Need Help with these 2 SAT Math Questions

Test Preparation SAT Preparation
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<p>Here are pics of the 2 questions:
imgur:</a> the simple image sharer
imgur:</a> the simple image sharer</p>

<p>I guessed #19 right, it is C. However, I have no clue how to actually do it. As for #20, I'm completely lost. Help on how to solve these would be appreciated!</p>

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<p>For the first one, look at this thread:</p>

<p><a href=“http://talk.collegeconfidential.com/sat-preparation/1042482-hardest-math-question-ever.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/1042482-hardest-math-question-ever.html&lt;/a&gt;&lt;/p&gt;

<p>The second one is a common SAT game: foil out (x+y)^2 and then look at what you have…x^2 + 2xy + y^2…or if you prefer: x^2 + y^2 + 2 xy …and they give you both x^2 + y^2 and xy in terms of a.</p>

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<p>Hey pcKeller, appreciate your help. It’s so clear, I can’t believe I didn’t understand in the first place. I understand now, thanks buddy!</p>

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<p>Wait…the answer to #20 is 3a-20, right?</p>

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<p>yes…</p>

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<p>Just finished today’s practice test. Got 2270, I’m so happy! It’s 120 points higher than my normal score. </p>

<p>How do I do this question?: [imgur:</a> the simple image sharer](<a href=“http://imgur.com/KHKcZ]imgur:”>http://imgur.com/KHKcZ)</p>

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<p>Is the answer to this problem 4<em>3</em>2=24?
If it is, my way of solving was following: the lead band is always first. So it doesn’t change in any situation, right? Then forget about this band. Now you have only 4 bands which are left. In the first position we can place 4 bands, in the second - 3, the third - 2, and in the last position - 1 band. We know the lead band always remains in its position. Thus we get the answer - 4!=4<em>3</em>2*1=24.</p>

<p>I’m sorry for my English, it’s not my native language. I’m also preparing to the SAT so it will be really helpful if you show me my mistakes.</p>

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<p>Yes, the answer is 24. </p>

<p>I didn’t quite understand your explanation of how 4 bands go first, 3 go second, so on. Can you please apologize? Maybe I’m missing something obvious here.</p>

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<p>It’s OK.</p>

<p>Do you know how to solve the next problem: “There are Five books on a shelf. How many different ways to arrange the books?”</p>

<p>The answer for this type of question is always x!, where x is a number of quantity of something. In my example it is 5 books, so the answer is 5!=5<em>4</em>3<em>2</em>1</p>

<p>Your problem is changed a little, but the main idea remains.
You have 5 bands. One of them was already chosen as a lead band, so you can eliminate it from band list. It doesn’t really matter, which spot this band will get, 1st, 2nd,…5th, because in all possible situation this band alwais remains in its only spot.
Thus, your problem became the problem from my example(about books). You have 4 bands, 4 spots and you have to arrange them. This is the way you get the answer - 4!=4<em>3</em>2*1</p>

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<p>The thing I’m confused about is what determines whether it is 5! or simply 5x5? </p>

<p>I know I have used the 5x5 method on probability questions much more than 5!. Possibly so much so that I’m failing to tell the difference. How will I know whether I have to do 5x5 rather than 5!?</p>

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<p>Where do you use this method (5x5)? Could you give me an example of question?</p>

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<p>[A</a> permutation is an arrangement in which order matters. The formula to calculate a permutation is…](<a href=“File Not Found [404 Error]”>A permutation is an arrangement in which order matters. The formula to calculate a permutation is...)</p>

<p>i found here a good explanation of arrangements</p>

<p>Actually, there is a good explanation of combinations and advanced arrangements. You should read it too.
If something will still be unclear - just ask me. I will try to help</p>

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<p>I got it. The difference is, that bands are not equal each other. They are different.
For instance, you have 2-digit code and 5 different numbers (1,2,3,4,5, for example). Also we know that this numbers can repeat.
Possible codes: 12,13,24 etc, but 11,22,33,44,55 are also possible options.
In this problem answer will be, like you said, 5x5 (you can put 5 numbers to the first place, and 5 numbers to the other)</p>

<p>Let’s modify this problem. Now numbers CANNOT repeat.
It means we have next possible codes: 12,24,35,11 etc, but only NOT 55,33,44,11,22.
The answer: 5*4
Because you can put all of 5 numbers to the first place, but only 4 (we remove 1 number which repeats number in the first place) to the second.</p>

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<p>AH, GOT IT! Thanks!</p>

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<p>For #19, you treat this question as in the form of a 30-60-90 degrees triangle.

  1. The side adjacent to 30 degrees is x/2(square root 3)
  2. The side adjacent to 60 degrees is x/2
  3. Find the slope of the slope, which is rise over run.
  4. The slope would be x/2, which is the rise, divided by x/2(square root 3), which is the run
  5. As a result, you’ll get 1/(square root 3), which is the same as x/(square root 3), which is C.</p>

<p>For # 20, you factor out (x+y)^2 FIRST

  1. (x+y)(x+y)= x^2+2xy+y^2
  2. Using x^2+2xy+y^2, you can work out each statement step by step
  3. Take x^2+y^2 from the factored-out equation and make it equals to a. And make 2xy equals to 2(a-10)
  4. Add the substituted equation a+2a-20 and you’ll get the result, 3a-20, which is answer choice D.</p>
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<p>For the band question, here’s a simple way to solve it.
You have 5 bands to be put in 5 places.
First place was chosen, so you have 1 choice for this palce (1)
Second place has 4 options left to be put in (4)
Third place has 3 options (3)
Then 2 then 1
Then multiply all of the possible options:
1x4x3x2x1 = 24</p>

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<p>kevin101996: Could you please explain #19 a little better. I am so confused on that one! Many say that just use tan of the angle to find the slope but I want to understand the other method as well. How did you get the sides of the triangles?</p>

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<p>Maybe this picture will help you understand better</p>

<p><a href=“http://mathworld.wolfram.com/images/eps-gif/30-60-90Triangle_1000.gif[/url]”>http://mathworld.wolfram.com/images/eps-gif/30-60-90Triangle_1000.gif&lt;/a&gt;&lt;/p&gt;

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<p>Ah yes that certainly helps, thanks a lot!</p>