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# find the number of nickels

A collection of dimes, nickels and quarters amounts to \$11.95. If there are 173 coins in all and there are twice as many dimes as there are quarters, find the number of nickels.

1) Write down all of the variables:

dimes = d

nickels=n

quarters=q

Then write down the facts.

d=2q (there are twice as many dimes as quarters)

d+n+q=173 (there are 173 coins in all)

d(.1)+n(.05)+q(.25) = 11.95 (the dimes, nickels and quarters amounts to \$11.95)

There are three separate equations and three variables.  You can solve them as a system of equations.  It is probably simplest to use substitution.

for:

d+n+q=173

We already know that d=2q so we can substitute 2q for d in the second equation.

2q+n+q=173

you can simplify this to

n=173-3q

Now we have substitutions for 2 of the three variable that can be plugged into the third equation.

2q=d

n=173-3q

for

d(.1)+n(.05)+q(.25) = 11.95

substitution yeilds

2q(.1)+(173-3q)(.05)+q(.25)=11.95

multiply both sides of the equation by 100 to eliminate the decimal.

20q+(173-3q)(5)+q(25)=1195

20q+(865-15q)+q(25)=1195

30q+865=1195

30q=330

q = 11

now take your value for q and find the values of n and d

d=2q

d=2(11)

d=22

d+n+q=173

22+n+11=173

n=140

number of dimes=22

number of quarters=11

number of nickels=140

d(.1)+n(.05)+q(.25) = 11.95

22(.1)+140(.05)+11(.25)=11.95

2.2+7+2.75=11.95

11.95=11.95

We can reasonably assume that we are correct in saying that there are 140 nickels.

Q= # quarters; D= #of dimes; N= # of nickels

Set up the following simultaneous equations:

1.      Q +      D +     N = 173  (number of coins)

2.  .25Q + .10D + .05N = \$11.95 (value of coins)

We have a problem because there are three variables, but only two equations. Remember, to solve simultaneous equations, the number of equations must be equal to or greater than the number of variables.

There isn't enough information to enable us to create a third equation, but there is information that will allow us to reduce the number of variables from three to two. The problem tells us that there are twice as many dimes as quarters, which allows us to set up the equality D = 2Q. This allows us to substitute 2Q for D in the two equations:

1. Q + 2(Q) + N = 173 (number of coins)

2. .25Q + .10(2Q) + .05N = \$11.95 (value of coins)

This reduces to:

1. 3Q   +      N = 173 (number of coins)

2. .45Q + .05N = \$11.95 (value of coins)

We now have two equations and two variables, so we can proceed to solve the simultaneous equations. In this case, multiply the first equation by -.05, leading to the following:

1. -.15Q  - .05N = -8.65

2.   .45Q + .05N = \$11.95

.30Q = 3.3  =>  Q=11 quarters; since we know D = 2Q, then D = 2*11 = 22 dimes

Finally, we know that 173 = Q + D + N. Solving for N:  N= 173 - Q - D; N= 173 - 11- 22 = 140 nickels.

d+n+q = 173

d = 2q ......(1)

=> n = 173 - 3q ......(2)

10d + 5n + 25q = 1195 ......(3)

(1), (2) -> (3): 10(2q) + 5(173-3q) + 25q = 1195

Solve for q,

q = 11

d = 22

n = 140