Please Help: Three SATI Math Questions

Test Preparation
1

<p>Hey guys!</p>

<p>I am preparing for the December 4th SATI next Saturday, and I have done a bunch of the "Real SAT" practice tests. Each time I get 750+ Verbal but all my Math scores are 680 or 690, I can't seem to break into the 700s! The thing is, looking back, I seem to get the same sort of questions wrong, and if I could get just one of them right I'd be in the 700s.</p>

<p>So maybe some of you math whizzes can help me, please?</p>

<p>I asked my friend Sam about it and he said they are called "permutations" and something that starts with a D. He said it's really easy if you know how to do it with your calculator. Here are some questions I'm talking about:</p>

<hr>

<p>(This one is a Quantitative Comparison)</p>

<p>For all positive intergers j and k, let j[]k be defined to be the sum of the k consecutive intergets beginning with j. For example, 9[]4 = 9 + 10 + 11 + 12.</p>

<p>Column A: 100[]99
Column B: 99[]100</p>

<p>(This one is a Student-Produced Response)</p>

<p>Tim wrote a seven-digit phone number on a piece of paper. He later tore the paper accidentally and the last two digits were lose. What is the maximum number of arrangements of two digits, using digits 0 through 9, that he could use in attempting to reconstruct the correct phone number?</p>

<hr>

<p>They are the ones that really get me! I know you're not supposed to count them all out, there has to be a "trick." =&lt;/p>

<p>Thank you so much,
Richard</p>

<p>PS: Here's another one I couldn't get. It's a quantitative comparison: </p>

<p>Given: 1 < s + t < r</p>

<p>Column A: r + s + t / r - s - t
Column B: 0</p>

<p>I put in sample numbers and got that A was a negative number so I put B, but it says it is A!</p>

2

<p>B is greater coz:</p>

<p>A = 100 + 101 + .....+ 198
B = 99 + 100 + 101 + ....+198
= 99 + A</p>

<hr>

<p>100
there are hundred possibilities for the last two digits....00 to 99</p>

<hr>

<p>you cant get a -ve number on A
r - (s+t) cant be -ve coz r > s+t
r + s + t cannot be -ve coz all three are greater than 1</p>

3

<p>ahtsisab,</p>

<p>I'm not sure I follow you on the first one. How did you come up with the number 198?</p>

<p>The second one - So, there is always one more than the number of inclusive numbers? Like 0-10 has 11 numbers, and 0-100 has 101 numbers, so 0-99 would be 100 numbers? Hah... never knew that, makes sense...</p>

<p>The third one - if r is 1, s is 1, and t is 3 wouldnt r - s - t be 1 - 1 - 3 which would be -3? How did you get r - (s+t)?</p>

<p>Is there any sort of equation you can use if you have a long pattern like where you multiply each successive number by 5 and add 3 and you want the 35th number?</p>

4

<p>Quantative Comparison:
For all positive intergers j and k, let j[]k be defined to be the sum of the k consecutive intergets beginning with j. For example, 9[]4 = 9 + 10 + 11 + 12.</p>

<p>Column A: 100[]99
Column B: 99[]100</p>

<p>The answer is B.
Explanation: If there are 99 and numbers and you start with 100, 100 + 101 + 102... + 198 gives you your 99 numbers. If there are 100 numbers and you start with 99, 99 + 100 + 101.... + 198. 100 numbers is from 99 to 198, hence Column B is the same as Column A, except for the fact that you add 99 to Column B, making it greater.</p>

5

<p>You can use smaller numbers too in the same pattern.</p>

<p>3[]2 = 3 + 4 = 7
2[]3 = 2 + 3 + 4 = 9</p>

<p>B</p>

6

<p>Given: 1 < s + t < r</p>

<p>Column A: r + s + t / r - s - t
Column B: 0</p>

<p><quote> The third one - if r is 1, s is 1, and t is 3 wouldnt r - s - t be 1 - 1 - 3 which would be -3? How did you get r - (s+t)? </quote></p>

<p>You cannot have r = 1. It says <b>1 < s + t < r</b>. If r = 1, it already isn't greater than 1. </p>

<p>So lets try numbers like s = -1, t = 3. That way s + t = 2. And lets say r = 3. r is greater than s + t now.</p>

<p>Therefore r + s + t / r - s - t is equal to (3 + -1 + 3) / (3 - -1 - 3)
= (3 - 1 + 3) / (3 + 1 + 3)
= 5 / 7 which is greater than B
Answer is A.</p>

<p>Just remember to read carefully, s + t and r must be greater than 1.</p>

7

<p>Oops, I meant s is 1, t is 1, and r is 3. Making s+t 2. Then wouldnt 3-1-1... ok whoops, I get it now.. :)</p>

<p>Thanks everyone, you've been helpful.</p>

8

<p>hell i count them out and i got two 800s</p>

9

<p>Given: 1 < s + t < r</p>

<p>Column A: r + s + t / r - s - t
Column B: 0</p>

<p>This problem is very simple if you use simple algebra.</p>

<p>Notice the inequality 1 < s + t < r
This inequality is key to solving the problem. With a bit of algebraic manipulation we get r-s-t>0. for the top part we get that s+t>1. If this is the case then r>1, thus the result has to be greater than 0. No need to plug in :)</p>

10

<p>Tim wrote a seven-digit phone number on a piece of paper. He later tore the paper accidentally and the last two digits were lose. What is the maximum number of arrangements of two digits, using digits 0 through 9, that he could use in attempting to reconstruct the correct phone number?</p>

<p>I get 2880 for this one.</p>

11

<p>2880 is wrong</p>

12

<p>I see my mistake dwintz. I think it is actually 100. because you can have the last two digits as 00 all the way through 99, which would render 100.</p>