2006-10-11 17:55:05 · 6 answers · asked by ourdoorsangler 1 in Science & Mathematics ➔ Mathematics
how did you figure or what formula did you use to get 56 tables. its a math question on a paper i have. Thanks everyone.
2006-10-11 18:09:28 · update #1
First, the maximum THEORETICALLY possible is 56. Since the area of the room is equal to the area of 56.74 tables, it is impossible to get a 57th table in.
So we have to find a configuration that works for 56 tables.
Fortunately, the person who came up with the dimensions for the problem had a method in mind:
First, put 4 tables lengthwise (end to end) across the width (52') of the room. This takes up 51' out of the 52' width.
Then do this again 4 times, so that you have a rectangle of 20 tables whose dimensions are 4 tables lengthwise and 5 tables widthwise.
The 5 table widths take up 28'9" of the length of the room, leaving 51'3" available out of the 80' length of the room. This permits 4 tables to be put in lengthwise. And we can set up 9 rows of 4 tables (9 table-widths take up 51'9", which doesn't quite use up all the width of the room. These 9 rows of 4 tables account for another 36 tables, for a total of 56 tables.
And we know (see the first paragraph above) that 56 is the most tables that could possibly fit into the room (without stacking, and without worrying about whether anyone could play pool).
2006-10-11 19:21:41 · answer #1 · answered by actuator 5 · 0⤊ 0⤋
54 pool tables, if they are all push in close together.
Put the width of the pool tables along the width of the room and you will fit in 9 of them. The length of the pool table go along the length of the room and you will fix in 6 pool tables.
9 x 6 = 54.
2006-10-12 01:18:47 · answer #2 · answered by Brenmore 5 · 0⤊ 0⤋
Assuming no stacking, and the tables packed as close as possible, and the floor is strong enough, about 56 tables will fit. Why do you want to know ?
2006-10-12 01:05:14 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Hmmmm...do you want anybody to be able to play pool? If not, is stacking legal? How tall is the room?
Insufficient information to answer.
2006-10-12 00:57:51 · answer #4 · answered by Beckee 7 · 1⤊ 0⤋
That would all depend on the spectator space....are you also trying to figure out how much space a player needs?
Or is this just to pack them in.
2006-10-12 01:05:53 · answer #5 · answered by Anonymous · 0⤊ 0⤋
9. Don't forget the elbow room
2006-10-12 01:03:25 · answer #6 · answered by profitmessenger 2 · 0⤊ 0⤋