Mrs. Yang spends $3.00 a day on cookies for her family at a local supermarket . When the price of a cookie increased by 10 cents , the number of cookies she bought decreased by 1. What was the original price , in dollars , of a cookie ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Define Variables
Let \( N \) be the number of cookies Mrs. Yang originally buys, and let \( P \) be the original price of one cookie in dollars.

### Step 2: Set Up the Equations
Since Mrs. Yang spends $3.00 a day on cookies, we can express this as:
\[ N \times P = 3 \]
From this, we can express the price of one cookie as:
\[ P = \frac{3}{N} \]

### Step 3: Account for Price Increase
When the price of a cookie increases by 10 cents (which is $0.10), the new price becomes:
\[ P + 0.10 = \frac{3}{N} + 0.10 \]
At this new price, the number of cookies she buys decreases by 1, so she buys \( N - 1 \) cookies. The total amount spent remains $3.00, leading to the equation:
\[ (N - 1) \left( P + 0.10 \right) = 3 \]

### Step 4: Substitute for \( P \)
Substituting \( P \) into the equation gives:
\[ (N - 1) \left( \frac{3}{N} + 0.10 \right) = 3 \]

### Step 5: Expand and Simplify
Expanding the left side:
\[ (N - 1) \left( \frac{3}{N} + 0.10 \right) = (N - 1) \frac{3}{N} + (N - 1) \times 0.10 \]
This simplifies to:
\[ \frac{3(N - 1)}{N} + 0.10(N - 1) = 3 \]
Multiplying through by \( N \) to eliminate the fraction:
\[ 3(N - 1) + 0.10N(N - 1) = 3N \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ 3N - 3 + 0.10N^2 - 0.10N = 3N \]
This simplifies to:
\[ 0.10N^2 - 3 - 0.10N = 0 \]

### Step 7: Multiply through by 10
To eliminate decimals, multiply the entire equation by 10:
\[ N^2 - 30 - N = 0 \]
Rearranging gives:
\[ N^2 - N - 30 = 0 \]

### Step 8: Factor the Quadratic
Factoring the quadratic equation:
\[ (N - 6)(N + 5) = 0 \]
This gives us two potential solutions:
\[ N = 6 \quad \text{or} \quad N = -5 \]
Since \( N \) must be a positive number, we take \( N = 6 \).

### Step 9: Calculate the Original Price
Now substituting \( N \) back to find the original price \( P \):
\[ P = \frac{3}{N} = \frac{3}{6} = 0.5 \]

### Final Answer
Thus, the original price of one cookie is:
\[ \boxed{0.50} \text{ dollars or 50 cents.} \]