(SAT1 Practice Test Question: Last Question in Set)
A man is laying tiles in his bathroom. He uses five colors of tile: red, blue, yellow, green, and pink, and lays them down in pairs of different colored tiles. If he uses each possible color pair (regardless of order, so red-blue is the same as blue-red) exactly 3 times, how many total tiles are there in his bathroom?
Edited on behalf of Ergot.
30
0. He's laying tiles not tires.
Quote from: rabbit on December 03, 2006, 02:00:58 PM
30
Close. Pay special attention to the wording of the problem.
60. Labels are important. 30 pairs = 60 tiles.
Quote from: Ergot on December 03, 2006, 05:10:28 PM
60. Labels are important. 30 pairs = 60 tiles.
Correct.
The solution:
(http://latex.sidoh.org/?render=%5C%5CThere%5C%20are%5C%20%5Cfrac%7B5%5Ctimes4%7D%7B2%7D%20%3D%2010%5C%20combinations.%20%0D%0A%5C%5C%0D%0A%5C%5CThe%5C%20person%5C%20desires%5C%20three%5C%20of%5C%20each%5C%20pair%2C%20which%5C%20results%5C%20in%5C%203%5Ctimes10%20%3D%2030%5C%20pairs.%20%0D%0A%5C%5C%0D%0A%5C%5CThere%5C%20are%5C%20two%5C%20tiles%5C%20per%5C%20pair%2C%5C%20so%5C%20there%5C%20are%5C%202%5Ctimes30%20%3D%2060%5C%20tiles.)
What was the original wording, that was edited on behalf of ergot?
Quote from: Joex86] link=topic=8055.msg101336#msg101336 date=1165191407]
What was the original wording, that was edited on behalf of ergot?
It was a typo. He probably had: ". . .how many total tires are there in his bathroom?"
I have 2 tires in my bathroom ^^.
Wow, you obviously have no life.
Quote from: Ender on December 07, 2006, 05:17:20 PM
Wow, you obviously have no life.
Wow, you obviously have no tires in your bathroom.
I believe that the answer Ender put as correct it wrong.
Take this as a summarized example. There are 4 tile types: 1,2,3 & 4. They are laid down in pairs: 1-2 3-4 in row one, and 4-3-2-1 in row two. (see below.)
(http://static.flickr.com/120/316766090_2c38f962a1_o.jpg)
However, 1-2 3-4, 4-3, & 2-1 are not the only pairs there. As shown by the lines in the diagram, there are really 10.
1-2, 2-3, 3-4
4-3, 3-2, 2-1
1 2 3 4
| | | |
4 3 2 1
Therefore, the way used to calculate the answer I believe is false.
Quote from: Ender on December 03, 2006, 05:26:14 PM
Correct.
The solution:
(http://latex.sidoh.org/?render=%5C%5CThere%5C%20are%5C%20%5Cfrac%7B5%5Ctimes4%7D%7B2%7D%20%3D%2010%5C%20combinations.%20%0D%0A%5C%5C%0D%0A%5C%5CThe%5C%20person%5C%20desires%5C%20three%5C%20of%5C%20each%5C%20pair%2C%20which%5C%20results%5C%20in%5C%203%5Ctimes10%20%3D%2030%5C%20pairs.%20%0D%0A%5C%5C%0D%0A%5C%5CThere%5C%20are%5C%20two%5C%20tiles%5C%20per%5C%20pair%2C%5C%20so%5C%20there%5C%20are%5C%202%5Ctimes30%20%3D%2060%5C%20tiles.)
Yes, there are 10 combinations. However the person want three of each pair. When you put 2 pair of tile together in a square:
1 - 2
| |
3 - 4
you get 4 color pairs out of these two pairs of tiles. Taking this into account I do not believe that 60 is the correct answer.
Here is my answer and work (references to figure 1 and 2 refer to the figures at the bottom of the post):
First, I figured out how many possible pairs their were. Like Ender, (5*4)/2, 10 combinations.
Following that I calculated how many pairs, connections between adjacent tiles, would be needed in order to have 3 set of each of the 10 combinations. That is 30 connections.
By changing the dimensions around until i got the 30 connections necessary (see figure 1), I then added the sum of the tiles needed to create that figure 30 connections.
As seen in figure 2, the total number of tiles needed is 40.
EDIT: I counted all the tiles in figure 2 twice, the correct answer therefore is 20.
I believe the correct answer to this problem is
40 20.
(http://static.flickr.com/112/316786859_77892ab000_o.jpg)
He lays them down in pairs. The pairs are not adjacent to eachother (he puts them down after he works out each color pair) and does not form new pairs. I'm sorry that this was unclear. In this context, 30 is the correct answer.