A horse can consume as much grass in 3 days as a deer can in 7 days. Similarly, a cow can consume as much grass in 4 days as a horse can in 9 days. If it takes the deer and the cow a combined total of 309 days to graze a particular field, how many days would it take for the horse, deer, and cow together to graze the same field?

Let's break down the problem step by step to find how many days it would take for the horse, deer, and cow together to graze the field.

### Step 1: Define the Rate of Grazing for Each Animal
Let:
- The rate of grazing for the deer be \( D \) (in terms of the fraction of the field consumed per day).
- The rate of grazing for the horse be \( H \).
- The rate of grazing for the cow be \( C \).

### Step 2: Set Up Relationships Based on the Problem Statement
1. **Horse and Deer**: A horse can consume as much grass in 3 days as a deer can in 7 days.  
   This means the amount of grass a horse consumes in 3 days is equal to the amount of grass a deer consumes in 7 days.
   \[
   3H = 7D
   \]
   From this, we can express \( H \) in terms of \( D \):
   \[
   H = \frac{7D}{3}
   \]

2. **Cow and Horse**: A cow can consume as much grass in 4 days as a horse can in 9 days.  
   This means the amount of grass a cow consumes in 4 days is equal to the amount of grass a horse consumes in 9 days.
   \[
   4C = 9H
   \]
   Substitute the value of \( H \) from the previous equation:
   \[
   4C = 9 \left(\frac{7D}{3}\right)
   \]
   Simplify this:
   \[
   4C = \frac{63D}{3} = 21D
   \]
   From this, we can express \( C \) in terms of \( D \):
   \[
   C = \frac{21D}{4}
   \]

### Step 3: Combine Deer and Cow Grazing Time
The deer and the cow together take 309 days to graze the entire field:
\[
309 \times (D + C) = 1
\]
This equation means the total fraction of the field consumed by the deer and cow in 309 days is 1 (the entire field).

Substitute \( C \) in terms of \( D \):
\[
309 \times \left( D + \frac{21D}{4} \right) = 1
\]

Combine the terms inside the parentheses:
\[
D + \frac{21D}{4} = \frac{4D + 21D}{4} = \frac{25D}{4}
\]

Substitute this into the equation:
\[
309 \times \frac{25D}{4} = 1
\]

Solve for \( D \):
\[
\frac{7725D}{4} = 1
\]
\[
7725D = 4
\]
\[
D = \frac{4}{7725} = \frac{4}{7725} \approx 0.000517
\]

### Step 4: Find Rates of Horse and Cow
Now, using the expressions we found earlier:
\[
H = \frac{7D}{3} = \frac{7 \times 4}{3 \times 7725} = \frac{28}{23175} \approx 0.001208
\]
\[
C = \frac{21D}{4} = \frac{21 \times 4}{4 \times 7725} = \frac{84}{30900} = \frac{28}{10300} \approx 0.002718
\]

### Step 5: Combined Grazing Rate of Horse, Deer, and Cow
Now, find the combined rate of all three animals:
\[
H + D + C = \frac{28}{23175} + \frac{4}{7725} + \frac{28}{10300}
\]

Find a common denominator (let’s use \(23175 \times 10300 = 238792500\)) and calculate the sum:
\[
H + D + C \approx 0.004443
\]

### Step 6: Total Time to Graze the Field
Now, find the time it takes for all three to graze the entire field:
\[
\text{Time} = \frac{1}{H + D + C} \approx \frac{1}{0.004443} \approx 225 \text{ days}
\]

So, it would take approximately **225 days** for the horse, deer, and cow together to graze the entire field.

Would you like a detailed explanation of any particular step, or do you have further questions?

### Related Questions
1. What is the individual grazing rate for each animal?
2. How long would it take for the horse alone to graze the field?
3. How long would it take for the deer alone to graze the field?
4. How does the grazing rate change if the cow's rate doubles?
5. What if another animal with the same grazing rate as the deer is added?

### Tip
When solving problems involving rates and time, setting up equations that reflect the relationships given in the problem helps break down the steps systematically.

Solve the work rate problem involving a horse, deer, and cow grazing a field. This detailed solution walks through the algebraic steps to determine how many days it would take for all three animals together to graze the field. Suitable for high school students (Grades 9-12).

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