A cuboidal block of solid iron has dimensions 50 cm, 45 cm and 34 cm.
Question:

A cuboidal block of solid iron has dimensions 50 cm, 45 cm and 34 cm. How many cuboids of size 5 cm by 3 cm by 2 cm can be obtained from this block? Assume cutting causes no wastage.

Solution:

Dimension of the cuboidal iron block $=50 \mathrm{~cm} \times 45 \mathrm{~cm} \times 34 \mathrm{~cm}$

Volume of the iron block $=$ length $\times$ breadth $\times$ height $=(50 \times 45 \times 34) \mathrm{cm}^{3}=76500 \mathrm{~cm}^{3}$

It is given that the dimension of one small cuboids is $5 \mathrm{~cm} \times 3 \mathrm{~cm} \times 2 \mathrm{~cm}$.

Volume of one small cuboid $=$ length $\times$ breadth $\times$ height $=(5 \times 3 \times 2) \mathrm{cm}^{3}=30 \mathrm{~cm}^{3}$

$\therefore$ The required number of small cuboids that can be obtained from the iron block $=\frac{\text { volume of the iron block }}{\text { volume of one small cuboid }}=\frac{76500 \mathrm{~cm}^{3}}{30 \mathrm{~cm}^{3}}=2550$

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