The Gold Coin

Geometry Level 2

The radius of a cylindrical gold coin is 0.7 0.7 c m cm and thickness is 2 2 m m mm . How much gold will be required to prepare 10 10 such coins?

308 308 c u . m m cu.mm 3080 3080 c u . m m cu.mm 30800 30800 c u . m m cu.mm

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1 solution

Ram Gautam
Mar 8, 2015

A coin is cylindrical

Hence, V o l u m e O f A C y l i n d e r = π × R a d i u s 2 × H e i g h t Volume\quad Of\quad A\quad Cylinder=\pi { \times Radius }^{ 2 }\times Height

Radius = 0.7cm i.e. 7 mm Height = 2 mm

V o l u m e O f C o i n = 22 7 × 7 × 7 × 2 = 308 m m 3 Volume\quad Of\quad Coin=\frac { 22 }{ 7 } \times 7\times 7\times 2=308{ mm }^{ 3 }

Hence Volume Of 10 Coins = 308 × 10 = 3080 m m 3 308\times 10=3080{ mm }^{ 3 }

1080 or 3080? @Ram Gautam

Anuj Shikarkhane - 6 years, 3 months ago

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Yeah! 3080! :P

Ram Gautam - 6 years, 3 months ago

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@Ram Gautam 108X10=3080?

Mehul Arora - 6 years, 3 months ago

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@Mehul Arora Typo Bro! :P

Ram Gautam - 6 years ago

You should try these questions - QuEsTiOnS

Sakanksha Deo - 6 years, 3 months ago

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