The length and the breadth of a conference hall are in the ratio $7 : 4$ and its perimeter is $110 m$ . Find:
(1) area of the floor
(2) number of tiles, each a rectangle of size $25 c m \times 20 c m$ , required for flooring of the hall.
(3) the cost of the tiles at the rate of $R s . 1400$ per hundred tiles.

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Solution

Length $= 7 x$
Breadth $= 4 x$
Perimeter $= 110$ m
Perimeter $= 2$ (length $+$ breadth)
$110 = 2 (7 x + 4 x)$
$110 = 2 \times 11 x$
$110 = 22 x$
$x = \frac{110}{22}$
$x = 5$
Therefore, length $= 35 m$
Breadth $= 20 m$
Area of the hall $= l \times b$
$= 35 \times 20 m^{2}$
$= 700 m^{2}$
Now,
Area $= 700 m^{2} = 700 \times 10000 = 7 \times 10^{6} c m^{2}$
Area of 1 tile $= 25 \times 20 = 500 c m^{2}$
No. of tiles required = ar. of hall $\div$ ar. of 1 tile
$= \frac{7000000}{500}$
$= 14000$
Cost of tiles at rate of $R s . 1400$ per hundred tiles $= \frac{14000}{100} \times 1400$
$= R s . 196000$

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The length and the breadth of a conference hall are in the ratio 7:4 and its perimeter is 110m. Find: (1) area of the floor (2) number of tiles, each a rectangle of size 25cm×20cm, required for flooring of the hall. (3) the cost of the tiles at the rate of Rs.1400 per hundred tiles.