<p>@MITer94 Thank you, much appreciated. I’m taking SATs soon for the very first time in oct, and i’m trying to understand how to approach the questions.</p>
<p>Another:
A positive integer is said to be “tri-factorable” if it is the product of three consecutive integers. How many positive integers less than 1,000 are there? @MITer94 is there a method? </p>
<p>And Another: (These are all stdent response questions) From a jar containing 50 pieces of candy, of which 25 ared and 25 are green, Ari has taken 3 red and 4 green pieces. He takes an additional 13 pieces from the jar. What is tbe least number of these additional pieces that must be red in order fo rAri to have more red candies than green cadies among all the pieces he has taken?</p>
<p>@ashario
For the tri-factorable one, just try stuff. The smallest tri-factorable integer is 6 (=1<em>2</em>3), and the tri-factorable integers are 1<em>2</em>3, 2<em>3</em>4, 3<em>4</em>5, …, 9<em>10</em>11. (9<em>10</em>11 = 990). The product of any integers with 0 is 0, which is not positive, and the product of three negative integers is negative. There should be 9, if I’m not mistaken.</p>
<p>If Ari takes 7 red and 6 green, he will have an equal number of red/green candies. Therefore he needs 8 red candies.</p>
<p>@MITer94 Thanks. Could you help me with 5 and 6 on section 5 please. <a href=“Box”>https://app.box.com/s/0fh44vcdtl72a9szut8p</a></p>
<p>5 is just a straightforward application of 45-45-90 triangles. Try solving it again…</p>
<ol>
<li>If x is inversely proportional to y, then x = k/y for some constant k, and x^2 = k^2 / y^2. This means that 1/x^2 = y^2 / k^2, so 1/x^2 is directly proportional to y^2.</li>
</ol>
<p>I didn’t get this one at all. how??? #27
thanks</p>
<p>@emily1212 #27 on which test/section?</p>
<p>@MITer94 thanks I get 6 but 5 i still don’t know what to do. I know its a 45 45 90 but isn’t the diagonal supposed to be side radical 2?</p>
<p>@sat2014 No. The <strong>ratio</strong> of the side lengths is 1:1:sqrt(2).</p>
<p>@MITer94 yes so how is the digonal 4 then? sorry I’m still lost :(</p>
<p>4 = s*sqrt(2) where s is the side length. What is s equal to?</p>
<p>While I am sure MITer knows this, it is not a bad idea to learn and memorize of couple of additional relationships for the simple geometry the SAT is testing. </p>
<p>In this case, a student who knows or remembers that given the lengths of the diagonals in a square, the area is half the product of the diagonals can rely on a short formula. In so many words, the answer is 1/2 * 4^2 or 8. </p>
<p>Obviously, that is the same as labeling the side s and the diagonal d and applying the Pyth Theorem</p>
<p>s^2 + s^2 = d^2</p>
<hr>
<p>Fwiw, another approach is to draw the second diagonal and realize that the area is none other than the sum of 4 right triangles with sides of 2. Hence, each one is 2x2/2 or 2. Four triangles of 2 = 8. </p>
<p>PS The diagonals of a square intersect at their middle point. That gives you the value of 2. </p>
<p>Thanks guys for helping. need some more help please. <a href=“Box”>https://app.box.com/s/5lq25lhegnhno6e8ujgl</a> questions 16 17 and 20 please. 16 i guessed right but that’s because i’m very good at guessing. still want to know how to do those 3 problems. Thank You. section 4 by the way</p>
<p>@sat2014</p>
<p>16) If the y-coordinates are 6, then we can get the x-coordinates by solving 6=x^2. P has a the negative x-coordinate and Q the positive one. Solving 6=x^2, gives x= +/- sort(6), so the distance is 2*sqrt(6). C</p>
<p>17) By the Pythagorean theorem, DE^2=AD^2+EA^2. 25x^2=9x^2+EA^2, so EA^2=4x. Now the volume=length<em>width</em>height=(AD)(EA)(GH)=(3x)<em>(4x)</em>(6x)=72x^3=576, x=2. A</p>
<p>20) The angle of the shaded arc is 180-x. The radius of the circle is 6/2=3. The area of the shaded region is pi<em>3^2</em>(180-x)/360=pi.180-x=40. x=140 degrees. C</p>
<p>
</p>
<p>All correct, but remember how the SAT loves the 3-4-5 right triangle. You should be on the lookout for any diagonal that has a value of 5. The 3 x 4 x 6 = 72 should be an easy step. Then you have to deal with the x^3.</p>
<p>Also, you should always take a look at the values of the answers: Since x will be cubed, you will deal with values of the cubes of 2 3 5 8 9. Those numbers are 8, 27, and more than 125. Obviously, even you missed the base of 3 x 4 and picked 3 x 5, you would have a 90 (and not 72) and a value MUCH greater than 576 when adding the cubed x… And if all that fails, you could also notice that 576 is not divisible by 27 or 125. That really leaves 2 as the only possible answer.</p>
<p>All in all, this problem can (and should) be solved in just 5 to 10 seconds by merely looking at the numbers and … remaining calm at the end of a section! </p>
<p>Thanks for the help guys. <a href=“Box”>https://app.box.com/s/5lq25lhegnhno6e8ujgl</a> section 8 problems 11 and 13 i’m stuck. please help. thank you</p>
<p>i had trouble understanding this one…
(DEF has measure 180 - 50 - 60 = 70. BEF has measure 70/2 = 35. FAD has measure 50/2 = 25. BCF has measure 60/2 = 30. AFC has measure 180 - 30 - 50 = 100. CFE has measure 180 - 100 = 80. Finally, x = 180 - 80 - 35 = 65.).
thanks</p>
<p>For the 11, you can approach it in different ways.</p>
<ol>
<li>You could solve (x + 2y)^ 2 and find that the answer is </li>
</ol>
<p>3 + ? + 64 (the ? is not relevant) </p>
<p>and see that there is only one answer with 67 + something! Of course, the purists might say that there could be a different answer that is also correct. Yet, at that stage of the test, the SAT will prefer one that is a tad more complex but not as complex as playing with the negative in the middle.</p>
<p>So there is the second approach</p>
<p>(x + 2y)^ 2 = x^2 + 4xy + 4y^2</p>
<p>Again you have the values shown above. You also have the 4xy and that is broken down as 4.4 (easy as 4 * sqrt 16 ) and sqrt of 3. They are really spotting you the answer! ;)</p>
<p>For 13, just plug the points into the equation.</p>
<p>Starting with the point (a,0), we have 0 = 2 - a^3. So a^3 = 2.</p>
<p>Now (-a,b): we have b = 2 - (-a)^3 = 2 + a^3 = 2 + 2 = 4.</p>