How do you solve this math word problem?

Question

You find 38 coins consisting only of nickels, dimes and quaters, 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 quater, for a total of $410. How many dimes did you find?

Answer 1

Assign a letter variable to the number of each type of coin (e.g. N=nickle, D=dime, Q=quarter).

Then write three formulas to express the three facts stated in the problem. (I'll do the first one for you: N+D+Q=38.)

a) Next solve one equation for N. Simplify as much as possible.
b) Replace N in the next equation with the answer to (a).
c) Solve the equation in (b) for Q. Simplify as much as possible.

By now it should be easy to determine the value of one of the variables. Plug it into the other equations and you should be able to find the other two quickly. If you publish your first steps here, I'll help you along, but I will not do the whole problem for you. That would be too much like doing your homework for you.

Answer 2

You find 38 coins consisting only of nickels, dimes and quarters

> n + d + q = 38 (1)

with a face value of $4.75.

> .05n + .1d + .25q = 4.75
> Multiply by 20 for scaling.
> n + 2d + 5q = 95 (2)
> Subtract equation (1) to eliminate n's
> d + 4q = 57 (3)

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.

> 8n + 6d + 22q = 410
> Divide by 2 for scaling.
> 4n + 3d + 11q = 205 (4)

How many dimes did you find?

> Multiply equation (2) by 4 and subtract equation (4) to eliminate n terms.
> 5d + 9q = 175 (5)
> Get rid of the q terms by multiplying equation (5) by 4 and equation (3) by -9, and then summing:
> 11d = 187
> d = 17

Seventeen dimes.

Answer 3

Here is how you solve it, using the facts to make 3 equations for the three variables. After that youuse the typical reduction of the equations substitutiong to get two variables out.

Number of nickels = N
Number of dimes = D
Number of quarters = Q

Total coins N + D + Q = 38 for one equation.

Total original value 5N + 10D + 25Q = 475 for second.

Total offered value 800N + 600D + 2200Q = 41000 for third

Now just do your algebra elimination with pairs of these equations and the dimes D will come out easily.

Answer 4

Let:
a = number of nickels
b = number of dimes
c = number of quarters

Equations from the problem:

a + b + c = 38 ... (1)

0.05 a + 0.1 b + 0.25 c = 4.75 ... (2)

8a + 6b + 22c = 410
or, in simplified form:
4a + 3b + 11c = 205 ... (3)

From (1):
c = 38 - a - b ... (4)

Inserting (4) into (3):
4a + 3b + 11c = 205
4a + 3b + 11(38 - a - b) = 205
4a + 3b + 418 - 11a - 11b = 205
-7a - 8b = 205 - 418
-7a - 8b = -213
-7a = -213 + 8b
a = (213 - 8b) / 7 ... (5)

Inserting (5) into (4):
c = 38 - a - b
c = 38 - (213 - 8b) / 7 - b
c = [266 - 213 + 8b - 7b] / 7
c = (53 + b) / 7 ... (6)

Inserting (6) and (5) into (2):
0.05 a + 0.1 b + 0.25 c = 4.75
0.05[(213 - 8b) / 7] + 0.1 b + 0.25[(53 + b) / 7] = 4.75

Multiplying both side by 7:
0.05(213 - 8b) + 0.7 b + 0.25(53 + b) = 33.25
10.65 - 0.4b + 0.7b + 13.25 + 0.25b = 33.25
0.55b = 9.35
b = 17 dimes

Therefore, from (5) and (6):
a = 11 nickels
c = 10 quarters

Answer 5

take a nickles, b dimes and c quarters. Then setup three equations with three unknowns.
a+b+c=38
0.05*a+0.1*b+0.25*c=4.75
8*a+6*b+22*c=410

After solving for b obtain the numner of dimes

Answer 6

Let n,d,q = the number of those coins
38=n+q+d
4.75=0.05n+0.25q+0.1d
410=8n+22q+6d

You have 3 unknowns and 3 equations. Solve for "d"