How many of each coin does she have?

Question

Alice has a total of 29 coins composed of only nickels, dimes and quarters. She is one more nickel than the sum of the number of the other two coins. if the total value of the coins is $3.35, how many of each type of coin does she have? Right and solve a system of equations, state your answer in a sentence.

Answer 1

Let n, d, and q represent the number of nickels, dimes, and quarters.

n + d + q = 29 and n + 1 = d + q ⇒ n = d + q - 1 are two equations to start with.

(d + q - 1) + d + q = 29 ⇒ 2d + 2q = 28 ⇒ d + q = 14, so n = 15.

You know you have 15 nickels, and each nickel is 5 cents, a dime is 10 cents, and a quarter is 25 cents,• so we have 15•5 + d•10 + q•25 = 335, or 10d + 25q = 260.

We also know that d + q = 14 ⇒ d = 14 - q. Plugging this in to the previous equation we get

(14 - q)•10 + 25q = 260 ⇒ 140 - 10q + 25q = 260 ⇒ 15q = 140 ⇒ q = 8. Since d = 14 - q, d = 6.

So we have 8 quarters, 6 dimes, and 15 nickels. 8•25 + 6•10 + 15•5 = 200 + 60 + 75 = 335.

Answer 2

Hi Lucy M.

15 nickels, 6 dimes and 8 quarters

Total number of coins 29

Total value of coins $.3.35

Let nickels = x

And we given that nickels is 1 more than the sum of the other two coins

x = 1 + y + z

Let dimes = y

Let quarters = z

Equation 1 for the total number of coins

x + y + z = 29

Equation 2 for the total value

0.05x + 0.10y + 0.25z = 3.35

In Equation 1 we can make the given substitution for x

(1 + y + z) + y + z = 29

1 + y + z + y + z =29

2y + 2z + 1 = 29

2y + 2z = 29 - 1

2y + 2z = 28

Divide both sides of the equation by 2

y + z = 14

Since

x = 1 + y + z = 1 + 14

x = 15

Substitute this value for x in Equation 2

0.05(15) + 0.10y + 0.25z = 3.35

0.75 + 0.10y + 0.25z = 3.35

0.10y + 0.25z = 3.35 -0.75

0.10y + 0.25z = 2.60

Now we have two equations in two unknowns y and z

y + z = 14

0.10y + 0.25z = 2.60

Multiply (0.10y + 0.25z = 2.60) by negative 4 and combine it with y + z = 14 to eliminate z

-0.40y - z = -10.4

y + z = 14

This leaves

0.60y = 3.60

y = 6

Substitute this value into the original Equation 1 to find z

x + y + z = 29

15 + 6 + z = 29

21 + z = 29

z = 29 - 21

z = 8

We check all these values in the original Equation 2

0.05(15) + 0.10(6) + 0.25(8) = 3.35

0.75 + 0.60 + 2.00 = 3.35

And we check in the original Equation 1

x + y + z = 29

15 + 6 + 8 = 29

29 = 29

Answer 3

Alice has a total of 29 coins composed of only nickels, dimes and quarters. She is one more nickel than the sum of the number of the other two coins. if the total value of the coins is $3.35, how many of each type of coin does she have? Right and solve a system of equations, state your answer in a sentence.

Let N = # of nickels, D = # of dimes, and Q = # of quarters.

(1) Alice has a total of 29 coins composed of only nickels, dimes and quarters: --> N + D + Q = 29

(2) She is one more nickel than the sum of the number of the other two coins--> N = (D + Q) + 1

if the total value of the coins is $3.35, how many of each type of coin does she have?

(3) Since total = $3.35, it follows that 0.05N + 0.1D + 0.25Q = 3.35

Using (1) and (2) via substitution, (D + Q + 1) + D + Q = 29 --> 2D + 2Q + 1 = 29 or 2(D+Q) = 28

Therefore, (D + Q) = 14 (4)

Using (2) and (3) via substitution, 0.05 (D+Q+1) + 0.1(D) + 0.25Q = 3.35

0.05D + 0.05Q + 0.05 + 0.1D + 0.25Q = 3.35

0.15D + 0.3Q = 3.3 (5)

Solving for (4) and (5):

0.15*(4): D + Q = 14--> 0.15D + 0.15Q = 2.1

(5): 0.15D + 0.3Q = 3.3

Subtracting 0.15*(4) from (5): 0.15Q = 1.2 or Q = 8. Therefore, (D+Q) = 14-->(D +8) = 14--> D = 6

Since N + D + Q = 29, then N + 8 + 6 = 29--> N = 29 -(14) = 15.

Answer: Alice has fifteen nickels, six dimes, and eight quarters.

Check: 15*0.05 + 6*.1 + 8*.25 = 3.35