Question

A rectangular floor which measures 15 m*8 m is to be laid with tiles measuring 50 cm*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?

Answer 1

AREA OF THE FLOOR = 15*8 m^2
AREA OF ONE TILE = 0.5*.25 m^2
NO. OF TILES REQUIRED = 120 / 0.125 = 960 tiles
NOW,
AFTER PUTTING COVER, IT WILL LEAVE TWO RECTANGLES OF 13*1 AND TWO RECTANGLES OF 8*1
THEREFORE TOTAL AREA LEFT = 26 + 16 = 42
fraction of floor uncovered = 120/42 = 20/7

Answer 2

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

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