SOLUTION: Julie bought donuts for a party - 1/6 of the donuts were jelly donuts, 1/3 of the donuts were cinnamon donuts, and 5/6 of the remainder were glazed donuts. Julie bought 20 glazed d

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Julie bought donuts for a party - 1/6 of the donuts were jelly donuts, 1/3 of the donuts were cinnamon donuts, and 5/6 of the remainder were glazed donuts. Julie bought 20 glazed d      Log On


   

Question 587573: Julie bought donuts for a party - 1/6 of the donuts were jelly donuts, 1/3 of the donuts were cinnamon donuts, and 5/6 of the remainder were glazed donuts. Julie bought 20 glazed donuts. How many donuts did Julie buy in all?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = the number of donuts she bought
+x%2F6+ = number of jelly donuts
+x%2F3+ = cinnamon donuts
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To find the fraction of donuts that are left,
So far there are +1%2F6+%2B+2%2F6+=+%283%2F6%29%2Ax+ and
+%283%2F6%29%2Ax+=+%281%2F2%29%2Ax+
So 1/2 of the donuts are left
+%285%2F6%29%2A%281%2F2%29%2Ax+ are glazed donuts and
%285%2F6%29%2A%281%2F2%29%2Ax+=+20+
+%285%2F12%29%2Ax+=+20+
+x+=+%2812%2F5%29%2A20+
+x+=+48+
She bought 48 donuts
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check:
+48%2F6+=+8+ = number of jelly donuts
+48%2F3+=+16+ = cinnamon donuts
+8+%2B+16+=+24+
There are +48+-+24+=+24+ donuts left
+%285%2F6%29%2A24+=+20+ are glazed donuts
There are 4 donuts of unknown type
OK