Question

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?

Answer 1

44 nickels and 22 dimes

Answer 2

follow my example in
http://www.jiskha.com/display.cgi?id=1441601720

Answer 3

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

Answer 4

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