SOLUTION: Twelve coins consisting of nickels, dimes, and quarters is worth $1.20. The number of dimes is two more than twice the number of nickels. How many nickels are there? thank you for

Algebra.Com
Question 949124: Twelve coins consisting of nickels, dimes, and quarters is worth $1.20. The number of dimes is two more than twice the number of nickels. How many nickels are there?
thank you for your help. even just a set up of the problem and i can solve it. been working on it for a couple hours and im going crazy now!
Found 3 solutions by josgarithmetic, macston, MathTherapy:
Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!
n, nickel count
d, dime count




Answer by macston(5194)   (Show Source): You can put this solution on YOUR website!
N=no. of nickels; D=no. of dimes=2N+2; Q=no. of quarters
N+D+Q=12 Substitute for D
N+2N+2+Q=12 Subtract 3N from each side.
2+Q=12-3N Subtract 2 from each side.
Q=10-3N
$0.05N+$0.10D+$0.25Q=$1.20 Substitute for D and Q
STOP HERE AND TRY THE REST ON YOUR OWN.
$0.05N+$0.10(2N+2)+$0.25(10-3N)=$1.20
$0.05N+$0.20N+$0.20+$2.50-$0.75N=$1.20 Combine like terms.
$-0.50N+$2.70=$1.20 Subtract $1.20 from each side
-$0.50N+$1.50=0 Add $0.50N to each side.
$1.50=$0.50N Divide each side by $0.50
3=N ANSWER 1: There are 3 nickels
D=2N+2=2(3)+2=6+2=8 ANSWER 2: There are 8 dimes.
Q=10-3N=10-3(3)=1 ANSWER 3: There is 1 quarter
CHECK
$0.05(3)+$0.10(8)+$0.25(1)=$1.20
$0.15+$0.80+$0.25=$1.20
$1.20=$1.20

Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!
Twelve coins consisting of nickels, dimes, and quarters is worth $1.20. The number of dimes is two more than twice the number of nickels. How many nickels are there?
thank you for your help. even just a set up of the problem and i can solve it. been working on it for a couple hours and im going crazy now!
Let number of nickels be N, and number of quarters, Q

Then number of dimes = 2N + 2

We then get the following coin-count equation: N + 2N + 2 + Q = 12

3N + Q = 10________Q = 10 - 3N --------- eq (i)



We then get the following value equation: .05N + .1(2N + 2) + .25Q = 1.2

.05N + .2N + .2 + .25Q = 1.2________.25N + .25Q = 1 ----- eq (ii)



.25N + .25(10 - 3N) = 1 ------- Substituting 10 - 3N for Q in eq (ii)

.25N + 2.5 - .75N = 1

.25N - .75N = 1 - 2.5

- .5N = - 1.5

N, or number of nickels = , or  

RELATED QUESTIONS


A donation box contains 49 coins consisting of nickels, dimes, and quarters. The number... (answered by ewatrrr)

SOLVE USING DETERMINANTS and CRAMER'S RULE

A coin collector has 41 coins consisting of  (answered by pjalgeb)

70. Nickels, dimes, and quarters. Bernard has 41 coins consisting of nickels, dimes, and... (answered by macston)

A collection of nickels, dimes and quarters is worth $5.30.  There are two more dimes... (answered by ikleyn)

a cashier has 25 coins consisting of nickels dimes and quarters with a value of $4.90.... (answered by lwsshak3)

Rita has a collection of 105 coins consisting of nickels, dimes, and quarters. The number  (answered by nerdybill)

Rita has a collection of 105 coins consisting of nickels , dimes and quarters. The number  (answered by rwm)

Rita has a collection of 105 coins consisting of nickels, dimes, and quarters. The number  (answered by stanbon)

Rita has a collection of 108 coins consisting of nickels, dimes, and quarters. The number  (answered by FrankM)