Question.
A godown measures $40 \mathrm{~m} \times 25 \mathrm{~m} \times 15 \mathrm{~m}$. Find the maximum number of wooden crates each measuring $1.5 \mathrm{~m} \times 1.25 \mathrm{~m} \times 0.5 \mathrm{~m}$ that can be stored in the godown.
Solution:
The godown has its length $\left(l_{1}\right)$ as $40 \mathrm{~m}$, breadth $\left(b_{1}\right)$ as $25 \mathrm{~m}$, height $\left(h_{1}\right)$ as $15 \mathrm{~m}$, while the wooden crate has its length $\left(t_{2}\right)$ as $1.5 \mathrm{~m}$, breadth $\left(b_{2}\right)$ as $1.25 \mathrm{~m}$, and height $\left(h_{2}\right)$ as $0.5$ $\mathrm{m}$.
Therefore, volume of godown $=l_{1} \times b_{1} \times h_{1}$
$=(40 \times 25 \times 15) \mathrm{m}^{3}$
$=15000 \mathrm{~m}^{3}$
Volume of 1 wooden crate $=I_{2} \times b_{2} \times h_{2}$
$=(1.5 \times 1.25 \times 0.5) \mathrm{m}^{3}$
$=0.9375 \mathrm{~m}^{3}$
Let n wooden crates can be stored in the godown.
Therefore, volume of n wooden crates = Volume of godown
$0.9375 \times n=15000$
$\mathrm{n}=\frac{15000}{0.9375}=16000$
Therefore, 16,000 wooden crates can be stored in the godown.
The godown has its length $\left(l_{1}\right)$ as $40 \mathrm{~m}$, breadth $\left(b_{1}\right)$ as $25 \mathrm{~m}$, height $\left(h_{1}\right)$ as $15 \mathrm{~m}$, while the wooden crate has its length $\left(t_{2}\right)$ as $1.5 \mathrm{~m}$, breadth $\left(b_{2}\right)$ as $1.25 \mathrm{~m}$, and height $\left(h_{2}\right)$ as $0.5$ $\mathrm{m}$.
Therefore, volume of godown $=l_{1} \times b_{1} \times h_{1}$
$=(40 \times 25 \times 15) \mathrm{m}^{3}$
$=15000 \mathrm{~m}^{3}$
Volume of 1 wooden crate $=I_{2} \times b_{2} \times h_{2}$
$=(1.5 \times 1.25 \times 0.5) \mathrm{m}^{3}$
$=0.9375 \mathrm{~m}^{3}$
Let n wooden crates can be stored in the godown.
Therefore, volume of n wooden crates = Volume of godown
$0.9375 \times n=15000$
$\mathrm{n}=\frac{15000}{0.9375}=16000$
Therefore, 16,000 wooden crates can be stored in the godown.