<p>Interesting problem, and also much more elaborate than the old SAT. Since this is a grid-in, we probably do not have the luxury to find a quick and dirty shortcut. </p>
<p>Anyhow, here is a way to solve it. I'll put you on the right track</p>
<ol>
<li><p>To solve it you need to remember that (a-b)^2 = a^2 -2ab + b^2</p></li>
<li><p>Then, look at what was given by the ETS maniacs... You need to find a way to reduce one of the functions. The value of t=0 seems interesting because of the zero. Plug in the function for a height of 6:</p></li>
</ol>
<p>6 = c-(d-0)^2 or c = 6 + (d-0)^2 or C = 6 + d^2 <= remember this one</p>
<ol>
<li>Now that we have C, we can plug it in the function for 106:</li>
</ol>
<p>106 = (6+d^2) - (d-10)^2 ... play around a bit and you'll find that d = 10. </p>
<ol>
<li>Now that you have c AND d, you can simply write the function for t = 1. The answer is indeed 70.</li>
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<p>It takes a bit of manipulation. As usual, make sure not to miss any clues. In this case, the t=0 was an important clue and not an afterthought.</p>
<p>Well...c-d^2=6 (initially means t=0; plug in 0 for t and that equation is equal to 6). At any other point, you get h(t)=c-(d-10)^2=c-d^2+8dt+16t^2=6+8dt-16t^2. You can then plug in (2.5, 106) so 106=6+8d(2.5)-16(2.5)^2.
100=20d-100, or d=10. Plug that back in; h(x)=6+8(10)t-16t^2; plug 1 in for t and you get h(1)=6+80(1)-16(1)=70.</p>
<p>Not exactly sure how to explain this but I'll try. So alice goes 10 steps to the left and Carrine 10 steps to the right. Then they said that the distance between them is equal to only 17 of Alice's steps. So what I did was I said that first alice went back 10 steps and started in her original position. Then her 7 steps (that she still has since she already used 10) equaled to Corinnes 10 steps so the length of 1 step compared to corinnes is 10/7.</p>