Confusing Questions...for me atleast.

Test Preparation
1

<p>So I started reviewing over some practice test carefully and in my math section
I came across a couple problems which I do not know either how to get the answer or why the answer is different than what I thought it to be.Help me please =) !
[Please do not scrtinize me for not knowing the answer] </p>

<p>The Questions are as following and are found in the CB's The SAT Preparation Booklet for 2008-2009:</p>

<p>1)What is the greatest possible area of a triangle with one side of length 7 and another side of length 10?
A)17 B)34 C)35 D)70 E)140 {now i know its 35 but why?} </p>

<p>[Section 6 #5 in actual booklet It has a circle incribe with 2 triangle i cant draw it here but ill post the questions anyway]</p>

<p>2)In the figure above, triangle ABC is inscribed in the circle with center O and diameter AC. If AB = AO what is the degree measure of triangle ABO ?
A)15 B)30 C)45 D)60 E)90 (It says C but idk how any ideas ?)</p>

<p>(again theres like 3 segments which make 3 angle that meet at a certain point. Hopefully you guys will recognize it. I cant draw it but heres the question Section 6 # 7 btw)</p>

<p>3) In the figure above, segment AB, segment CD,& segment EF intersect at P. If r=90, s=50, t=60, u=45, and w=50, what is the value of x?
A045 B)50 C)65 D)75 E) Cannot e determined by info given</p>

<p>4) If 6<|x-3|<7 and x<0, what is one possible value of |x|? (i guess like 3.1 and the answer was any number greater than 3 but less than 4 so i got it right but how?)(Section 6 #16)</p>

<p>5)What is the product of the smallest prime number that is greater than 50 and the greatest prime number that is less than 50? (Section 6 #17)</p>

<p>(number line) -8 a b c d e f 10</p>

<p>6) On the number line above, the tick marks are equally spaced and their coordinates are shown. Of these coordinates, which has the smallest positive value? (Section 8 #5)
A)a B)b C)c D)d E)e</p>

<p>7) Two spheres, one with radius 7 and one with radius 4, are tangent to each other. If P is any point on one sphere and Q is any point on the other sphere, what is the maximum possible length of segment PQ?
A)7 B)11 C)14 D)18 E)22</p>

<p>I cant illustrate these problem i think the image is needed but they are on Section 8 numbers 9,12,& 14. If anyone knows them please let me know.</p>

<p>8) Let @x be defined as x + 1\x for all nonzero integer x. If @x= t, where t is n integer, which of the following is a possible value of t ?
A)1 B)0 C)-1 D)-2 E)-3</p>

<p>Those are it again there in the booklet mostly every receives and hopefully you guys can help me understand the answers to these questions.</p>

<p>thanks !</p>

2

<ol>
<li>bh(1/2) = 35</li>
<li>x<0, the answer was any number greater than 3 but less than 4. what?</li>
<li>53*47= 2491</li>
</ol>

<p>your format is really annoying to look at. also, how can you expect people to help without diagrams? not everyone has the booklet</p>

3

<p><a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools;

4

<p>yeah thats what it said reater than 3 but less than 4</p>

5

<ol>
<li><p>Consider a right triangle with base 10 and height 7: this clearly has area 35. And other triangle is going to have that angle at which the length-10 and length-7 side meet either greater or less than 90 degrees - you can represent this range of triangles in comparison to your original one by drawing a semi-circle of radius 7 with center where the two specified sides meet on the original drawing: any other triangle with side lengths 10 and 7 will be congruent to one with the same 10-side and some other 7-side starting at that same center and continuing to some point on the circle. Notice that any of these pairings which do not meet at right angles result in a smaller height than 7, so a smaller overall area=1/2<em>base</em>height.</p></li>
<li><p>Notice that BO is a radius of the circle, as is AO, so AO=BO=AB and this triangle is equilateral. Then It should be that angle ABO is 60 degrees, not 45. The test books (sometimes even the exam grading!) are not always correct.</p></li>
<li><p>Unfortunately, I do not have this book and so would need more careful description of where all the letters are placed to make sense of this question.</p></li>
<li><p>As x is negative (x<0), so is x-3 (x<0 -> x-3<0-3=-3<0), so we need only consider the cases in which |x-3|=3-x. Then as 6<|x-3|=3-x, 6+x<3-x+x=3, and x=-6+6+x<-6+3=-3: x<-3. As 7>|x-3|=3-x, 7+x>3-x+x=3, and x=-7+7+x>-7+3=-4: x>-4. As -3>x>-4, 3<|x|<4.</p></li>
<li><p>For 2-digit numbers, you only need to check whether the number in question is divisible by 2, 3, 5, or 7 to determine whether it has prime factors or is instead itself prime (as the smallest composite number which does not have 2, 3, 5, or 7 as factor uses the next smallest prime, 11<em>11=121>99 - all other composite numbers without 2, 3, 5, or 7 as factors are even larger than 11</em>11). 2 is easy - is the number even? 5 similarly, does it end in a 0 or 5? 3 is a little trickier, do the digits of the number add up to a multiple of 3? 7 has some ridiculously silly rule but for small numbers like this I think it’s just easiest to remember some multiples of 7 from your times tables. Then we can find these primes!</p></li>
</ol>

<p>Least prime >50: (in both of these we only consider odd numbers, as of course evens are divisible by 2 and so not prime)

  • 51 is divisible by 3: 5+1=6=3<em>2
  • 53 is not divisible by 3: 5+3=8. Doesn’t end in 0 or 5. 7</em>7=49 and 7*8=56 so 53 is not divisible by 7 - this is our prime!</p>

<p>Greatest prime <50:

  • 49 = 7<em>7 is not prime.
  • 47 is not divisible by 3: 4+7=11. It doesn’t end in 0 or 5. 7</em>6=42 and 7*7=49 so 47 is not divisible by 7 - this is our prime!</p>

<p>Then simply multiply 47*53=2491.</p>

<ol>
<li><p>The distance represented is |10-(-8)|=10+8=18, and is divided into 7 even pieces -8 to a, a to b, b to c, c to d, d to e, e to f, and f to 10. Then each piece is of length 18/7. a=-8+18/7=-(5 3/7); b=a+18/7=-(5 3/7)+18/7=-(2 6/7); c=b+18/7=-(2 6/7)+18/7=-(2 6/7)+(2 4/7) is just shy of positive, so d=c+18/7 will be your first positive coordinate and so smallest such.</p></li>
<li><p>Notice that the maximum value they suggest is 22. Notice also that this is achievable by points positioned along the line through the centers of the two spheres: the diameter of the first sphere along this line contributes 7+7=14, while the diameter of the second along this line contributes 4+4=8, summing totally to 22. As the longest distance has to then be greater than or equal to 22 and 22 is the highest suggested value, 22 must be the answer they seek.</p></li>
<li><p>As x is an integer, and all suggested values of t are integers, consider only those x such that x+1/x is an integer: namely, as x is already an integer, those where 1/x is, 1 and -1. 1+1/1=2, which is not a listed answer. However, -1+1/-1=-1±1=-2, showing d as a possible answer.</p></li>
</ol>

<p>Let me know if you have any further questions about these!</p>

6

<ol>
<li>The sum of interior angles of any triangle is 180, and vertically opposite angles are equal - knowing this, we find some of the unknown angles. As s is 50 and t is 60, angle BPF (and so angle EPA) is 70. As u is 45 and w is 50, angle APD (and so angle CPB) is 85. Then the remaining angle CPE=FPD must sum to 360 in the circle or 180 along one line, so angle CPE is 180-85-70=25. Then as CPE is 25 and r is 90, x must be 180-90-25=65.</li>
</ol>

7

<p>i didnt really get your explanations on 4,6,8.</p>

8

<p>Have pm’d you this as well, but will post here as you may not be the only one interested.</p>

<ol>
<li>Basically, I am solving for x a bunch of times, narrowing down the inequalities that pin its absolute value between 3 and 4.</li>
</ol>

<p>The first thing to notice is that 6<|x-3|<7 means both 6<|x-3| and |x-3|<7 are true.
Generally with solving absolute values, we then have to consider the cases both
~ if x-3 is positive, in which case |x-3| = x-3, or
~ if x-3 is negative, in which case |x-3| = -(x-3) = -x+3 = 3-x
However, as the question also stipulates that x<0, I know x-3 is negative: a negative number minus some positive amount is even ‘more’ negative. Then we only have to consider case 2: |x-3| = 3-x.</p>

<p>Plugging this equality into the first half of the inequality, we see that
as |x-3| = 3-x,
6 < |x-3| implies 6 < 3-x
(is that a better way of phrasing it? Let me know if this step does not make sense - it’s difficult to word these things online sometimes)
I then add x to both sides of the equation and subtract 6 from both sides,
ending up with x < 3-6,
so x < -3.</p>

<p>Similarly with the other half of the inequality,
as |x-3| = 3-x and |x-3| < 7,
we find 3-x < 7.
Then adding x to both sides and subtracting 7 from both sides,
I find 3-7 < x,
so -4 < x.</p>

<p>As -4 < x and x < -3, x is between -4 and -3.
As the question is asking about the absolute value of x, though, we look at the absolute values of this solution set:
x being between -4 and -3 translates to |x| being between 3 and 4.</p>

<ol>
<li>As the ticks are all evenly spaced, I divide the interval between -8 and 10 evenly:
The distance between -8 and 10 is |-8 - 10| = |-18| = 18,
and there are 7 intervals.
~ if there was just the one interval from -8 to 10, it would be of distance 18.
~ if the line looked like -8 a 10, there would be two intervals: one between -8 and a, and one between a and 10. Then as the total distance is 18, and the spaces are all equal, each would be of distance 18/2=9, so as a is 9 more than -8 (similarly, 9 less than 10), a = 1.
~ (last example before the real one) if the line looked like -8 a b 10, there would be three intervals: -8 to a, a to b, and b to 10. Then as the total distance is 18 and the spaces are all equal, each would be of distance 18/3=6.
So as a is 6 more than -8, a = -8+6 = -2,
and as b is 6 more than a, so 6 more than -2, b = 4.
~~ (actual problem) the line looks like -8 a b c d e f 10, so there are 7 intervals.
Then as total distance is 18, each equal interval is of distance 18/7 = 2 4/7 (but this does not simplify to an integer, like the nice earlier examples, as 7 does not evenly divide 18).
So a is 2 4/7 more than -8: a = -8 + (2 4/7) = -6 + 4/7 = -(5 3/7).
b is 2 4/7 more than a: b = -(5 3/7) + (2 4/7) = -(3 3/7) + 4/7 = -(2 6/7).
c is 2 4/7 more than b: c = -(2 6/7) + (2 4/7) = -6/7 + 4/7 = -2/7.
d is 2 4/7 more than c: d = -2/7 + (2 4/7) = 2 2/7.
We can similarly solve for e and f, but as we are looking for the first positive-valued letter and both e and f will be greater than d, we can simply stop here.
d is the smallest of those letters along this line with positive value.</li>
</ol>

<p>Looking at the problem in the book, I see that it is phrased differently than you’ve stated it here - the one in the booklet going -8 a b c d e 10, not -8 a b c d e f 10. This leaves you with 6 intervals, much easier as 6 divides 18. From my explanation of how to do it with 1, 2, 3, or 7 intervals, though, can you figure out how to find the correct answer with 6 intervals now? For the record, d would be the correct answer if there were 7 intervals.</p>

<ol>
<li>For @x to be an integer, x and 1/x have to sum to an integer. Can you figure out why this only happens when 1/x is an integer? (if not, let me know and I will try to explain further)
For x=1, 1/x = 1/1 = 1 is an integer! Notice that for any integer n greater than 1, 1/n is not an integer (1/2 is not, 1/3 is not, 1/4 is not…). So we try @1 and see if it is on the list of multiple-choice answers:
@1 = 1 + 1/1 = 1 + 1 = 2, unfortunately no on the list.
That’s ok though, because while we’ve rules out now all integers greater than 1, we still have to consider negative integers.
for x=-1, 1/x = 1/-1 = -1 is an integer. Notice similarly that for any integer n less than -1, 1/n is not an integer (1/-2=-1/2 is not, 1/-3=-1/3 is not…) So we try @-1:
@-1 = -1 + 1/-1 = -1 + -1 = -2, so -2 is a possible value of @x. As it is on the list of multiple-choice answers, I maintain that answer d) -2 is correct.</li>
</ol>