<p>@Greatakosah </p>
<p>Sometimes when you perform division of whole numbers, the divisor doesn’t go into the dividend a whole number of times; there will be something left over. For example, 5 goes into 22 4 times but there is a 2 “left over” from the division. This is called the “remainder”. So the question is asking how much of 2^2014 will be left after 11 has been taken away from it as many times as possible.</p>
<p>Hi, I need help with some math SAT problems. My goal is to get a 650 in math. Section 2 I got 16/20 correct. I got 17 wrong and skipped 16 19 and 20. If someone could please explain how to do them, I’d appreciate it <a href=“http://www.scribd.com/doc/110664351/SAT-October-2010”>http://www.scribd.com/doc/110664351/SAT-October-2010</a></p>
<p>For number 20, (x+y)^2 = x^2 + y^2 + 2xy = a + 2(a - 10) = a + 2a - 20 = 3a - 20, choice (D).</p>
<p>For number 19, just look at the special 30, 60, 90 triangle given to you. If you let the side opposite the 30 degree angle have length 1, then the adjacent side has length sqrt(3). So the slope of the line is rise/run = 1/sqrt(3). So the answer is choice ©.</p>
<p>For 17, start by drawing the median from angle A to base BC. Since the triangle is isosceles, this median is also an altitude and angle bisector. Therefore, you get two 3, 4, 5 right triangles. So the area is (1/2)(6)(4) = 12, choice (D).</p>
<p>For 16, us the formula Total = X + Y - Both + Neither. So 78 = 56 + 42 - Both + 7 = 105 - Both. </p>
<p>So Both = 105 - 78 = 27, choice ©.</p>
<p>Note that I just gave one solution for each of these problems. There are other ways to solve them. I’ll leave it to others to suggest other methods.</p>
<p>@DrSteve thanks for helping. For 19 and 20, I couldn’t finish the problems. For 16, that formula works great. 17 I;m still kinda lost on how you got the 3,4,5 triangles</p>
<p>Back to 17: Since the length of BC is 6, the median splits it into two pieces of length 3. You get two right triangles. Just look at one of them (either one). It has a leg of length 3, and a hypotenuse of 5. If you remember the Pythagorean triple, then you know the other leg has length 4 - otherwise just use the Pythagorean Theorem to get sqrt(5^2 - 3^2) = sqrt(25 - 9) =sqrt(16) = 4.</p>
<p>Once you finish drawing this picture you should see that you have a triangle with base 6 and height 3 (BC is the base, and the median that we added in is the height). So the area is (1/2) bh = (1/2)(6)(4) = 12.</p>
<p>Does that help?</p>
<p>@DrSteve yeah i flipped my paper around and saw that the 6 is split into 3 and when you have 3 and 5 4 is the missing piece and also the height and 3x4=12. Thanks. Got to get hard questions right more often. My weaknesses are sequences, geometry and functions. Everything else I excel at</p>
<p>For the geometry problem, it helps to pay attention to the mention of the figure not being to scale. In this case, it did NOT say that, This means you could use your own little self-made ruler and measure the value of 4! That is the trust your eyes approach! </p>
<p>One another approach is to realize that the answers are (more than probably) a multiple of 3. The small values of A - B -C only leave D and E. And E is not a multiple of 3. </p>
<p>PS The 3-4-5 is really the chosen approach, but I offered you a couple of other approaches. </p>
<p><a href=“Box”>https://app.box.com/s/nr6n326vo8c2s6o1sdlx</a> section 6. skipped 7. got 8 wrong. 14 don’t have a scientific calculator with me so i could type in permutation. 16. i put 15 for the answer and the answer was between 10-14 :(. 17 and 18 i also got wrong. please help if you can</p>
<p>@sat2014 </p>
<p>7) Draw a picture. It won’t matter where exactly you draw your lines so long as you follow the constraints because all configurations will have the same lines perpendicular and parallel to each other (because, for example if we’re given a line t, if we draw two perpendiculars, p, and q, they will be parallel). Look at the picture you draw to conclude that I and II are right but III is wrong. B.</p>
<p>8). If we know that a and b are two positive numbers, a>b is equivalent to a^2>b^2. Let’s think about why this is the case (on the SAT you don’t want to do this of course). To get from a>b to a^2>b^2, do: a>b means that a<em>a>a</em>b>a<em>b>a</em>b-b(a-b)>b^2 (this is of course more work than you’ll need to do on the SAT, but why not be careful now while practicing?). To get from a^2>b^2 to a>b with the knowledge that a and b are positive, note that a^2-b^2=(a-b)(a+b)>0. For the product of two numbers to be positive, they must have the same sign. a+b is positive, so a-b is too. So a-b>0, a>b. </p>
<p>Armed with the knowledge that a>b is equivalent to a^2>b^2 when a and b are positive, we can square both sides to get 3x>5y or x>5y/3. B.</p>
<p>14) No need for a calculator here. There are four positions left for the four remaining bands. There are four possibilities for who can take the second spot, but then only three spots for the next spot after the second is taken, etc. Proceeding this way gives us that the total number of possible orders is 4!=4<em>3</em>2*1=24.</p>
<p>16) In an acute triangle, if the shorter sides are a and b and the longest side is c, then a^2+b^2<c^2. this="" gives="" us="" that="" bc^2="">100. Now by what we did for question 8, we know that this means BC>10. But by the triangle inequality, we know that AB+AC>BC, so BC is smaller than 14. Hence your range. </c^2.></p>
<p>Here’s why the rule for obtuse triangles works: In a right triangle with the same legs, then a^2+b^2=c^2 holds (Pythagorean Theorem). But in the obtuse triangle, the legs are the same length by construction, but the hypotenuse is now longer. So the ‘=’ in the Pythagorean theorem turns into a ‘<’.</p>
<p>17) Plugging in gives us g(m+2)=(m+2)(m+1)=(2+2)(2+1)=12. m=2.</p>
<p>18) Let n be the number of 9th grade students assigned a locker in hall B. Then n/(155+25+n)=1/5. 5n=180+n. n=45. So the total number of students assigned a locker in hall B is 45+100+70+75=190.</p>
<p>@RandomHSer Thanks. I still don’t understand 7 because I don’t know what exactly to draw. 8 is confusing as is 17 and 18. got 14 and 16 but that’s it :(</p>
<p>@RandomHSer got 18! used a proportion get 45. 7 i would definately skip on the SAT. not even bother answering those type of question. 8 and 17 still not sure exactly how to attack</p>
<p>@sat2014 </p>
<p>For 7 draw the lines that they describe in the problem with the conditions of which are perpendicular to which. For 8, if you don’t care for my long-winded explanation, the point is that you can just square both sides. What still confuses you about 17?</p>
<p>Edit: Careful, 45 is the number of 9th graders that have a locker in hall B, not the total number of students with a locker there. This trap is probably a key reason why it’s so late in the test.</p>
<p>@RandomHSer I don’t understand what to plug in for 17. and 8 if i square it, would it become 3x>5y and then i basically plug in values for x and y to find the answer? Any tips on how to get a 650? I know you need to get 44 out of 54 correct. I’m trying to improve how i attack the hard problems</p>
<p>@RandomHSer I tried plugging in for 18 but still don’t understand how E is the correct answer :(</p>
<p>@sat2014</p>
<p>Once you have 3x^2>5y^2, divide by 3 to get x^2>5y^2/3 which is E. There’s no magic formula to getting a good score. Everyone has their own individual weaknesses that they need to fix. Personally, I found my math errors were always the “easy” questions because I didn’t check them/read them carefully, but of course everyone is different. If you’ve identified your weaknesses and are working on them (which it seems you’re doing) , you’re on the right track.</p>
<p>@RandomHSer i see it now thank you. 17 I still don’t see what you plug in for the function. My biggest weakness if functions as a whole as well as hard geometry. I used to make mental mistakes on the easy but now. i double check my answer to eliminate it so i can spend time on the harder questions.</p>
<p>Ok, I probably misled you at first since I did I forgot to square the x and y at first haha. For 17, if g(x)=x(x-1), then g(m)=(m+2)(m+2-1)=12, which you can solve.</p>