For all real numbers s,t,u and v such that s+t +u = 29 and s < v, which of the following statement is true ?

Options

A. s+t+vlt29
B. t+u+v gt29
C. s+t+v=29
D. s+t+v=29

Answer

To solve the problem, we need to analyze the given conditions and check which of the statements is true based on those conditions.

### Step-by-Step Solution:

1. **Write down the given equation and inequality:**
We have two conditions:
- Equation: \( s + t + u = 29 \)
- Inequality: \( s < v \)

2. **Express \( s \) in terms of \( t \) and \( u \):**
From the equation \( s + t + u = 29 \), we can isolate \( s \):
\[ s = 29 - t - u \]

3. **Substitute \( s \) into the inequality:**
Now, we substitute \( s \) in the inequality \( s < v \):
\[ 29 - t - u < v \]

4. **Rearrange the inequality:**
To isolate \( v \), we can rearrange the inequality:
\[ 29 < v + t + u \]
This can be rewritten as:
\[ t + u + v > 29 \]

5. **Identify the correct option:**
Now, we compare our derived inequality \( t + u + v > 29 \) with the options given in the question:
- Option 1: \( s + t + v < 29 \)
- Option 2: \( t + u + v > 29 \) (This matches our derived inequality)
- Option 3: \( s + t + v = 29 \)
- Option 4: \( s + t + v > 29 \)

The correct statement is **Option 2: \( t + u + v > 29 \)**.

### Conclusion:
Thus, the true statement among the options provided is:
**Option 2: \( t + u + v > 29 \)**.