tim has nickles,dimes, & quarters in his pocket. he has twice as many dimes as nickles. there are 63 coins in all. the coins in his pocket have a total value of $5.75. how many of each coin does he have
Don't let all the words scare you. Just write it down as math.
d=2n
n+d+q=63
5n+10d+25q=575
now crank it out.
To solve this problem, we can set up a system of equations based on the given information.
Let's denote the number of nickels as "n," the number of dimes as "d," and the number of quarters as "q."
From the information given in the problem, we have the following:
1. Tim has twice as many dimes as nickels, so we can write d = 2n.
2. There are 63 coins in total, so we have n + d + q = 63.
3. The total value of the coins is $5.75. Since a nickel equals $0.05, a dime equals $0.10, and a quarter equals $0.25, we can write 0.05n + 0.10d + 0.25q = 5.75.
Now we have a system of equations:
Equation 1: d = 2n
Equation 2: n + d + q = 63
Equation 3: 0.05n + 0.10d + 0.25q = 5.75
We can solve this system of equations using various methods, such as substitution or elimination.
Let's use substitution to eliminate one variable. Since Equation 1 tells us that d = 2n, we can substitute 2n for d in Equation 2 and Equation 3:
Equation 2 becomes: n + 2n + q = 63 --> 3n + q = 63
Equation 3 becomes: 0.05n + 0.10(2n) + 0.25q = 5.75 --> 0.05n + 0.20n + 0.25q = 5.75 --> 0.25n + 0.25q = 5.75
Now we have a system of two equations with two variables:
Equation 4: 3n + q = 63
Equation 5: 0.25n + 0.25q = 5.75
We can solve this system using substitution or elimination. Let's use substitution.
From Equation 4, we can rewrite it as q = 63 - 3n.
Now, substitute q = 63 - 3n into Equation 5:
0.25n + 0.25(63 - 3n) = 5.75 --> 0.25n + 15.75 - 0.75n = 5.75 --> -0.5n = -10 --> n = 20
Using n = 20, substitute it back into q = 63 - 3n:
q = 63 - 3(20) = 63 - 60 = 3
From Equation 1, we know that d = 2n:
d = 2(20) = 40
Therefore, Tim has 20 nickels, 40 dimes, and 3 quarters.