Basic but tricky concept of DS

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Basic but tricky concept of DS

by sui generis » Sun Nov 13, 2011 12:03 am
I am presenting the problem in the simpler algebraic version so as to emphasize on the concept.

Q1) What is x ?

(1) (x-2)(x-3) = 0
(2) (x-3)(x-4) = 0

I want to know the fundamental approach for DS questions.

Soln:
St (1) . insufficient as x could be either 2 or 3
Similarly St (2) . insufficient as x could be either 3 or 4
St (1 & 2) . x = 2 or 3 AND x =3 or 4
how to proceed next ?


Similarly another question testing the same concept which I have correctly solved:
Q2) Is pq =1 ?

(1) p2q = p (it is p squared times q = p)
(2) q2p = q (it is q squared times p = q)

Soln:

St1. p(pq-1) = 0 ; either p=0 or pq =1 ; => pq = 0 or 1 so not sufficient
St2. q(pq-1) = 0 ; either q=0 or pq =1 ; => pq = 0 or 1 so not sufficient
St1&2: p(pq-1) = 0 AND q(pq-1) =0
pq=1 OR p=q=0
=> pq = 1 or 0 therefore insufficient.
Answer : E


My question rather doubt is how to approach question 1 in the lines of reasoning of question 2 ?
Source: — Data Sufficiency |

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by shankar.ashwin » Sun Nov 13, 2011 12:36 am
When you combine the 2 statements you should make sure the solutions satisfy both equations.

If you have 3 values of x combing the 2 statements, 2,3 and 4.
value of x = 4 would not hold true in statement 1
similarly, value of x = 2 will not hold true in statement 2.

SO x can only take '3' to satisfy both, you can't have all 3 possibilities as the conditions would not hold true. If you still get multiple answer, a finite solution cannot be determined.
sui generis wrote:I am presenting the problem in the simpler algebraic version so as to emphasize on the concept.

Q1) What is x ?

(1) (x-2)(x-3) = 0
(2) (x-3)(x-4) = 0

I want to know the fundamental approach for DS questions.

Soln:
St (1) . insufficient as x could be either 2 or 3
Similarly St (2) . insufficient as x could be either 3 or 4
St (1 & 2) . x = 2 or 3 AND x =3 or 4
how to proceed next ?


My question rather doubt is how to approach question 1 in the lines of reasoning of question 2 ?

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by sui generis » Sun Nov 13, 2011 12:46 am
Thanks bro, got it.

OA here is C where x =3.

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by zooki » Sun Nov 13, 2011 9:55 am
for Q2) Is pq =1 ?

(1) p2q = p (it is p squared times q = p)
(2) q2p = q (it is q squared times p = q)

Can someone please explain why the answer is NOT C?

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by neelgandham » Sun Nov 13, 2011 10:36 am
zooki wrote:for Q2) Is pq =1 ?

(1) p2q = p (it is p squared times q = p)
(2) q2p = q (it is q squared times p = q)

Can someone please explain why the answer is NOT C?
I reckon the answer is C
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by sui generis » Sun Nov 13, 2011 10:39 am
The answer to the discussed question is E. Here is how:

Taking Statement 1&2 together we get: p(pq-1) = 0 AND q(pq-1) =0
the both equations are only true in two cases:
case 1: pq=1
case 2: p=q=0

Now lets try to find value of the pq (as asked in question) in the above two cases

case 1: pq =1
case 2: pq =0

as there is no unique value of pq so both statements are not sufficient.

Hence E.

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by mankey » Sun Nov 13, 2011 10:40 am
Please explain this one.

Thanks.

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by shankar.ashwin » Sun Nov 13, 2011 10:48 am
Q2) Is pq =1 ?

(1) p^2*q = p (it is p squared times q = p)
(2) q^2*p = q (it is q squared times p = q)


The 2 conditions work for both p=q=1 (or) p=q=0.

Should be E IMO

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by neelgandham » Sun Nov 13, 2011 11:25 am
neelgandham wrote:
zooki wrote:for Q2) Is pq =1 ?

(1) p2q = p (it is p squared times q = p)
(2) q2p = q (it is q squared times p = q)

Can someone please explain why the answer is NOT C?
I reckon the answer is C
Oops I am a perfect example of 'glazed-eyes'. I read it as Can someone please explain why the answer is NOT E:lol:.

It is indeed E
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