Okay well here is is the problem
Randy has $2.55 in dimes and quarters. He has eight more dimes than quarters. How many of each coin does he have?
2006-11-30 03:09:49 · 2 answers · asked by swweeeetttt 1 in Education & Reference ➔ Primary & Secondary Education
13 dimes, 5 quarters. You know that there has to be an odd number of quarters because the amount ends in a 5 and there are no nickels. Since the number is only $2.55, you know there's probably only 3 or 5 quarters, and 5 works.
2006-11-30 03:26:16 · answer #1 · answered by Justin B 2 · 0⤊ 0⤋
Let the number of quarters he has be x.
Let the number of dimes he has be (8 + x) since he has 8 more dimes than quarters.
since the total amount of money is made up of the total amount of dimes and quarters, the equation is:
[(8 + x)*$0.10] + $0.25x = $2.55
$0.80 + $0.10x + $0.25x = $2.55
$0.35x = $2.55 - $0.80
$0.35x = $1.75
x = $1.75 / $0.35
x = 5
The number of quarters is 5.
The number of dimes is (8+x) = (8+5) =13
2006-11-30 03:32:32 · answer #2 · answered by reut 4 · 0⤊ 0⤋