John has 11 more quarters than dimes. He has a total of $7.65. How many quarters and dimes does he have?
Q = 11 + D
25Q + 10D = 765
Substitute 11+D for Q in second equation and solve for D. Insert that value into the first equation and solve for Q. Check by inserting both values into the second equation.
To solve this problem, we can set up a system of equations with the information given. Let's assume that the number of dimes John has is represented by "x", and the number of quarters he has is represented by "x + 11" since he has 11 more quarters than dimes.
The value of x dimes is 0.10x dollars, and the value of (x + 11) quarters is 0.25(x + 11) dollars.
We are told that the total value of the coins is $7.65. So we can write the equation:
0.10x + 0.25(x + 11) = 7.65
To simplify the equation, distribute 0.25 to (x + 11):
0.10x + 0.25x + 2.75 = 7.65
Combine like terms:
0.35x + 2.75 = 7.65
Subtract 2.75 from both sides of the equation:
0.35x = 7.65 - 2.75
0.35x = 4.90
Divide both sides of the equation by 0.35:
x = 4.90 / 0.35
x ≈ 14
Therefore, John has 14 dimes.
To find the number of quarters, substitute x back into the expression x + 11:
14 + 11 = 25
Therefore, John has 25 quarters.
In conclusion, John has 14 dimes and 25 quarters.