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At the MP Donut Hole Factory, Niraek, Theo, and Akshaj are coating spherical donut holes in powdered sugar. Niraek's donut holes have radius 6 mm, Theo's donut holes have radius 10 mm, and Akshaj's donut holes have radius 18 mm. All three workers coat the surface of the donut holes at the same rate and start at the same time. Assuming that the powdered sugar coating has negligible thickness and is distributed equally on all donut holes, how many donut holes will Niraek have covered by the first time all three workers finish their current donut hole at the same time?

 Nov 24, 2020

Best Answer 

 #1
avatar+11086 
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At the MP Donut Hole Factory, Niraek, Theo, and Akshaj are coating spherical donut holes in powdered sugar. Niraek's donut holes have radius 6 mm, Theo's donut holes have radius 10 mm, and Akshaj's donut holes have radius 18 mm. All three workers coat the surface of the donut holes at the same rate and start at the same time. Assuming that the powdered sugar coating has negligible thickness and is distributed equally on all donut holes, how many donut holes will Niraek have covered by the first time all three workers finish their current donut hole at the same time?

 

Hello Guest!

 

\(\frac{N\cdot Parts}{6^2mm^2}+\frac{T\cdot Parts}{10^2mm^2}+\frac{A\cdot Parts}{18^2mm^2}\\ =\frac{N}{36}+\frac{T}{100}+\frac{A}{324}\\ =\frac{N}{9}+\frac{T}{25}+\frac{A}{81}\\ = \frac{2025N+729T+225A}{18225}\cdot \frac{Parts}{mm^2}\)

 

In the period in which Akshaj 225 and Theo 729 coated donut holes,

Niraek coated 2025 donut holes.

laugh  !

 Nov 25, 2020
edited by asinus  Nov 25, 2020
 #1
avatar+11086 
+3
Best Answer

At the MP Donut Hole Factory, Niraek, Theo, and Akshaj are coating spherical donut holes in powdered sugar. Niraek's donut holes have radius 6 mm, Theo's donut holes have radius 10 mm, and Akshaj's donut holes have radius 18 mm. All three workers coat the surface of the donut holes at the same rate and start at the same time. Assuming that the powdered sugar coating has negligible thickness and is distributed equally on all donut holes, how many donut holes will Niraek have covered by the first time all three workers finish their current donut hole at the same time?

 

Hello Guest!

 

\(\frac{N\cdot Parts}{6^2mm^2}+\frac{T\cdot Parts}{10^2mm^2}+\frac{A\cdot Parts}{18^2mm^2}\\ =\frac{N}{36}+\frac{T}{100}+\frac{A}{324}\\ =\frac{N}{9}+\frac{T}{25}+\frac{A}{81}\\ = \frac{2025N+729T+225A}{18225}\cdot \frac{Parts}{mm^2}\)

 

In the period in which Akshaj 225 and Theo 729 coated donut holes,

Niraek coated 2025 donut holes.

laugh  !

asinus Nov 25, 2020
edited by asinus  Nov 25, 2020

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