SOLUTION: you have at least $30.00in change consisting of dimes and quaters. Write an inequality that shows the differant number of coins

Question 42046: you have at least $30.00in change consisting of dimes and quaters. Write an inequality that shows the differant number of coins
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
You have at least $30.00 in change consisting of dimes and quaters.
What's the least number of coins you can have? If you could have all quarters,
that would be the least number, but you must have some dimes.
All quarters would be $30/.25 = 120 coins.
You can't take away 1 quarter to get change in dimes. You must take away 2
quarters, and get 5 dimes change.
120 coins - 2 quarters + 5 dimes = 123
That's the low end of the inequality.
The most coins you could have would be all dimes.
That would be $30/.10 = 300, but you must have quarters. Dimes don't
divide evenly into 1 quarter, but there are 5 dimes in 2 quarters.
300 coins - 5 dimes + 2 quarters = 297

for exactly $30 worth of dimes and quarters
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