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

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Answer 1 · 2016-11-06T08:13:48

All the coins are dimes and none are nickels.

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.