A room is 18m long 4m wide, It's floor is to be covered with rectangular tiles of size 9cm by 8cm. Find the cost of tiles at 5.25 per tile.
Answers
ANSWER :-
- Cost of tiles used is 525.
GIVEN :-
- Floor of room has dimensions 18m × 4m.
- Rectangular tiles has dimensions 9cm × 8cm.
- Cost of one tile is 5.25.
TO FIND :-
- The total cost of tiles.
TO KNOW :-
★ Area of rectangle = length × breadth
★ 1m = 100cm
SOLUTION :-
Firstly we will convert the dimensions of room to cm as dimensions of tiles are given in cm.
As , 1m → 100cm
Dimensions of room becomes 1800cm × 400cm
Let 'n' numbers of tiles can be fit in the room .
So, Area of all tiles will be equal to Area of room
→ n × Area of one tile = Area of room
→ n × 9 × 8 = 1800 × 400
→ n × 72 = 7200
→ n = 7200/72
→ n = 100
Hence , 100 tiles are used to cover the floor of room.
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1 tile costs 5.25.
100 tiles will cost → 5.25 × 100
Hence , Total cost of tiles that are used is 525.
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THINGS TO REMEMBER :-
- Make sure that all the values are in same units . In this question , dimensions of tiles were given in cm and that of room in m.
MORE TO KNOW :-
★ Area of square = side²
★ Area of triangle = (1/2) × bh
★ Area of circle = πr²
★ Area of semi-circle = πr²/2
★ Area of trapezium = h(sum of parallel sides/2)
★ Area of parallelogram = b × h
Question:-
A room is 18m long 4m wide, It's floor is to be covered with rectangular tiles of size 9cm by 8cm. Find the cost of tiles at 5.25 per tile.
Given:-
• Length of room = 18m
• Width of room = 4m
• Length of tile = 9cm
• Breadth of tile = 8cm
To find :-
Cost of tiles
Using Formula :-
Area of rectangle = l × b
Solution :-
Length of room = 18m = 1800cm
Breadth of room = 4m = 400cm
Area of room = l × b
= 1800 × 400
= 7200cm².
Area of room = 7200cm².
Length of tile = 9cm
Breadth of tile = 8cm
Area of tile = l × b
= 9 × 8
= 72cm²
Area of tile = 72cm²
No. of tiles = Area of room / Area of tile
= 7200cm²/72cm²
= 100
No. of tiles = 100
Cost of 1 tile = 5.25
Cost of 100 tiles = 5.25 × 100
= Rs.525
Therefore,
Cost of tiles = Rs.525