Answer:
12 bats
Step-by-step explanation:
Let us number the bat caves from 1 to 45 and divide them into 5 parts:
a) 1
b) 2 to 29 ([tex]7\times 4=28[/tex] bats)
c) 30
d) 31 to 44 ([tex]7\times2=14[/tex] bats)
e) 45
It is given that any 7 caves in a row has 77 bats. Hence, the total number of bats in part b and d = ( [tex]4\times77[/tex] ) + ( [tex]2\times77[/tex] ) = 462 bats.
There remains ( 490 - 462 ) = 28 bats in cave number 1, 30 and 45.
Now, our question demands the maximum possible number of bats in cave 30.
Minimum number of bats in a cave is 2. So we shall put 2 bats in the last cave, which gives us [tex]7\times2=14[/tex] bats in the first cave.
Therefore, the greatest possible number of bats in cave number 30 = ( 28 - 2 - 14 ) = 12 bats.
