IIT JEE , AIEEE Forum - Mathematics , Algebra

tushartuteja

A tiled floor of a room has dimensions m*m sq.m.The dimensions of the tile used are n*n sq.m.All tiles used are green tiles except diagonal tiles which are red. After some time some green tiles are removed to form an alternate red and green pattern. How many green tiles are removed?(m is not equal to n and the total number of tiles are odd).

i need the solution with correct answer

for_succes

Is the answer
1/2*m/n .....
if that's rite i wud giv my argument.......

tushartuteja

your answer is wrong for_success

the answer is

[(m-2n)-2n^2]/2n²

neeraj_agarwal_1990

the answer is
[(m-2n)-2n2]/2n2

avinash.sharma

As the alternate colours arrangement is possible only for odd tiles and in a row the number of tiles are Ö(m²/n²) = m/n

Now put the m/n = 3+{(r-1) *2} -------------------------(1)

It is a simple AP r^th term formula with 3 as first term and 2 is the common difference, which yields the value of r.

Now total replacements are

= 4 * [(r-1)+(r-2)+(r-3)+..........+1]

Summation of the (r-1) terms where first term is 1 and last term is (r-1) and common difference is 1

= 4*[ {(r-1) * (1+r-1)}/2

= 2*(r-1) * (r) ---------------------- (2)

m/n	3	5	7	9	11	13
r by (1)	0	2	3	4	?	?
Total replacements by (2)	0	4	12	24	?	?

You should try to solve ??? marked fields for practice.

Note : I have taken both the diagonals having red tiles initially. But if you want to say that only one diagonal has red tiles than you should add (m/n)-1 in the final result [Total replacements by (2) ] to get the total replaced tiles.

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