SOLUTION: Sherry bought 15 plain and glazed doughnuts for $4.65. Each plai doughnut cost.10cents less than a glazed doughnut. How many of each did she buy?

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Sherry bought 15 plain and glazed doughnuts for $4.65. Each plai doughnut cost.10cents less than a glazed doughnut. How many of each did she buy?      Log On


   

Question 1054839: Sherry bought 15 plain and glazed doughnuts for $4.65. Each plai doughnut cost.10cents less than a glazed doughnut. How many of each did she buy?
Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +n+ = number of plain doughnuts
+15+-+n+ = number of glazed doughnuts
Let +c+ = cost of a glazed doughnut
+c+-+10+ = cost of a plain doughnut
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(1) +%28+c+-+10+%29%2An+%2B+c%2A%28+15+-+n+%29+=+465+ ( in cents )
(1) +c%2An+-+10n+%2B+15c+-+c%2An+=+465+
(1) +15c+-+10n+=+465+
(1) +15c+=+10n+%2B+465+
(1) +c+=+%28+2%2F3+%29%2An+%2B+31+
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I'll try +n=3+
(1) +c+=+2+%2B+31+
(1) +c+=+33+
and
+c+-+10+=+23+
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(1) +%28+c+-+10+%29%2An+%2B+c%2A%28+15+-+n+%29+=+465+
(1) +23%2A3+%2B+33%2A12+=+465+
She bought 3 plain doughnuts
She bought 12 glazed doughnuts



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Sherry bought 15 plain and glazed doughnuts for $4.65. Each plai doughnut cost.10cents less than a glazed doughnut. How many of each did she buy?
Number of plain can be any number between 0 and 15, exclusive. The same goes for the glazed. They need to total 15 though. Something more

specific relating to one type can determine a specific number of each type.

It can range from 1 plain for .216667, and 14 glazed for .316667 each, to 14 plain at .303333 each, and 1 glazed for .4033333.

I believe this problem doesn't make much sense.