# There were 80 nickels and dimes on a table. It was worth $8.00. How many of each coin were on the table? ##### 1 Answer 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 $N \left(.05\right) + D \left(.1\right) = 8$- 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$$\left(80 - D\right) \left(.05\right) + D \left(.1\right) = 8$$4 - .05 D + .1 D = 8$$4 + .05 D = 8$$.05 D = 4$$D = 80\$

So all the coins are dimes and none are nickels.