a childs bank contains $6.30 in dimes and quarters. there are twice as many dimes as quarters. how many of each kind of coin are in the bank?
let d = #dimes and q=#quarters.
The total is worth $6.30.
eqn 1 is 0.10*d + 0.25*q=$6.30
eqn 2 is 2*#quarters = #dimes or
2q=d.
Two equatins and two unknowns. Solve for q and d.
Post your work if you get stuck.
in a jar of coins there is 15 more dimes than nickels. If the total value of the coins in the jar is $6.90 how many of each coin is in the jar?
Any help for this?
We have two equations:
Equation 1: 0.10d + 0.25q = $6.30
Equation 2: 2q = d
We can substitute Equation 2 into Equation 1 to solve for q:
0.10(2q) + 0.25q = $6.30
0.20q + 0.25q = $6.30
0.45q = $6.30
q = $6.30 / 0.45
q = 14
Now, we can substitute the value of q into Equation 2 to find d:
2(14) = d
d = 28
So, there are 28 dimes and 14 quarters in the bank.
To solve this problem, we can set up a system of equations and use substitution to find the values of d (number of dimes) and q (number of quarters).
Equation 1: 0.10d + 0.25q = 6.30
Equation 2: 2q = d
Let's start with equation 2 because it gives us a relationship between the number of dimes and quarters. We can substitute the value of d in equation 1 with 2q:
0.10(2q) + 0.25q = 6.30
0.20q + 0.25q = 6.30
0.45q = 6.30
Now, we can solve for q by dividing both sides of the equation by 0.45:
q = 6.30 / 0.45
q ≈ 14
We have found that there are approximately 14 quarters in the bank. Let's substitute this value back into equation 2 to find the number of dimes:
2q = d
2(14) = d
d = 28
Therefore, there are 28 dimes and 14 quarters in the child's bank.