Hard Math Problems! HELP!

Test Preparation SAT Preparation
1

<h2>1. </h2>

<h2>Number of Telephones in One Residential Area</h2>

<p>Telephones Per Home l Number of Homes l
l
0 ------------------------- 2 l
1 ------------------------- 9 l
2 ------------------------ 18 l
3 ------------------------- 4 l
4 ------------------------- 2 l</p>

<h2> 5 ------------------------- 1 l</h2>

<p>(Sorry if the chart looks a bit confusing.. I did my best to recreate it)</p>

<p>A survey revealed the data shown in the table above. What fraction of the homes in the survey had exactly 1 telephone?</p>

<p>a) 1/36 b) 1/9 c) 1/6 d) 1/4 e) 1/3
*ANSWER: D</p>

<ol>
<li>The areas of the bottom, the side, and the front of a rectangular box are r, s, t squares inches, respectively. What is the volume of the box, in cubic inches?</li>
</ol>

<p>a) (rst)^3 b)(rst)^2 c) rst d)Square Root of (rst) e) Cube root of (rst)
*ANSWER: D</p>

<ol>
<li>A train 200 meters long was traveling at a constant rate of 20 meters per second through a tunnel. It took 1 minute and 10 seconds from the time that the front of the train entered the tunnel until the time that the back of the train left the tunnel. What is the length of the tunnel, in meters?</li>
</ol>

<p>a) 1000 b) 1200 c) 1400 d) 1600 e)2000
*ANSWER:B</p>

<ol>
<li><p>kx^a = x^a+1
In the equation above, k, x, and a are positive integers greater than 1. What is the value of k-x?</p></li>
<li><p>A wheel made 3,000 revolutions while traveling 40,000pi inches in a straight line along the ground. What is the radius, in inches, of the wheel?</p></li>
<li><p>In a biology class of s students, there are m microscopes available. If the instructor assigns one microscope to each student, 6 more microscopes will be needed. If the instructor had twice as many microscopes available and assigned one microscope to each student, 6 microscopes would be left over. What is the value of s?</p></li>
</ol>

<p>Please explain thoroughly and clearly as much as possible (Explaining why?). Thank you!</p>

2

<ol>
<li>9/(2+9+18+4+2+1)= d .25</li>
<li>area is s^2 so to find a side would be the sqrt of the area or sqrt(s)
since the area of the sides are r, s, and t, the sides are sqrt(r), sqrt(s), sqrt(t)</li>
</ol>

<p>volume is height x width x length or sqrt(rst)</p>

<ol>
<li><p>the distance it took for the train (head in and butt out) would be 1400, but since the time includes the extra time it took the butt to come out, you have to subtract 200 from the total distance (i have no clue how to explain it, someone else help? lol) so you get 1200.</p></li>
<li><p>um are there any parenthesis?</p></li>
<li><p>you are talking about circumference.
3000 revolutions traveling 40000(pi) is 40/3(pi) distance a revolution.
2(pi)r=40/3(pi)
2r=40/3
r=40/6 in</p></li>
<li><p>you got yourself 2 equations
s=m+6 (you need another 6 microscopes uin order for each kid to get a microscope)
s=2m-6 (too many microscopes)</p></li>
</ol>

<p>since both =s, combine them</p>

<p>m+6=2m-6
12=m</p>

<p>so there are “12” microscopes, substitute into an equation and you will get 18 kids.</p>

3

<p>No parenthesis for number 4. Also, what makes you think that squaring an area will get you the side of a rectangular box?</p>

4

<p>mmmmhm, thinking a little more i remember that a square IS a rectangle, so unless it states something else, you can always assume. And being the forgetful person that I am, I used process of elimination for #2.</p>

<p>a. can’t be right (obviously too big)
b. wrong (obviously too big)
c. too big</p>

<p>That left D and E. In my PSAT, I came across a similar problem and I chose E which I got wrong. So now I choose D. lol
((x+1)/x)-x works for #4 but I don’t remember how to solve it o_o</p>

5

<p>Hmm interesting… square is a rectangle? Never thought of that… Also doing the process of elimination will be quite harder than it looks. If you do not know the right value how is it possible to just assume if one is too big for an answer or too small for an answer. Thanks anyways, I will look more into question #2.</p>

6

<ol>
<li><p>36 homes were surveyed, 9 of which had only one phone.
9/36 = 1/4</p></li>
<li><p>Let l = length, w = width, h = height of box. Then r = lw, s = wh, t = lh
(r)(s)(t) = (lw)(wh)(lh)
(r)(s)(t) = l^2 * w^2 * h^2
sqrt(l^2 * w^2 * h^2) = sqrt (rst)
lwh = sqrt (rst)</p></li>
</ol>

<p>Since length x width x height = volume, sqrt (rst) is the volume)</p>

<ol>
<li>Let’s look at everything from the back of the train’s perspective. At t = 0 seconds (when the front is just entering the tunnel), the back is 200 m away from the tunnel entrance. The back of the train travels the 200 m to the tunnel entrance and then through the tunnel. At t = 70 seconds (1 min 10 sec), the back has just left the tunnel. So the total distance traveled by the train in that time is 200 m + the tunnel length. Use d = rt now</li>
</ol>

<p>Let x = tunnel length</p>

<p>x + 200 meters = (20 meters/second)(70 seconds)
x + 200 meters = 1400 meters
x = 1200 meters</p>

<p>I’ll do the next few in the next post</p>

7

<ol>
<li>3000 revolutions is equal to 3000 circumferences of the circle. 3000 circumferences = 40,000 pi inches so</li>
</ol>

<p>3000 (circumference) = 40,000 pi
3000 (2 pi r) = 40,000 pi
6000 r = 40,000
r = 6 and 2/3 inches</p>

<ol>
<li>Translate the English into math</li>
</ol>

<p>If the instructor assigns one microscope to each student, 6 more microscopes will be needed. </p>

<p>s - 6 = m (there are more students than microscopes)</p>

<p>If the instructor had twice as many microscopes available and assigned one microscope to each student, 6 microscopes would be left over. </p>

<p>s + 6 = 2m (there are more microscopes than students)</p>

<p>Subtract the equations</p>

<p>s + 6 = 2m
-(s - 6 = m)
12 = m</p>

<p>s - 6 = m
s - 6 = 12
s = 18</p>

8

<ol>
<li>kx^a = x^a+1
In the equation above, k, x, and a are positive integers greater than 1. What is the value of k-x?</li>
</ol>

<p>Is it kx^a=(x^a)+1 or kx^a=x^(a+1)?</p>

9

<p>When the OP said there were no paranthesis in #4, he was right b/c there wouldn’t be any paranthesis in the actual problem on paper. When the problem is being typed out, there needs to be clarification.</p>

<p>kx^a = x^a + 1 is probably not right because
kx^a - x^a = 1
x^a(k - 1) = 1</p>

<p>That last equation would imply that either x^a or k - 1 is less than 1, which is not possible for either with the given conditions.</p>

<p>if the problem is
kx^a = x^(a + 1) then it’s much easier</p>

<p>kx^a - x^(a + 1) = 0
kx^a - x*x^a = 0
x^a (k - x) = 0
k - x = 0</p>

10

<p>My bad there was a parenthesis. Dang I totally got stuck on the part from x^(a+1) to x*x^a… I was not thinking clearly haha. Thank you jamesford.</p>

11

<p>This problem is begging for you to make up numbers…</p>

<p>Make up dimensions for a rectangular solid: say 3 by 4 by 6 or whatever you like.</p>

<p>Now find the dimensions of each side:</p>

<p>r = 3x4 = 12
s = 4x6 = 24
t = 3x6 = 18</p>

<p>The volume is 3x4x6 = 72.</p>

<p>Now check each answer choice.</p>

<p>And before anyone argues that the algebra is quicker, I’ll say: yes it is, if you know exactly what to do. But OP didn’t and neither do lots of other people. Making up #s is a mindless trick that works in many settings.</p>

12

<p>These questions look familiar – are they from a previous SAT test?</p>

13

<p>I think they’re from a practice PSAT test. I remember #3.</p>