Katie had 21 nickels, dimes and quarters worth $2.25. If she had twice as many nickels as quarters, how many of each type of coin did Katie have?

Sure, let's solve this step by step.

Let's denote: N = number of nickels Katie has D = number of dimes Katie has Q = number of quarters Katie has

From the problem, we know that:

1. N + D + Q = 21 (total number of coins)
2. N = 2Q (twice as many nickels as quarters)
3. 0.05N + 0.10D + 0.25Q = 2.25 (total value of the coins)

We can substitute N from equation 2 into equations 1 and 3:

2Q + D + Q = 21 and 0.05(2Q) + 0.10D + 0.25Q = 2.25

This simplifies to: 3Q + D = 21 and 0.35Q + 0.10D = 2.25

We can solve these two equations simultaneously to find the values of D and Q.

Let's multiply the second equation by 10 to make the calculations easier: 3.5Q + D = 22.5

Now we can subtract the first equation from this new equation: 3.5Q + D - (3Q + D) = 22.5 - 21 0.5Q = 1.5

Solving for Q gives us Q = 3.

Substitute Q = 3 into the first equation: 3*3 + D = 21 D = 21 - 9 = 12

Finally, substitute Q = 3 into the second equation: N = 2*3 = 6

So, Katie has 6 nickels, 12 dimes, and 3 quarters.