Alex plans to cover a rectangular floor of dimensions 9.5 meters and 11.5 meters using tiles. Two types of square shaped tiles are available in the market. A tile with side 1 meter costs 100 dollars and a tile with side 0.5 meters costs 30 dollars . The tiles can be cut if required. What will be the minimum cost of covering the entire floor with tiles? (Assume cost of cutting to be 0)

My approach:-

Firstly I noticed that a 1x1 tile could be divided into 4 (0.5x0.5) tiles. Now we know the cost of 1x1 tile is 100, and cost of 4 (0.5 x 0.5) tiles would be 30 * 4 =120 , enter image description here

therefore it would be economical to use as many 1x1 tiles as possible in case of covering, so I ended up with the following diagram, having complete 1x1 tiles used for area = 11 * 9 = 99 m$^2$ , enter image description here, there are (11 * 9) tiles of 1x1 dimension used and each tile would cost 100 , so therefore the cost of these 99 tiles would be 9900 dollars.

Now for the remaining 0.5 m gaps instead of fitting (0.5 x 0.5) tiles , what I did was to use the 1x1 tile only and cut it into tiles of (1x0.5) and fill it in the spaces left enter image description here

Number of (1x0.5) tiles required in the very last column = 9 Number of (1x0.5) tiles required in the very bottom row = 11 therefore total number of (1x0.5) tiles = 20 , and these can be obtained from 10 (1x1) tiles therefore a total of (10*100) dollars more = 1000 dollars , so my answer would become 9900+1000=10900 dollars , but the answer is 10930 dollars

Official Answer given:

Area of floor = 109.25 m$^2$

Cost of covering 109 m$^2$ (with 1x1 tiles) = 10900 dollars

Cost of covering 0.25 m$^2$ (with 0.5x0.5 tile)= 30 dollars

Total cost = 10930 dollars

Why is my approach wrong ?

  • 4
    $\begingroup$ Your approach works, you just missed the corner in the bottom right of your diagram that's 0.5m by 0.5m. That's where the official answer's extra 0.25 square meters comes from. It's cheapest to cover that with a 0.5m square for \$30 more. $\endgroup$
    – Amaan M
    Commented Dec 28, 2021 at 23:31
  • $\begingroup$ yes , thank you :) , do you think the official solution is correctly presented, I am confused as why they took 109 m^2 to be covered first ? $\endgroup$
    – Fin27
    Commented Dec 29, 2021 at 21:15
  • 2
    $\begingroup$ I think it's fine the way they presented. The idea is that since you can cut up the $1 m^2$ tiles as needed, and they're cheaper than the $0.5 m \times 0.5 m$ tiles, you can buy $109$ tiles that are $1 m^2$ each, and cut them/arrange them as needed to fit the $109 m^2$ of the floor. Then, for the additional $0.25 m^2$ left over, you could either buy another $1 m^2$ tile for \$100 and cut out the area needed, or you could buy a $0.5 m \times 0.5 m$ tile for \$30, and the latter is obviously the cheaper way to get the remainder. $\endgroup$
    – Amaan M
    Commented Dec 29, 2021 at 21:52

1 Answer 1


Your approach nearly solves the puzzle. It is correct that using almost all $1*1$ tiles is optimal. However, you neglected the issue of the final corner. Using your method, one corner would have a gap of $\dfrac12$m $*\dfrac12$m. This is not a problem that is difficult to solve. As you have covered $109$m$^2$ with the larger tiles already, there are no leftovers. So using a smaller tile would perfectly fill in the gap required, and only cost an additional \$30. This is why your total comes up short of the actual answer.


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