Question

The length of a room is 50 per cent more than its breadth. The cost of carpeting the room at the rate of Rs38.50m^2 is Rs924 and the cost of papering the walls at Rs.3.30m^2 is Rs.214.50. If the room has one door of dimensions 1m * 2m and two windows each of dimensions 1m * 1.5m, find the dimensions of the room.

Answer 1

Step-by-step explanation:

SOLUTION

Given :

The length of a room is 50% more than its breadth.

The cost of carpeting the room at the rate of ₹ 38.50.

The cost of papering the walls at ₹ 3.30 m² is ₹ 214.

One dimension of the room = 1 m × 2 m.

Second dimension of the room = 1 m × 1.5 m.

Find :

The dimensions of the room.

SO :

Let the breadth of the room be x meters.

As the length of the room is 50% more than its breadth,

$length \: of \: the \: room = (1 + \frac{50}{100} ) \: of \: x \: meters \: = \frac{3}{2} xmeters.$

$= > the \: area \: of \: the \: floor \: of \: the\: room \: = ( \frac{3}{2} x \times x) {m}^{2} = \frac{ {3x}^{2} }{2} {m}^{2} .$

Cost of carpeting the room at the rate of ₹38.50 m² = ₹(3x²/2 × 38.5).

According to given :

$= > \frac{ {3x}^{2} }{2} \times 38.5 = 924$

$= > {x}^{2} = \frac{924 \times 2}{3 \times 38.5}$

$= > {x}^{2} = 16$

$= > x = \sqrt{16}$

$= > x = 4.$

Hence :

the length of the room = 3/2x meters = 3/2 × 4 meters = 6 meters

and its breadth = x meters = 4 meters.

Let the height of the room be h meters.

Then :

The surface area of all the walls of the room

=> 2(length + breadth)  × height

=> 2(6 + 4) × h m²

=> 20h m².

Area of one door = 1 × 2 m² = 2m²,

area of two windows =  2(1 × 1.5) m² = 3m².

Surface area to be prepared  

=> surface area of walls - surface area of door and windows

=> (20h - 2 - 3) m²

=> (20h - 5) m².

Cost of preparing the walls of the rate of ₹3.30 m² = ₹(20h - 5) × 3.30.

According to given :

         (20h - 5) × 3.30 = 214.50

=> 20h - 5 = 214.50/3.30 = 65

=> 20h = 65 + 5

=> 20h = 70

=> h = 70/20

=> h = 3.5

Hence :

The height of the room = 3.5 meters.

Answer 2

Answer:

Step-by-step explanation:

SOLUTION

Let the breadth of the room be x meters.

As the length of the room is 50% more than its breadth,

$(1 + \frac{50}{100} )of \: x \: meter \: = \frac{3x}{2} {m}^{2} \\ the \: area \: of \: the \:floor \: of \: the \: room = ( \frac{3x}{2} \times x) {m}^{2} \\ acoording \: to \: the \: question \: \\ = \frac{3 {x}^{2} }{2} \times 38.5 = 924 \\ {x}^{2} = \frac{924 \times 2}{3 \times 38.5} {x}^{2} = 16 \\ x = \sqrt{16} \\ x = 4$

Cost of carpeting the room at the rate of ₹38.50 m² = ₹(3x²/2 × 38.5).

Hence :

the length of the room = 3/2x meters = 3/2 × 4 meters = 6 meters

and its breadth = x meters = 4 meters.

Let the height of the room be h meters.

Then :

The surface area of all the walls of the room

=> 2(length + breadth) × height

=> 2(6 + 4) × h m²

=> 20h m².

Area of one door = 1 × 2 m² = 2m²,

area of two windows = 2(1 × 1.5) m² = 3m².

Surface area to be prepared

=> surface area of walls - surface area of door and windows

=> (20h - 2 - 3) m²

=> (20h - 5) m².

Cost of preparing the walls of the rate of ₹3.30 m² = ₹(20h - 5) × 3.30.

According to given :

(20h - 5) × 3.30 = 214.50

=> 20h - 5 = 214.50/3.30 = 65

=> 20h = 65 + 5

=> 20h = 70

=> h = 70/20

=> h = 3.5

Hence :

The height of the room = 3.5 meters.