How many tiles with dimensions $5 c m$ and $12 c m$ will be required to fit a region with length $100 c m$ and breadth $96 c m$ respectively?

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Solution

Area of region $= l e n g t h \times b r e a d t h$
$= 100 c m \times 96 c m = 9600 c m^{2}$
Area of one tile $= l e n g t h \times b r e a d t h$
$= 5 c m \times 12 c m = 60 c m^{2}$
Number of tiles $= \frac{Area of region}{Area of one tile} = \frac{9600}{60} = 160$
Number of tiles $= 160$

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Q. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are 100 cm and 144 cm respectively?

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How many tiles with dimensions 5 cm and 12 cm will be required to fit a region with length 100 cm and breadth 96 cm respectively?

Area of region =length×breadth =100cm×96cm=9600 cm2Area of one tile=length×breadth =5cm×12cm=60 cm2Number of tiles=Area of regionArea of one tile=960060=160Number of tiles=160

How many tiles with dimensions $5 c m$ and $12 c m$ will be required to fit a region with length $100 c m$ and breadth $96 c m$ respectively?

Area of region $= l e n g t h \times b r e a d t h$
$= 100 c m \times 96 c m = 9600 c m^{2}$
Area of one tile $= l e n g t h \times b r e a d t h$
$= 5 c m \times 12 c m = 60 c m^{2}$
Number of tiles $= \frac{Area of region}{Area of one tile} = \frac{9600}{60} = 160$
Number of tiles $= 160$