A bought some carrots and apples for $24.80. A carrot and an apple $0.90 altogether. She bought more carrots than apples. The cost of the extra number of carrots was $6.80. How many apples did A buy?

1C+1A=.90

nC+mA=24.80
(n-m)C=6.80
nC-MC=6.80
subtract second and last equation
0C+M(A+C)=18.00

But A+C=.90
M=18/.9=20
M is the number of apples

Let's assume the number of apples bought is 'x' and the number of carrots bought is 'y'.

We know that the cost of one carrot and one apple is $0.90 altogether, so we can write the first equation as:
0.90 = x + y

We also know that A bought more carrots than apples, so the cost of the extra number of carrots is $6.80, which can be written as:
6.80 = (y - x) * 0.90

Since the total cost of the carrots and apples is $24.80, we can write another equation as:
24.80 = (x * 0.90) + (y * 0.90)

Now, we can solve this system of equations to find the values of x and y. Let's start by solving the first equation for x:
x = 0.90 - y

Substituting this value of x in the second equation, we get:
6.80 = (y - (0.90 - y)) * 0.90
6.80 = (2y - 0.90) * 0.90
6.80 = 1.8y - 0.81
6.80 + 0.81 = 1.8y
7.61 = 1.8y
y = 7.61 / 1.8
y ≈ 4.228

Since the number of apples can only be a whole number, we will round it to the nearest whole number. Therefore, A bought approximately 4 apples.

To find the number of carrots, we can substitute this value of y in the first equation:
0.90 = x + 4.228
x = 0.90 - 4.228
x ≈ -3.328

Since the number of carrots cannot be a negative value, we will conclude that A did not buy any carrots.

In summary, A bought approximately 4 apples and did not buy any carrots.

To solve this problem, we can use a system of equations. Let's define a few variables:

Let's say the cost of a carrot is C, and the cost of an apple is A.
Let's also say the number of carrots A bought is X, and the number of apples is Y.

From the given information, we can derive the following equations:

Equation 1: C + A = 0.90 (The total cost of a carrot and an apple is $0.90)
Equation 2: C * X + A * Y = 24.80 (The total cost of the carrots and apples is $24.80)
Equation 3: X - Y = 6.80 (The cost of the extra carrots is $6.80)

Now, let's solve the system of equations:

Equation 1: C + A = 0.90
Since we know that C + A = 0.90, we can solve this equation to find the value of A:

A = 0.90 - C

Now, substitute this value of A into Equation 2:

C * X + (0.90 - C) * Y = 24.80

Equation 3: X - Y = 6.80
Rearrange this equation to solve for X:

X = 6.80 + Y

Now, substitute the value of X into Equation 2:

C * (6.80 + Y) + (0.90 - C) * Y = 24.80

Simplify this equation:

6.80C + CY + 0.90Y - CY - CY = 24.80

Combine like terms:

-4CY + 0.90Y + 6.80C = 24.80

Now, we have a system of two equations:

Equation 1: -4CY + 0.90Y + 6.80C = 24.80
Equation 2: C + A = 0.90

From here, we can use substitution, elimination, or other methods to solve for the values of C and Y. Once we have the values, we can find the value of A by substituting the value of C and Y into A = 0.90 - C.

Unfortunately, I cannot give you the exact values of C, Y, and A without all the numerical values in the problem. However, you can follow these steps to find the solution to the problem.