Help Wanted: 2 SAT Math Questions!

Test Preparation SAT Preparation
1

<p>1)
t^2 - k^2 < 6
t + k > 4</p>

<p>If t and k are positive integers in the inequalities above and t > k, what is the value of t?</p>

<p>A 1
B 2
C 3
D 4
E 5</p>

<p>2) Any 2 points determine a line. If there are 6 points in a plane, no 3 of which lie on the same line, how many lines are determined by pairs of these 6 lines? </p>

<p>A 15
B 18
C 20
D 30
E 36</p>

<p><em>the ans for 1) is C and the answer for 2) is A</em></p>

<p>Can someone kindly explain how to do these questions, because the explanation given by collegeboard is far from clear.</p>

2

<p>Hey!</p>

<p>For question 1: </p>

<p>t^2-k^2= (t+k)(t-k)</p>

<p>(t+k)(t-k)<6.</p>

<p>Since (t+k)> 4, and t-k cannot equal 0, (t+k)= 5. (it CANNOT equal six because even if t-k was 1, then the expression would equal six, but it must be LESS than six).</p>

<p>so, 5(t-k) < 6.</p>

<p>This expression holds if t-k is 1.</p>

<p>Use system of equations: t+k= 5. t-k=1.</p>

<p>2t=6. t=3.</p>

<p>I am not sure about #2…but good luck!</p>

3

<p>

</p>

<p>I don’t think I’m going to explain this too well, so I’m going to leave this to some other… smart people.</p>

<p>

</p>

<p>The answer is 15. You made a typo.
6 choices for the first point
5 choice for the second point</p>

<p>6 * 5 = 30 possibilities
However, AB is same as BA (in a line), so 30/2 = 15</p>

4

<p>For number 2, if you have a fx-570ES calculator, you can solve this question in less than 10 seconds.</p>

<p>Simply press 6, then “Shift”, followed by the “divide” sign on the calculator.
It should like this: 6C2</p>

<p>This is a long equation (which if you want me to explain for you, just ask), that is wrapped up in one letter, and written on the calculator as such. It is used to give the number of different combinations of y (in this case, 2), that can be found in x (in this case, 6) different units.</p>

<p>For example, how many different combinations of 3, can we make out of 10 pencils with different colors? 10C3 = 120 different combinations.</p>

<p>6C2 = 15. As JJ said, you made a typo. The answer is 15.</p>

5

<p>For the second question, if there are six points, shouldn’t there be five possible lines for the first and then 4 for the second?</p>

6

<p>@ Morton, yes.</p>

<p>But if you choose A as the first choice, and B as the second choice.
And you choose B as the first choice, and A as the second choice.
Then you’ve made the same combination. AB = BA. So, 6 x 5 = 30, and since we have 2 of each combination, we divide the result by 2. 30/2 = 15.</p>

<p>Or 6C2 as I explained in the post above for people who are more comfortable with calculators.</p>

7

<p>Here is a pic of what I am trying to explain in my question above: <a href=“http://i41.■■■■■■■.com/2a9y2vs.png[/url]”>http://i41.■■■■■■■.com/2a9y2vs.png&lt;/a&gt;&lt;/p&gt;

<p>So, when we count all the lines, there are 13 lines altogether…right?</p>

8

<p>

</p>

<p>Yes. You can do that manually.
Let there be points A, B, C, D, E and F.
For point A, there are 5 possible lines (AB, AC, AD, AE, and AF)
For point B, there are 4 possible lines (BC, BD, BE, and BF) Keep in mind that AB is excluded since lines AB and BC are considered the same line.
For point C, there are 3 possible lines (CD, CE, and CF) (CA and CB excluded)
For point D, there are 2 possible lines (DE and DF) (DA, DB, and DC excluded)
For point E, there is 1 possible line (EF) (EA, EB, EC, and ED excluded)
For point F, there is no possible line. (FA, FB, FC, FD, and FE excluded)</p>

<p>5 + 4 + 3 + 2 + 1 = 15 possible lines.</p>

<p>OR</p>

<p>You can do this:
A line is created when two points are connected.
For the first point on the line, there are 6 possible points.
For the second point on the line, there are 5 possible points.</p>

<p>6*5 = 30.</p>

<p>However, in the process, you counted AB and BA. You need to exclude them.</p>

<p>30/2! = 15 possible lines.</p>

9

<p>

</p>

<p>You actually get 15. The OP made a typo.</p>

10

<p>More precise drawing: <a href=“http://i43.■■■■■■■.com/ftenup.png[/url]”>http://i43.■■■■■■■.com/ftenup.png&lt;/a&gt;&lt;/p&gt;

<p>I see 13…</p>

<p>EDIT: WHOOPS! I see 15. I counted wrong! My bad LOL</p>

11

<p>You can also apply this counting method to SirWanksalot’s example.

</p>

<p>There are 10 possible choices for the first pencil you pick.
There are 9 possible choices for the second pencil you pick.
There are 8 possible choices for the third pencil you pick.</p>

<p>10 * 9 * 8 = 720 possibilities</p>

<p>However, in the process, you counted ABC, ACB, BAC, BCA, CAB, and CBA as different combinations. You must divide the total possibilities by 6 or 3!.</p>

<p>720 / 3! = 120 possibilities.</p>

<p>You really don’t have to learn anything about combinations and permutations if you have a strong foundation in combinatorics(counting). Try to understand where your combinations/permutations formulas are derived from…</p>

12

<p>Exclamation points are rude. I know it’s 3; you don’t have to shout.</p>

<p>;)</p>

13

<p>wow! thanks for all the feedback guys, this concept of permutations is still a fairly new one to me but your explanations have really helped.</p>

<p>you just gotta love CC!!</p>