Mary has nickels, dimes and quarters worth $12:85. If the number of dimes and the number of quarters is more than twice the number of dimes, how many more dimes are there.
Well, well, well! It seems Mary has been busy playing with her coins. Let's use our mathematical clown calculator to solve this puzzle!
Let's assume the number of dimes is "x", the number of quarters is "y", and the number of nickels will be "N". We are given that the value of all these coins combines to $12.85.
So, we can set up an equation:
0.10x + 0.25y + 0.05N = 12.85
Now, we also know that the number of dimes and the number of quarters combined is more than twice the number of dimes. In other words,
y + x > 2x
Well, we clowns love to simplify things, so let's rewrite this inequality:
y > x
Now, let's solve this clownish puzzle and find out how many more dimes there are!
But, before we continue, a little clown joke to lighten the mood:
Why did the clown go to school?
He wanted to improve on his whacky formulas!
Now, back to business. Let's simplify the equations and solve them:
0.10x + 0.25y + 0.05N = 12.85
0.05N = 12.85 - 0.10x - 0.25y
Since we know that y > x, we can assume y = x + k, where k is a positive integer representing the number of additional dimes.
Substituting y = x + k into the equation, we have:
0.05N = 12.85 - 0.10x - 0.25(x + k)
Simplifying further:
0.05N = 12.85 - 0.10x - 0.25x - 0.25k
0.05N = 12.85 - 0.35x - 0.25k
Now, we need to admit something. Clowns are not always great with numbers (we prefer making balloon animals instead). So, we'll need a little help from a human mathematician to solve this equation.
But hey, here's a clown riddle to keep you entertained while the mathematician crunches the numbers:
Why don't scientists trust atoms?
Because they make up everything!
Once the mathematician has solved the equation, they'll be able to determine how many more dimes there are.
But remember, laughter is the best medicine, so keep smiling and enjoying those silly coin puzzles!
Let's assume the number of nickels is x, the number of dimes is y, and the number of quarters is z.
The value of x nickels is 5x cents.
The value of y dimes is 10y cents.
The value of z quarters is 25z cents.
According to the given information, we know that the total value of all the coins is $12.85, which is equivalent to 1285 cents.
So, we can now set up an equation based on the given information:
5x + 10y + 25z = 1285 ----(1)
We are also given that the number of dimes (y) and quarters (z) is more than twice the number of dimes:
y + z > 2y ----(2)
Now, let's solve for y in equation (2):
z > 2y - y
z > y
From equation (2), we can see that z is always greater than y.
To find the number of dimes that are more, we need to find the difference between the number of dimes and quarters: z - y.
However, we still need to solve the system of equations to find the actual values of x, y, and z.
To solve this problem, let's use algebraic equations and create a system of equations.
Let's represent the number of nickels as 'n', the number of dimes as 'd', and the number of quarters as 'q'.
We can start by writing the equations based on the given information:
1. The total value of the coins is $12.85:
0.05n + 0.10d + 0.25q = 12.85
2. The number of dimes and quarters combined is more than twice the number of dimes:
q + d > 2d
Now, let's simplify the second equation:
q > d
We also know that the total value of the coins is in cents. Therefore, we can multiply the total value of each coin by 100 to get the equation without decimals:
5n + 10d + 25q = 1285 (equation 1)
q > d (equation 2)
To find the number of additional dimes, we need to determine the values of 'n,' 'd,' and 'q.' Let's solve the system of equations using logical elimination:
From equation 2, we know that 'q' is greater than 'd.' We'll start by assuming that 'q' is equal to 'd+1.'
q = d + 1
We'll substitute this value in equation 1:
5n + 10d + 25(d+1) = 1285
5n + 10d + 25d + 25 = 1285
5n + 35d + 25 = 1285
5n + 35d = 1260 (equation 3)
Now, we'll look for a combination of 'n' and 'd' that satisfies equation 3. We can start with the assumption that there are no nickels (n = 0):
35d = 1260
d = 36
Now, let's substitute the values of 'd' and 'n' into equation 1 to find 'q':
5(0) + 10(36) + 25q = 1285
360 + 25q = 1285
25q = 925
q = 37
So, when there are no nickels (n = 0), dimes (d) = 36, and quarters (q) = 37.
To find the number of additional dimes, we can subtract the assumed value of dimes (36) from the actual value of dimes (d):
Additional dimes = Actual dimes - Assumed dimes
Additional dimes = 36 - 36
Additional dimes = 0
Therefore, there are no additional dimes.
In conclusion, there are no additional dimes when the number of nickels is assumed to be 0.
5n + 10d + 25q = 12.85
d + q > 2d