<p>Ok, this question is really bugging me, I thought I had it figured out but apparently not.</p>
<p>here goes:</p>
<p>In the figure above, PQRS is a rectangle. The area of (triangle)RST is 7 and PT=(2/5)PS. What is the area of PQRS?</p>
<p>I made a simple drawing of the figure too</p>
<p>
<a href="http://img.photobucket.com/albums/v701/StoneImmaculate/satq.jpg%5B/IMG%5D">http://img.photobucket.com/albums/v701/StoneImmaculate/satq.jpg
</a></p>
<p>I know that the area of the square if a line is drawn from T straight up would be 14(obviously) I thought the other fact meant that that square is 3/5 of the total area which apparently isn't true because it doesn't add up right.</p>
<p>The correct answer is 70/3. According to CB the difficulty is 5 on this one.</p>
<p>Drawing a perpendicular line from T will create a rectangle with the area 3/5(PS)(PQ).</p>
<p>As you figured out, 3/5(PS)(PQ) = 14.</p>
<p>Solve for (PS)(PQ), the area of PQRS, to get 70/3.</p>
<p>cool, thanks for the help</p>
<p>got another couple if someone wants to help</p>
<p>If j, k, and n are consecutive integers such that 0<j<k<n and the units digit of the product jn is 9 what is the units digit of k?</p>
<p>and the last one if even tougher I think, I'm posting it as an image b/c it has lots of superscripts in it.</p>
<p>
<a href="http://img.photobucket.com/albums/v701/StoneImmaculate/satq1.jpg%5B/IMG%5D">http://img.photobucket.com/albums/v701/StoneImmaculate/satq1.jpg
</a></p>
<p>thanks</p>
<p>Huh? I would have NEVER thought of drawing a line perpendicular to T. If I don't don't make these extra lines, will I not do as good on the math section? I did it a separate way actually, but are those ways of just making those lines very useful and almost essential to a good score?</p>
<p>PT=(2/5)PS, so
ST=(3/5)PS
(1/2)RS<em>ST=7=area of Triangle
Substute ST with 3/5 PS
(3/10)RS</em>PS=7
RS*PS=(70/3)</p>
<p>I still don't undestand the idea of drawing a line from T to what, the (2/5) point of QR? How does that make a square? </p>
<p>The JKN one looks tough, but how many integers have a units digit of 9? Well, infinite. Hmmm.. J<em>N could also very well be a number like 1293848948349999999. Well I would say that J and N are odd #s if the units digit is 9, so K=even. So now we have even</em>odd, so the end digit should be even. So at least now we've made it into a multiple choice question(2-8 even). It could be 0, but I wouldn't guess that if I had to guess. I would guess I would guess 2 maybe, but that is quite a hard one.</p>
<p>Stoneimmaculate, you're killing me here! Sorry, but plug in your own #s for 19 because plugging in/coming up with your own #s is the best way to answer some hard questions; it's the # 1 strategy you want to know if you're trying to answer these questions. I sometimes question why CB puts these questions when we can just come up with our own #s, but that's the advantage of MC. Let's say X=1, Y=2. So you get 60. C is the first one that comes out to be 60. I did some algebra first though by realizing that you factor (2x)^y, so the it'll be (2x)^y *(something -1(has to be 1 because it's the same term)), so I would only try C, D, E.</p>
<ol>
<li><p>the numbers are 9, 10, 11 (9*11 = 99); so the units number of k would be 0</p></li>
<li><p>you just factor out (2x)^y and because you subtract exponents when you divide, you must subtract y from 3y. there for the answer would be c.</p></li>
</ol>
<p>did that make sense?</p>
<p>
[quote]
I still don't undestand the idea of drawing a line from T to what, the (2/5) point of QR?
[/quote]
</p>
<p>Drawing the perpendicular line forms a triangle identical to RST.</p>
<p>Are you supposed to find out that the #s are 9, 10, 11, or is there an algebraic way of finding out? Like the relation between if the end digit is 7 or 5 and what the middle digit is.</p>
<p>there are a few ways of figuring out the three numbers:
1. first of all, with the SATs, you can be pretty sure that they would not choose a product (or a set of numbers) that are extremely high. Therefore, I just went through non-prime numbers that end in 9 (9, 39, 49, 69, 99...) which had factors two numbers apart - because I did not think there would be that many.</p>
<pre><code> 2. If you think about it, the factors of any number that ends in 9 must themselves end in 1 and 9 or 3 and 3. because no numbers that both end in 3 are two apart from one another, you know that j and n must end in 1 and 9. Therefore, because k must be in between them, it would have to end in 0.
</code></pre>
<p>I was going through 9-99 until I realized that the # could be something like 194839384999999 or some other crazy number. Were these questions on actual SATs(to the OP) though? Just curious.</p>
<p>jn=99=9*11.</p>
<p>==>k=10. so units digit of k=0.</p>
<p>I just went through all the numbers ending with 9 from 19 onwards, and hey presto, only 99 fulfilled the condidion!</p>
<p>cool, thanks for all the help guys</p>
<p>RahoulVA, yes these are actual questions from a test in 2005 ranked 5/5 in difficulty</p>
<p>I like guessing.</p>
<p>For the area of the rectangle, I simply assumed that PT=2, and TS=3. I then calculated the value of RS, using the given area of the triangle. 7=(.5)(3)(RS).
RS = 4.67. Then, it's an easy multiplication: (4.67)(5) = 70/3. </p>
<p>I agree, the integers are 9,10, and 11.</p>
<p>The other question: the answer is C (I think). This time it's not so much about guessing than it is about seeing the answer choices. Factor for 2x to the yth power. Compare with the other answer choices. Careeeful with E, though!</p>
