A barrel of a fountain pen, …
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Sia ? 6 years, 6 months ago
We have,
Volume of a barrel {tex}= \left( \frac { 22 } { 7 } \times 0.25 \times 0.25 \times 7 \right) \mathrm { cm } ^ { 3 }{/tex}= = 1.375 cm3
Volume of ink in the bottle = {tex}\frac { 1 } { 5 } \text { litre } = \frac { 1000 } { 5 } \mathrm { cm } ^ { 3 } = 200 \mathrm { cm } ^ { 3 }{/tex}
{tex}\therefore{/tex} Total number of barrels that can be filled from the given volume of ink = {tex}\frac { 200 } { 1.375 }{/tex}
So, required number of words ={tex}\frac { 200 } { 1.375 } \times 330 = 48000{/tex}
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