Answer:
71
Step-by-step explanation:
Let's denote the initial number of cakes as \( x \).
1. **After Mississauga takes 1/4 of the cakes:**
- Mississauga takes \( \frac{1}{4} \) of \( x \) cakes, which is \( \frac{x}{4} \).
- Cakes remaining after Mississauga: \( x - \frac{x}{4} = \frac{3x}{4} \).
2. **After Miss Molly, Aunt Ena, and Miss Feldman take 1/4 of the remaining cakes:**
- They take \( \frac{1}{4} \) of \( \frac{3x}{4} \) cakes, which is \( \frac{3x}{16} \).
- Cakes remaining after them: \( \frac{3x}{4} - \frac{3x}{16} = \frac{9x}{16} \).
3. **After Mr. Alvarez takes half of the remaining cakes:**
- Mr. Alvarez takes \( \frac{1}{2} \) of \( \frac{9x}{16} \) cakes, which is \( \frac{9x}{32} \).
- Cakes remaining after Mr. Alvarez: \( \frac{9x}{16} - \frac{9x}{32} = \frac{18x}{32} - \frac{9x}{32} = \frac{9x}{32} \).
Given that 20 cakes remain at the end of the day, we have:
\[ \frac{9x}{32} = 20 \]
To find \( x \), multiply both sides by \( \frac{32}{9} \):
\[ x = 20 \times \frac{32}{9} \]
\[ x = \frac{640}{9} \]
\[ x = 71 \frac{1}{9} \]
Since \( x \) must be a whole number (you can't have a fraction of a cake), we round \( 71 \frac{1}{9} \) down to the nearest whole number.
Therefore, the initial number of cakes that Miss Grady had was \( \boxed{71} \).
