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H Harsh Kankaria
Given, Dimensions of the bigger box = , Dimensions of smaller box =   We know, Total surface area of a cuboid =   Total surface area of the bigger box =   Area of the overlap for the bigger box =  Similarly,   Total surface area of the smaller box =   Area of the overlap for the smaller box =  Since, 250 of each box is required,  Total area of carboard required =  Cost of  of the cardboard...

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H Harsh Kankaria
Given, dimensions of the greenhouse =  (ii) Tape needed for all the 12 edges = Perimeter = Therefore,  of tape is needed for the edges.
Given, Edge of the cubical box =  Dimensions of the cuboid =  (ii) The total surface area of the cubical box =  The total surface area of the cuboidal box =  Clearly, the total surface area of a cuboidal box is greater than the cubical box. Difference between them = 
Given, Edge of the cubical box =  Dimensions of the cuboid =  The lateral surface area of the cubical box =  The lateral surface area of the cuboidal box =  Clearly, Lateral surface area of the cubical box is greater than the cuboidal box. Difference between them = 

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H Harsh Kankaria
Given, dimensions of the brick =  We know, Surface area of a cuboid =  The surface area of a single brick =   Number of bricks that can be painted =  Therefore, the required number of bricks that can be painted = 100
Given, Perimeter of rectangular hall =  Cost of painting the four walls at the rate of Rs 10 per  = Rs 15000 Let the height of the wall be   Area to be painted =  Required cost =  Therefore, the height of the hall is 
Given, Dimensions of the room =  Required area to be whitewashed = Area of the walls + Area of the ceiling =  Cost of white-washing per  area = Cost of white-washing area = Therefore, the required cost of whitewashing the walls of the room and the ceiling is 

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H Harsh Kankaria
Given, dimensions of the plastic box Length,  Width,  Depth,  We know, area of the sheet required for making the box is  (ii) Cost for   of sheet = Rs 20  Cost for  of sheet =  Required cost of the sheet is 

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H Harsh Kankaria
Given, dimensions of the plastic box Length,  Width,  Depth,  (i) The area of the sheet required for making the box (open at the top)= Lateral surface area of the box. + Area of the base.   =  The required area of the sheet required for making the box is   
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