45 dimes and 93 nickels were in bank
<em><u>Solution:</u></em>
Let "n" be the number of nickels
Let "d" be the number of dimes
We know that,
value of 1 nickel = $ 0.05
value of 1 dime = $ 0.10
<em><u>Given that There are three more than twice as many nickels as there are dimes</u></em>
Number of nickels = 3 + 2(number of dimes)
n = 3 + 2d ---- eqn 1
<em><u>Also given that coin bank that excepts only nickels and dimes contains $9.15</u></em>
number of nickels x value of 1 nickel + number of dimes x value of 1 dime = 9.15
0.05n + 0.10d = 9.15 ---- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
Substitute eqn 1 in eqn 2
0.05(3 + 2d) + 0.10d = 9.15
0.15 + 0.1d + 0.10d = 9.15
0.2d = 9
<h3>d = 45</h3>From eqn 1
n = 3 + 2(45)
n = 3 + 90 = 93
<h3>n = 93</h3>Thus 45 dimes and 93 nickels were in bank