3 simple questions. Please don’t make fun of me. Thanks. If you have a box with dimensions 5, 6, 5, the volume is 5 x 6 x 5 = 150. how many small boxes can you divide this big box into perfectly and what will be the dimensions of those little boxes. There could be more than one set of such boxes. They must all have the same dimensions. If I do a job in s minutes, and you do a job in r minutes than I can find how long it will take both of us by using the formula: 1/s + 1/r = 1/t. But lets say that we do 5 of these jobs. Then what percentage of the job did you do? Is there a formula to figure this out? Finally, I know that x:y :: 3:4 implies that x/y = ¾. What does x:y:z :: 2:3:4 imply?
needhelp Wrote: ------------------------------------------------------- > 3 simple questions. Please don’t make fun of me. > Thanks. > > If you have a box with dimensions 5, 6, 5, the > volume is 5 x 6 x 5 = 150. how many small boxes > can you divide this big box into perfectly and > what will be the dimensions of those little boxes. > There could be more than one set of such boxes. > They must all have the same dimensions. Let the small boxes be of 1 x 1 x1 demension, then you could fit 150 exactly > If I do a job in s minutes, and you do a job in r > minutes than I can find how long it will take both > of us by using the formula: 1/s + 1/r = 1/t. But > lets say that we do 5 of these jobs. Then what > percentage of the job did you do? Is there a > formula to figure this out? With the formula you given, it would just be (1/r)/(1/t)… > Finally, I know that x:y :: 3:4 implies that x/y = > ¾. What does x:y:z :: 2:3:4 imply? 2/3/4?
needhelp Wrote: ------------------------------------------------------- > 3 simple questions. Please don’t make fun of me. > Thanks. > > If you have a box with dimensions 5, 6, 5, Then it will be easy to hit the bottom but hard to f#$& up the edges.
- 150 1 x 1 x 1 boxes 2. 3. x/y = 2/3, y/z = 3/4, x/z = 2/4 or 1/2
The first question asks what are the other possibilities other than the 1x1x1. is there another set of boxes? and if so how can you determine that?
1 x 2 x 1 75 of them.
needhelp Wrote: ------------------------------------------------------- > The first question asks what are the other > possibilities other than the 1x1x1. is there > another set of boxes? and if so how can you > determine that? you could have 2 * 3 * 2 boxes or 3 * 2 * 3 boxes if they didn’t have to be oriented the same way.
There are a lot answers to #1, you probably missed mentioning the whole question
Ugghhhhh… I made up those dimensions dudes. My question is if you have a box of dimensions x,y,z, how do you determine how it cab divided into smaller boxes and what are the dimensions of such boxes. geniuses.
If the smaller boxes are the same shape then 1 * 1.2 * 1 = 125 boxes. That assumes the minimum length of a dimension is 1.
needhelp Wrote: ------------------------------------------------------- > 3 simple questions. Please don’t make fun of me. > Thanks. > > Finally, I know that x:y :: 3:4 implies that x/y = > ¾. What does x:y:z :: 2:3:4 imply? 1/6
well damn you could say the dimension of the small boxes are .1 x .1 x .1 for 1500, or .01 x. 01 x .01 for 15000
needhelp Wrote: ------------------------------------------------------- > 3 simple questions. Please don’t make fun of me. > Thanks. > > If I do a job in s minutes, and you do a job in r > minutes than I can find how long it will take both > of us by using the formula: 1/s + 1/r = 1/t. But > lets say that we do 5 of these jobs. Then what > percentage of the job did you do? Is there a > formula to figure this out? > 5(1/s) + 5(1/r) = 5(1/t) For your percentage it would be (5(1/s))/(5(1/t))*100
Why dont you post the actual questions instead of your hard to understand abridged versions?
well, what its asking you is how many ways to make a volume of 150 with uniform shapes. I’m going to make an assumption here: That if you change the orientation of a box, you’ve actually changed the dimensions (width becomes depth, etc) so changing orientations is not allowed. The given one is 1 unit * 5 wide * 5 deep * 6 high = 150 Assuming you can’t split units, the opposite case (maximum boxes) would be 150 units * 1 wide * 1 deep * 1 high = 150 So, what you need to figure out is, without breaking outside the 5x5x6 dimensions, how many different configurations can you come up with? You need to focus on the integer factors, since you can’t split the unit (I’m assuming this, you never said it though) and you can’t change orientation. So, the 5 sides can be 1 or 5, and the 6 side can be 1, 2, 3, 6 So 2 possibilities for width * 2 possibilities for height * 4 possibilities for depth = 16 possible box shapes (1 of which is the big box). Without looking at depth, we know we have 4 possible LxW shapes (1x1, 1x5, 5x1, 5x5). Now we figure out how many boxes for each of these LxW possibilities… You’re an actuary, so I assume you can take it from here.
dlpicket Wrote: ------------------------------------------------------- > well, what its asking you is how many ways to make > a volume of 150 with uniform shapes. > I’m going to make an assumption here: That if you > change the orientation of a box, you’ve actually > changed the dimensions (width becomes depth, etc) > so changing orientations is not allowed. > > The given one is > > 1 unit * 5 wide * 5 deep * 6 high = 150 > > Assuming you can’t split units, the opposite case > (maximum boxes) would be > > 150 units * 1 wide * 1 deep * 1 high = 150 > > So, what you need to figure out is, without > breaking outside the 5x5x6 dimensions, how many > different configurations can you come up with? You > need to focus on the integer factors, since you > can’t split the unit (I’m assuming this, you never > said it though) and you can’t change orientation. > > > So, the 5 sides can be 1 or 5, and the 6 side can > be 1, 2, 3, 6 > > So 2 possibilities for width * 2 possibilities for > height * 4 possibilities for depth = 16 possible > box shapes (1 of which is the big box). Without > looking at depth, we know we have 4 possible LxW > shapes (1x1, 1x5, 5x1, 5x5). Now we figure out how > many boxes for each of these LxW possibilities… > > > You’re an actuary, so I assume you can take it > from here. This is more in line with what I was looking for.
the funny thing is that i am using this book called GRE math bible which has questions categorized as easy, medium, hard, very hard… i get stumped at the easy ones and cant believe how easy the hard ones are.
X is number of boxes X = 5/L * 5/W * 6/H X * L * W * H = 150 So I guess you are summing the set of possible X. I get 432 minus the 1 big box = 431 little boxes