Question

the length and the breadth of a conference hall are in the ratio 7 : 4 and its perimeter is 110 m number of tiles, each a rectangle of size 25×20 cm , required for flooring of the hall​

Answer 1

Answer:

Hey mate ur answer is :

Step-by-step explanation:

Ratio in length and breadth = 7 : 4

Perimeter = 110 m

∴ Length + Breadth = 110/2 = 55 m

Sum of ratios = 7 + 4 = 11

∴ Length = (55 × 7)/11 = 35 m

And breadth = (55 × 4)/11 = 20 m

(i) Area of floor = l × b

= 35 × 20 = 700 m2

(ii) Size of tiles = 25 cm × 20 cm

= (25 × 20)/(100 × 100)

= (1/20) m2

∴ Number of tiles =

= Area of floor/Area of one tile

= (700 × 20)/1 = 14000

(iii) Cost of tiles = ₹ 1400 per 100 tiles

∴ Total cost = (14000 × 1400)/100

= ₹ 196000

Answer 2

Answer:

14000 tiles

Step-by-step explanation:

Let, the length of the hall be 7x and the breadth be 4x

Perimeter of hall - 110m

perimeter of rectangle - 2(l+b)

So,

110 = 2(7x+4x)

110 = 14x + 8x

110 = 22x

x = 110/22

x = 5

So, length of hall = 35

breadth of hall = 20

Area of hall = length X breadth

= 35 X 20

= 700msq

length of tile = 25cm

= 0.25m

breadth of tile = 20cm

= 0.2m

area of one tile = 0.25 X 0.2

0.05ms

No. of tiles for flooring = area of hall / area of one tile

= 700/0.05

= 14000 tiles