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The paint in a certain container is sufficient to paint an area equal to 9.375 ${m^2}$. How many bricks of dimensions 22.5 cm x10 cm x7.5 cm can be painted out of this container.

Answer
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Hint : We need to find the area that can be painted by the container by using total surface area and then the number of bricks can be painted out of the container. Container is in the form of a cuboid. So the total surface area of the container will be the same as the cuboid.

Given,
Dimensions of the brick 22.5 cm x10 cm x7.5 cm i.e.., l=22.5 cm, b=10cm, h=7.5 cm. Total surface area of the container is given as 9.375 ${m^2}$.As we know that the total surface area formula i.e.., 2(lb+bh+hl). Now let us find the total surface area of the brick by substituting the values of l, b, and h, we get
$
   \Rightarrow T.S.A = 2(lb + bh + hl) \\
   \Rightarrow T.S.A = 2((22.5*10) + (10*7.5) + (7.5*22.5)) \\
   \Rightarrow T.S.A = 2(225 + 75 + 168.75) \\
   \Rightarrow T.S.A = 2(468.75) \\
   \Rightarrow T.S.A = 937.5{\text{c}}{{\text{m}}^{\text{2}}} \\
 $
Let us convert the total surface area in the terms of ${m^2}$ as the area total area painted by the container is given in ${m^2}$i.e..,
\[ \Rightarrow T.S.A = 937.5{\text{c}}{{\text{m}}^{\text{2}}} = \dfrac{{937.5}}{{10000}}{\text{ }}{{\text{m}}^2} = 0.09375{\text{ }}{{\text{m}}^2}\]
Now, let us consider ‘x’ as the number of bricks. Therefore, by dividing ‘total area painted by the container’ with ‘total surface area of brick’ the number of bricks (x) can be computed i.e..,
$
   \Rightarrow {\text{x = }}\dfrac{{{\text{total area painted by the container}}}}{{{\text{total surface area of the bricks}}}} \\
   \Rightarrow {\text{x}} = \dfrac{{9.375}}{{0.09375}}{\text{ }}{{\text{m}}^2} \\
   \Rightarrow {\text{x}} = 100{\text{ }}{{\text{m}}^2} \\
 $
Therefore, 100 bricks can be painted out with the given amount of paint.

Note: Make a note that since the brick is considered as the cuboid the total surface area of the brick is calculated using the formulae of the total surface area of the cuboid.