A dollar was changed into 16 coins consisting of just nickels an dimes. how many coins of each kind were in the change?

Answer 1

This is an algebra problem that requires two equations to be solved simultaneously. The two key pieces of information are that there are 16 coins and that they add to 1 dollar. For the first equation, let's suppose we have X nickels and Y dimes. Therefore we may say:

X + Y = 16

For the second equation we apply what we know about the values of nickels and dimes. A nickel is worth 5 cents. So the value of X number of nickels is 5*(X). The value of Y dimes is 10*(Y). Notice that the units of this value is in cents. Our given information says dollar, so it now makes sense to think of this dollar as 100 cents to match our equation. Now we may say:

5X + 10Y = 100

These two equations can be solved simultaneously by add/subtracting them or solving for a variable and substituting. I think add/subtracting is cooler but everyone knows substitution:

Rearrange X+Y=16 to

X=16-Y

Then substitute for X in the other equation:

5(16-Y)+10Y=100

Distribute, group terms, subtract, and divide both sides:

(80-5Y)+10Y=100

80+5Y=100

5Y=20

Y=4

Therefore we have 4 dimes, or 40 cents. The total is a dollar so we must have 60 cents in nickels. The only way to have that is with 12 coins. These numbers make sense: they total 16 coins and 1 dollar.

QED.