SOLUTION: Alan and Mike had $800 altogether. 1/4 of Alan’s money was $65 more than 1/5 of Mike’s money. How much more money than Mike did Alan have?

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Alan and Mike had $800 altogether. 1/4 of Alan’s money was $65 more than 1/5 of Mike’s money. How much more money than Mike did Alan have?      Log On


   

Question 1184819: Alan and Mike had $800 altogether. 1/4 of Alan’s money was $65 more than 1/5 of Mike’s money. How much more money than Mike did Alan have?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
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Alan and Mike had $800 altogether. 1/4 of Alan’s money was $65 more than 1/5 of Mike’s money.
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Give variables and translate into equations.

n, alan
k, mike
system%28n%2Bk=800%2Cn%2F4=65%2Bk%2F5%29

Do what you need. Question is find k-n.

Answer by greenestamps(13215) About Me  (Show Source):
You can put this solution on YOUR website!


Probably easier if you use a single variable....

x=Alan's amount
800-x=Mike's amount

1/4 of Alan’s money was $65 more than 1/5 of Mike’s money:

%281%2F4%29x=%281%2F5%29%28800-x%29%2B65

I would multiply through by the least common denominator (20) to make the rest of the work easier.

5x=4%28800-x%29%2B1300

I'll let you finish from there....