Asish was given Rs. 158 in denominations of Rs 1 each. He distributes these in diff bags, such that ne sum of money of denomination betn 1 and 158 can be given in bags. The min no. of such bags reqd
10 / 17 / 15 / none

• 1,2,4,8,16,32,64,31 coins in 8 different bags.
  min 8 bags
  so none of these
• 12 years agoHelpfull: Yes(37) No(39)
• ans:none
  soln: 8 bags required
  
  Let the Bags be marked as A,B,C,D - -- - - - so on.
  Let the first bag [ A] amt. be 1. second bag [ B] amt. be 2. These two bags cover 1,2 & 3 { 3= A+B }
  Hence Bag C will hold 4. Now combinations of A,B,C can manage any amt. from 1 to 7. So D becomes 8.
  These Nos.form a GP. [ Geometric progression ] having 1 as first term and 2 as common ratio.
  The sum og GP is ==== a*( r^n -1) / (r - 1) - - - - - where a=1 , r= 2 and n = No. of terms.
  This should be equated to 158.
  
  If u put n=7 it gives result as 127. So any sum from 1 to 127 can be managed with 7 Bags.
  They are 1,2,4,8,16,32,64. The 8th bag will be obviously 128. These 8 Bags can manage from 1 to 255.
  [ Includes 1 to 158 easily].
  
  As the difference 158-127 is 31. we cannot put eighth bag as 31. It will not help to give sum of 129.
  
  So only 8 bags are required.
• 10 years agoHelpfull: Yes(10) No(2)
• 1+ 2+3+5+ 7+9+11+ 13+15+17+ 19+ 21+ 23+12=158
• 7 years agoHelpfull: Yes(1) No(3)
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• 7 years agoHelpfull: Yes(0) No(12)