SOLUTION: A square floor is fitted with rectangular tiles of perimeter 220cm. Each row (tile lengthwise) carries 20 less tiles than each column (tile breadth wise) if the length of the floo

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Question 1161306: A square floor is fitted with rectangular tiles of perimeter 220cm. Each row (tile lengthwise) carries 20 less tiles than each column (tile breadth wise) if the length of the floor is 9.6m, calculate the dimensions of the tiles
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The easiest way I see for solving the problem is by trial and error, aided by logical reasoning.

(1) Since the square floor of side length 9.6m = 960cm is exactly covered by the tiles, the length and width both have to be factors of 960cm

(2) The perimeter of each tile is 220cm, so the sum of the length and width is 110cm

There are not many combinations of tile dimension and number of tiles that have a product of 960cm; and there is only one case where the sum of two of the possible dimensions is 110cm:
    tile   # of
  dimension  tiles
  ----------------
      96     10
      80     12
      64     15
      60     16
      48     20
      40     24
      32     30
      30     32
      24     40
      20     48
      16     60
    ...

The only pair of dimensions that give a perimeter of 220cm are 80cm and 30cm.

And those dimensions satisfy the condition that the number of tiles required lengthwise (960/80 = 12) is 20 less than the number required breadth wise (960/30 = 32); so all the conditions of the problem are satisfied.

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Below is a formal algebraic solution I came up with....

There might well be other formal algebraic processes that find the solution more easily than the one I found.

Let x be the length (cm) of each tile; then, since the perimeter of the tile is 220cm, the width is 110-x.

The number of tiles required to cover the floor lengthwise is 960/x; the number required breadth wise is 20 more than that, or (960/x)+20.

The product of the width of the tiles and the number of tiles required to cover the floor breadth wise is 960:

%28110-x%29%28%28960%2Fx%29%2B20%29+=+960
105600%2Fx%2B2200-960-20x+=+960
105600%2Fx-20x+=+-280
20x-280-105600%2Fx+=+0
20x%5E2-280x-105600+=+0
x%5E2-14x-5280+=+0
%28x-80%29%28x%2B66%29+=+0
x+=+80 or x+=+-66

Obviously we want the positive answer -- which gives us tile dimensions of 80cm and 30cm.

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.

            There is an elegant way to construct an equation and solve the problem. 


Let w be the width of a tile, in centimeters (the same for all tiles).

Then the length of the a tile is 110-x centimeters (since 110 = 220%2F2 is half of the perimeter).


The number of tiles in each row (i.e. longwise) is  960%2F%28110-x%29. 

The number of tiles in each column              is  960%2Fx. 


Second number is 20 more that the first one.  

It gives you this equation

    960%2Fx - 960%2F%28110-x%29 = 20.


It is your basic equation.


To solve it, first cancel the common factor 20 in both sides

    48%2Fx - 48%2F%28110-x%29 = 1.


Next multiply both sides by x*(110-x).  You will get

    48*(110-x) - 48*x = x*(110-x).


Simplify and reduce to the standard form quadratic equation

    48*110 - 48x - 48x = 110x - x^2

    x^2 - 206x + 5280 = 0.


At this point, you can solve this quadratic equation EITHER using the quadratic formula OR factoring

    (x-30)*(x+176) = 0.


This equation gives the only positive solution x= 30 centimeters.


ANSWER.  The dimensions of a tile are 30 cm (width) by (110-30) = 80 cm (length).

Solved.