Let's break down the problem step by step:
1. A got \(33.33\%\) of the total marks of \(300\), which is \(0.3333 \times 300 = 100\) marks.
2. B's marks are one fourth more than A's marks. So, B got \(100 + \frac{1}{4} \times 100 = 125\) marks.
3. C got \(15\) marks less than B, which means C got \(125 - 15 = 110\) marks.
4. D got \(20\) marks more than the average marks of A, B, and C. First, let's find the average marks of A, B, and C:
\[ \text{Average marks} = \frac{(100 + 125 + 110)}{3} = \frac{335}{3} \]
Now, D got \(20\) marks more than this average, which is:
\[ \text{D's marks} = \frac{335}{3} + 20 \]
Now, let's calculate D's marks:
\[ \text{D's marks} = \frac{335}{3} + 20 = \frac{335 + 60}{3} = \frac{395}{3} \approx 131.67 \text{ marks} \]
Now, we need to determine how many among A, B, C, and D got more than \(40\%\) of the total marks.
The total marks are \(300\).
40% of \(300\) is \(0.4 \times 300 = 120\) marks.
From our calculations:
- A got \(100\) marks
- B got \(125\) marks
- C got \(110\) marks
- D got approximately \(131.67\) marks
So, B, C, and D got more than \(40\%\) of the total marks. Therefore, the answer is \( \boxed{3} \).
#studyharder
