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Bunuel
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GMAT 2: 760 Q51 V40
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We are told that p(2) = 41 and p(5) = 26 and
p(t) = rt -5(t)^2 + b

We can set some equations given the information and the function.

(1) 41 = 2t - 5(2)^2 + b -> 41 = 2r - 20 + b

(2) 26 = 5t - 5(5)^2 + b -> 26 = 5r - 125 + b

Subtract eq. 2 from eq.1 to get rid of b and find value of r.

15 = -3r + 105
3r = 90
r = 30

Plug 30 into either equation to find value of b

41 = 2(30) - 20 + b
41 = 60 - 20 + b
41 = 40 + b
b = 1

Now that we have a value for r and b, we can input
p(4) into our function.

p(t) = rt - 5(t)^2 + b
p(4) = (30)(4) - 5(4)^2 + 1
p(4) = 120 - 80 + 1
p(4) = 41

Option D
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My half a second approach:

We are given that p(2)=41 and p(5)=26 and asked value of p(4) would not go too far from 26 and 41.
Thus, possible values are (option B)26, (option C)39, and (option D)41


If I draw the graph of p(t)=rt-5t^2+b and check for the value, I realize that p(2) and p(4) would touch the same point of the given parabola.
Therefore, option D i.e. 41 is the correct option.
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