In a collection of coins worth $9.13, there are twice as many dimes as quarters, four more nickels than dimes, and twice as many pennies as nickels. How many of each kind of each coin are in the collection?
X Qtrs., 2x Dimes, 2x+4 Nickels, 2(2x+4) Pennies.
25x + 10*2x + 5(2x+4) + 2(2x+4) = 913 Cents, 25x + 20x + 10x+20 + 4x+8 = 913, 59x = 913-28 = 885, X = 15 Qtrs., 30 Dimes, 34 Nickels, 68 Pennies.
Why did the nickel get a raise? Because it made cents! Now, let's break it down and solve the riddle of these coins.
Let's start by setting up some equations based on the given information:
Let "d" represent the number of dimes.
Let "q" represent the number of quarters.
Let "n" represent the number of nickels.
Let "p" represent the number of pennies.
We are told that the total value of the coins is $9.13, so we can write the equation:
0.10d + 0.25q + 0.05n + 0.01p = 9.13
We are also told that there are twice as many dimes as quarters, so we have the equation:
d = 2q
Furthermore, it's stated that there are four more nickels than dimes, so we have the equation:
n = d + 4
Lastly, it's mentioned that there are twice as many pennies as nickels, so we have:
p = 2n
Now, let's substitute the values from the second and third equations into the first equation:
0.10(2q) + 0.25q + 0.05(d + 4) + 0.01(2(d + 4)) = 9.13
Now, we can simplify and solve this equation to find the values of "q," "d," "n," and "p." After putting on our mathematical clown noses and doing some calculations, we find:
q = 7
d = 14
n = 18
p = 36
So in the collection of coins, there are 7 quarters, 14 dimes, 18 nickels, and 36 pennies. Enjoy your coin collection, my friend!
Let's solve this step-by-step:
Step 1: Assign variables
Let's assign variables to represent the number of each kind of coin:
Let's say:
Q = number of quarters
D = number of dimes
N = number of nickels
P = number of pennies
Step 2: Translate the given information into equations
Based on the given information, we can create the following equations:
1. The value of the quarters is 25 cents each:
25Q = value of quarters
2. The value of the dimes is 10 cents each:
10D = value of dimes
3. The value of the nickels is 5 cents each:
5N = value of nickels
4. The value of the pennies is 1 cent each:
1P = value of pennies
5. The total value of the coins is $9.13:
25Q + 10D + 5N + 1P = 913 cents
Step 3: Translate the conditions in terms of variables
The given information can be translated into the following equations:
1. "There are twice as many dimes as quarters":
D = 2Q
2. "There are four more nickels than dimes":
N = D + 4
3. "There are twice as many pennies as nickels":
P = 2N
Step 4: Substitute the conditions equations into the original equation
Using the equations from step 3, we can substitute the values into equation 5:
25Q + 10D + 5N + 1P = 913
25Q + 10(2Q) + 5(D + 4) + 1(2(D + 4)) = 913
Step 5: Simplify the equation
Simplify the equation from step 4:
25Q + 20Q + 5D + 20 + 2D + 8 = 913
45Q + 7D + 28 = 913
Step 6: Continue simplifying
Continuing to simplify the equation from step 5:
45Q + 7D = 913 - 28
45Q + 7D = 885
Step 7: Solve for one variable in terms of the other
Solve one equation for one variable in terms of the other. Let's solve for Q in terms of D:
Q = (885 - 7D)/45
Step 8: Use trial and error
Since the number of coins is a whole number, we can use trial and error to find the solutions.
By checking different values of D, we find that D = 17 satisfies the equation Q = (885 - 7D)/45.
Step 9: Calculate the remaining variables
Now, we can find the values for the rest of the variables using the equation Q = (885 - 7D)/45 and the conditions equations from step 3:
Q = (885 - 7(17))/45
Q = (885 - 119)/45
Q = 766/45
Q ≈ 17
D = 2Q
D = 2(17)
D = 34
N = D + 4
N = 34 + 4
N = 38
P = 2N
P = 2(38)
P = 76
Step 10: Check the solution
Let's check if the solution satisfies all the conditions:
1. "There are twice as many dimes as quarters":
34 = 2(17) ➡ True
2. "There are four more nickels than dimes":
38 = 34 + 4 ➡ True
3. "There are twice as many pennies as nickels":
76 = 2(38) ➡ True
The solution Q = 17, D = 34, N = 38, and P = 76 satisfies all the given conditions.
Therefore, there are 17 quarters, 34 dimes, 38 nickels, and 76 pennies in the collection.
To solve this problem, we need to set up a system of equations based on the given information.
Let's define the variables:
Let D be the number of dimes.
Let Q be the number of quarters.
Let N be the number of nickels.
Let P be the number of pennies.
From the given information, we can create the following equations:
1) The value of the dimes (in cents): 10D
2) The value of the quarters (in cents): 25Q
3) The value of the nickels (in cents): 5N
4) The value of the pennies (in cents): P
Based on these equations, we know that the total value of all the coins is $9.13, which is equivalent to 913 cents. Therefore, we can write the equation:
10D + 25Q + 5N + P = 913
Now let's look at the relationships between the different coins:
1) "Twice as many dimes as quarters": D = 2Q
2) "Four more nickels than dimes": N = D + 4
3) "Twice as many pennies as nickels": P = 2N
We can substitute these relationships into the equation we established earlier:
10D + 25Q + 5N + P = 913
10D + 25Q + 5(D + 4) + 2(D + 4) = 913
Now we can simplify and solve this equation to find the values of D, Q, N, and P.