SAT Math Problems

Test Preparation SAT Preparation
1

<p><a href=“http://k.minus.com/jNTK6XfJPunYF.png[/url]”>http://k.minus.com/jNTK6XfJPunYF.png&lt;/a&gt;&lt;/p&gt;

2

<p><a href=“http://k.minus.com/jTkd0suTDStkA.png[/url]”>http://k.minus.com/jTkd0suTDStkA.png&lt;/a&gt;&lt;/p&gt;

3

<p>First question: Six
Second question: 2</p>

<p>Please confirm, so I may elaborate.</p>

4

<p>You are right on both :). Please explain - especially the first one. </p>

<p>I don’t know why I posted the second one; 2 is the only answer that makes sense.</p>

5

<p>@IceQube - the answer is 6 because the only other house which can be painted grey is the one which is in the bottom row in the extreme right. All others cannot be painted grey.
So the answer is 7-1=6</p>

6

<p>Y X X Y X</p>

<p>X X Y X X</p>

<p>Let X be unpainted houses, and let Y be painted houses. The question is straight-forward from here, but you just have to read it carefully. No two houses next to each other on the same side can be painted the same color. That’s the first condition. No two opposite (different sides) houses can be painted the same color either. Let’s sketch a new line up now. Y is going to be painted houses, and Z is going to be houses that cannot be painted.</p>

<p>Y X X Y X -> Y Z Z Y Z</p>

<p>X X Y X X -> Z Z Y Z X</p>

<p>We basically applied both conditions. No two houses next to each other were painted the same color, on either side of each house (Y). And no houses opposite to the painted houses (Y) could be painted either. After applying all these conditions, the only remaining house that can be painted is the one of the bottom right. Elaborating on this question is rather difficult, because I actually need to be able to draw on the drawing itself to illustrate. Let me know if you need further explanation.</p>

<p>For question two: AC is 6, BC is 3. CP perpendicular to AC. Therefore, triangle CPB is right-angle. In any right-angle triangle, the hypotenuse must be greater than any one of the other sides, in which case, the only value smaller than 3 given is 2.</p>

7

<p>Oh crap!!! I see what I did wrong on the 1st problem. I overcounted :mad:. </p>

<p>The houses to the left and to the right of the gray houses cannot be gray, so I counted 5 houses that way.</p>

<p>In addition, the house across from each gray house cannot be painted gray, so I counted an additional 3 houses. </p>

<p>I added 5 and 3 and got 8. Unfortunately, that wasn’t an answer choice, so I picked the largest number, which was 6, and I still got the question right.</p>

8

<p>Now you know. :)</p>

<p>Good luck.</p>

9

<p>Can someone explain the second one without just saying that it’s the only one that looks reasonable?</p>

10

<p>In the second one, you have two right triangles, one with a hypotenuse of 6 (AC) and one with a hypotenuse of 3 (BC). </p>

<p>The legs of a right triangle cannot exceed the length of the hypotenuse. Otherwise, the Pythagorean Theorem would yield negative numbers.</p>

<p>hypotenuse^2 - legA^2 = legB^2</p>

<p>3^2 - 4^2 = -7 This is not possible. </p>

<p>2 is the only answer that works.</p>

<p>3^2 - 2^2 = 5.</p>