<p>How would you do these 2 SAT math problems:</p>
<p>1) If the length of AB is 5 and the length of BC is 6, which of the following could be the of AC?</p>
<p>(10, 12, 14, 15, 16)</p>
<p>2)</p>
<p><a href="http://artpad.art.com/?iu3feh3psz4%5B/url%5D">http://artpad.art.com/?iu3feh3psz4</a></p>
<p>On the number line above, the tick marks are equally spaced. Which of the lettered points represents y?</p>
<p>(A, B, C, D, E)</p>
<p>Thank you!</p>
<p>IS 1) 10?</p>
<p>5+6=11</p>
<p>x<11 ?</p>
<p>I got it...because 5+6>11.</p>
<p>Yes, the first one is 10. But the art pad thing won't load for the second one :-/</p>
<p>its 10 because in a triangle one side cant be bigger than the other 2 added together or else there would be a hole</p>
<p>Yep, the third side rule says that:
The third side must be between the difference of the first two sides and the sum of the first two sides.
If .5 were an option you could count that out because c has to be greater than 6-5=1 :)</p>
<p>I hope that isn't TOO small :-X</p>
<p>For the second one it is E. The spaces are .5 apart from each other. </p>
<p>What I did is make x+y = -2
therefore (x+y)/2 = -1 </p>
<p>from this you can figure out that the spaces are .5 apart because there are 2 dots between x+y and (x+y)/2. You can also infer that x = -3.</p>
<p>From there, -3 + y = -2
y = 1</p>
<p>Count out the spaces from (x+y)/2 and you can see that E is 1. Therefore E is y.</p>
<p>This should work with other values as long as you keep in mind that x, x+y, and (x+y)/2 all must be negative values. You can see this because the only way for a number divided by a whole number to be greater than the original numerator it must be negative. IE, no fraction in denominatior.</p>
<p>How would you approach a question like this:</p>
<p>"How many positive integers less than 1000 are multiples of 5 and are equal to 3 times an even integer?"</p>
<p>and </p>
<p>"In art class there were just enough staplers, rulers, and glue bottles, so that every 2 students had to share a stapler, every 3 tudents had to share a ruler, and every 4 students had to share a glue bottle. If the sum of the number of all the supplies used by the class was 65, how many students are in the class?"</p>
<p>Sort of, what mind-set do you go through on these free-response questions.</p>
<p>For the first one:
Only multiples of 30 fulfill those requirements.
Question basically says the number has to be a multiple of 5 and 3 and an even integer.
5x3x2 = 30. I chose 2 because its the smallest even integer, which means that we will get every possible answer.
Then just divide 1000 by 30, which equals 33.333, disregard the remainder so the answer is 33.</p>
<p>Second one:
If every 2 students have to share a stapler, there is only enough staplers for 1/2 the class. So number of staplers (s) = .5 (number of students x). s = .5x. By teh same logic, g = .25x and r = 1/3x. g+s+r = 65, substitute and solve. </p>
<p>For the way to approach these, I think you just have to rewrite the questions in terms you understand. As soon as you look at a question, kind of go over in your head what things mean...i.e. realizing that there was only 1/2 the number of staplers as students if 2 students had to share 1 stapler. Just try to write whatever equations you can. If you can't, think about it logically like in the first question.</p>
<p>Thank you so much evilmonkey!</p>
<p>I was on the right track for the 2nd one...and I was close on the first one. Thanks for the tips, very insightful! I'll put those tips in my pocket now! :)</p>
