Question

Mr Smith has a bag A. Bag A has "n" numbers of bag B. Each bag B has " n " numbers of bag C. Each bag C has " n" number of one rupee coins in it. If one bag B is removed from bag A, then the total number of coins left in bag A is.....
1) n(n-1)(n+1)
2) n(n-1)
3) n2 (n+1)
4) n2(n-1)

Answer 1

Suppose "n" = 10

So, bag A has 10 Bags of B

Every Bag of B will have 10 Bags of C.

So, Total number of C bags = 10 * 10 = 100 bags of C

Every Bag of C will have 10 coins.

So, Total coins = 10 * 100 = 1000 coins

Now, 1 Bag B is removed which has 10 bags of C and each bag of C has 10 coins. So, coins removed = 10 * 10 =100 coins

Remaining coins = 1000 - 100 = 900

Now verify options by putting "n" = 10. The option which gives the value as 900 will be the correct answer.

Option 1 = 10(9)(11) = 990 (wrong)

Option 2 = 10(9) = 90 (wrong)

Option 3 = 10 square * (11) = 1100 (wrong)

Option 4 = 10 square * (9) = 900 (right)

So, Correct answer is option 4.