In each of questions two quantities I and II are given compare both quantities choose correct option and give your answer accordingly
Quantity I:the average of three numbers b,c,d is 1 more than average of a,b,c the average of a and d is 19.5 value of a
QuantityII:21

Options

A. `quantityIltquantityII`
B. `quantity I gtquantityII`
C. `quantityIleqquantityII`
D. `quantityIgeqquantityII`

Answer

To solve the problem, we need to analyze the two quantities provided and derive the value of \( a \) based on the information given.

### Step-by-Step Solution:

1. **Understanding the Averages**:
- The average of three numbers \( b, c, d \) is given to be 1 more than the average of \( a, b, c \).
- Mathematically, this can be expressed as:
\[ \frac{b + c + d}{3} = \frac{a + b + c}{3} + 1 \]

2. **Rearranging the Equation**:
- Multiply both sides by 3 to eliminate the fraction:
\[ b + c + d = a + b + c + 3 \]
- Simplifying this gives:
\[ d = a + 3 \]

3. **Average of \( a \) and \( d \)**:
- We know that the average of \( a \) and \( d \) is 19.5:
\[ \frac{a + d}{2} = 19.5 \]
- Multiplying both sides by 2:
\[ a + d = 39 \]

4. **Substituting for \( d \)**:
- From the earlier step, we have \( d = a + 3 \). Substitute this into the equation \( a + d = 39 \):
\[ a + (a + 3) = 39 \]
- This simplifies to:
\[ 2a + 3 = 39 \]

5. **Solving for \( a \)**:
- Subtract 3 from both sides:
\[ 2a = 36 \]
- Divide by 2:
\[ a = 18 \]

6. **Comparing Quantities**:
- Now we have found that \( a = 18 \).
- Quantity I (value of \( a \)) is 18.
- Quantity II is 21.
- Therefore, we compare:
\[ 18 < 21 \]

### Conclusion:
- Since \( a = 18 \) is less than 21, we conclude that Quantity I is less than Quantity II.

### Final Answer:
- The correct option is that Quantity I is less than Quantity II.

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