Question

joe has a piggy bank with $8.90 split upon nickels, dimes, and quarters. The piggy bank contains 76 coins in all. The number of dimes is equal to the sum of quarters and nickels. How many of each coin does he have ?

Answer 1

Answer:

There are 38 dimes, 22 nickels and 16 quarters.

Step-by-step explanation:

Let n, d and q represent the # of nickels, dimes and quarters respectively.

Then n + d + q = 76

The value of a nickel is $0.05; that of a dime is $0.10, and that of a quarter is $0.25.

Thus, the value of n nickels is $0.05n (and so on).

The total value of the coins is $0.05n + $0.10d + $0.25q = $8.90.

d = n + q allows us to eliminate d.

First, n + d + q = 76 becomes n + (n + q) + q = 76, and second:

$0.05n + $0.10(n + q) + $0.25q = $8.90.  Here we have succeeded in eliminating d from two different equations, and now we have these two different equations in two unknowns (n and q), which is solvable.

Simplifying both equations, we get:

2n + 2q = 76 and

5n + 10n + 10q + 25q = 890, or 15n + 35q = 890

Let's use the substitution method of solving linear equations:

Rewrite 2n  + 2q = 76 as n + q = 38, or n = 38 - q.  Substituting this result into the second equation, we get:

15(38 - q) + 35 q = 890, or

  570 - 15q + 35q = 890, or

   570 + 20 q = 890.  Then 20q = 890 - 570 = 320, and q = 320/20 = 16.

There are 16 quarters.  Thus, the number of nickels is n = 38 - 16 = 22.

Finally, since n + d + q = 76, 22 + d + 16 = 76, or:

22 + d = 60, or d = 60 - 22 = 38.

There are 38 dimes, 22 nickels and 16 quarters.