Question

A certain number of cakes is distributed equally among Alex, Bruce, Clive and Daniel. Alex gives away \((\frac{3}{4})^{th}\) of his cakes equally among the other three people. Later Bruce gives half of his cakes to Daniel. What fraction of the total number of cakes does Daniel have in the end?

• A. \(\frac{5}{16}\)
• B. \(\frac{7}{16}\)
• C. \(\frac{15}{32}\)
• D. \(\frac{17}{32}\)

Answer 1

The Correct Option is C

Distribution of Cakes 

The total number of cakes is distributed equally among four people: Alex, Bruce, Clive, and Daniel.

Let the total number of cakes be \( C \).

Each person initially gets:

\[ \frac{C}{4} \]

Step 1: Alex Gives Away \( \frac{3}{4} \) of His Cakes Equally

Alex gives away:

\[ \frac{3}{4} \times \frac{C}{4} = \frac{3C}{16} \]

This amount is divided equally among Bruce, Clive, and Daniel:

\[ \frac{3C}{16} \div 3 = \frac{3C}{48} = \frac{C}{16} \]

So, Daniel receives an additional \( \frac{C}{16} \).

Step 2: Bruce Gives Half of His Cakes to Daniel

After receiving \( \frac{C}{16} \) from Alex, Bruce's total cakes become:

\[ \frac{C}{4} + \frac{C}{16} = \frac{4C}{16} + \frac{C}{16} = \frac{5C}{16} \]

Bruce gives half of his cakes to Daniel:

\[ \frac{1}{2} \times \frac{5C}{16} = \frac{5C}{32} \]

Step 3: Compute Daniel's Total Cakes

Initially, Daniel had:

\[ \frac{C}{4} = \frac{4C}{16} = \frac{8C}{32} \]

After receiving \( \frac{C}{16} = \frac{2C}{32} \) from Alex:

\[ \frac{8C}{32} + \frac{2C}{32} = \frac{10C}{32} \]

After receiving \( \frac{5C}{32} \) from Bruce:

\[ \frac{10C}{32} + \frac{5C}{32} = \frac{15C}{32} \]

Thus, the fraction of total cakes Daniel has in the end is:

\[ \frac{15}{32} \]

Final Answer:

Option (C) \( \frac{15}{32} \)