Question

Plz help me (All my points if correct)

Answer 1

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 ($7\times 4=28$ bats)

c) 30

d) 31 to 44 ($7\times2=14$ 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 = ( $4\times77$ ) + ( $2\times77$ ) = 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 $7\times2=14$ bats in the first cave.

Therefore, the greatest possible number of bats in cave number 30 = ( 28 - 2 - 14 ) = 12 bats.