SAT math questions -- help!--

<p>Ok, I’m working on SAT math and need some help:</p>

<li>A total of k passengers went on a bus trip. Each of the n buses that were used to transport the passengers could seat a maximum of x passengers. If one bus had 3 empty seats and the remaining buses were filled, what is the relationship among n, x, and k?</li>
</ol>

<p>I stick with my answer- k=nx + 3x (k/x = n+3)</p>

<li>In the xy-plane, line L passes through the origin and is perpendicular to the line 4x + y= k, where k is a constant. If the two lines intersect at the point (t, t+1), what is the value of t?</li>
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<p>No idea here.</p>

<p>I really don’t understand how I get the hard ones but not the mediums…</p>

<p>Thanks for the help.</p>

<p>K 2nd one took me a while but I got them :S</p>

<p>1rst:
was pretty straightforward. Just put words into numbers, so:
k = nx if you dont apply "if one bus had 3 empty seats and the remaining buses were filled". Then you just subtract 3, so k = nx -3 = TEH ANSWER :D</p>

<p>2nd:
Took me too long, if this was on SAT i'd be screwed, but at least I got it :S</p>

<p>Line L is y = (1/4)x + 0. I say 1/4, b/c the other line (which it has to be perp. to) is y = -4x + k. Regardless of the k value, Line L would always be perp. to it. So then I just plugged in random k values and then calculate the co-ordinates of the interception, until I found 2 pairs of values that worked for (t, t+1). The X value of the interception would have to be 4 times as large as the Y value b/c Line L is y = (1/4)x meaning for Y =1, X must be 4 in order for the equation to be true. Just try it: any X value on the L line is always 4 times the Y value.</p>

<p>So after much trial and error, I found that t = -1.3 recurring. That + 1 = -0.3 recurring. (-1.33... , -0.33...) is on Line L. Also, notice how -1.3 recurring is 4 times the size of 0.3 recurring.
hope it helps :D
I'm sure theres an actual method of solving this instead of using primitive trial and error like I did...</p>

<p>For the 2nd prob.
you solved methodically till y=1/4.x
therfore 4y = x
now t= x and t+1 = y
i.e. y=x+1
Now we have 2 eqns. </p>

<ol>
<li> 4y=x</li>
<li> y=x+1
solving these two simulataneously we get</li>
</ol>

<p>y=-1/3
x=-4/3
i.e. t = -4/3</p>

<p>ahh cool thx for demonstrating that that helps :D</p>

<p>good. thank you.</p>

<p>I like both solutions for #2 because they get you to the correct answer obviously. After getting to 4y = x, however, I would have finished the problem differently:</p>

<p>Remember that the point (t, t+1) is the point of intersection. This means that it has to work when plugged into either linear equation. Forget<br>
4x +y = k because is has the k. Simply plug (t, t+1) into 4y = x as follows and solve for t:</p>

<p>4(t + 1) = t
4t + 4 = t
3t = -4
t = -4/3</p>

<p>I think reducing the problem to one equation with one unknow is a little quicker that solving the system of equations.</p>

<p>bah I missed that line question because I couldn't figure out to move the y = -4x</p>

<p>So bad >_<</p>