silverturtle challenge:Math Questions

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1

<p>I couldn't solve these SAT math questions. Please explain how the answer was derived. Thank you</p>

<p>40% Transportation
30% Food
10% Misc.
20% Hotel Room</p>

<p>1) The graph above shows the distribution of Tom's $240 trip expenses. The amount Tom paid for the hotel room was only part of the total hotel room cost, because he shared the cost of the room equally with 3 room. What was the total cost of the hotel room?</p>

<p>A 30
b 80
c 144
d 192
e 240</p>

<p>2) On a square gameboard is divided into n rows of n squares each, k of these squares lie along the boundary of the gameboard. Which of the following is a possible value for K?</p>

<p>a 10
b 25
c 34
d 42
e 52</p>

<p>3) Also question 12 on page 485 test number 2 section 8 in math.</p>

<p>4) If n is a positive integer and 2^(n) + 2^(n+1) = k, what is 2^(2n+2) in terms of k?</p>

<p>5) Question 4 section 2 page 453, test number 2.</p>

<p>6) 5,6,5,6,7,5,n,6</p>

<pre><code>For the numbers listed above, the only mode is 5 and the median is 6. Each of the following could be the value of n EXCEPT
</code></pre>

<p>A 6
B 7
C 8
D 9
E 10</p>

<p>7) Question 18 page 475 section 2 test number 2.</p>

<p>8) From a jar containing 50 pieces of candy, of which of 25 are red and 25 are green, Ari has taken 3 red and 4 green pieces. He takes an additional 13 pieces from the jar. What is the least number of these additional pieces that must be red in order for Ari to have more red cadies than green candies among all the pieces he has taken?</p>

<p>9) A positive integer is said to be "tri-factorable" if it is the product of three consecutive integers. How many positive integers less than 1000 are tri factorable? </p>

<p>Thank you for all the help CC members! :)</p>

2

<p>(1) Including Tom, 4 people shared the hotel room. Therefore, each person paid 240<em>.20 = $48. The total cost for the hotel room was 48</em>4 = $192.</p>

3

<p>Thank you, that was actually very simple then wasn’t it? I was thinking in more complex ways which only confused me.</p>

4

<p>2). 52
The board is nxn, so n^2. The perimeter is n+n+n-2+n-2, so 4n-4.
When you apply this to every possibility, the only one that come out with a whole number answer is 52, because 52+4 is divisible by 4.</p>

5

<p>9). Think of the upper limit of the three numbers to find this solution. 10x10x10 equals 1000, so we know it must be near there. 9x10x11 equals 990, so that is the highest one, because the next integer up would be well over 1000. Then count up to 9. 1,2,3,4,5,6,7,8,9. 9 solutions, because the numbers that are the product of the of each of these integers are less than 1000. I started with the first integer in each series, because they are consecutive. 1 is the minimum to start at, and after the starting integer of 9, the answers are above 1000.</p>

<p>Nine numbers</p>

6

<p>Or, more simply for #2, the perimeter of this particular square must be a multiple of 4, which only 52 is.</p>

7

<p>^Right. I was just trying to get to the reasoning of why that was true.</p>

8

<p>And I agree with every word of it.</p>

9

<p>Is # 6 reproduced correctly?</p>

<p>I see 3 5’s and 3 6’s, yet the problem says 5 is the only mode. And 5 is clearly not a permissible value of n, because that would make the median 5.5, and also because there are supposed to be LOTS of possible values of n.</p>

<p>Is it just my middle-aged eyes, or is there a problem here?</p>

10

<h1>8) After Ari has taken 13 more pieces from the jar, he will have taken a total of 20 pieces: 3 + 4 + 13 = 20. In order for more than half the pieces to be red, at least 11/20 must be red. He as already taken 3 reds out of the jar; he’ll need to take at least 8 more reds among the 13.</h1>

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<h1>6. (It’s copied wrong, there are supposed to be four fives; I just took this test today. [760, I wish I would have read the last two questions correctly {Read your #1 as if he shared it with 2 others I.E 3 total, and then on the one before it I misread which sides were larger than each other.} as I would have gotten an 800. >.<] )</h1>

<p>The answer would have to be A - 6, because if you have an additional 6 the data set will have an additional mode (Both six and five), and it’s stated that it only has one mode (five).</p>

12

<h1>4 seems hard.</h1>

<p>2^n + 2^(n+1) = k</p>

<p>Since 2^(n+1) = 2<em>(2^n), rewrite 2^n as (1/2)</em>[2^(n+1)].</p>

<p>(1/2)*[2^(n+1)] + 2^(n+1) = k</p>

<p>Factor out 2^(n+1).</p>

<p>[2^(n+1)]<em>(1/2 + 1) = k
[2^(n+1)]</em>(3/2) = k
2^(n+1) = (2/3)k or 2k/3</p>

<p>Next, note that 2^(2n+2) = 2^[2*(n+1)], or [2^(n+1)]^2.</p>

<p>If 2^(n+1) = 2k/3, then
[2^(n+1)]^2 must equal (2k/3)^2, or (4k^2)/9.</p>

<p>So 2^(2n+2) = (4k^2)/9.</p>

<p>Sheesh! This took me WAY too long! Where did you get this question?</p>

13

<p>Thanks, Adam. I suspected as much.</p>

14

<p>Thank you for the help, all questions came from the BB test 2.</p>

<p>and Sikosky sorry for the mistake here is the questions fixed</p>

<p>6) 5,6,5,6,7,5,5,n,6</p>

<p>For the numbers listed above, the only mode is 5 and the median is 6. Each of the following could be the value of n EXCEPT</p>

<p>A 6
B 7
C 8
D 9
E 10</p>

15

<p>6) The answer should be choice A, 6. If n were 6, there would be four 5’s and four 6’s in the set of numbers. That would mean the mode is both 5 and 6. However, the problem clearly states that the mode is (and only is) 5.</p>

16

<p>Where did you get the n-2 component from?</p>

17

<p>@SATnoob, why are you wasting your precious time typing and asking those questions, while you can just log onto your CB account & press either the Official SAT Study Guide First Edition or the OSSG Second Edition for answers explanations. They are available for free!! ;)</p>

18

<p>" #4 seems hard.</p>

<p>2^n + 2^(n+1) = k"</p>

<p>This problem is exactly the kind that is best attacked by making up #s. Pick an n value, see what k is. Then, find the answer based on the n value you chose. Then, put your k value into each answer choice…</p>

<p>For example, if n=5, 2^5 + 2^6 = 32 + 64 = 96 = k.</p>

<p>Then, 2^(2n+2) = 2^12 = 4096</p>

<p>Now put k = 96 into each answer choice. Pick the one that also comes out to 4096.</p>

<p>Whether you self-prep, take a course or work with a tutor, this technique should definitely be in your collection.</p>

<p>And,btw, this is NOT the method you will see at the college board site.</p>

19

<p>for #4 just factor out 2^n
you will get 4k/3 for answer</p>

20

<p>^Yes, that is how you solve it, but your solution is wrong.</p>

<p>2^n + 2^(n+1) = k
2^n(1+2^1) = k //Factor out 2^n from both terms
3<em>2^n = k //Simplify 1+2^1=3
9</em>2^(2n) = k^2 //Square both sides
9/4*2^(2n+2) = k^2 //Raise the exponent from 2^(2n) to 2^(2n+2) to get your desired exponent, which is the same thing as multiplying by 2^2=4. So you divide by 4 outside this expression to balance it out.
2^(2n+2)=4k^2/9 //Solve for your final expression</p>