Question

Daniel is painting the wall and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10m and 7 respectively. From each can of paint, 100m² of area is painted. How many cans of paint will he need to paint the entire room?​

Answer 1

5

Explanation:

Given,

Length = 15m

Breadth = 10m

Height = 7m

Area painted by 1 can = 100m²

To find,

Number of cans of paint will be needed to paint the entire room.

Procedure,

$no. \: of \: cans = \frac{area \: of \: the \: hall}{area \: painted \: by \: 1 \: can}$

Since, only wall and ceiling is painted, therefore we need to subtract the bottom area of the hall from the total surface area of the hall.

$total \: surface \: area \: of \: cuboid = 2(lb + bh + hl)$

$= 2(15 \times 10 + 10 \times 7 + 7\times 15)$

$= 2(150 + 70 + 105)$

$= 2(325)$

$= 650 {m}^{2}$

Therefore, total surface area of the cuboid is 650m².

$bottom \: area = l \times b$

$= 15m \times 10m$

$= 150 {m}^{2}$

Therefore, bottom area of the cuboid is 150m²

$area \: to \: be \: painted = 650m {}^{2} - 150 {m}^{2} \\ = 500m {}^{2}$

Now,

$no. \: of \: cans = \frac{500}{100} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 5$

Therefore, 5 cans are required to paint the entire room.

Answer 2

Answer:

5 cans

Step-by-step explanation:

Hint: We need to determine area painted by Daniel of hall which is given by the sum of area of four walls and the area of ceiling which is equal to where l is the length of the hall breadth be b and height be h of the cuboidal hall.

Complete step-by-step solution -

Given, Daniel is painting the walls and ceiling of a cuboidal hall, the length of cuboidal hall is 15m, breadth of cuboidal hall is 10m and height of cuboidal hall is 7m.Question

Let the length be l, breadth be b and height be h of the cuboidal hall then l=15m, b=10m, h=7m

Since the hall is cuboidal in shape so, the area painted by Daniel of hall= Area of the cuboid, which is equal to the area of four walls + area of ceiling substituting the values of the length l breadth b and height h of the cuboidal hall in the above formula we get,

area painted by Daniel of hall . Now we are given that from each can of paint, 100m of area is painted implies 100m square of area is painted by 1 can 1m square of area is painted by can

Therefore, 500 m square of area is painted by Daniel in can.

Hence, we got that Daniel will need 5 cans to paint the room.

Note: The possibility of the error in the question is that you can ignore adding the area of the ceiling as well, which is wrong because Daniel needed to paint both the walls and the ceiling. So you need to add the area of ceiling as well which is nothing but length multiplied by breadth of the hall.