Sat Math Questions Hard!!! Omg

<p>I really dont understand these can someone explain me how to do these problems </p>

<h2>GRID INS</h2>

<li><p>If (3x + y)/y = 6/5, what is the value of x/y?</p></li>
<li><p>A postive 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></li>
</ol>

<h2>MC</h2>

<li><p>In thee xy-coordinate planes, line l and q are perpendicular. If line l contains the points (o,o) and (2,1) and line q contains the points (2,1) and (o,T), what is the value of T?</p></li>
<li><p>A box contains wood beads, red glass beads, and blue glass beads. The number of glass beads is 4 times the number of wood beads. If one bead is to be chosen at random from the box, the probability that a red glass will be chosen is 3 times the probability that a blue glass bead will be chosen. If there are 12 red glass beads in the box, what is the total number of beads in the box.</p></li>
</ol>

<ol>
<li>1/15
I could be wrong, but I solved it like this:
(3x + y)/y = 6/5
(3x)/y + 1 = 6/5
(3x)/y = 1/5
Multiply each side by 1/3 to cancel the 3 in the numerator:
x/y = 1/15</li>
</ol>

<p>Lalanikcols has # 12 right, just manipulate the problem a little if you are lost. #16 is easy, if you know how to use your calculator. Take each #, starting at 1, and raise it to the power of 3. So 1=1, 2=8, 3=27, 4=64, 5=125, and you get the idea. Eventually you'll get to a # where it goves over 1000. For #20, the slow of line L=1/2, so the slope of Line T=-2. So slope=y1-y2/x1-x2. So you say that -2=T-1/0-2 and you just solve for T. Now #8 is cruel, but I say that because I remember that from the Bluebook and I Could have sworn that there were 5 answer choices, and sometmies solving questions like that ist he best way when presented with answer choices. I did that problem by looking at each answer choice and seeing if it worked. If not, use variables and make them related to each other. Translate words into math. </p>

<p>W=wood
R=red glass
B=blue glass</p>

<p>R+B=4W</p>

<p>R/R+W+B=3(B/R+B+W)</p>

<p>Now R+B=4W, and there are two places where I see R+B. So substitute</p>

<p>R/5W=3(B/5W). Divide by 5W to get R=3B. Now we know that there are 12 Red, so there are 4 Blue. So then you can substite both of those back into R+B=4W. So 12+4=4W. So 16=4W, so w=4. So 12+4+4=20, is that right?</p>

<p>i dont understand in the problem number 12 where does the 1 come from ?</p>

<p>(3x + y)/y = 6/5
is the same as saying (3x)/y + y/y = 6/5
y/y = 1</p>

<p>20) T should be 5
then 20 for the last one</p>

<p>for the tri one, answer should be 9.</p>

<p>I kinda did for 16 like this but my way is tedious
1.2.3, 2.3.4, 3.4.5, 4.5.6, 5.6.7..... 9.10.11</p>

<p>The answer is 9 integers
I listed till no 11 and took 3 numbers at a time,</p>

<p>In number 16 the highest tri factorable number less then 1000 is 990 because 9<em>10</em>11 equals 990. So the only possibilities are (1,2,3) (2,3,4) (3,4,5) (4,5,6) (5,6,7) (6,7,8) (7,8,9) (8,9,10) and (9,10,11), all the multiplications give a number smaller then 1000. For number 20, L has a slope of 1/2 so line q has a slope of -2/1,which is the same as 4/-2 and now we can do 2-2/1+4 and get a point of coordinates (0,5) therefore T equals 5.</p>

<p>^^Just a little faster:
Since 10^3=1000, 10x11x12>1000 - not good.
9x10x11=990 - works.
From 1x2x3 to 9x10x11 - 9 products.</p>

<p>The algebra looks like this...</p>

<p>for a number to be "tri factorable" it has to be a product of 3 consecutive numbers so just set it up like this</p>

<p>a<em>(a+1)</em>(a+2) ---simplify---> a^3+3a^2+2a</p>

<p>Solve for a when that formula equals 1000, or just plug it into your calculator and see when it goes above 1000.</p>

<p>jay,
that is a cool method!</p>

<p>jay way :) :
(a-1) a (a+1) = a^3 - a.
This is why you can use a^3 for an estimate.</p>

<p>I don't get what your saying gcf lol</p>

<p>(a-1) a (a+1) = a^3 - a....why isnt it a^3+3a^2+2a</p>

<p>^just like you said jay, three consecutive integers can be expressed as
a, a+1, a+2
or
a-1, a, a+1
(or even a+1000, a+1001, a+1002 ----- but why?)
where a is an integer.
If you use the second option:
(a-1) a (a+1) = a (a^2 - 1) = a^3 - a ----- slightly more compact, that's all (well, I could not pass on a chance to show off lol).</p>

<p>Adam's way:'</p>

<p>solve((3x + y)/y = 6/5 and x/y=z,z)</p>

<p>ti-89, mathematica etc</p>

<p>ohh lol yeah that's pretty clever</p>