Question

How many silver coins of diameter \[5cm\] and thickness \[4mm\] have to be melted to prepare a cuboid of \[12cm \times 11cm \times 5cm\] dimension?

Answer 1

Hint: Assume n numbers of the silver coins to be needed, now since they are melted and reformed into cuboid, total volume of the silver coins will be equal to the volume of the cuboid.

Complete step-by-step answer:

Given diameter of the silver coin \[d = 5cm\] and thickness \[h = 4mm = 0.4cm\]
Since, \[r = \dfrac{d}{2}\]
\[
  r = \dfrac{5}{2} \\
  \therefore r = 2.5cm \\
\]
Consider the volume of the one silver coin \[{V_S}\]
i.e. \[{V_S} = \pi {r^2}h\]
\[
  {V_S} = \pi {\left( {2.5} \right)^2}0.4 \\
  {V_S} = \dfrac{{22}}{7}\left( {2.5 \times 2.5} \right)0.4 \\
  {V_S} = \dfrac{{22}}{7}\left( {6.25} \right)0.4 \\
  {V_S} = \dfrac{{22}}{7} \times 2.5 \\
  \therefore {V_S} = \dfrac{{22}}{7} \times 2.5{\text{ }}c{m^2} \\
\]
Given, length of cuboid \[l = 12cm\]
           breadth of cuboid \[b = 11cm\]
           height of cuboid \[h = 5cm\]
Volume of the cuboid \[{V_C} = lbh\]
\[
  {V_C} = 12cm \times 11cm \times 5cm \\
  {V_C} = 12 \times 11 \times 5{\text{ }}c{m^3} \\
  {V_C} = 660{\text{ }}c{m^3} \\
  \therefore {V_C} = 660{\text{ }}c{m^3} \\
\]
Now, since volume of cuboid is equal to volume of one silver coin
\[
  {\text{Number of coins = }}\dfrac{{{\text{Volume of cuboid}}}}{{{\text{Volume of 1 silver coin}}}} \\
  {\text{ = }}\dfrac{{{V_C}}}{{{V_S}}} \\
  {\text{ = }}\dfrac{{660}}{{\dfrac{{22}}{7} \times 2.5}} \\
  {\text{ = }}\dfrac{{660 \times 7}}{{22 \times 2.5}} \\
  {\text{ = }}\dfrac{{4620}}{{55}} \\
  {\text{ = 84}} \\
\]

 Thus, \[84\] silver coins can be made by melting cuboids of given dimensions.

Note: In these types of problems leave the common terms undisturbed in the values so that simplifications can be done easier. Also note that thickness of the silver coin is equal to the height of the silver coin.