Alberto has decided to paint his dining room. Paint comes in 1 litre cans. The paint in one can will cover an area of approximately 24 square metres. The dining room is 4 m x 6 m x 3.5 m high. There is just one window which is in one of the long walls and is 1.5 m x 2 m.

All of the walls, door and ceiling are to be painted with the same type of paint.

Approximately 20% of the wall area to be painted is wood which will need a second coat of paint.

What is the minimum number of cans of paint that Alberto should buy to have sufficient to complete the room?

Determine what they are trying to ask, and what you will need to solve it

Eliminate any non-essential information

Draw a picture, graph, or equation

In moments of high stress like exam taking, always work with the paper they give you to avoid careless mistakes.

Solve.

First we need to distinguish between what needs to be painted and what will not. Notice that we need to paint ‘walls, door and ceiling are to be painted…” and it does not include the floor. Also note that we need to subtract the area of one window. So we can start by finding the area of the 4 walls and the ceiling:

Ceiling = 4m x 6m = 24 m^2

Short walls = (2)[(4m)(3.5m)] = 28 m^2

Long walls = (2)[(6m)(3.5m)] = 42 m^2

Total area wall and ceiling area = 24m + 28m + 42m = 94 m^2

Area of the window:

(1.5m )(2m) = 3 m^2

Now we need to find the area that will have the second coat.

Total wall area = short wall surface area + long wall surface area

Total wall area = 42 m^2 + 28 m^2

Total wall area = 70 m^2

20% Total wall area = (20%)(70 m^2)

20% Total wall area = 14 m^2

Surface Area needed to painted = $94 m^2 - 3 m^2 + 14 m^2

Surface Area needed to be painted = 105 m^2

Now we can figure out how many cans of paint we need:

1 can = 24 m^2

We have 105 m^2 that needs to be painted, and so this will require 5 cans.

4 cans is not enough because (4 cans)(24m^2) = 96m^2

(5 cans)(24m^2) = 120m^2

\fcolorbox{red}{grey!30}{Therefore the answer is C and we need $5$ cans of paint.}