eaakbari wrote:IMO A
I tend to use calculus to solve this type (a very basic concept of calculus though)
Info -
(x) = k(x - a)(x - b)
Simplifying
(x) = kx^2 -kx(a+b) +abx
not quite right, simplifying we get P(x)= k*x^2-k(a+b)*x+kab
d(x) = 2kx - k(b+a)
equating to zero for maximum and dividing equation by k (constant)
0 = 2x - (b+a)
x = (b+a)/2 is when x is maximum
Really? Have you passed calculus or failed it?
P`(x)=2kx-k(a+b)=0, 2kx=k(a+b), 2x=a+b, x=(a+b)/2 <== This is the extremum point (where function P may get minimum or maximum value, we don't know yet).
Take the second derivative d^2P/dx^2=P``(x) or 2k. Now depending on the sign of k we get close to our answer. If k<0, we will have maximum value in the point x=(a+b)/2 and by substituting this into P(x) we get P(x) maximum.
The condition in the question says that k<0, hence P is maximized in the point x=(a+b)/2.
(I) Clearly implies that a & b are equal to 20 and 1020
Hence we have a & b values, hence Suff
Not right again. St(1) P(20) = P(1,020) = 0 supplies as with information (remember our function is parabolic from k*x^2-k(a+b)*x+kab, where a,b,k are constants and k cancelled out leaves us with x^2-(a+b)x+ab, similar to GMAT Foiler) that there's some in-between point between symmetric coordinates of x on the cross-over of y=P(x)=0. That is the half of interval [20;1020] will be our maximum point. Hence our function will be maximzed at x=(1020-20)/2=500. By plugging this into our previously found (derived) extremum point when x=(a+b)/2 we obtain a+b=1000.
Now to finalize our answer we would need individual values for a, b and k. Because to find P(500)=k*500^2-k(1000)*500+kab we are missing exactly a,b,k.
Since in St(1) we are given P(20)=0 and P(1020)=0 we may attempt to set several equations to find the three unknown variables (a,b,k).
P(20)=k*20^2-k(a+b)*20+kab=0 and P(1020)=k*1020^2-k(a+b)*1020+kab=0. We know that k is not 0 and we are left with 20^2-(a+b)*20+ab=1020^2-(a+b)*1020+ab. Cancelling ab on both sides we get, 20^2-(a+b)*20=1020^2-(a+b)*1020 or 1040*(a+b)=1020^2-20^2=(1020-20)*(1020+20)=1000*1040 and a+b=1000
I have gone through so much tedious explanation with calculations to show that St(1) is not sufficient, i.e. St(1) In Not Right.
St(2) P(120) = 9,000 Cannot be right as well, as we are put to find the values of a,b,k again. So St(2) alone is not sufficient.
Combining both statements(1&2), we still have only two equations on the side of St(1) which are equivalent (we have seen that) and one equation on the side of St(2). So three variables and two equations. Unless we introduce a new parameter, this may not solvable. The parametric equation will not indicate one unique answer, rather the set of values underpinned by parameter (GMAT wants one, unique answer)
Answer is E (the question is not solvable with any statement taken alone, neither combined).
