Andrew has a combination of 8 quarters, dimes and nickels that add up to a value of 95 cents. How many of each coin does Andrew have?
2 quarters = $.50
3 dimes = $.30
3 nickels = $.15
So, if 2 quarters equals .50 cents, and 3 dimes equals .30 cents, and 3 nickels add up to .15 cents, add them all together. If you add: .50
+.30
+.15
--------
.95 cents
Then, there's your answer.
q=2
d=3
n=3
76 pennys and 95 nickels and 50 quarters 310 dimes
To solve this problem, we can set up a system of equations. Let's use the variables Q, D, and N to represent the number of quarters, dimes, and nickels, respectively.
Given that Andrew has a total of 8 coins, we can write the equation:
Q + D + N = 8
The value of the coins is given as 95 cents, so we can write another equation for the value:
25Q + 10D + 5N = 95
Now we have a system of two equations:
Q + D + N = 8
25Q + 10D + 5N = 95
To solve this system, we can use substitution or elimination. Let's use substitution.
From the first equation, we can rewrite it as Q = 8 - D - N, and substitute this into the second equation:
25(8 - D - N) + 10D + 5N = 95
Expanding this equation, we get:
200 - 25D - 25N + 10D + 5N = 95
Combining like terms, we have:
215 - 15D - 20N = 0
Rearranging this equation, we get:
15D + 20N = 215
Now we have a new equation. We can simplify the coefficients by dividing both sides by 5:
3D + 4N = 43
Now we have a system of equations:
Q + D + N = 8
3D + 4N = 43
To solve for the values of D and N, we can use trial and error or guess-and-check method.
Let's start with a few possible values for D and N:
D = 1, N = 10
D = 2, N = 9
D = 3, N = 8
D = 4, N = 7
D = 5, N = 6
D = 6, N = 5
D = 7, N = 4
D = 8, N = 3
D = 9, N = 2
D = 10, N = 1
Let's substitute these values into the first equation to check if they satisfy the conditions:
1) Q + D + N = 8
Q + 1 + 10 = 8 (D = 1, N = 10)
Q + 11 = 8
Q = -3
Since Q is negative, this combination is not valid.
2) Q + D + N = 8
Q + 2 + 9 = 8 (D = 2, N = 9)
Q + 11 = 8
Q = -3
Similarly, this combination is invalid.
We can continue this process for the other values of D and N until we find a combination that satisfies both equations.