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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 8 mm, and Akshaj's donut holes have radius 10 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?

mathtoo  Mar 17, 2018

Best Answer 

 #1
avatar+6945 
+2

These problems easily confuse me but here's my attempt....

 

surface area of N's donut holes   =   4 pi * (6 mm)2   =   144pi   mm2

surface area of T's donut holes   =   4 pi * (8 mm)2   =   256pi   mm2

surface area of A's donut holes   =   4 pi * (10 mm)2   =   400pi   mm2

 

When N finishes  1  donut, T will have finished  \(\frac{144}{256}\)  donuts and A will have finished  \(\frac{144}{400}\)  donuts.

When N finishes  1  donut, T will have finished  \(\frac{9}{16}\)  donuts and A will have finished  \(\frac{9}{25}\)  donuts.

 

When N finishes  2  donuts, T will have finished  2*\(\frac{9}{16}\)  donuts and A will have finished  2*\(\frac{9}{25}\)  donuts.

When N finishes  2  donuts, T will have finished  \(\frac98\)  donuts and A will have finished  \(\frac{18}{25}\)  donuts.

 

When N finishes  x  donuts, T will have finished  \(\frac{9x}{16}\)  donuts and A will have finished  \(\frac{9x}{25}\)  donuts.

 

We need the smallest integer  x  such that  \(\frac{9x}{16}\)  is an integer and  \(\frac{9x}{25}\)  is an integer.

 

For a fraction to be an integer, all the factors in the denominator must be in the numerator.

For  9x/16  to be an integer,  2*2*2*2  needs to be factors of  x  .

For  9x/25  to be an integer,  5*5  needs to be factors of  x .

 

So  x  must be  2*2*2*2*5*5   =   400

 

When Niraek finishes 400 donut holes, Theo will have finished  400*9/16 = 225  donut holes, and Akshaj will have finished  400*9/25 = 144  donut holes.

hectictar  Mar 18, 2018
Sort: 

2+0 Answers

 #1
avatar+6945 
+2
Best Answer

These problems easily confuse me but here's my attempt....

 

surface area of N's donut holes   =   4 pi * (6 mm)2   =   144pi   mm2

surface area of T's donut holes   =   4 pi * (8 mm)2   =   256pi   mm2

surface area of A's donut holes   =   4 pi * (10 mm)2   =   400pi   mm2

 

When N finishes  1  donut, T will have finished  \(\frac{144}{256}\)  donuts and A will have finished  \(\frac{144}{400}\)  donuts.

When N finishes  1  donut, T will have finished  \(\frac{9}{16}\)  donuts and A will have finished  \(\frac{9}{25}\)  donuts.

 

When N finishes  2  donuts, T will have finished  2*\(\frac{9}{16}\)  donuts and A will have finished  2*\(\frac{9}{25}\)  donuts.

When N finishes  2  donuts, T will have finished  \(\frac98\)  donuts and A will have finished  \(\frac{18}{25}\)  donuts.

 

When N finishes  x  donuts, T will have finished  \(\frac{9x}{16}\)  donuts and A will have finished  \(\frac{9x}{25}\)  donuts.

 

We need the smallest integer  x  such that  \(\frac{9x}{16}\)  is an integer and  \(\frac{9x}{25}\)  is an integer.

 

For a fraction to be an integer, all the factors in the denominator must be in the numerator.

For  9x/16  to be an integer,  2*2*2*2  needs to be factors of  x  .

For  9x/25  to be an integer,  5*5  needs to be factors of  x .

 

So  x  must be  2*2*2*2*5*5   =   400

 

When Niraek finishes 400 donut holes, Theo will have finished  400*9/16 = 225  donut holes, and Akshaj will have finished  400*9/25 = 144  donut holes.

hectictar  Mar 18, 2018
 #2
avatar+186 
+1

Very, very nice, detailed solution, hectictar! Bravo!smiley

mathtoo  Mar 18, 2018

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