Hardest SAT question ever (math)

Test Preparation SAT Preparation
1

<p>So i was going thru the Blue book today and found this weird question. I tried to solve it but i was not able to, AFTER 10 minutes of trying.
Heres the Question:</p>

<p>h(t)=c-(d-4t)^2
At time t=0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds was given by the function h above, in which c and d are positiveconstants. If the ball reached its maximum height of 106 feet at time t=2.5, what was the height of the ball, in feet, of the ball at time=1?</p>

<p>Please help :(</p>

<p>edit: the answer is 70 if that helps.</p>

2

<p>this is actually not that hard…
ok, so you have points, (0, 6) and (2.5, 106)</p>

<p>plug them into the equation and you get 6=c-d^2 and 106=c-(d^2)+20d-100</p>

<p>at this point, you can either substitute for c or do something to cancel out one of the variable, but either way you get c= 106 and d=10. </p>

<p>now plug 1 into the equation, so you have:
h(1)= 106- (10-4(1))^2 = 70</p>

3

<p>^exactly</p>

<p>system of equations</p>

4

<p>Well the problem gives you a few things to start out with. So you know at time 0, the height is 6. So lets put that into the h(t)=c-(d-4t)^2 equation.</p>

<p><a href=“0”>code</a> = c -(d-0)^2… We know h(0) = 6 so
6 = c - d^2


</p>

<p>Know lets see the other part of the problem they gave...
At time 2.5, height = 106
So lets put that info into h(t)=c-(d-4t)^2 equation</p>

<p>

h(2.5) = c - (d-10)^2
h(2.5) = c - (d^2-20d+100)… distribute the negative
h(2.5) = c - d^2 +20d -100
We know from before that (c-d^2 = 6)
So H(2.5) = 6 + 20d -100. We also know h(2.5) = 106
106 = 6 + 20d -100… Solve through
d = 10


&lt;/p&gt;

&lt;p&gt;Okay know we know d. Since we know one of the variables we can solve for the other. 
Go back to the 6 = c -d^2. Plug in for  d and you get c = 106.&lt;/p&gt;

&lt;p&gt;Finally, do h(1) using our found values of c and d
h(1) = 106 - (10-4)^2
h(1) = 106 - 36
h(1) = 70&lt;/p&gt;
5

<p>Got it, but its still preety hard.</p>

6

<p>You can also do it another way, though it is not as practical. You can take the derivative of the function dt/dh(t). You have the maxima of t=2.5 so you know that
h’(2.5) = 0.
h(t)=c-d^2+8dt-16t^2
h’(t)=8d-32t
0=8d-32(2.5)
8d=80
d=10
Then you solve for c by pluging in (0,6) and d=10 into h(t) to get c=106. From there you can solve for x.</p>

7

<p>This is a poorly written SAT question that would probably never make it onto an actual test (BB was written before the new SAT came out). It’s not “difficult” in the problem solving sense, it’s just too long. It would make a great SAT Math Level 1 problem, though, which is where I suspect it was lifted from. </p>

<p>General rule of thumb: a well-trained/sharp SAT’er should be able to solve any given SAT math problem and bubble in the answer in no more than 30 seconds; the computations need to be easy!</p>

8

<p>Here is a pretty short way.</p>

<p>The function is (positive constant #) - (positive varying #).</p>

<p>The biggest this function will be is when the (positive varying #) is zero (making c the maximum).</p>

<p>So, c = 106, and (d-4t) = 0 at t=2.5, so d=10.</p>

<p>h(1) = 106 - (10-4)^2 = 70.</p>

9

<p>it honestly doesn’t take THAT long to solve. There is a definite advantage for those who have physics and calculus here though.</p>

10

<p>^ Yes, and that alone makes it inappropriate. Attacking it with the SAT-level toolset is not the most efficient way to solve the problem. This totally removes a lot of its dependence on intelligence :P</p>

11

Xx

12

Rewrite h(t) in vertex form: h(t)=-16(t-d/4)^2 + c
vertex=(2.5, 106)=(d/4,c). Hence d=10, c=106.
h(1)=106-(10-4)^2 =106-36=70.

Vertex form of the parabola is in the skill base for both old and new SAT’s. No physics knowledge is required.