There are 27 nickels, dimes, and quarters in the drawer with a value of $3. How many coins of each type are there if there are three times as many nickels as there are dimes?

Answer 1

Let n = number of nickels

Let d = number of dimes

let q = number of quarters

n + d + q = 27

Each nickel is 5 cents, dime is 10 cents and quarter is 25 cents so if you have 3 quarters it's 3*25 = 75 cents and $3 = 300 cents

So we can put together another equation in those terms

5n + 10d + 25q = 300

We also know that there are 3x as many nickels as there are dimes so 3d = n, and we can substitute that into our equations and solve:

3d + d + q = 27 ⇒ 4d + q = 27, q = 27 - 4d

5(3d) + 10d + 25(27-4d) = 300

3d + 2d + 5(27-4d) = 60

5d + 135 - 20d = 60

75 = 15d

5 = d

So we know there are 5 dimes.

Since n = 3d, n = 15, hence there are 15 nickels.

Putting those into the original equation

n + d + q = 27

15 + 5 + q = 27

q = 7

So there are 7 quarters.

15 nickels, 5 dimes and 7 quarters

Checking:

15*5 + 5*10 + 7*25 = 300

75 + 50 + 175 = 300

300 = 300 check.

Answer 2

15 Nickels, 10 Dimes and 7 Quarters

that sums to 27 coins

5(15) + 10(5) + 25(7) = 300= $3

75+50 + 175 =300

5N+10D+25Q =300 divide by 5

N+2D +5Q = 60

N+ D + Q =27, subtract

D+ 4Q = 33

N=3D

N+2D +5Q =60

3D +2D +5Q =60

5D +5Q =60

D + Q = 12

D+4Q = 33

D +Q =12

3Q = 21

Q=7

D=12-7 = 5

N=3(5) = 15