Question

Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15m, 10m and 7m respectively. From each can of paint, 100m of area is painted and how many cans of paint will she need to paint the room.

Answer 1

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 \[=\text{ }2\left( l+b \right)h\text{ }+\text{ }lb\], 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.
The given details can be illustrated in a figure as,

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
Then the area painted by Daniel of hall \[=\text{ }2\left( l+b \right)h\text{ }+\text{ }lb\]
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 \[=\text{ }2\left( 15\text{ }+\text{ }10 \right)7\text{ }+\text{ (}15)(10)\]

\[\Rightarrow \]area painted by Daniel of hall \[=350\text{ }+\text{ }150\]

\[\Rightarrow \]area painted by Daniel of hall \[=500{{m}^{2}}\]

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
\[\Rightarrow \]1 m square of area is painted by \[\dfrac{1}{100}\]can
Therefore, 500 m square of area is painted by Daniel in \[\dfrac{500}{100}=5\]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.