Number of dimes were 81 and number of quarters were 47
<em><u>Solution:</u></em>
Let "d" be the number of dimes
Let "q" be the number of quarters
We know that,
value of 1 dime = $ 0.10
value of 1 quarter = $ 0.25
<em><u>Given that There are 128 coins in all</u></em>
number of dimes + number of quarters = 128
d + q = 128 ------ eqn 1
<em><u>Also given that collection of dimes and quarters is worth $19.85</u></em>
number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85
0.1d + 0.25q = 19.85 -------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
From eqn 1,
d = 128 - q -------- eqn 3
<em><u>Substitute eqn 3 in eqn 2</u></em>
0.1(128 - q) + 0.25q = 19.85
12.8 - 0.1q + 0.25q = 19.85
12.8 + 0.15q = 19.85
0.15q = 7.05
<h3>q = 47</h3>Therefore from eqn 3,
d = 128 - q
d = 128 - 47
<h3>d = 81</h3>Thus number of dimes were 81 and number of quarters were 47
