A can X contains 399 litres of petrol and a can Y contains 532 litres of diesel. They are to be bottled in bottles of equal size so that whole of petrol and diesel would be separately bottled. The bottle capacity in terms of litres is an integer. How many different bottle sizes are possible?

The correct option is 4.

To find the number of different bottle sizes possible, let us analyze the conditions given in the problem:
1. Can X contains 399 litres of petrol.
2. Can Y contains 532 litres of diesel.
3. The petrol and diesel are to be bottled separately in bottles of equal size.
4. The bottle capacity (let it be $d$ litres) must be an integer.

For the petrol to be bottled completely without any waste, the bottle capacity $d$ must be a factor of 399. Similarly, for the diesel to be bottled completely, $d$ must be a factor of 532.
Therefore, the bottle capacity $d$ must be a common factor of both 399 and 532.

Let us find the prime factorization of both numbers to determine their common factors:
For 399:
399 is divisible by 3:
$399=3\times 133$
133 is divisible by 7:
$133=7\times 19$
So, the prime factorization of 399 is:
$399=3\times 7\times 19$

For 532:
532 is even, so it is divisible by 2:
$532=2\times 266$
266 is divisible by 2:
$266=2\times 133$
And 133 is divisible by 7:
$133=7\times 19$
So, the prime factorization of 532 is:
$532=2^2\times 7\times 19$

Now, we find the Highest Common Factor (HCF) of 399 and 532 by taking the lowest power of the common prime factors:
The common prime factors are 7 and 19.
$\text{HCF}(399,532)=7\times 19=133$

Any common factor of 399 and 532 must also be a factor of their HCF, which is 133.
The factors of 133 are:
1, 7, 19, and 133.

These factors represent the possible integer capacities (in litres) for the bottles:
- 1-litre bottles
- 7-litre bottles
- 19-litre bottles
- 133-litre bottles

Counting these possibilities, we find there are exactly 4 different bottle sizes possible.