System of Equations Word Problem

Question

A change machine contains nickels, dimes, and quarters. There are a total of 99 coins in the machine, and the value of those coins totals $12.40. There are five times as many dimes as nickels in the machine. Set-up and solve a system of equations to find the number of each type of coin in the machine.

Answer 1

Here's what we know:

n + d + q = 99 (there are 99 coins total)

.05n + .1d + .25q = 12.40 (the dollar amount is $12.25)

d = 5n ((there are 5 times as many dimes as nickels)

 

We can start by putting in our variable value:

n + 5n + q = 99

6n + q = 99

 

.05n + .1(5n) + .25q = 12.40

.05n + .5n + .25q = 12.40

.55n + .25q = 12.40

 

Does this now look familiar to previous problems? Anyways, we can isolate q and use that to solve via substitution:

q = 99 - 6n

 

.55n + .25(99 - 6n) = 12.40

.55n + 24.75 - 1.5n = 12.40

-.95n + 24.75 = 12.40

-.95n = -12.35

n = 13

 

d = 5 * 13

d = 65

 

q = 99 - 6 * 13

q = 99 - 78

q = 21

 

check:

13 + 65 + 21 = 99

99 = 99

 

.05 * 13 + .1 * 65 + .25 * 21 = 12.4

.65 + 6.5 + 5.25 = 12.4

12.4 = 12.4