Qestion in description

Question

A jar contains nickels, dimes, and quarters. There are 105 coins with a total value of $8.40. If there are 3 more than twice as many dimes as quarters, find how many of each coin are in the jar.

Answer 1

Let N = number of nickels

Let D = number of dimes

Let Q = number of quarters

 

There are 105 coins  ==>  N + D + Q = 105

 

The total value is $8.40  ==>  .05N + .10D + .25Q = 8.40

 

There are 3 more than twice as many dimes as quarters  ==>  D = 2Q + 3 

 

Substitute the value for D from the 3rd equation in to the 1st equation to eliminate D and solve for N 

 

N + (2Q + 3) + Q = 105

N + 3Q + 3 = 105

N = 102 - 3Q

 

Substitute this value for N and the value for D in to the 2nd equation and solve for Q

 

.05(102 - 3Q) + .10(2Q + 3) + .25Q = 8.40

5.10 -.15Q + .20Q + .30 + .25Q = 8.40

5.40 +.30Q = 8.40

.30Q = 3.00

Q = 3.00 / .30 = 10

 

So D = 2Q + 3 = 23

and N = 102 - 3Q = 72

 

So there are 72 nickels, 23 dimes and 10 quarters 

 

Answer 2

That was a great problem to solve.