Lizzie has 30 coins that total $4.80. All of her coins are dimes, D, and quarters, Q. Algebraically determine how many of each t
1 answer:
Lizzie has 18 dimes and 12 quarters
<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
Given that LIzzie has 30 coins
number of dimes + number of quarters = 30
d + q = 30 ---- eqn 1
Also given that the coins total $ 4.80
number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 4.80
0.1d + 0.25q = 4.8 ------ eqn 2
Let us solve eqn 1 and eqn 2
From eqn 1,
d = 30 - q ---- eqn 3
Substitute eqn 3 in eqn 2
0.1(30 - q) + 0.25q = 4.8
3 - 0.1q + 0.25q = 4.8
0.15q = 1.8
<h3>q = 12</h3>From eqn 3,
d = 30 - 12
<h3>d = 18</h3>Thus she has 18 dimes and 12 quarters
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