n = nickels
q = quarters
d = dimes
There are a total of 30 coins.
n + d + q = 30
There are twice as many quarters as dimes means:
q = 2 d
n + d + q = 30
n + d + 2 d = 30
n + 3 d = 30
1 nickel = $0.05
1 dime = $0.10
1 quarter = $0.25
A value is $3.30
0.05 ∙ n + 0.1 ∙ d + 0. 25 ∙ q = 3.30
0.05 n + 0.1 d + 0. 25 ∙ 2 d = 3.30
0.05 n + 0.1 d + 0. 50 d = 3.30
0.05 n + 0.60 d = 3.30
Divide both sides by 0.05
n + 12 d = 66
Now you must solve system:
n + 3 d = 30
n + 12 d = 66
___________
First equation:
n + 3 d = 30
Subtract 3 d to both sides
n = 30 - 3 d
Second equation:
n + 12 d = 66
Subtract 12 d to both sides
n = 66 - 12 d
n = n
30 - 3 d = 66 - 12 d
Add 12 d to both sides
30 + 9 d = 66
Subtract 30 to both sides
9 d = 36
d = 36 / 9 = 4
n = 30 - 3 d = 30 - 3 ∙ 4 = 30 - 12 = 18
q = 2 d = 2 ∙ 4 = 8
18 nickels
4 = dimes
8 = quarters
Proof:
q = 2 d
8 = 2 ∙ 4
n + d + q = 18 + 8 + 4 = 26 + 4 = 30
0.05 ∙ n + 0.1 ∙ d + 0. 25 q =
0.05 ∙ 18 + 0.1 ∙ 4 + 0. 25 ∙ 8 =
0.9 + 0.4 + 2 = 1.3 + 2 = 3.30
SOLVING SYSTEMS:
30 coins having a value of $3.30 consists of nickels, dimes and quarters. If there are twice as many quarters as dimes, how many coins of each kind were there?
-Thank you!
1 answer
1 answer