If the numbers between 01 to 65 which will be divisible by 4 are taken and then if the number present in the units places and tens places is swapped, post which they are written in ascending order, then which of the following number will be at 10th place from the last ?

Note that 01 when swapped will become 10 and so on. 

The total number of multiples of 4 is 16. (04, 08, 12, ...., 64)

The numbers, which are multiples of 4, have their unit digit 0/2/4/6/8

The multiples of 4, with their unit digit as 8, are (08,28,48), and the multiples of 4, with their unit digit as 6, are (16, 36, 56).

The multiples of 4, with their unit digit as 4, are (04, 24, 44, 64), and the multiples of 4, with their unit digit as 2, are (12, 32, 52), the multiples of 4, with their unit digit as 0, are (20, 40, 60).

It is given that the number present in the units places and tens places is swapped, post which they are written in ascending order, then which of the following numbers will be at 10th place from the last => (16+1-10) = 7th from the beginning.

In ascending order (after swapping the digits), the least value will be those numbers that had '0' as their unit place. After that, the numbers had '2' as their unit place, and so on.

7th from the beginning = (3 numbers that had 0 as their unit place) + (3 numbers that had 2 as their unit place) + (1st number that had '4' as their unit place)

Hence, the first number that had 4 as their unit place now becomes (04 => 40). Hence, 40 is the correct answer.