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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) |
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Answer» Each bag C contains= n coins thus, n no. of bag C contain = n × n= n² Now n no. of bag C = 1 bag B therefore 1 bag B contains =n² coins If a bag B is removed from bag A, final no. of bag B in bag A = (n-1) [as there were n no. of bag B in bag A initially]. thus (n-1) no. of bag B will contain= (n-1) × n² coins now bag A = (n-1) no. of bag B Thus coins left in a bag A is n²(n-1) [option 4] |
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