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Mar 18, 2018
 #1
avatar+9468 
+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.

Mar 18, 2018

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