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

The correct answer is (C): \(5544\)

The sum of possible even digit numbers in the form aabb is \(1100 + 1122 + 1144 1166 + 1188 + 2200 + 2222 + 2288 +\) .......\(9900 + 9922 + 9988\) i.e. (\(45\) numbers)

\(⇒ 1100+1100+1100+1100+1100+22+44+66+88+2200+2200+2200+2200+2200+22+44+66+88+…\)\(+9900+9900+9900+990+9900+22+44+66+88\)

\(⇒ 5(1100+2200+.....9900)+9(22+44+66+88)5×1100(1+2+....9)+9×22(1+2+3+4)\)

\(⇒ 5500(45) + 45 × 44 = 45(5544)\)

Hence mean = \(5544\)

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 \( \boxed{5544} \).