Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of $15 m$ , $10 m$ and $7 m$ respectively. From each Can of a paint $100 m^{2}$ of area is painted. How many Cans of paint will she need to paint the entire room?

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Solution

Given that,
Length of wall, $l$ $= 15 m$ ,
Breadth of wall, $b$ $= 10 m$
Height of wall, $h$ $= 7 m$
Area painted by one Can of paint $=$ $100 m^{2}$
To calculate:
Total number of cans required to paint entire area of the cuboidal hall.
We know that,
Total surface area of cuboid $= 2 (l b + b h + l h)$
Required area of cuboidal hall $=$ Total surface area of the room $-$ Base area of the room
$=$ $2 (l b + b h + l h)$ $- l b$
$= 2 (15 \times 10 + 10 \times 7 + 7 \times 15) - (15 \times 10) = 2 (150 + 70 + 105) - 150 = 650 - 150 = 500 m^{2}$
Now, Required number of cans $=$ $\frac{R e q u i r e d a r e a o f c u b o i d a l h a l l}{A r e a c o v e r e d b y o n e C a n} = \frac{500 m^{2}}{100 m^{2}} = 5$
Hence, $5$ Cans of paint will she need to paint the entire room.

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