Question

600 males and 492 females need to be transported from an island via sea. In each trip, number of males transported are same, and also number of females transported are same. The minimum number of passengers per trip is .

Answer 1

For minimum number of passengers on each trip, the total number of trips will be maximum.
Given that in each trip, number of males transported are same, and also number of females transported are same.

$\therefore$ We need to find the greatest common divisor of $600$ and $492.$

Using prime factorization,
$600=6\times 100=2\times 3\times 2\times 2\times 5\times 5$
$=2^3\times 3\times 5^2$
$492=2\times 2\times 3\times 41$
$=2^2\times 3\times 41$

$\therefore$ The common factors are $2,3,4,6,12.$
Hence, GCD $=2\times 2\times 3=12$

The maximum number of trips required is $12.$

$\therefore$ Number of male passengers in each trip
$=\frac{\text{Total number of male passengers}}{\text{Total number of trips}}$
$=\frac{600}{12}=50$

Similarly, number of female passengers in each trip
$=\frac{492}{12}$
$=41$

$\therefore$ Required number of passengers $=50+41=91$