Question

A gold biscuit of dimensions $5.9cm \times 3.2cm \times 3mm$ was used to form a hollow box of thickness $2mm$. If the external length and width of the box are $2cm$ and $1.5cm$, find the height of the box ?

Answer 1

Hint: Here, in this question, we are given a gold biscuit which is used to make a hollow box. Since, all the material of the gold biscuit goes into building the box, so the volume of the gold biscuit should be equal to the volume of the material used in making up the hollow box. The given question involves the concepts of mensuration.

Complete step by step answer:
In the given question, we are given the dimensions of the golden biscuit.
The length of the gold biscuit is $5.9cm = 59mm$
The breadth of the gold biscuit is $3.2cm = 32mm$
The height of the gold biscuit is $3mm$.
The volume of the cuboidal gold biscuit can be calculated using the formula $ = lbh$, where l, b and h represents the length, breadth and height of the biscuit.
So, the volume of the gold biscuit$ = \left( {59mm} \right)\left( {32mm} \right)\left( {3mm} \right) = 5664m{m^3}$
Now, this volume of gold is used to make the hollow box.
The external length of hollow box is $2cm = 20mm$.
The external breadth of hollow box is $1.5cm = 15mm$.
Let the height of the box be h.
Then, the volume of the outer box$ = \left( {20mm} \right)\left( {15mm} \right)\left( h \right) = 300h\,m{m^3}$
Also, we are given that the thickness of the hollow box is $2mm$.
Now, we know that inner dimensions and outer dimensions of a hollow box differs by $2$ times the thickness of the box.
So, the inner length of the hollow box is $20mm - 2\left( {2mm} \right) = 16mm$
Similarly, the inner breadth of the hollow box is $15mm - 2\left( {2mm} \right) = 11mm$
Similarly, the inner height will be $h - 2\left( 2 \right)\,mm = \left( {h - 4} \right)mm$
Then, the volume of the inner box$ = \left( {16mm} \right)\left( {11mm} \right)\left( {h - 4} \right) = 176\left( {h - 4} \right)\,m{m^3}$
So, the volume of hollow box can be calculated by subtracting the volume of inner box from the volume of outer box.
Hence, the volume of hollow box is $300h - 176\left( {h - 4} \right)\,m{m^3}$.
\[ \Rightarrow \left( {300h - 176h + 704} \right)\,m{m^3}\]
\[ \Rightarrow \left( {124h + 704} \right)\,m{m^3}\]
Now, equating the volume of material used in building the hollow box with the volume of the gold biscuit, we get,
\[\left( {124h + 704} \right)\,m{m^3} = 5664\]
Shifting all the constants to the right side of the equation, we get,
\[ \Rightarrow 124h = 5664 - 704\]
Simplifying the equation, we get,
\[ \Rightarrow 124h = 4960\]
Dividing both the sides of the equation by $124$, we get,
\[ \Rightarrow h = \dfrac{{4960}}{{124}} = 40mm\]
So, the height of the hollow box is $40mm$.

Note: One must know the formulae of volume of shapes in order to solve the given question. Inner dimensions and outer dimensions of a hollow box differs by $2$ times the thickness of the box. Care must be taken while doing the calculations in order to be sure of the answer.