ou find 38 coins consisting only of nickels, dimes, and quarters, with a face value of $4.75. However, the coins all date from 1926, and are worth considerably more than their face value. A coin dealer offers you $8 for each nickel, $6 for each dime, and $22 for each quarter, for a total of $410. How many dimes did you find? Type in your numerical answer only; do not type any words or letters with your answer.

17

Step-by-step explanation:

First of all let us assume the following:

Let the number of nickels = n

Let the number of dimes = d

Let the number of quarters = q

Note also the following conversion factors:

1 nickel = $0.05

1 dime = $0.1

1 quarter = $0.25

we are given the following information:

n + d + q = 38. . . . (1)   (38 coins consisting only of nickels, dimes, and quarters)

0.05n + 0.1d + 0.25q = 4.75 ( we can remove decimals by multiplying through by 100 to give)

5n + 10d + 25q = 475 . . . . (2) note that equation 2 can also be simplified by dividing through by 5, to give the following)

n + 2d + 5q = 95 . . . . . (2)

8n + 6d + 22q = 410 . . . . (3)

Next we will look for a way to eliminate two of the unknowns in each equation to have just one unknown that we can use to solve.

we have three equations

n + d + q = 38. . . . . . . . (1)

n + 2d + 5q = 95 . . . . . (2)

8n + 6d + 22q = 410 . . .(3)

Next we will look for a way to eliminate two of the unknowns in each equation to have just one unknown that we can use to solve.

we can eliminate n by subtracting equation (1) from (2) {eq (2) - eq (1)}

(n - n) + (2d - d) + (5q - q ) = (95 - 38)

= d + 4q = 57 . . . . . . (3)

Next we can multiply eqn (1) through by 8, to make it look like eqn (3), the we can subtract eqn (1) from eqn (3) as follows:

eqn (1) × 8 = 8 × (n + d + q = 38)

8n + 8d + 8q = 304 . . . . (1b) ; subtracting this from eqn (3) we have:

(8n - 8n) + (6d - 8d) + (22q - 8q) = (410 - 304)

= -2d + 14q = 106 . . . . (4)

Now, we have two simplified equations that we can work with:

d + 4q = 57 . . . . . . (3)

-2d + 14q = 106 . . . . (4)

from eqn (3), making 'd' the subject of the formular:

d = 57 - 4q ; replacing the value of d in eqn 4 with this value, we have:

-2(57 - 4q) + 14q = 106

-114 + 8q + 14q = 106

-114 + 22q = 106

22q = 106 + 114 = 220

22q = 220

∴ q = 220 ÷ 22 = 10

∴ q = 10

Next let us solve for 'd' by replacing the value of q with 10 in eqn (3)

d + 4q = 57 . . . . . . (3)    (but q = 10)

d + 4(10) = 57

d + 40 = 57

d = 57 - 40 = 17

d = 17

finally to solve for n, let us input the values of q and d into eqn (1)

n + d + q = 38. . . . (1)  (q = 10 ; d = 17)

n + 17 + 10 = 38

n + 27 = 38

∴ n = 38 - 27 = 11

Therefore the number of each coin are as follows:

nickel = 11

dime = 17

quarters = 10

the solution is quite lengthy, but if you don't understand it at first look, please go through it again.