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

#1**+2 **

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

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

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

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

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

#1**+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 mm^{2}

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

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

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