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anonymous

  • 4 years ago

Fool's problem of the day, A tiled floor has dimensions \(m \times m \) sq.m. The dimensions of the tile used are \(n\times n\) sq.m. All the tiles used are green tiles except the diagonals tiles which are red. After some years some green tiles are replaced by red tiles to form an alternate red and green tile pattern.If \(m \neq n \) and total number of tiles are odd then how many green tiles are removed?

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  1. anonymous
    • 4 years ago
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    PS: http://finance.groups.yahoo.com/group/ascent4cat/message/2900 http://finance.groups.yahoo.com/group/ascent4cat/message/2915 These solutions are incorrect.

  2. anonymous
    • 4 years ago
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    I am not so sure. is it \(\large m(\frac{m}{n*2} - \frac{3}{2})\)

  3. anonymous
    • 4 years ago
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    I am sorry Ishaan :(

  4. anonymous
    • 4 years ago
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    Umm m is the dimension of the tiled floor, n the dimension of a tile. So, total number of tiles in a column or row (since it's a square) must be \(\frac{m}{n}\). Hmm then we have boundary of red tiles and we wish to create an alternate pattern of Red and Green (Horrible color combination btw). The number of red tiles would be always 1 greater than Green ones after the completion of alternate pattern.(initially we have two tiles in red of each row or column) So, in one row \(\large\frac{\frac{m}{n} -2 -1}{2}\) must be green. We have m-2 column not m sorry. Hence, the answer should be \(\large \frac{\frac{m}{n} - 3}{2} (m-2)\) Am I doing it right?

  5. anonymous
    • 4 years ago
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    Oh Wait.. I did something again umm It should be \[\left(\frac{\frac{m}{n} -1}{2}\right) ( m-2)\]

  6. anonymous
    • 4 years ago
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    Oh no I did something again \[\frac{\frac{m}{n} - 1}{2}, \text{ is the number of green tiles} \implies \text{number of red tiles must be } \frac{m}{n}- 2 - \frac{m}{2n} - \frac{1}{2} \] \[\frac{m}{n}- 2 - \frac{m}{2n} - \frac{1}{2} = \frac{m}{2n} - \frac{3}{2}\] So number of tiles to be replaced are \(\left( \frac{m}{2n} - \frac{3}{2} \right) (m-2)\)

  7. anonymous
    • 4 years ago
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    I can't believe I took so much time to solve a CAT question and still don't know if it's right

  8. anonymous
    • 4 years ago
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    Sorry again it's not what I got, I am getting confused again and again \[\left( \frac{m}{2n} - \frac{3}{2} \right) (\frac{m}{n}-2)\] Because m/n is the number of rows and column not m

  9. anonymous
    • 4 years ago
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    Doesn't seem correct, if you want I can give you the answer.

  10. anonymous
    • 4 years ago
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    Impossibru I checked my logic for even number of rows and column

  11. anonymous
    • 4 years ago
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    Hmm lemme check again

  12. anonymous
    • 4 years ago
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    For Odd this time

  13. anonymous
    • 4 years ago
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    Mere working for small m,n doesn't mean a conclusive proof.

  14. anonymous
    • 4 years ago
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    The correct answer is \( \frac{1}{2} n^2 \left((m-2 n)^2-n^2\right)\)

  15. anonymous
    • 4 years ago
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    \[\left( \frac{m}{2n} - \frac{1}{2} \right) (\frac{m}{n}-2)\] Umm but you don't have n in the denominator? How do you get rows and column?

  16. anonymous
    • 4 years ago
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    \[\frac{1}{2}* \frac{1}{n^2} (m-n)(m-2n) = \frac{1}{2n^2} ( m^2 - 2mn -nm + 2n^2)\]This is all I am getting umm looks like I will have to check again

  17. anonymous
    • 4 years ago
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    \[\left( \frac{m}{2n} - \frac{3}{2} \right) (\frac{m}{n}-2)\] \[\frac{1}{2}\left( \frac{m}{n} - \frac{3}{1} \right) (\frac{m}{n}-2)=\frac{1}{2}* \frac{1}{n^2} (m-3n)(m-2n) = \frac{1}{2n^2} ( m^2 - 2mn -3nm + 6n^2)\] Lets take the case of a floor where m = 99 and n = 1, total rows = columns = 99. 97 tiles in each row are green, 2 are red. We need an alternate pattern which consists of 50 red tiles and 49 green. Since 2 were already red 48 tiles were replaced in each row and there are 99-2 columns on the floor (first and last columns are completely red so no replacement). total tiles to be replaced = 48* (99-2) = 48*97 By my formula I get the same \[\frac{1}{2}(99-3)(99 -2) = 48*97=4656\] And with yours (foolformath's solution) \(\frac{1}{2} n^2 \left((m-2 n)^2-n^2\right)\), I am getting \[\frac{1}{2}* 1* ((99-2)^2 - 1) = 4704\] What did I miss? Please tell me where I am wrong?

  18. anonymous
    • 4 years ago
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    Oh I get it Diagnol's tiles are red, I thought the floor was bounded by red tiles. Silly me I should learn to read questions properly

  19. anonymous
    • 4 years ago
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    \[f(tiles)= \left\{\begin{array}{r}\left( \frac{1}{2}\right)\left( \frac{m}{2n} - 3 \right) \left(\frac{m}{n}-2\right) + \frac{m}{2n} -\frac{3}{2}, \text{if} &\frac{m}{n} \in \{(4k-3), k \in \mathbb{N}\} \\ \left(\frac{1}{2}\right)\left( \frac{m}{2n} - 3 \right) \left(\frac{m}{n}-2\right) + \frac{m}{2n} - \frac{1}{2}, \text{if}& \frac{m}{n} \in \{(4k -1), k \in \mathbb{N}\} \end{array} \right.\] \(f(tiles)\) is the function for number of tiles lol I know I have made it so complicated but it works! :-D lol

  20. anonymous
    • 4 years ago
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    I can't stop laughing lol I made it so bad and yet it works.

  21. anonymous
    • 4 years ago
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    Nah it doesn't works I committed some error again I will check out this problem later

  22. anonymous
    • 4 years ago
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    work*

  23. anonymous
    • 4 years ago
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    \[f(tiles)= \left\{\begin{array}{r}\left( \frac{1}{2}\right)\left( \frac{m}{n} - 3 \right) \left(\frac{m}{n}-2\right) + \frac{m}{2n} -\frac{3}{2}, \text{if} &\frac{m}{n} \in \{(4k-3), k \in \mathbb{N}\} \\ \left(\frac{1}{2}\right)\left( \frac{m}{n} - 3 \right) \left(\frac{m}{n}-2\right) + \frac{m}{2n} - \frac{1}{2}, \text{if}& \frac{m}{n} \in \{(4k -1), k \in \mathbb{N}\} \end{array} \right.\] Correction* Note. I am testing it

  24. anonymous
    • 4 years ago
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    I have it FINALLY! \[\frac{1}{2} \left(\left(\frac{m}{n}-2\right)\left(\frac{m}{n}-3\right) + \frac{m}{n} -3\right) = \frac{1}{2} \left(\frac{m}{n} -3\right) \left( \frac{m}{n} -1\right)\]

  25. anonymous
    • 4 years ago
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    Yay! I got it! The closed form! Yay!

  26. anonymous
    • 4 years ago
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    Took me time, pathetic :-/

  27. anonymous
    • 4 years ago
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    Hehe, should i post my solution?

  28. anonymous
    • 4 years ago
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    Yeah, you should

  29. anonymous
    • 4 years ago
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    Okay, after few hours :)

  30. anonymous
    • 4 years ago
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    Done! :-)

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