A collection of nickels dimes and quarters consists of 50 coins with a total of $5.50. If there are two times as many dimes as quarters find the number of each type of coins?

quarters --- x

dimes ----- 2x
nickels --- 50 - 3x

equation based on "value" of those coins:

25x + 10(2x) + 5(50-3x) = 550
25x + 20x + 250 - 15x = 550
30x = 300
x=10

there are 10 quarters, 20 dimes and 20 nickels

post it.

Let's assume the number of quarters is "x".

Since there are two times as many dimes as quarters, the number of dimes is "2x".
The number of nickels can be calculated by subtracting the total number of quarters and dimes from 50, as follows:

Number of nickels = 50 - (x + 2x)
= 50 - 3x

Now we can calculate the total value of the coins. The value of a quarter is $0.25, the value of a dime is $0.10, and the value of a nickel is $0.05. The total value can be calculated as follows:

Total Value = (Value of quarters * Number of quarters) + (Value of dimes * Number of dimes) + (Value of nickels * Number of nickels)
= ($0.25 * x) + ($0.10 * 2x) + ($0.05 * (50 - 3x))
= $0.25x + $0.20x + $0.05(50 - 3x)
= $0.25x + $0.20x + $2.50 - $0.15x
= $0.30x + $2.50

Since the total value of the coins is given as $5.50, we can set up the following equation:

$0.30x + $2.50 = $5.50

Now we can solve for x:

$0.30x = $5.50 - $2.50
$0.30x = $3.00
x = $3.00 / $0.30
x = 10

Therefore, the number of quarters is 10.
The number of dimes is 2x, which is 2 * 10 = 20.
The number of nickels is 50 - (10 + 20) = 20.

So, there are 10 quarters, 20 dimes, and 20 nickels.

To solve this problem, we can set up a system of equations using the given information.

Let's assume:
x = number of quarters
y = number of dimes
z = number of nickels

From the problem, we know:
x + y + z = 50 (equation 1)
0.25x + 0.10y + 0.05z = 5.50 (equation 2)
y = 2x (equation 3)

Since we have three variables and three equations, we can solve them simultaneously.

First, let's substitute equation 3 into equations 1 and 2. We need to express both equations in terms of only x and z.

From equation 1:
x + y + z = 50 becomes
x + 2x + z = 50,
3x + z = 50 (equation 4)

From equation 2:
0.25x + 0.10y + 0.05z = 5.50 becomes
0.25x + 0.10(2x) + 0.05z = 5.50,
0.25x + 0.20x + 0.05z = 5.50,
0.45x + 0.05z = 5.50 (equation 5)

Now we have a system of two equations with two variables, x and z (equations 4 and 5). We can solve this system of equations to find the values of x and z.

Multiplying equation 4 by 0.45, we get:
0.45(3x + z) = 0.45(50),
1.35x + 0.45z = 22.50 (equation 6)

Subtracting equation 5 from equation 6, we have:
(1.35x + 0.45z) - (0.45x + 0.05z) = 22.50 - 5.50,
0.90x + 0.40z = 17.00 (equation 7)

Now we have a new equation (equation 7) with two variables, x and z. We can solve it for the values of x and z.

To eliminate decimals, we can multiply equation 7 by 10:
9x + 4z = 170 (equation 8)

Now we have a system of two linear equations:
3x + z = 50 (equation 4)
9x + 4z = 170 (equation 8)

We can solve this system of equations using substitution, elimination, or any other method. Solving it, we find that x = 8 and z = 26.

Since y = 2x (from equation 3), we get y = 2(8) = 16.

Therefore, there are 8 quarters, 16 dimes, and 26 nickels in the collection.