+545 The big sack contained $30 in nickels and dimes. If there were 500 coins in the sack, how many were nickels?
The big sack contained $30 in nickels and dimes. If there were 500 coins in the sack, how many were nickels?
Let the number of nickles=N
Let the number of dimes =D
N + D =500
.05N + .10D =$30
Solve the following system:
{D+N = 500 | (equation 1)
0.1 D+0.05 N = 30 | (equation 2)
Subtract 0.05 × (equation 1) from equation 2:
{N+D = 500 | (equation 1)
0 N+0.05 D = 5 | (equation 2)
Divide equation 2 by 0.05:
{N+D = 500 | (equation 1)
0. N+D = 100. | (equation 2)
Subtract equation 2 from equation 1:
{N+0 D = 400 | (equation 1)
0 N+D = 100 | (equation 2)
Collect results:
Answer: | N = 400 and D=100
+1904 \(0.1x+0.05y=30\)
\(x+y=500\)
Solve for x in the second equation
\(x+y=500\)
\(x=500-y\)
Subsitute x in th first equation and solve for y
\(0.1\times(500-y)+0.05y=30\)
\(50-0.1y+0.05y=30\)
\(50-0.05y=30\)
\(-0.05y=-20\)
\(y=400\)
Subsitute y in the second equation
\(x+400=500\)
\(x=100\)
There would be \(400\) nickels and \(100\) dimes totaling \(500\) coins with a face value of \($30.\)
