nickels --- x
dimes ---- x+10
quarters --- 2(x+10) = 2x + 20
now for the value of your coints
5x + 10(x+10) + 25(2x + 20) = 925
solve for x , easy to solve
dimes ---- x+10
quarters --- 2(x+10) = 2x + 20
now for the value of your coints
5x + 10(x+10) + 25(2x + 20) = 925
solve for x , easy to solve
Let's start by assigning variables to the number of nickels, dimes, and quarters Sarah has.
Let N be the number of nickels,
Let D be the number of dimes,
And let Q be the number of quarters.
According to the information given, we know three things:
1. The total value of the coins is $9.25.
2. Sarah has 10 more dimes than nickels.
3. Sarah has twice as many quarters as dimes.
First, let's convert everything to cents to make the math a bit easier.
The value of N nickels is 5N cents.
The value of D dimes is 10D cents.
The value of Q quarters is 25Q cents.
So from the first piece of information, we know that 5N + 10D + 25Q = 925 cents.
Now, let's work on the second and third pieces of information.
Sarah has 10 more dimes than nickels, so D = N + 10.
Sarah has twice as many quarters as dimes, so Q = 2D.
Now we can substitute these expressions into the first equation to solve for N, D, and Q.
5N + 10D + 25Q = 925.
Substituting D = N + 10 and Q = 2D:
5N + 10(N + 10) + 25(2(N + 10)) = 925.
Simplifying:
5N + 10N + 100 + 50N + 500 = 925.
Combining like terms:
65N + 600 = 925.
Subtracting 600 from both sides:
65N = 325.
Dividing both sides by 65:
N = 5.
Now we know there are 5 nickels.
Using D = N + 10, we find D = 5 + 10 = 15.
Using Q = 2D, we find Q = 2 * 15 = 30.
So Sarah has 5 nickels, 15 dimes, and 30 quarters.
Now, it's time to help Sarah become a successful coin collector!
Step 1: Assign variables.
Let's assume the number of nickels as "n", the number of dimes as "d", and the number of quarters as "q".
Step 2: Translate the given information into equations.
According to the problem, the value of all the coins is $9.25.
The value of a nickel is $0.05, the value of a dime is $0.10, and the value of a quarter is $0.25.
The equation for the total value of all the coins can be written as:
0.05n + 0.10d + 0.25q = 9.25 -- Equation 1
According to the problem, Sarah has 10 more dimes than nickels.
So, the equation for the number of dimes can be written as:
d = n + 10 -- Equation 2
According to the problem, Sarah has twice as many quarters as dimes.
So, the equation for the number of quarters can be written as:
q = 2d -- Equation 3
Step 3: Solve the system of equations.
To solve this system of equations, we will use the substitution method.
Substitute Equation 2 and Equation 3 into Equation 1:
0.05n + 0.10(n+10) + 0.25(2d) = 9.25
0.05n + 0.10n + 1.00 + 0.50d = 9.25
0.15n + 0.50d = 8.25 -- Equation 4
Now, substitute Equation 2 into Equation 4:
0.15n + 0.50(n+10) = 8.25
0.15n + 0.50n + 5.00 = 8.25
0.65n = 3.25
n = 5
Substitute the value of n into Equation 2:
d = n + 10
d = 5 + 10
d = 15
Substitute the value of d into Equation 3:
q = 2d
q = 2 * 15
q = 30
Step 4: Check the solution.
Let's check if our solution is correct by using the values of n = 5, d = 15, and q = 30 in Equation 1:
0.05n + 0.10d + 0.25q = 9.25
0.05(5) + 0.10(15) + 0.25(30) = 9.25
0.25 + 1.50 + 7.50 = 9.25
9.25 = 9.25
Since both sides of the equation are equal, our solution is correct.
Step 5: Answer the question.
Sarah has 5 nickels, 15 dimes, and 30 quarters.
Let's start by assigning variables:
Let the number of nickels be N
Let the number of dimes be D
Let the number of quarters be Q
We can now translate the given information into equations.
1) The total value of the coins is $9.25:
The value of N nickels is 5N cents.
The value of D dimes is 10D cents.
The value of Q quarters is 25Q cents.
The sum of the values is 5N + 10D + 25Q.
Therefore, the equation is:
5N + 10D + 25Q = 925 (because $9.25 is equal to 925 cents)
2) Sarah has 10 more dimes than nickels:
The number of dimes is 10 more than the number of nickels.
Therefore, the equation is:
D = N + 10
3) Sarah has twice as many quarters as dimes:
The number of quarters is twice the number of dimes.
Therefore, the equation is:
Q = 2D
Now we have a system of three equations:
1) 5N + 10D + 25Q = 925
2) D = N + 10
3) Q = 2D
We can solve this system of equations using substitution or elimination.
Let's use the substitution method to solve this system:
Substitute D in terms of N from equation (2) into equations (1) and (3):
1) 5N + 10(N + 10) + 25Q = 925
2) Q = 2(N + 10)
Simplify equation (1):
5N + 10N + 100 + 25Q = 925
15N + 25Q = 825
3N + 5Q = 165
Multiply equation (2) by 5:
5Q = 10N + 50
Substitute the value of Q from equation (2) into equation (1):
3N + 5(10N + 50) = 165
3N + 50N + 250 = 165
53N = 165 - 250
53N = -85
N = -85/53 (rounded to the nearest whole number)
Since the number of coins cannot be negative, we know that N must be equal to 0.
Therefore, the number of nickels (N) is 0.
Substitute the value of N back into equation (2):
D = N + 10
D = 0 + 10
D = 10
Substitute the value of D back into equation (3):
Q = 2D
Q = 2(10)
Q = 20
So, Sarah has 0 nickels, 10 dimes, and 20 quarters.