Question

The mean of all 4 -digit even natural numbers of the form 'aabb', where a >0, is

• A. 5050
• B. 4466
• C. 5544
• D. 4864

Answer 1

The Correct Option is C

To find the mean of all 4-digit even natural numbers of the form 'aabb' where a > 0, follow these steps:

1. The number 'aabb' can be expressed as: 1000a + 100a + 10b + b = 1100a + 11b.
2. Since the number should be even, the last digit must be even. Thus, b can be 0, 2, 4, 6, 8.
3. a has possible values from 1 to 9 (since it's a 4-digit number and a > 0).
4. Calculate the sum of all possible numbers:

Total sum = Σ(1100a + 11b)
Where a ranges from 1 to 9 and b is 0, 2, 4, 6, 8.

Total sum = 1100Σa + 11Σb.

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 a: 1100 × 45 = 49500.
Sum for b: 11 × 20 = 220.
Total sum = 49500 × 5 + 220 × 9 = 247500 + 1980 = 249480.

Total number of numbers = 9 × 5 = 45.
The mean = 249480 ÷ 45 = 5544.

Therefore, the mean of all 4-digit even natural numbers of the form 'aabb' is 5544.

Answer 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 \( \boxed{5544} \).