Question 148698: Mike has 19 coins (Nickels, Dimes and quarters) in his piggy bank with a total value of $2.35. There are 2 more dimes than nickels. How many of each type of coins does mike have?
I know I have to make 3 equations
N + D + Q= 2.35
D= N +2 ( I think)
Not sure on the third one.
Found 2 solutions by oscargut, jojo14344:
Answer by oscargut(2103)   (Show Source):
You can put this solution on YOUR website! 1 Nickel =0.05 N= number of nickels
1 Dimes =0.1 D= number of dimes
1 quarter =0.25 Q=number of quarters
Mike has 19 coins then
N+D+Q=19
total value of $2.35 then
0.05N+0.1D+0.25Q=2.35
There are 2 more dimes than nickels then
D=N+2
3 equations are:
N+D+Q=19
0.05N+0.1D+0.25Q=2.35
D=N+2
using the third eq in 1 and 2
N+N+2+Q=19
0.05N+0.1(N+2)+0.25Q=2.35
2N+Q=17 then Q=17-2N
0.15N+0.25Q=2.15
0.15N+0.25(17-2N)=2.15
0.15N+4.25-0.5N=2.15
-0.35N=-2.1
N=2.1/0.35=6
Q=17-2(6)=5
D=19-5-6=8
Answer:
N=6,D=8,Q=5
Answer by jojo14344(1513)  (Show Source):
You can put this solution on YOUR website! You're on the right track amigo, and we'll help you along the way.
To continue, we mark the following eqns okay,
> There are 2 more dimes than nickels,  -------> eqn 1
> Since the total coins is 19, so  -------> eqn 2
> Then, when we multiply each coin to their value and add them up we get $2.35,
to show,  ---------> eqn 3
.
We substitute "D" from eqn 1 to eqn 2 we get,  , and from this eqn we get  ----> eqn 4
Now we substitute value of "D" from eqn 1, also of "Q" from eqn 4 to eqn 3:
 -------------->  , # of nickels
For # of dimes go back eqn 1,
D=6+2= 8
For # of quarters go back eqn 2,
Q=19-8-6= 5
In doubt? go back to eqn 3,
Thank you,
Jojo
|
|
|
