Sufficient Together with (2) r≠-s, we have q=1.So the answer is C

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

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19 Sep 2012, 10:16

Ans is C. stmt 1) qr + qs = r + s => q(r+s) = (r+s) => q=1 if (r+s) not equal to 0 OR q does not have unique value if (r+s)=0 stmt 2) r not equal to -s => (r+s) not equal to 0 Hence, to find unique value of q stmt 1) & 2) both necessary

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

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19 Sep 2012, 10:38

I will choose Answer C:

Statement 1: qr + qs = r + s q(r + s) = r + s

most of the people prefer to divide both sides by (r + S) and hence conclude that q = 1. As per my understanding, GMAT does not prefer that way because it eliminates one solution. (if one does as this, one will miss (r + s) = 0 as answer)

As per my understanding, q(r + s) = r + s q(r + s) - (r + s) = 0 (r + s)(q - 1) = 0

either (r + s) = 0 or (q - 1) = 0. if (r + s) = 0 then q can have any value. statement 1 insufficient.

Statement 2: r is not equal to -s does not provide any information about value of q statement 2 is insufficient.

statement 1 and 2: from statement 1, we know that either (r + s) = 0 or (q - 1) = 0 but from statement 2, we know that r NE -s, so (r + s) NE 0

substitute this in statemet 1. we will get (q - 1) = 0

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

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19 Sep 2012, 10:46

Ans : C R+c should not be 0 in the 1st equation to get a value for Q This is confirmed by 2nd statement. " R+c not= 0" We need both statements together so ans is C

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

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19 Sep 2012, 11:41

C. - (2) is not sufficient since irrelevant. BD out - (1) q (r+s) = r + s (q-1)(r+s)= 0 => Either (q-1) = 0 or (r+s) = 0. Not sufficient in case of (r+s) = 0. A out - Both (1)&(2), from (2) we have r+s <>0, combine with (1) we have q = 1: sufficient

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

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19 Sep 2012, 20:18

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 _________________

" Make more efforts " Press Kudos if you liked my post

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

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19 Sep 2012, 20:48

1) qr + qs = r + s thus, q (r+s) = (r+s). Now you can't divide since r+s=0. If r+s is not 0, then q=1. Hence insufficient 2) r+s is not = 0. no information about q.

Both combined - we can say that q = 1. Thus, answer is C

Last edited by prep on 19 Sep 2012, 20:51, edited 1 time in total.

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

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19 Sep 2012, 21:09

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

A will only be the answer if statement (1) is sufficient to provide one exact answer to the question, what is the value of q? _________________

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

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19 Sep 2012, 22:29

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. _________________

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

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19 Sep 2012, 22:33

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.

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

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20 Sep 2012, 00:15

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Ans:C Statement1:given qr+qs=r+s =>q(r+s)=r+s =>q=1 only when r not equal to -s If r=-s,then q is undefined Hence Statement1 is insufficient Statement2:given r not equal to -s No information about q Hence Statement2 is insufficient

By combining both statements 1 and 2 q=1 when r not equal to -s Hence C is the correct answer

Last edited by sdpp143 on 20 Sep 2012, 00:24, edited 1 time in total.

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