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To solve question 5(b), we need to understand the setup of the problem. Here's the step-by-step breakdown:

### Problem Restatement:
You have a grain silo that is a combination of a cylinder and two identical shallow cones. Initially, the silo is filled to the top, and then some barley is removed such that the level of grain drops by 1.1 meters. You need to find the percentage of grain that was removed.

### Given Information:
- The total height of the silo: \(7.8 \, \text{m}\) 
- The height of the conical section: \(1.8 \, \text{m}\)
- The height of the cylindrical section: \(7.8 - 2(1.8) = 4.2 \, \text{m}\)
- The radius of the silo: \(2.2 \, \text{m}\)
- The grain level drops by \(1.1 \, \text{m}\).

### Steps to Solve:

#### 1. **Volume of the Full Silo:**
   The silo consists of a cylindrical part and two conical sections.

   - **Volume of the cylinder**: 
     \[
     V_{\text{cyl}} = \pi r^2 h = \pi \times (2.2)^2 \times 4.2
     \]
     \[
     V_{\text{cyl}} \approx \pi \times 4.84 \times 4.2 = \pi \times 20.328 \approx 63.9 \, \text{m}^3
     \]
   - **Volume of each cone**:
     \[
     V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times (2.2)^2 \times 1.8
     \]
     \[
     V_{\text{cone}} = \frac{1}{3} \times \pi \times 4.84 \times 1.8 \approx \pi \times 2.904 \approx 9.12 \, \text{m}^3
     \]
     For two cones:
     \[
     V_{\text{total\_cones}} = 2 \times 9.12 = 18.24 \, \text{m}^3
     \]
   - **Total volume of the silo**:
     \[
     V_{\text{silo}} = V_{\text{cyl}} + V_{\text{total\_cones}} = 63.9 + 18.24 = 82.14 \, \text{m}^3
     \]

#### 2. **Volume of Grain Removed:**
   When the grain level drops by 1.1 meters, we only consider the cylindrical portion.

   - **New height of the grain**:
     \[
     h_{\text{new}} = 4.2 - 1.1 = 3.1 \, \text{m}
     \]
   - **Volume of the cylinder with new height**:
     \[
     V_{\text{new}} = \pi \times (2.2)^2 \times 3.1 \approx \pi \times 4.84 \times 3.1 \approx \pi \times 15.004 \approx 47.1 \, \text{m}^3
     \]
   - **Volume of grain removed**:
     \[
     V_{\text{removed}} = V_{\text{cyl}} - V_{\text{new}} = 63.9 - 47.1 \approx 16.8 \, \text{m}^3
     \]

#### 3. **Percentage of Grain Removed:**
   \[
   \text{Percentage} = \frac{V_{\text{removed}}}{V_{\text{silo}}} \times 100 = \frac{16.8}{82.14} \times 100 \approx 20.45\%
   \]

### Final Answer:
Approximately **20.45%** of the grain was removed when the level dropped by 1.1 meters.

Would you like more details or have any questions? Here are some related questions:

1. What would the percentage be if the level dropped by 2 meters instead?
2. How does the radius of the silo affect the percentage of grain removed?
3. What would be the volume of grain removed if only 10% was taken out?
4. How would the calculation change if the conical sections had different heights?
5. Can you calculate the volume if the height of the cylindrical section was 5 meters?
6. What is the relationship between the radius and the total volume of the silo?
7. How would the problem change if the entire silo was in the shape of a cylinder?
8. What would happen if the silo was only half-filled initially?

**Tip:** When dealing with composite shapes, break them down into simpler parts (like cylinders and cones) to simplify volume calculations.

Learn how to calculate the percentage of grain removed from a silo with a detailed mathematical solution. This problem involves finding volumes of cylindrical and conical sections to determine the percentage of grain removed when the grain level drops by 1.1 meters.

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