Suppose there is a pile of quaters, dimes, an pennies with a total is $1.06,
How much of each coin can be present without being able to change for a dollar? If there are multiple selection of coins that will work
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Would you rather have $200 base 10 or $1000 base 5? What would be the base for the $200 do that it would be the better of the two choices
To answer this question, we can set up a system of equations to represent the given scenario. Let's assume the number of quarters is q, the number of dimes is d, and the number of pennies is p.
1. The first equation represents the total value of the coins:
0.25q + 0.10d + 0.01p = 1.06
2. The second equation represents the total number of coins:
q + d + p = ?
Since the question asks for the maximum amount of each coin without being able to change for a dollar, we want to find the largest possible values for q, d, and p.
Let's solve this step-by-step:
Step 1: Multiply equation 1 by 100 to get rid of decimals:
25q + 10d + p = 106
Step 2: Subtract equation 2 from equation 1:
25q + 10d + p - (q + d + p) = 106 - ?
Simplifying:
24q + 9d = 106 - ?
From this point, we need more information about the total number of coins (equation 2). Without knowing the exact number, we cannot determine the specific values for q, d, and p that would make it impossible to change for a dollar.
To find out how much of each coin can be present without being able to change for a dollar, we can use a systematic approach. Let's assume the number of quarters, dimes, and pennies as follows:
Let's denote the number of quarters as Q, dimes as D, and pennies as P.
1. First, let's set up the equations based on the given information:
- For the number of quarters, the value is $0.25, so the equation is: 0.25Q
- For the number of dimes, the value is $0.10, so the equation is: 0.10D
- For the number of pennies, the value is $0.01, so the equation is: 0.01P
2. The total value of the coins is given as $1.06, so the equation becomes:
0.25Q + 0.10D + 0.01P = 1.06
3. The problem is asking for the number of each coin that can be present without being able to change for a dollar, which means the total value must be less than or equal to $1.00.
4. Now, we need to find integer solutions for Q, D, and P that satisfy the equation mentioned earlier and have a total value less than or equal to $1.00.
We can use trial and error or a more systematic approach, such as using loops or a table, to check different combinations of coins until we find a solution(s) that meets the criteria.
Here is one possible solution:
Q = 1 (1 quarter)
D = 4 (4 dimes)
P = 6 (6 pennies)
If we substitute these values into the equation:
0.25(1) + 0.10(4) + 0.01(6) = 0.25 + 0.40 + 0.06 = 0.71
We can see that the total value is less than $1.00 (specifically, $0.71).
So, one possible combination is 1 quarter, 4 dimes, and 6 pennies, where the total value is $0.71, and it cannot be changed for a dollar.