A miniature golf course charges different prices for adults and children. On Saturday, 50 adults and 50 children played, and the golf course earned $800. On Sunday, 65 adults and 75 children played, and the golf course earned $1,100. How much does the golf course charge for adults

Answer 1

Let's denote the price charged for adults as \( a \) and the price charged for children as \( c \).

From the information provided, we can set up the following equations based on the number of adults and children playing on Saturday and Sunday.

Saturday's equation (50 adults and 50 children): \[ 50a + 50c = 800 \] Simplifying this equation by dividing everything by 50: \[ a + c = 16 \quad \text{(1)} \]

Sunday's equation (65 adults and 75 children): \[ 65a + 75c = 1100 \] We can simplify this equation by dividing everything by 5: \[ 13a + 15c = 220 \quad \text{(2)} \]

Now we have a system of equations:

1. \( a + c = 16 \)
2. \( 13a + 15c = 220 \)

We can solve this system. From equation (1), we can express \( c \) in terms of \( a \): \[ c = 16 - a \quad \text{(3)} \]

Now, we substitute equation (3) into equation (2): \[ 13a + 15(16 - a) = 220 \] Expanding this gives: \[ 13a + 240 - 15a = 220 \] Combine like terms: \[ -2a + 240 = 220 \] Subtracting 240 from both sides: \[ -2a = -20 \] Dividing by -2 gives: \[ a = 10 \]

Now we can find \( c \) using equation (3): \[ c = 16 - 10 = 6 \]

Thus, the prices charged by the miniature golf course are:

• Adult price \( a = 10 \)
• Child price \( c = 6 \)

Therefore, the golf course charges $10 for adults.

Answer 2

Xavior took a total of 124 quarters and dimes to trade in for cash at the bank. He got exactly $25 back. How many quarters did he have?

Answer 3

To solve for the number of quarters and dimes Xavior had, we can set up a system of equations.

Let:

• \( q \) = number of quarters
• \( d \) = number of dimes

From the problem, we gather the following information:

1. The total number of coins (quarters and dimes) is 124: \[ q + d = 124 \quad \text{(1)} \]

2. The total value of the coins is $25. Since quarters are worth $0.25 and dimes are worth $0.10, this gives us the equation: \[ 0.25q + 0.10d = 25 \quad \text{(2)} \]

To eliminate the decimals in equation (2), we can multiply the entire equation by 100: \[ 25q + 10d = 2500 \quad \text{(3)} \]

Now, we can use equations (1) and (3) to solve the system. From equation (1), we can express \( d \) in terms of \( q \): \[ d = 124 - q \quad \text{(4)} \]

Now substitute equation (4) into equation (3): \[ 25q + 10(124 - q) = 2500 \] Expanding the equation gives: \[ 25q + 1240 - 10q = 2500 \] Combining like terms: \[ 15q + 1240 = 2500 \] Subtracting 1240 from both sides: \[ 15q = 1260 \] Dividing by 15: \[ q = 84 \]

Now that we have \( q \), we can find \( d \) using equation (4): \[ d = 124 - 84 = 40 \]

In conclusion, Xavior had 84 quarters and 40 dimes.