SOLUTION: A had some chocolate cookies and gave {{{1/2}}} to B. A also had some strawberry cookies and gave {{{1/2}}} to B. A ate 32 chocolate cookies and B ate 14 strawberry cookies. B had

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: A had some chocolate cookies and gave {{{1/2}}} to B. A also had some strawberry cookies and gave {{{1/2}}} to B. A ate 32 chocolate cookies and B ate 14 strawberry cookies. B had      Log On


   

Question 1184678: A had some chocolate cookies and gave 1%2F2 to B. A also had some strawberry cookies
and gave 1%2F2 to B. A ate 32 chocolate cookies and B ate 14 strawberry cookies. B had 1%2F8 as many strawberry cookies are chocolate cookies left and A had 1%2F5 as many strawberry
cookies as chocolate cookies left. How many strawberry cookies did A have at first?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A had some chocolate cookies and gave 1%2F2 to B.
A also had some strawberry cookies and gave 1%2F2 to B.
let c = original number of Chocolate cookies
let s = number of straw berries
then
A and B both have .5c and .5s
:
A ate 32 chocolate cookies and B ate 14 strawberry cookies.
A then had: (.5c-32) and .5s, while B has: .5c and (.5s-14)
:
B had 1%2F8 as many strawberry cookies as chocolate cookies left
.5s-14 = 1%2F8(.5c)
get rid of the fraction, mult by 8, rearrange fractions for elimination
8(.5s-14) = .5c
4s - 112 - .5c = 0
4s -.5c = 112
:
and A had 1%2F5 as many strawberry cookies as chocolate cookies left.
.5s = 1%2F5(.5c-32)
multiply by 5
5(.5s) = .5c - 32
2.5s - .5c = -32
:
4s -.5c = 112
2.5s - .5c = -32
-------------------subtraction eliminates c, find s
1.5s = 144
s = 144/1.5
s = 96 strawberry cookies A had at first
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