Question

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

Answer 1

Given : 4 -digit even natural numbers of the form 'aabb', where a >0

To Find : The mean  

Solution:

aabb

even natural numbers  Hence b can be 0  , 2 , 4 , 6 , 8

a can be 1 , 2 ............,  9

1100  , 1122 , 1144 , 1166 , 1188

2200 , 2222 , 2244 ,2266 , 2288

9900 , 9922 , 9944 , 9966 , 9988

Total 45 numbers

unit digit has 0 , 2 , 4 , 6 , 8  - each 9 times

= 9 ( 0 + 2 + 4 + 6 + 8)

= 180

Tens digits also same Hence 180 x 10  = 1800

Hundreds digit    1 to  9  each 5 times

= 5 * 9 * 10/2  = 225  

225 x 100 = 22500

Thousand digits also same

225 x 1000 = 225000

225000 + 22500 + 1800 + 180

= 249480

Mean = 249480/45 = 5,544

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

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Answer 2

Answer:

5544

Step-by-step explanation:

Note that first number would be 1100 and last one 9988. Just take the average of these two. (All numbers are equally spaced)

$x=\frac{1100+9988}{2}$

$x=5544$

Thus average is 5544