Lana is calibrating a scale. The scale calibrates at 50g. When she puts a nickel on the scale, it weighs 4.94g. A nickel weighs 5g. If a fifty cent piece weighs 11.17g, a quarter 5.71g, a dime 2.26g, a nickel at 4.94g, and a penny at 2.52g, how many of each can she have to make 50.00g?

To find out how many of each coin Lana can have to make 50.00g, we can set up a system of equations. Let's assign variables to represent the number of each coin:

Let
N = number of nickels
Q = number of quarters
D = number of dimes
P = number of pennies

We know the weights of each coin:

1 nickel weighs 5g
1 quarter weighs 5.71g
1 dime weighs 2.26g
1 penny weighs 2.52g

And we need to find the combination of coins that adds up to 50.00g, so the equation is:

5N + 5.71Q + 2.26D + 2.52P = 50.00

We also know that the total number of coins must be 50, so we can write another equation:

N + Q + D + P = 50

Now, we have a system of two equations:

5N + 5.71Q + 2.26D + 2.52P = 50.00
N + Q + D + P = 50

To solve this system of equations, we can use the method of substitution or elimination. However, since this is a relatively small system, let's use the method of substitution.

From the second equation, we can rewrite N as:

N = 50 - Q - D - P

Now substitute this expression for N into the first equation:

5(50 - Q - D - P) + 5.71Q + 2.26D + 2.52P = 50.00

Simplifying this equation will give us the value of Q in terms of D and P:

250 - 5Q - 5D - 5P + 5.71Q + 2.26D + 2.52P = 50.00

Simplifying further, we get:

-2.74Q - 2.74D - 2.48P = -200

Dividing through by -2.74, we have:

Q + D + P ≈ 72.99

Since the number of coins cannot be a decimal or fraction, we can start by assuming Q, D, and P are integers. We want the sum of Q, D, and P to be as close to 73 as possible without going over.

One way to achieve this is to set Q = 29, D = 24, and P = 21. This gives us:

29 quarters (29 * 5.71g) = 165.59g
24 dimes (24 * 2.26g) = 54.24g
21 pennies (21 * 2.52g) = 52.92g

Adding up the weights, we have:

165.59g + 54.24g + 52.92g = 272.75g

Since this is greater than 50.00g, our assumption of Q = 29, D = 24, and P = 21 is incorrect. We need to adjust the numbers.

To get closer to 73 without going over, let's try Q = 28, D = 24, and P = 20:

28 quarters (28 * 5.71g) = 159.88g
24 dimes (24 * 2.26g) = 54.24g
20 pennies (20 * 2.52g) = 50.40g

Adding up the weights, we have:

159.88g + 54.24g + 50.40g = 264.52g

Since this is still greater than 50.00g, we need to adjust the numbers again.

Continuing this process of trial and error while maintaining the condition that Q + D + P should be as close to 73 as possible without going over, we find that the combination Q = 24, D = 23, P = 26 satisfies the equation:

24 quarters (24 * 5.71g) = 137.04g
23 dimes (23 * 2.26g) = 51.98g
26 pennies (26 * 2.52g) = 65.52g

Adding up the weights, we have:

137.04g + 51.98g + 65.52g = 254.54g

Since this is less than 50.00g, we need to increase the total weight. To do this, we can adjust the number of nickels.

Let's calculate the weight of 50 nickels:

50 nickels (50 * 5g) = 250g

Adding the weight of nickels to the current total:

254.54g + 250g = 504.54g

Since this is greater than 50.00g, we can subtract the weight of the extra nickels to achieve the desired weight.

Subtracting:

504.54g - 50.00g = 454.54g

So, Lana can have 24 quarters, 23 dimes, 26 pennies, and no nickels to make the total weight 50.00g.