Question 1179642: 1.the classroom is 20 feet long and 30 feet wide.the principal decided that the tiles would look attractive in that class.if each tile is 24 inches long and 36 inches wide,how many tiles are needed to fill the classroom?
2.a rectangle is 4 times as long as it is broad.if the lenth is increased by 4 inches and the width is decreased by 1 inch,the area would be 60 squares inches.what were the dimensions of the original rectangle?
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! 1.the classroom is 20 feet long and 30 feet wide.the principal decided that the tiles would look attractive in that class.if each tile is 24 inches long and 36 inches wide,how many tiles are needed to fill the classroom?
Area of floor = L * W 30 fT820 ft = 600 ft^2
Area of tile = 24 inch * 36 in.
= 3ft*2ft
=6ft2
Number of tiles = area of floor/ area of tile
= 600/6
You continue
2.a rectangle is 4 times as long as it is broad.if the lenth is increased by 4 inches and the width is decreased by 1 inch,the area would be 60 squares inches.what were the dimensions of the original rectangle?
a rectangle is 4 times as long as it is broad.
let breadth be x
length will be 4x
if the length is increased by 4 inches(4x+4) and the width is decreased by 1 inch,the (x-1)area would be 60 squares inches.
(x-1)(4x+4)=60
4x^2+4x-4x-4=60
4x^2=64
x^2=16
x= 4
Breadth = 4 in
length = 16 in
Answer by ikleyn(53742)  (Show Source):
You can put this solution on YOUR website! .
1.the classroom is 20 feet long and 30 feet wide. the principal decided that the tiles would look attractive
in that class. if each tile is 24 inches long and 36 inches wide, how many tiles are needed to fill the classroom?
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In his post, @mananth solved this problem comparing the area of one single tile with the area of the classroom.
Although in this case it leads to correct answer, nevertheless it is not a correct way to solve
and to construct arguments. In other words, his solution is badly designed.
Below I place my correct and accurate solution.
With the dimension of the room in one direction of 20 feet, 20/2 = 10 tiles can be placed in this direction.
With the dimension of the room in the other direction of 30 feet, 30/2 = 15 tiles can be placed in this direction.
Hence, 10 x 20 = 200 tiles are needed to cover the floor of the classroom. ANSWER
Now the problem is solved correctly and accurately.
Why it is necessary to look in each dimension separately ?
To guarantee that the integer number of tiles fits in each dimension.
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