Question

The length and breadth of a conference hall are in the ratio 7 : 4 and its perimeter is 110 m. Find:

(i) Area of the floor of the hall.

(ii) number of tiles, each a rectangle of size $25cm\times 20cm,$ required for flooring of the hall.

(iii) The cost of the tiles at the rate of Rs 1,400 per hundred tiles.

Answer 1

Ratio in length and breadth =7 : 4

Perimeter =110 m

$\therefore$ Length +Breadth $=\frac{110}{2}=55m$

Sum of ratios =7+4=11

$\therefore$ Length $=\frac{55\times 7}{11}=35m$ and breadth =\frac{55 \times 4}{11}=20 m

(i)Area of floor $=l\times b=35\times 20=700m^2$

(ii)Size of tile =25 cm $\times$ 20 cm

$=\frac{25\times 20}{100\times 100}$

$=\frac{1}{20}{m}^2$

$\therefore$ Number of tiles

$=\frac{\text{Area of floor}}{\text{Area of one tile}}$

$=\frac{700\times 20}{1}=14000$

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

$\therefore$ Toal cost $=\frac{14000\times 1400}{100}$

=Rs 196000