With nothing but nickels and Pennie’s in his pocket, Rupert can afford 77 cents worth of candy. If he has 21 coins in his pocket how many coins does he have of each type?

Answer 1

Let the number of nickels be \( n \) and the number of pennies be \( p \).

We can set up two equations based on the information given:

1. The total number of coins: \[ n + p = 21 \]

2. The total value of the coins in cents: \[ 5n + p = 77 \] (since each nickel is worth 5 cents and each penny is worth 1 cent).

Now, we can solve this system of equations.

First, we can express \( p \) from the first equation: \[ p = 21 - n \]

Next, we can substitute this expression into the second equation: \[ 5n + (21 - n) = 77 \]

Now, simplify the equation: \[ 5n + 21 - n = 77 \] \[ 4n + 21 = 77 \]

Subtract 21 from both sides: \[ 4n = 56 \]

Now, divide by 4: \[ n = 14 \]

Now, substitute \( n \) back into the equation for \( p \): \[ p = 21 - n = 21 - 14 = 7 \]

So, Rupert has:

• \( n = 14 \) nickels
• \( p = 7 \) pennies

In conclusion, Rupert has 14 nickels and 7 pennies.