Barbara has $4.40 worth of change in nickels and dimes. If she has 2 times as many nickels as dimes, how many of each type of coin does she have?
44 nickels and 22 dimes
follow my example in
http://www.jiskha.com/display.cgi?id=1441601720
let N = # of Nickels
let D = # of Dimes
Take Note: 1N = .20 dollar and
1D = ,10 dollar k?
now
(# of coins) x (Amt$ per coin) = Amt$
also Amt$ of Nickel + Amt$ of Dimes = TotalAmt$
and Amt$ of Nickel = 0.2 x N = 0.2N
also Amt$ of dimes = 0.1 x D = 0.1D
therefore:
.2N + .1 D = 4.40 (let this be our Eqn**1)...
but as stated in the Problem, that 2 times as many nickels as dimes, and in equation form, we get:
2 x N = D or 2N=D (let this be our Eqn**2.
Now Substitute Eqn**2 into Eqn**1
so that we have:
.2N + .1 x (2N) = 4.40
then we have now a new equation with only one variable, N:
therefore, reducing the new eqn, it becomes:
.2N + .2N = 4.40
so that .4N =4.40
Solving for N, we get
N= 4.40/.4
N=11. (meaning the quantity is 11 Nickel coins)
and solving for D: (use Eqn**2)
2N=D so, 2(11)=D and D= 22
therefore, 22 Dime coins
Now Check: (using Eqn **1)
.2(11) + .1(22)
= 2.2 + 2.2 = 4.40
Checked!
therefore,
There are 11 Nickel Coins and
there are 22 Dime Coins!
QED~! Yhehyyy Solved!
fro Teacher Jon
let N = # of Nickels
let D = # of Dimes
Take Note: 1N = .20 dollar and
1D = ,10 dollar k?
now
(# of coins) x (Amt$ per coin) = Amt$
also Amt$ of Nickel + Amt$ of Dimes = TotalAmt$
and Amt$ of Nickel = 0.2 x N = 0.2N
also Amt$ of dimes = 0.1 x D = 0.1D
therefore:
.2N + .1 D = 4.40 (let this be our Eqn**1)...
but as stated in the Problem, that 2 times as many nickels as dimes, and in equation form, we get:
2 x N = D or 2N=D (let this be our Eqn**2.
Now Substitute Eqn**2 into Eqn**1
so that we have:
.2N + .1 x (2N) = 4.40
then we have now a new equation with only one variable, N:
therefore, reducing the new eqn, it becomes:
.2N + .2N = 4.40
so that .4N =4.40
Solving for N, we get
N= 4.40/.4
N=11. (meaning the quantity is 11 Nickel coins)
and solving for D: (use Eqn**2)
2N=D so, 2(11)=D and D= 22
therefore, 22 Dime coins
Now Check: (using Eqn **1)
.2(11) + .1(22)
= 2.2 + 2.2 = 4.40
Checked!
therefore,
There are 11 Nickel Coins and
there are 22 Dime Coins!
QED~! Yhehyyy Solved!
fro Teacher JonSihay
To solve this problem, let's assign variables to the unknowns. Let's say the number of dimes Barbara has is D, and the number of nickels she has is N.
We are given two pieces of information:
1. Barbara has $4.40 worth of change.
2. She has 2 times as many nickels as dimes.
The value of a nickel is $0.05, while the value of a dime is $0.10.
Using the information provided, we can set up two equations:
Equation 1: The total value of the nickels and dimes is $4.40.
0.05N + 0.10D = 4.40
Equation 2: Barbara has 2 times as many nickels as dimes.
N = 2D
Now, we have a system of equations. We can substitute the value of N from Equation 2 into Equation 1:
0.05(2D) + 0.10D = 4.40
0.10D + 0.10D = 4.40
0.20D = 4.40
D = 4.40 / 0.20
D = 22
Now that we know D (the number of dimes), we can substitute this value back into Equation 2 to find N (the number of nickels):
N = 2D
N = 2(22)
N = 44
So Barbara has 22 dimes and 44 nickels.