Question

With nothing but nickels and pennies 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?


Use the guess and check method to solve this problem.

With nothing but nickels and pennies 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?


16 nickels and 5 dimes


15 nickels and 6 dimes


13 nickels and 8 dimes


14 nickels and 7 pennies

Answer 1

To solve the problem using the guess and check method, we start by defining the two types of coins Rupert has: nickels (N) and pennies (P).

We know:

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

2. The total number of coins is 21:
\[
N + P = 21
\]

Now we can use these two equations to set up our checks. First, we can express P in terms of N using the second equation:
\[
P = 21 - N
\]

Now we substitute this expression for P into the first equation:
\[
5N + (21 - N) = 77
\]
Simplifying gives:
\[
5N + 21 - N = 77
\]
\[
4N + 21 = 77
\]
\[
4N = 77 - 21
\]
\[
4N = 56
\]
\[
N = 14
\]

Now, substituting \(N=14\) back into the equation for \(P\):
\[
P = 21 - N = 21 - 14 = 7
\]

Thus, Rupert has 14 nickels and 7 pennies.

Let's confirm this:
- Value from nickels: \(14 \times 5 = 70\) cents
- Value from pennies: \(7 \times 1 = 7\) cents
- Total value: \(70 + 7 = 77\) cents

The total number of coins is \(14 + 7 = 21\) coins.

So, we have the correct solution: **14 nickels and 7 pennies.**