. A floor which measures 15 mx 8 m to be laid with tiles measuring 50 cm by 25 cm Find the number
of tiles required. Further, if a carpet is laid on the floor so that a space of 1 m exists between its edges
and the edges of the floor, what fraction of the floor is uncovered?
Answers
Step-by-step explanation:
We know, 50cm=0.5m and
25cm=0.25m
Tiles along the length of the room=0.5/15
=30 tiles
Tiles along the length of the room=0.2/58
=32 tiles
Total number of tiles required =30×32
=960 tiles
To leave 1m between the carpet and wall on all sides, the carpet needs to be 2m shorter in each dimension.
Therefore,
(15−2)m(8−2)m
=(13×6)m
=78m2
Therefore, the room area is 15×8
=120m2
The carpet area is 78m2
The fraction of the area uncovered is
=120/120−78
=120/42
=20/7
Consider ABCD as a rectangular field of measurement 15m × 8m
Length = 15 m
Breadth = 8 m
Here the area = l × b = 15 × 8 = 120 m2
Measurement of tiles = 50 cm × 25 cm
Length = 50 cm = 50/100 = ½ m
Breadth = 25 cm = 25/100 = ¼ m
So the area of one tile = ½ × ¼ = 1/8 m2
No. of required tiles = Area of rectangular field/Area of one tile
Substituting the values
= 120/ (1/8)
By further calculation
= (120 × 8)/ 1
= 960 tiles
Length of carpet = 15 – 1 – 1
= 15 – 2
= 13 m
Breadth of carpet = 8 – 1 – 1
= 8 – 2
= 6 m
Area of carpet = l × b
= 13 × 6
= 78 m2
We know that
Area of floor which is uncovered by carpet = Area of floor – Area of carpet
Substituting the values
= 120 – 78
= 42 m2
Fraction = Area of floor which is uncovered by carpet/ Area of floor
Substituting the values
= 42/120
= 7/20