Hi Sir thanks for the explanation. I used visualization. for testing C, I drew to check how many parabolas are possible that intersect y axis at -20 and have y min at -36. There were 2: one with vertex in the Q3 and one with vertex in Q4. So eliminated option C since p+q can be +ve or -ve
Option B says minimum value of y is -36. So let's take -35=(x-p)(x-q) Either (x-p) or (x-q) is negative. Let's assume, x-p0? We can't say for sure. If we combine statements 1 and 2 We know the curve cuts at -20. So at x=0 y=-20 -20= pq and x>p and x
Hi Sir thanks for the explanation. I used visualization. for testing C, I drew to check how many parabolas are possible that intersect y axis at -20 and have y min at -36. There were 2: one with vertex in the Q3 and one with vertex in Q4. So eliminated option C since p+q can be +ve or -ve
Excellent Kinshuk!
well at least I know I won't score 750 on GMAT............even if I can reason this out, no way I can do this in 2 minutes
Can you use calculus , the first derivation to find out the x value?? Reply if you are noted ..
It's strange that I got this right, but still can't score above 620 on the GMAT. And I got it right using the same logic used by wizako.
Option B says minimum value of y is -36.
So let's take -35=(x-p)(x-q)
Either (x-p) or (x-q) is negative.
Let's assume, x-p0? We can't say for sure.
If we combine statements 1 and 2
We know the curve cuts at -20. So at x=0 y=-20
-20= pq and x>p and x
That's exactly what my situation is too