SOLUTION: A jar of dimes and quarters contained $10.95 There were at 93 coins in the jar. How many dimes? thanx

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A jar of dimes and quarters contained $10.95 There were at 93 coins in the jar. How many dimes? thanx      Log On


   

Question 40010: A jar of dimes and quarters contained $10.95 There were at 93 coins in the jar. How many dimes?
thanx
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Let d be the number of dimes and q will be quarters.
Their values will be 10d and 25q respectively. So we have
d + q = 93
10d + 25q = 1095 (talking about cents now)
We can do a linear sum of these by multiplying the first equation by 25 and subtracting...so we get
10d + 25q = 1095
-(25d + 25q = 2325)
and we have
-15d = -1230
d = 82
Thus there are 82 dimes and 11 quarters.