Question

the length and breadth of the prayer hall of a school 18m and 1200 respectively sqaretiles of 12 cm are to be fixed on its floor find the number of tiles required find the cost of fixing tiles at the rate of₹8 per m

Answer 1

Area of the hall=lxb
=18x1200
=21600m sq.
Area of one tile=sxs
=12x12
=144 cm sq.
or 144/100 m sq.
So,number of tiles needed to be fixed =21600m sq./144/100m sq.
=2160000/144
=15000 tiles.
So, the cost of fixing the tiles=(8x15000)Rs
=120000Rs.