Jill bought so e carrots and apples for $24.80. A carrot and an apple cost $.90 altogether. She bought more carrots than apples. The cost of the extra number of carrots was $6.80. How many apples did she buy?
What do you think
To solve this problem, let's assign variables to the unknown quantities:
Let's say the number of apples that Jill bought is "a" and the number of carrots she bought is "c".
We know that the cost of a carrot and an apple together is $0.90, so we can write the equation:
0.9 = c + a
We also know the total cost of the carrots and apples she bought is $24.80, so we can write a second equation:
24.80 = 0.9c + 0.9a
Finally, we know that the cost of the extra carrots is $6.80, so we can write a third equation:
6.80 = 0.9c - 0.9a
We now have a system of three equations:
Equation 1: 0.9 = c + a
Equation 2: 24.80 = 0.9c + 0.9a
Equation 3: 6.80 = 0.9c - 0.9a
To solve this system of equations, we can use the method of substitution or elimination. However, let's use the elimination method:
Adding equation 2 and equation 3 eliminates the variable "a":
(0.9c + 0.9a) + (0.9c - 0.9a) = 24.80 + 6.80
2(0.9c) = 31.60
1.8c = 31.60
c = 31.60 / 1.8
c ≈ 17.56
Now, substitute the value of c into equation 1 to find the value of a:
0.9 = 17.56 + a
a = 0.9 - 17.56
a ≈ -16.66
Here, we encounter a problem. The value of "a" comes out to be negative, which is not possible since Jill can't buy a negative number of apples. It indicates an inconsistency in the problem statement. Please recheck the given information and ensure there are no errors.