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The mean of all $4-$digit even natural numbers of the form $aabb\text{’},$ where $a>0,$ is

1. $5050$
2. $4466$
3. $5544$
4. $4864$

Given that, $4 \; \text{-digit}$ even natural numbers of the form $aabb\text{’}, a>0.$

Now,

• $\underline{a} \quad \underline{a} \quad \underline{b} \quad \underline{b}$
• $1 \quad 1 \quad {\color{red} {0}} \quad {\color{red} {0}}$
• $1 \quad 1 \quad {\color{blue} {2}} \quad {\color{blue} {2}}$
• $1 \quad 1 \quad {\color{purple} {4}} \quad {\color{purple} {4}}$
• $1 \quad 1 \quad {\color{green} {6}} \quad {\color{green} {6}}$
• $1 \quad 1 \quad \underbrace{{\color{magenta} {8}} \quad {\color{magenta} {8}}}_{5}$
• $\vdots \quad \;\vdots \quad \;\vdots \quad \vdots$
• $9 \quad 9 \quad {\color{red} {0}} \quad {\color{red} {0}}$
• $9 \quad 9 \quad {\color{blue} {2}} \quad {\color{blue} {2}}$
• $9 \quad 9 \quad {\color{purple} {4}} \quad {\color{purple} {4}}$
• $9 \quad 9 \quad {\color{green} {6}} \quad {\color{green} {6}}$
• $9 \quad 9 \quad \underbrace{{\color{magenta} {8}} \quad {\color{magenta} {8}}}_{5}$

Total possibilities $= 9 \times 5 = 45$

$\boxed{\text{Average} = \frac{\text{Sum of possibilities}}{\text{Number of possibilities}}}$

Sum of possibilities $= 5 \times (1100 + 2200 + \dots + 9900) + 9 \times (00 + 22 + 44 + 66 + 88)$

$\qquad \qquad = 5 \times 100 \times 11 \times (1 + 2 + 3 + \dots + 9) + 9 \times 11 \times ( 0 + 2 + 4 + 6 + 8)$

$\qquad \qquad = 5500 \times 45 + 9 \times 11 \times 20$

$\text{Average} = \frac{ 5500 \times 45 + 9 \times 11 \times 20}{45}$

$\qquad \qquad = \frac{5500 \times 45}{45} + \frac{9 \times 11 \times 20}{45} = 5500 + 44 = 5544$

$\therefore$ The required mean (average) is $5544.$

$\textbf{Short Method:}$

First , $4 \; \text{digit}$ even numbers is $aabb = 1100$

Last $4 \; \text{digit}$ even number in $aabb = 9988$

$\therefore$ The mean (average) $= \frac{1100+9988}{2} = \frac{11088}{2} = 5544.$

Correct Answer$: \text{C}$

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