How many of each type of coin does Jamie have?

Question

Jamie has a collection of 78 nickels,dimes,and quaters worth $12.40.If the number of quaters is doubled,the value becomes $22.15.

Answer 1

How many of each type of coin does Jamie have? Jamie has a collection of 78 nickels,dimes,and quarters worth $12.40.If the number of quarters is doubled,the value becomes $22.15.

Detailed Solution:

Given/Known: nickels = n, dimes = d, quarters = q

n + d + q = 78 ==> Equation 1

0.05n + 0.1d + 0.25q = 12.40 ==> Equation 2

0.05n + 0.1d + 0.25(2q) = 22.15 <== (Double the quarters to get $22.15)

0.05n + 0.1d + 0.5q = 22.15 ==> Equation 3

In TI-84 Plus/TI-83 Plus:

1) Create (3 x 4) Matrix A and enter the coefficients of Equation 1, Equation 2, Equation 3 (Steps: 2nd Matrix ==> Edit ==> 3 x 4 ==> enter the coefficients)

2) Solve the Matrix: rref(A) (Steps: 2nd Matrix ==> Math ==> rref(A))

Solution: Numbers of: (nickels, dimes, quarters) = (n, d, q) = (25, 14, 39) <== Final Answer

Check the three Equations:

Equation 1: n + d + q = 78 ==> 25 + 14 + 39 = 78

Equation 2: 0.05n + 0.1d + 0.25q = 12.40 ==> 0.05(25) + 0.1(14) + 0.25(39) = 12.40

Equation 3: 0.05n + 0.1d + 0.5q = 22.15 ==> 0.05(25) + 0.1(14) + 0.5(39) = 22.15

Answer 2

n + d + q = 78 (Eq. 1)

5n + 10d + 25q = 1240 (Eq. 2)

5n + 10d + 2(25q) = 2215 (Eq. 3)

5n + 10d + 50q = 2215 (Eq. 3 expanded)

Subtracting equation 2 from equation 3,

25q = 975

q = 39

Substituting into equations 1 and 2,

(1)

n + d + 39 = 78

n + d = 39

d = 39 - n

(2)

5n + 10d + 25(39) = 1240

5n + 10d + 975 = 1240

5n + 10d = 265

Substituting from eq 1 into eq 2

5n + 10(39 - n) = 265

390 - 5n = 265

5n = 125

n = 25

d = 39 - n

d = 39 - 25

d = 14

n = 25

d = 14

q = 39 (doubled for the second amount to 78)

Answer 3

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

n+d+q=78

.05n+.10d+.25q = 12.40

if doubling the number of quarters gives a total of $22.15, then the increase = 22.15-12.40 = 9.75

.25q = 9.75

q = 39

n+d+39=78 --> n+d = 39 --> d = 39-n

Now

.05n + .10d + .25(39) = 12.40

.05n + .10d = 2.65

Substituting for d into this equation

.05n+.10(39-n) = 2.65

.05n + 3.90 - .10n = 2.65

-.05n = -1.25

n = 25

d = 39-25 = 14

Originally, Jamie had 25 nickels, 14 dimes and 39 quarters for a total of 78 coins.