Adam has dimes in quarters in his bucket there are 28 points at all and they are worth $4.60 altogether how many of each coin does he have

Answer 1

This problem can be represented as a system of equations. Since we know the total number of coins, we can say that the number of dimes plus the number of quarters equals the total number of coins or

D + Q = 28

Since we know the total value of the coins, we know that the value of a dime times the number of dimes we have plus the value of a quarter times the number of quarters we have will equal this total value so

0.10D + 0.25Q = 4.60

Thus, the system of equations is:

D + Q = 28

0.10D + 0.25Q = 4.60

We can solve this system by elimination or substitution. With the decimals in the second equation, it may be in our interest to use elimination to get rid of the decimals and only deal with whole numbers. We want to get rid of one of the variables so we will multiply the first equation by 10 and the second equation by 100:

10(D + Q = 28)

100(0.10D + 0.25Q = 4.60)

This gives us:

10D + 10Q = 280

10D + 25Q = 460

Note that both equations now have 10D so subtracting the two equations will eliminate that variable leaving us with an equation with one variable:

10D + 10Q = 280

- 10D + 25Q = 460

-15Q = -180

We can cancel the negative sign on each side to get:

15Q = 180

and now we can simplify the equation to solve for Q:

Q = 12

We now know that Q = 12 is the solution for Q that solves both equations, we can now plug this value in for Q in either equation to solve for D, let's use the first equation as it does not involve any decimals:

D + 12 = 28

D = 16

We now know that we have 12 quarters and 16 dimes. To ensure that this is the proper solution, we can plug both values into the other equation and make sure our answer is correct:

0.10(16) + 0.25(12) = 1.60 + 3.00 = 4.60

Since our answers for both variables satisfy both equations we know for certain that our answer was correct. There are 12 quarters and 16 dimes in the bucket.