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?

tertre Mar 22, 2018

#3**0 **

Then, why did the person who posted it the first time "mathtoo" praised the "solution" as very thorough and detailed answer? Do you mean "hectictar" didn't get right? I think, as CPhill said, " it is as good as it gets"!!

Guest Mar 22, 2018

#6**0 **

I think you should find the LCM of [6, 8, 10] =120, so you have:

120 / 6 =20 holes for N

120 / 8 =15 holes for T

120 / 10 =12 holes for A

Guest Mar 22, 2018

edited by
Guest
Mar 23, 2018

edited by Guest Mar 23, 2018

edited by Guest Mar 23, 2018

edited by Guest Mar 23, 2018

edited by Guest Mar 23, 2018

#7**+2 **

They're coating spherical donut holes, so...

SA of Niraek's donut hole = 144pi square millimeters

So to coat one donut hole, Niraek has to cover 144pi square millimeters with powdered sugar.

In the time Niraek covers 144pi square mm, Theo will have covered 144pi sq mm because they are coating at the same rate. But 144pi sq mm for Theo is not an entire donut hole. To coat one entire donut hole, Theo has to cover 256pi sq mm.

hectictar
Mar 22, 2018

#10**+1 **

**Mr. BB you are an enigma of chaotic stupidity!**

First, you post this dumb blarney:

*hectictar: You calculated the SA of a sphere. Wouldn't it be more accurate if you calculated it as SA of a circle, since it says "Assuming that the powdered sugar coating has negligible thickness and is distributed equally on all donut holes"??*

*At any rate, if you calculated them as circles, your numbers would be:*

*Niraek = 36pi*

*Theo = 64pi*

*Akshaj= 40pi*

*Since the LCM of [36, 64, 40] =2,880, therefore:*

*Niraek will have finished = 2,880 / 36 =80 donut holes*

*Theo will have finished = 2,880 /64 =45 donut holes*

*Akshaj will have finished= 2,880 /40 =72 donut holes.*

*Note: Somebody should check these numbers.** hectictar: what do you think of this approach?*

Then, after Hecticar answers, you replace the post with this:

*This doesn't lead anywhere !!*

**Deleting your question after specifically asking Hectictar to answer, and receiving an answer, is just bl****òó****dy rude. It looks like Hectictar is replying to empty air instead of a gasbag full of hot air. **

Now you’ve replaced the post with this BS:

*I think you should find the LCM of [6, 8, 10] =120, so you have:*

*120 / 6 =20 holes for N*

*120 / 8 =15 holes for T*

*120 / 10 =12 holes for A*

**(This is dumber blarney, because it’s not a square area)**

Mr. BB, you are failing, fast! The toxic blarney circulating in your system is accelerating your dementia! You should check in to a neurological treatment center ASAP. **They can’t save you, but maybe they will euthanize you and put you out of our misery. **

GA

GingerAle
Mar 23, 2018

#8**0 **

Well, even if you calculate them as SA of spheres, you don't have to use percentages to get the numbers you got:

The LCM of [144, 256, 400] =57,600. Therefore:

N =57,600 / 144 = 400 holes

T =57,600 / 256 = 225 holes

A =57,600 / 400 = 144 holes.

Which are the same numbers you got. But then, "tertre" says it is not right!!. So, what is, or where is the problem?

Guest Mar 23, 2018