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How many coins of diameter 1.75 cm and thickness 2 cm can be made by melting down a cuboid of sides 5.5 cm, 10 cm and 3.5 cm?​

Answer»

CORRECTION in the Question:

How MANY coins of DIAMETER 1.75 cm and thickness 2 mm can be made by melting down a cuboid of sides 5.5 cm, 10 cm and 3.5 cm?​

Solution:

Dimensions of the cylindrical coins to be made:

  • Diameter (d) = 1.75 cm.
  • Radius (r) = (1.75)/2 cm.
  • Height (H) = 2 mm → 2/10cm.

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Dimensions of the cuboid:

  • Assuming the sides are a, b & c;
  • a = 5.5 cm.
  • b = 10 cm.
  • c = 3.5 cm.

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Let "n" be the number of coins that have to be made.

Therefore, VOLUME of one cylindrical COIN multiplied by "n" is equal to the volume of the cuboid.

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\longmapsto \sf Volume \ of \ a \ Cylinder = \pi r^2h\\ \\ \\\longmapsto \sf Volume \ of \ a \ Cuboid = l \times b \times h\\

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Therefore,

\Longrightarrow \sf l \times b \times h = n \times \pi r^2 H\\ \\ \\\Longrightarrow \sf 5.5 \times 10 \times 3.5 = n \times \dfrac{22}{7} \times \Bigg( \dfrac{1.75}{2} \Bigg)^2 \times \dfrac{2}{10} \\ \\ \\ \\\Longrightarrow \sf 192.5 = n \times \dfrac{22}{7} \times \dfrac{1.75}{2} \times \dfrac{1.75}{2} \times \dfrac{2}{10} \\ \\ \\ \\\Longrightarrow \sf 192.5 = n \times \dfrac{11}{70} \times 1.75 \times 1.75 \\ \\ \\ \\

\Longrightarrow \sf \dfrac{192.5 \times 70}{11 \times 1.75 \times 1.75} = n \\ \\ \\ \\\Longrightarrow \sf \dfrac{17.5 \times 70}{1.75 \times 1.75} = n \\ \\ \\ \\\Longrightarrow \sf \dfrac{1225}{1.75 \times 1.75} = n \\ \\ \\ \\\Longrightarrow \sf \dfrac{700}{1.75} = n \\ \\ \\ \\\Longrightarrow \sf \dfrac{700}{1.75} \times \dfrac{100}{100} = n \\ \\ \\ \\\Longrightarrow \sf \dfrac{70000}{175} = n \\ \\ \\ \\\Longrightarrow \sf n = 400 \\ \\ \\

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Therefore, the number of coins that can be made is 400.


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