<p>Hmm am I posting too many problems? Is there a specific thread where I can post these? =/ </p>
<p>Anyways here goes:
There are g gallons of paint available to paint a house. After n gallons have been used, then, in terms of g and n, what percent of the paint has not been used?</p>
<p>a) 100n/g%</p>
<p>b) g/100n%</p>
<p>c) 100g/n%</p>
<p>d) g/(100(g-n))%</p>
<p>e) (100(g-n))/g%</p>
<p>Okay, here is the case. I've gotten this problem right the first time, however, I would like a solution from any of you ccers =] </p>
<p>Here's my solution:
Well since the total amount is expressed as g, the percent fraction obviously has to be over g(the gallons of paint available). Since n is the number of gallons used, the number of gallons not used would be g-n. </p>
<p>I'm confused about why you need to multiply it by 100. Is it because you need to convert a fraction in a percentage problem? Haha, I think I answered my own question but anyways...</p>
<p>I think 100 is put in front because a fraction is expressed in the form of a decimal and in order to make a percent (% is put in front) you need to multiply the fraction by 100 to move it two decimal places up? </p>
<p>ex: 1/4x 100 = 100/4= 25% same as .25 x 100 =25%? </p>
<p>Can anyone tell me if this works for all of these types of problems and is universal?? I don't know if I'm doing this right but I'm pretty sure =D</p>
<p>Another problem, I dunno if I start my own "I don't get it thread" or something xD Anyways, I just want to get things straight. Do you ever get a problem right but you still don't EXACTLY know why it is right or how you got it right? That is how I feel and too bad this BB doesn't give you solutions to most of the problems and I want them so badly!!! However, these questions are the most similar to real SAT questions. Just no solutions!!! arghh...</p>
<p>Anyways here goes xD:
HAHA I guessed this one but I think it is an educated guess.. =/</p>
<p>If x is an integer and 2<x<7, how many different triangles are there with sides of lengths 2,7, and x??</p>
<p>a) 1 b) 2 c)3 d)4 e)5</p>
<p>Correct answer a) 1</p>
<p>My solution: I thought about this a lot and I came up with a conclusion, there can be only 1 type of triangle with any give sides. for example 3,4,5 has only 1 type of triangle and it is right?? </p>
<p>I highly doubt that this is right.. Please, give me your solution to this problem thanks!!!</p>
<p>for the 1st one... is the answer e? if you have g total and n used... and you want to know the percent unused then that would be g-n divided by total, g and multiplied by 100 for percent... yeah i think your way is right :D
fo rhte 2nd one... there's this rule about triangle side lengths. the sum of any two sides of a triangle has to be greater than the third side. So... the only triangle that would work under htose conditions and this rule... is 2, 6, 7.</p>
<p>Well I came up with this: Can a right triangle be iscosoles or w/e? Can a iscocosle (lol) be a scalene triangle? I don't think any of these is possible. So I think that is why!</p>
<p>If 2lx+3l=4 and ly+1l/3=2, then lx+yl could equal each of the following EXCEPT:</p>
<p>a) 0
b) 4
c) 8
d) 10
e) 12</p>
<p>the correct answer is D and I think this is because the other numbers are divisble by 4 and 2 except 10?? lol Can anyone tell me a way of solving this numerically? Plug in numbers? solve for x and y? Please also state the way you think is most efficient. Thanks!</p>
<p>a triangle with more than two conbinations of sides? wouldn't that be a completley different triangle? a triangle with sides... 6,2,7 for ex will be the same triangle no matter where each side lenght is...
for your next problem...
isolate the absolut values: [x+3]=2 and [y+1]=6
then figure out the possible values for each variable: x=-1,-5 y=5,-7
then add each x with each y, take the absolute values of each sum
[-5+5]=0 [-5+1]=4 [-7+-1]=8 [-5+-7]=12</p>
<p>i don't think iscoceles can be scalene... because their definitions don't allow it... there are scalene right triangles (pythagorean triangle things, 30,60,90) and iscoceles right triangles (45,45,90)</p>
<p>Dang your hella smart. Yeah I never though of doing it that way! the absolute value problem. So I guess my method doesn't work, probably just a coincidence. </p>
<p>About the triangle, I meant can you have more than 2 different sides? like if two sides were given and you had to find x side, can there be more than 1 possibility? Can anyone think up of sides that would work?</p>
<p>yeah... you can have a 2,6,7 2,8,7 um... anything that works for that one rule i typed earlier.... as long as it fits the constraints placed by the problem.
i wish is was this good at the other two sections... ugh.</p>
<p>Hmm, okay. Haha I wish that I'm as good as you in ALL the sections. My psat score on math was in the 60s Bleh, without any studying/prepare though.</p>
<p>AznBoi:
About the triangle problem - draw the base AB of length 7, and let AC be the side of length 2. Then CB is the side of length x.</p>
<p>Start with angle CAB = 0 degrees, and slowly increase this angle through values 10, 20, 30.... degrees, all the way through 180 degrees. At CAB=0 degrees, x would be 7-2 = 5, but you won't have a triangle. As CAB increases, x can take on gradually increasing integer values of 6,7,8....but not 9, because angle CAB would be 180 degrees at that point, and again ABC would no longer be a triangle.</p>
<p>In summary, x can take on all integer values > 7-2 and < 7+2. The same logic would work for any pair of integers, but it is easier to visualize if the base AB is drawn as the greater length.</p>
<p>For any triangle with sides a,b,c: a+b>c a+c>b b+c>a The sum of the lengths of any two sides MUST be greater than the length of the third.</p>
<p>so, the only possible triangle is 2,7,6. (2,7,5) or (2,7,4) or (2,7,3) wont work because 2+5, 2+4, and 2+3 are not greater than 7. 2,7,7 isn't a choice because x must be less than 7</p>