English

The Floor of a Room is of Size 6 M X 5 M. Find the Cost of Covering the Floor of the Room with 50 Cm Wide Carpet at the Rate of Rs.24.50 per Metre. Also, Find the Cost of Carpeting the Same Hall

Advertisements
Advertisements

Question

The floor of a room is of size 6 m x 5 m. Find the cost of covering the floor of the room with 50 cm wide carpet at the rate of Rs.24.50 per metre. Also, find the cost of carpeting the same hall if the carpet, 60 cm, wide, is at the rate of Rs.26 per metre.

Sum
Advertisements

Solution

Area of floor of a room
= 6m x 5m
= 30m2 
Let the length of the carpet that is 50cm wide be l m.
∴ Area of carpet
= length x breadth
= l m x 50cm
= l x 0.5m2
Since area of carpet
= Area of floor of a room
⇒ l x 0.5 = 30

⇒ l  `(30)/(0.5)`
= 60m
∴ Cost of carpet at Rs.24.50per mate
= Rs.60 x 24.50
= Rs.1470
Let the length of the carpet that is 60cm wide be L m.
∴ Area of carpet
= length x breadth
= l m x 60cm
= l x 0.6m2
Since area of carpet
= Area of floor of a room
⇒ l x 0.6 = 30

⇒ l  `(30)/(0.6)`
= 50m
∴ Cost of carpet at Rs.26per mate
= Rs.50 x 26
= Rs.1300.

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Perimeter and Area - Exercise 24.2

APPEARS IN

Frank Mathematics Part 1 [English] Class 9 ICSE
Chapter 19 Perimeter and Area
Exercise 24.2 | Q 35

RELATED QUESTIONS

The area of a rectangular is 640 m2. Taking its length as x cm; find in terms of x, the width of the rectangle. If the perimeter of the rectangle is 104 m; find its dimensions.


 Trapezium given below; find its area.


A wire when bent in the form of a square encloses an area of 484 m2. Find the largest area enclosed by the same wire when bent to from:

  1. An equilateral triangle.
  2. A rectangle of length 16 m.

How many tiles, each of area 400 cm2, will be needed to pave a footpath which is 2 m wide and surrounds a grass plot 25 m long and 13 m wide? 


The figure given below shows the cross-section of a concrete structure. Calculate the area of cross-section if AB = 1.8 cm, CD = 0.6 m, DE = 0.8 m, EF = 0.3 m and AF = 1.2 m.


The length of a rectangular verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter.

(i) Taking x as the breadth of the verandah, write an equation in x that represents the above statement

(ii) Solve the equation obtained in (i) above and hence find the dimensions of the verandah. 


Using the information in the following figure, find its area.


Vertices of given triangles are taken in order and their areas are provided aside. Find the value of ‘p’.

Vertices Area (sq.units)
(0, 0), (p, 8), (6, 2) 20

Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC. Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.


When proving that a quadrilateral is a trapezium, it is necessary to show


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×