-
sui generis
- Senior | Next Rank: 100 Posts
- Posts: 89
- Joined: Wed Dec 29, 2010 4:07 am
- Thanked: 1 times
- Followed by:2 members
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 ?
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 ?
.
