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

Question

1370 views around the world

You can reuse this answer
Creative Commons License

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:

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

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

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

$N = 80 - D$ $N=80-D$

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

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

$4 + .05 D = 8$ $4+.05D=8$

$.05 D = 4$ $.05D=4$

$D = 80$ $D=80$

So all the coins are dimes and none are nickels.