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As I suspected, the fact that we know the values of two possible solutions does not determine the equation.

1) If x = 20, when P = 0, this means that 20 is one of the solutions of the equation. Also gives the equation 400d+20e+f=0, when substituting x and P in the equation. And that is all INSUFF

2) If P has the greatest value when x=25, this means that x is the x of the vertex. The x of the vertex has the formula -b/2a, considering (ax^2 + bx + c) Thus you have -e/2d = 25. As d<0 you have e = 50d And that is all INSUFF

However, if you put both stmts together, you can remember that the x of vertex is exactly between the two "zeros-solutions" of the equation. Thus, if one solution is 20, and the x of the vertex is 25, the other solution is 30.

So you might be attempt to think: Bingo! I know that when I sum the solutions I have the opposite of e, and when I multiply them I have f. But in fact, this is only valid when the term which multiply \(x^2\) is 1, or d=1.

In fact the right formula is (considering ax^2 + bx + c; x1 and x2 as the solutions) \(x1 + x2 = -b/a\) \(x1*x2 = c/a\)

This means that since the solutions are 20 and 30, the equation is x^2 - 50x + 600 Or, because d<0, -x^2 + 50x -600 But it could be: -2x^2 + 100x - 1200 -3x^2 + 150x - 1800 -4x^2 + 200x - 2400 . . .

When we read that f is a constant, this means that f is a number, but it can be -600, -1200, ... f is a constant means that f, at the end of the solution will be a number and not a equation depending on other variables.

So, even with the two stmts together we only find a ration of d to e to f

ANSWER E.

If you liked the explanation, consider a kudo I am trying to access those GMATClub tests!!!

hmmm, good for me that 2 days before the G-day I get to know I'm still lacking in few basic concepts

I always thought that if we increase the coefficients of x^2 and x along with C, they will give us a magnified graph, which will not be the same as previous one. Thanks for the explanation coelholds. Atleast now i know one more thing. You rightfully deserve kudos for this one

2)When x=25, P has the greatest value >> maxima = 0 >> 2dx+e = 0 (take derivative as slope = 0 @ maxima) >> 50d+e=0 >> insufficient

On combining both >> since d is -ve (refer figure attached) >>

Case I is ruled out

in case II >> only single root exists ie >> both are possible 20 or 25 = -d/2e >> insufficient in case III >> both roots are real >> so possiblity is at x=20 f(x) = 0 and at x = 25 f(x) = maximum >> insufficient Ans E

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Bhushan S. If you like my post....Consider it for Kudos

Don't worry about. Unless you are a 800 test taker you will always learn something new in the previous day of the test, or even in the day. That is my opinion.