A can X contains 399 liters of petrol and a can Y contains 532 liters 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 liters is an integer.
How many different bottle sizes are possible?

Correct Option (B)

To determine the number of different bottle sizes possible, the bottle capacity must be a common divisor of the volumes of petrol and diesel. Since the petrol (399 liters) and diesel (532 liters) are to be bottled separately in bottles of equal integer size, the bottle capacity must be a common factor of 399 and 532.

The possible bottle sizes are the factors of the Highest Common Factor (HCF) of 399 and 532.

• First, find the HCF of 399 and 532:
  • Prime factorization of 399: 3 × 7 × 19
  • Prime factorization of 532: 2² × 7 × 19
  • The common factors are 7 and 19.
  • Therefore, HCF(399, 532) = 7 × 19 = 133.
• Next, find all factors of the HCF, which is 133:
  • The factors of 133 are 1, 7, 19, and 133.
• Thus, there are 4 different possible bottle sizes (1 liter, 7 liters, 19 liters, and 133 liters).

Incorrect Options:

Options A (3), C (5), and D (6) are incorrect because the calculation of the common factors of 399 and 532 yields exactly four distinct values (1, 7, 19, 133). These options do not reflect the correct number of possible bottle sizes, which is determined by the total count of these common factors.