Question

There were 80 nickels and dimes on a table. It was worth $8.00. How many of each coin were on the table?

Answer 1

All the coins are dimes and none are nickels.

Explanation:

Let's let N be the number of nickels and D be the number of dimes. We know that:

`N+D=80` - this is for the actual number of coins

`N(.05)+D(.1)=8` - this is for the values of the coins

Let's solve the first equation for N then substitute into the second question:

`N=80-D`

`(80-D)(.05)+D(.1)=8`

`4-.05D+.1D=8`

`4+.05D=8`

`.05D=4`

`D=80`

So all the coins are dimes and none are nickels.