Qs from the full length SAT on collegeboard.com

Test Preparation
1

<p>well first of all heres the link
<a href="http://www.collegeboard.com/prod_downloads/sat/satguide/SAT_full_0405.pdf%5B/url%5D"&gt;http://www.collegeboard.com/prod_downloads/sat/satguide/SAT_full_0405.pdf&lt;/a&gt;&lt;/p>

<p>Could someone explain the reasoning behind </p>

<h1>20 in the 1st section math (I dont understand why its "C")</h1>

<h1>25 in the grid in math (its a rate problem, so if you could give me some advice in general about those it would be great)</h1>

<p>and</p>

<h1>6 in the last 10 question section (I have no clue how to solve these types of problems without writing them out.)</h1>

<p>Thanks</p>

2

<ol>
<li><pre><code>I.b is the y-coordinate and c is the x-coordinate. Since those two points are both on the axis, then b must be 0 and so must c, therefore the two are equal
II. As for this one, you can see that the point (e,f) is in the 2nd quadrant, meanint that the x-coordinate will be negative and the y-coordinate will be positive. Thus, e is negative and f is positive, so f must be greater than e
III. we kno that a is positive and that d is negative, but that does not mean they are equal, so that one may or may not be true.
</code></pre>

<p>only I and II are true</p></li>
<li><p>Diana ran a total of 700 meters in two laps of equal distance, so she ran 350 per lap. She also ran the first lap in 7 mps, so 350 divided by 7 would give us 50, which means she spent 50 seconds running the first lap. Similarly she ran the second lap going 5 mps. Thus 350 divided by 5 is equal to 70. So it took her a total of 70 plus 50 seconds to run the race. 700 meters over 120 seconds is your answer. This simplifies to 35 over 6.</p></li>
<li><p>Im not sure about this either. I did however notice a pattern. from 1 to 10 there are 11 digits. from 1 to 20 there are 31 digits. from 1 to 30 there are 51 digits. Thus every increment of 10 (like 1 to 10, 1 to 20, 1 to 30 etc.) adds a total of 20 digits (of course, because there are two digits in each number greater than or equal to 10). We can now easily find the 90th digit because 1 to 40 will yield 71 digits, and 1 to 50 will yield 91 digits. our 90th digit will be 5, because the 0 in 50 was our 91st digit.</p></li>
</ol>

<p>whew that was a lot of work, hope it helps!</p>

3

<h1>20: The answer is C b/c</h1>

<p>I: the coordinate for the (c,d) is the same as (0,?) so c = 0, and for (a,b) it is (?,0), so b = 0, so b = c
II: F is greater than e b/c f is positive y and e is negative x
so those two are true
and you can't tell III w/o exact values</p>

4

<p>ahh, darn, i was about to post the other two, but someone beat me too it</p>

5

<p>Thank you soooo much juzam_djinn and psquared. I cant believe i didnt notice that one about the coordinates.
The people on this board are so friendly, i really appreciate it.</p>

6

<p>This type of problem appears very often on the SAT and tends to trip many students. The good news is that the problems are rarely difficult. In fact, they are very easy, once you know how to look for the shortcuts. </p>

<p>There are a few variances in the problem statement and you still have to be careful not to misread the exact question. However, the typical problem will use the same distance BOTH ways and provide two separate speeds or rates. </p>

<p>What do you need to solve this? One very simple formula to memorize:</p>

<p>rate 1 * rate 2 * 2 /rate 1 + rate 2 (See no distance in formula)</p>

<p>Plug in numbers => 5<em>7</em>2/5+7 or 70/12 or 35/6, Bingo! </p>

<p>The only reason you need to reduce 70/12 is because you can only use 4 digits. Remember to always leave fractions in the grid in. It eliminates silly errors. </p>

<p>If this type of problem appears on a multiple choice, it is even easier to solve. Typically TCB will provide the first speed, the second speed, and the simple arithmetic average of the speeds as answers. Those are obviously impossible answers. It leaves you with two possible answers. But there is not even a need to guess. The correct answer is always the one between the lowest speed and the trick answer that is that average of the two speeds. In the case, the answer would show up as one between 5 and 6. Does not get any easier than this. </p>

<p>Enjoy! </p>

<p>PS There is nothing wrong with the proposed answer. However, it is important to know where to save time on the SAT. In this type of problem, the distance, being equal on the two legs, is not relevant. Knowing the formula is a SAFE time saver. </p>

<ol>
<li>Diana ran a total of 700 meters in two laps of equal distance, so she ran 350 per lap. She also ran the first lap in 7 mps, so 350 divided by 7 would give us 50, which means she spent 50 seconds running the first lap. Similarly she ran the second lap going 5 mps. Thus 350 divided by 5 is equal to 70. So it took her a total of 70 plus 50 seconds to run the race. 700 meters over 120 seconds is your answer. This simplifies to 35 over 6.</li>
</ol>

7

<p>Xiggi you are like an amazing test-taking machine. I wanted to ask you, the formula you gave which is 2rate1*rate2/(rate1+rate2), does that work for any average problem even when the distances are not the same?</p>

8

<p>Juzam, it does not work in the simplest form of the formula. But you should not worry about this on the old SAT. So far, TCB has kept things pretty simple by using the same distance for this type of problem. Remember, the longer the problem, the easier the solution is! </p>

<p>In this case, TCB wanted the students to start using their calculator. Imagine the mess if the distance would have been 695 meters instead of 700! They picked 700 meters on purpose. Again, nothing wrong just a loss of precious time. If the question was a MC, they would hope lots of students to pick 6 as the answer. Tricky devils!</p>

9

<p>Alright, another one from full length SAT on collegeboard.com ,</p>

<p><a href="http://www.collegeboard.com/prod_do...T_full_0405.pdf%5B/url%5D"&gt;http://www.collegeboard.com/prod_do...T_full_0405.pdf&lt;/a&gt;&lt;/p>

<h1>24 in the grid in math section 3. The figure has seven small circles contained in large circle. I wonder what's the best , safest, and preferably also quick way to solve.</h1>

10

<p>arghh, this is sooo frustrating!!!!</p>

11

<p><a href="http://www.collegeboard.com/prod_downloads/sat/satguide/SAT_full_0405.pdf%5B/url%5D"&gt;http://www.collegeboard.com/prod_downloads/sat/satguide/SAT_full_0405.pdf&lt;/a&gt;&lt;/p>

<p>please give answer explanations for Sec. 3 #24 and Sec. 6 #9. thanks</p>

12

<h1>24 in the grid in math section 3. The figure has seven small circles contained in large circle. I wonder what's the best , safest, and preferably also quick way to solve.</h1>

<p>Did you notice the little marks where circles touch each others? Notice how the marks in the center circle divide the circumference in 6 parts? </p>

<p>For this problem, you can trust your eyes, especially since the figure is of exact scale. Once you know that the gray area in the small circle is 1/6 of a small circle, look at the 2 other shaded areas. They each represent half circles. </p>

<p>So you have 1/2 + 1/2 + 1/6 or or 1 1/6 or 7/6</p>

13

<p>The central circle on #24 can be divided into 6 equal sections through use of the tangent points, so you know that angle apb is 60 degrees. From this it can be deduced that the shaded area of the central circle is 1/6 of the regular circle. The two other circles are bisected by their respective lines so the sum of their areas is equal to 1x that of circle radius r. 6/6 +1/6= 7/6</p>

14

<p>is th answer for sec 3 </p>

<p>7/6?</p>

15

<p>For the Sec. 6 #9, you will need to make a succession of easy and quick calculations of the value of each hypo of the 6 right triangles.</p>

<p>Playing on a^2+b^2=c^2, you will find the following values to be left in square root form. </p>

<p>value of c^2 (represents the hypo)
1. sqrt 6 for the top triangle
2. sqrt 17 for the left one
3. sqrt 10 for the bottom one
4. combine 2 and 3 to find sqrt 27 for the inside left</p>

<p>Now you can calculate the remaining values of the inside right triangle.
It will be ? + sqrt 6 = sqrt 27 => ? is sqrt 21. </p>

<p>Plug in x^2 + b^2 = (sqrt 21)^2 or simply 21. A is the answer. </p>

<p>This problem is solved quickly by filling all the values in the figure. Takes a lot longer to type all of this than to solve it! Enjoy!</p>

16

<p>darn it. and i was ready to post a nice elborate response too. gosh you guys are too fast. i'm going to race to solve the next on eonw. </p>

<p>Sec. 6 #9 here i come!</p>

17

<p>ugh. i got A too. but it took me longer. it was relatively easy but i really have to watch out for the calculator simple errors. does anyone else have some more? for some of us here (maybe just me) this is my SAT pratice</p>

18

<p>ok another one
same CB test, #25 in Sec. 1
How many positive integers less than 1,001 are divisible by either 2 or 5 or both?</p>

<p>A)400
B)500
C)540
D)600
E)700,</p>

<p>I got it right, but took some POE and guessing. Please show me a good and quick way to solve this permutation or whatever type problem</p>

19

<p>D because first how many divisible by 2?</p>

<p>1001/2= 500 Remainder 1, so 500</p>

<p>how many divisible by 5?</p>

<p>1001/5 = 200 remainder 1, so 200</p>

<p>then how many are overlaps, 2x5= 10..all 10's would be overlaps so
1001/10= 100 remainder 1, so 100</p>

<p>500+200-100= 600</p>