Bricks

Openings in perforated bricks occupy 10%, and brick has dimensions of 30 cm, 15 cm, and 7.5 cm. Calculate
a) the weight of a perforated brick if you know that the density of the full brick material is p = 1800 kg/m³ (1.8 kg/dm³)
b) the number of perforated bricks allowed the driver to load on the truck with a load capacity of 5 tons and avoid car overload.

Correct answer:

m =  5.4675 kg
n =  914

Step-by-step explanation:

$$\begin{gathered} a=30\text{cm}\to \text{dm}=30:10\text{dm}=3\text{dm} \\ b=15\text{cm}\to \text{dm}=15:10\text{dm}=1.5\text{dm} \\ c=7.5\text{cm}\to \text{dm}=7.5:10\text{dm}=0.75\text{dm} \\ V=a\cdot b\cdot c=3\cdot 1.5\cdot 0.75=\frac{27}{8}=3.375{\text{dm}}^3 \\ \rho =1.8{\text{kg/dm}}^3 \\ {m}_1=\rho \cdot V=1.8\cdot 3.375=\frac{243}{40}=6.075\text{kg} \\ q=100\%-10\%=1-\frac{10}{100}=0.9 \\ m=q\cdot m_1=0.9\cdot 6.075=5.4675\text{kg} \end{gathered}$$

$$\begin{gathered} M=5\text{t}\to \text{kg}=5\cdot 1000\text{kg}=5000\text{kg} \\ {n}_1=M/m=5000/5.4675\doteq 914.4947 \\ n=\lfloor n_1\rfloor =\lfloor 914.4947\rfloor =914 \end{gathered}$$

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