Respuesta :
Number of dimes were 81 and number of quarters were 47
Solution:
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 There are 128 coins in all
number of dimes + number of quarters = 128
d + q = 128 ------ eqn 1
Also given that collection of dimes and quarters is worth $19.85
number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85
[tex]d \times 0.10 + q \times 0.25 = 19.85[/tex]
0.1d + 0.25q = 19.85 -------- eqn 2
Let us solve eqn 1 and eqn 2
From eqn 1,
d = 128 - q -------- eqn 3
Substitute eqn 3 in eqn 2
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
q = 47
Therefore from eqn 3,
d = 128 - q
d = 128 - 47
d = 81
Thus number of dimes were 81 and number of quarters were 47
The number of dime coins is 81 and the number of quarter coins is 47.
Given to us,
Worth of all coins = $19.85,
Number of coins = 128 coins,
Value of $1 = 100 cent,
therefore, $19.85 = 19.85 x 100 = 1985 cent,
Value of dimes = 10 cent,
Value of quarter = 25 cent,
Assumption
Let the number of dime coins be x, therefore, the number of quarter coins is [tex](128-x)[/tex].
The total worth of dime coins,
the worth of dime coins = Value of dimes x number of dim coins
[tex]= 10 \times x[/tex]
[tex]= 10 x[/tex]
The total worth of quarter coins,
the worth of quarter coins = Value of quarter x number of quarter coins
[tex]= 25\times (128-x)\\= (3200-25x)\ cent[/tex]
Worth of all coins
Worth of all coins = worth of dime coins + worth of quarter coins
[tex]1985 = 10x + 3200 - 25x\\-1215 = -15x\\15x = 1215\\\\x = \dfrac{1215}{15}\\\\x = 81[/tex]
Number of coins
The number of dime coins is x, therefore, the number of dime coins is 81.
The number of dime coins is [tex](128-x)[/tex], therefore, the number of dime coins is [tex](128-81) = 47[/tex].
Hence, the number of dime coins is 81 and the number of quarter coins is 47.
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