Question

There is a mountain with 45 bat caves in a row. Every cave has at least 2 bats and there are 490 bats in all. Any 7 caves in a row contain exactly 77 bats. Suppose the first cave has 7 times more bats than the last cave. What is the greatest possible number of bats in the 30th cave?

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 number of bats in cave number 30 = $(28-2-14)=12$ bats.