If the arithmetic mean of 3a and 4b is greater than 50, and a is twice b, then the smallest possible Integer value of a is

Options

A. 20
B. 18
C. 21
D. 19

Answer

To solve the problem step by step, we will follow the given conditions and derive the required value of \( a \).

### Step 1: Write the condition for the arithmetic mean
The arithmetic mean of \( 3a \) and \( 4b \) is given by the formula:
\[ \text{Arithmetic Mean} = \frac{3a + 4b}{2} \]
According to the problem, this mean is greater than 50:
\[ \frac{3a + 4b}{2} > 50 \]

### Step 2: Multiply both sides by 2
To eliminate the fraction, we multiply both sides of the inequality by 2:
\[ 3a + 4b > 100 \]

### Step 3: Use the relationship between \( a \) and \( b \)
We are given that \( a \) is twice \( b \):
\[ a = 2b \]
Now, we can substitute \( a \) in terms of \( b \) into the inequality.

### Step 4: Substitute \( a \) in the inequality
Substituting \( a = 2b \) into \( 3a + 4b > 100 \):
\[ 3(2b) + 4b > 100 \]
This simplifies to:
\[ 6b + 4b > 100 \]
\[ 10b > 100 \]

### Step 5: Solve for \( b \)
Now, divide both sides by 10:
\[ b > 10 \]

### Step 6: Find the smallest integer value for \( b \)
Since \( b \) must be an integer, the smallest integer value for \( b \) that satisfies the inequality is:
\[ b = 11 \]

### Step 7: Calculate the corresponding value of \( a \)
Now, we can find \( a \) using the relationship \( a = 2b \):
\[ a = 2 \times 11 = 22 \]

### Conclusion
Thus, the smallest possible integer value of \( a \) is:
\[ \boxed{22} \]