nestor has nickels, dimes, and quarters amounting to $1.85. if he has twice as many dimes as quarters, and the number of nickels is two less than twice the number of dimes, how many quarters does he have?

5 n + 10 d + 25 q = 185

d = 2 q
n = 2 d -2 = 4 q - 2

so
5 (4q-2) + 10(2q) + 25q = 185

ok from there?

1.85= 0.05∗(4q−2)+0.25q+ 0.10∗2q1.85= 0.05∗(4q−2)+0.25q+ 0.10∗2q

1.85= 0.2q−0.1+0.25q+ 0.2q1.85= 0.2q−0.1+0.25q+ 0.2q
1.85+0.1= 0.65q1.85+0.1= 0.65q
1.95= 0.65q1.95= 0.65q
3= q

Well, let's solve this problem one giggle at a time, shall we? Since we don't know the exact number of quarters, let's represent it with a hilarious variable like "Q" (which also happens to be the letter that likes to skip the end of the line at school).

Now, we know that Nestor has twice as many dimes as quarters. So, if we let the number of quarters be "Q," then the number of dimes would be 2Q (cue dramatic drumroll).

Moving on, the number of nickels is two less than twice the number of dimes. Let's calculate that with a witty equation! If the number of dimes is 2Q, then the number of nickels would be (2Q x 2) - 2, which simplifies to 4Q - 2.

Now, time to add up the values! A nickel is worth 5 cents, a dime is worth 10 cents, and a quarter is worth 25 cents. So, we can write another joyful equation: 5(4Q - 2) + 10(2Q) + 25(Q) = 185 cents.

Simplifying that equation (and avoiding any summertime sadness), we get 20Q - 10 + 20Q + 25Q = 185. Combining like terms and ridding the equation of any mathematical gloom, we find 65Q - 10 = 185.

Solving for Q (the quarters I mean, not the famous British spy), we find Q = 3. So, Nestor has 3 quarters!

I hope this mathematical circus brought a smile to your face. If you have any more questions, keep 'em coming!

Let's assign variables to the unknown quantities:

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

According to the given information, we can form the following equations:

1. The total value of the coins is $1.85:
0.05N + 0.10D + 0.25Q = 1.85 (since each nickel is worth $0.05, each dime is worth $0.10, and each quarter is worth $0.25)

2. Nestor has twice as many dimes as quarters:
D = 2Q

3. The number of nickels is two less than twice the number of dimes:
N = 2D - 2

To solve the problem, we'll substitute the values from the second and third equations into the first equation:

0.05(2D - 2) + 0.10D + 0.25Q = 1.85

Now we can solve for Q.

To solve this problem, we can set up a system of equations based on the given information.

Let's represent:
- the number of nickels as "n",
- the number of dimes as "d", and
- the number of quarters as "q".

1. We know that the total value of all the coins is $1.85. So, we can write the first equation as:
0.05n + 0.10d + 0.25q = 1.85

2. The problem states that Nestor has twice as many dimes as quarters. We can express this relationship in an equation as:
d = 2q

3. It is also stated that the number of nickels is two less than twice the number of dimes. We can write this as:
n = 2d - 2

Now, let's solve the system of equations to find the value of "q" (the number of quarters).

Since we have two equations in terms of "d", we can substitute one equation into the other:

n = 2(2q) - 2
n = 4q - 2

Now, we'll substitute the value of "n" in equation 1:

0.05(4q - 2) + 0.10d + 0.25q = 1.85

Simplifying the equation:

0.20q - 0.10 + 0.10d + 0.25q = 1.85
0.45q + 0.10d = 1.95

Now, let's substitute the value of "d" from equation 2 into the equation above:

0.45q + 0.10(2q) = 1.95
0.45q + 0.20q = 1.95
0.65q = 1.95
q = 1.95 / 0.65
q = 3

Therefore, Nestor has 3 quarters.