The mean of all 4 -digit even natural numbers of the form 'aabb', where a >0, is
- 5050
- 4466
- 5544
- 4864
The Correct Option is C
Approach Solution - 1
- The number 'aabb' can be expressed as: 1000a + 100a + 10b + b = 1100a + 11b.
- Since the number should be even, the last digit must be even. Thus, b can be 0, 2, 4, 6, 8.
- a has possible values from 1 to 9 (since it's a 4-digit number and a > 0).
- Calculate the sum of all possible numbers:
Where a ranges from 1 to 9 and b is 0, 2, 4, 6, 8.
| Step | Calculation | Result |
|---|---|---|
| Sum of values of a | Σa where a = 1 to 9 | 45 |
| Sum of values of b | Σb where b = 0, 2, 4, 6, 8 | 20 |
| Number of terms in a | 9 | 9 |
| Number of terms in b | 5 | 5 |
Sum for b: 11 × 20 = 220.
Total sum = 49500 × 5 + 220 × 9 = 247500 + 1980 = 249480.
The mean = 249480 ÷ 45 = 5544.
Approach Solution -2
The first 4-digit even number in the form 'aabb' is \( 1100 \).
The last 4-digit even number in the form 'aabb' is \( 9988 \).
To find the mean, we use the formula:
\[ \text{Mean} = \frac{\text{Sum of all numbers}}{\text{Total number of numbers}} \]
So, the sum of all numbers is:
\[ \text{Sum} = \frac{(1100 + 9988)}{2} \]
\[ \text{Sum} = \frac{11088}{2} \]
\[ \text{Sum} = 5544 \]
And, the total number of numbers is:
\[ \text{Total number of numbers} = \frac{(9988 - 1100)}{100} + 1 \]
\[ \text{Total number of numbers} = \frac{8888}{100} + 1 \]
\[ \text{Total number of numbers} = 88 + 1 \]
\[ \text{Total number of numbers} = 89 \]
So, the mean of all 4-digit even numbers of the form 'aabb', where \( a > 0 \), is:
\[ \text{Mean} = \frac{5544}{89} = 62.337 \]
Therefore, the required mean is \( \boxed{5544} \).
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