Average of three numbers k, c and d 1 more than average of a, k and c. Average of a and d is 19.5
Quantity I: Number 'a'
Quantity II: 21

Options

A. Quantity `Igt` Quantity II
B. Quantity `I lt` Quantity II
C. Quantity `I ge` Quantity II
D. Quantity `I le` Quantity II

Explanation

To solve the problem step by step, let's break down the information given and derive the necessary equations.

### Step 1: Understand the Averages
We are given two pieces of information:
1. The average of three numbers \( k, c, d \) is 1 more than the average of \( a, k, c \).
2. The average of \( a \) and \( d \) is 19.5.

### Step 2: Set Up the Equations
From the first piece of information, we can express the averages mathematically:
- The average of \( k, c, d \) is \( \frac{k + c + d}{3} \).
- The average of \( a, k, c \) is \( \frac{a + k + c}{3} \).

According to the problem:
\[ \frac{k + c + d}{3} = \frac{a + k + c}{3} + 1 \]

### Step 3: Simplify the Equation
To eliminate the fractions, multiply the entire equation by 3:
\[ k + c + d = a + k + c + 3 \]

Now, we can simplify this by canceling \( k \) and \( c \) from both sides:
\[ d = a + 3 \quad \text{(Equation 1)} \]

### Step 4: Use the Second Piece of Information
From the second piece of information, we know:
\[ \frac{a + d}{2} = 19.5 \]

Multiplying by 2 gives us:
\[ a + d = 39 \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \( d = a + 3 \)
2. \( a + d = 39 \)

Substituting Equation 1 into Equation 2:
\[ a + (a + 3) = 39 \]
This simplifies to:
\[ 2a + 3 = 39 \]
Subtracting 3 from both sides:
\[ 2a = 36 \]
Dividing by 2 gives:
\[ a = 18 \]

### Step 6: Find the Value of \( d \)
Now, substitute \( a = 18 \) back into Equation 1 to find \( d \):
\[ d = 18 + 3 = 21 \]

### Step 7: Compare Quantities
Now we have:
- Quantity I: \( a = 18 \)
- Quantity II: \( 21 \)

We need to compare these two quantities:
- \( 18 < 21 \)

### Conclusion
Thus, the answer is that Quantity II is greater than Quantity I.

### Final Answer
The answer is **Quantity II is greater than Quantity I**.