there are a no. of chocolates in a bag.if they were equally divided among 14 children,there are 10 chocolates left.if they were to be equally divided among 15 children,there are 8 chocolates left.obviously,this can be satisfied if any multiple of 210 chocolates are added to the bag.What is the remainder when the minimum feasible number of chocolates in the bag is divided by 9

• let x be the total number of chocolates
  14*y1+10=x
  15*y1+8=x
  so the y1 is 2 and x=38
  and 210+38=248
  248/9=27 and remainder 5
  
• 9 years agoHelpfull: Yes(42) No(4)
• 14*2=28+left 10=38
  15*2=30+left 8=38
  210+38=248
  =248/9=27qu and 5 remainder
• 9 years agoHelpfull: Yes(9) No(1)
• i think that they are asking for wat will be the remainder wen min num of chocos left in the bag devided by 9. in this case ans should be 38%9=2.
  plz give ur feedback i may be wrong.
• 9 years agoHelpfull: Yes(8) No(5)
• how both the quotients are taken as y1? an number when divided by 14 and 15 will not give the same quotient ryt?
• 9 years agoHelpfull: Yes(3) No(3)
• let x be the total number of chocolates
  14*y1+10=x
  15*y1+8=x
  so the y1 is 2 and x=38
  now 38 % 9 = 2 Ans
  
• 9 years agoHelpfull: Yes(2) No(1)
• 14x+10=15y+8
  x,y are whole numbers so 0,1 not possible. x=y=2 by common sense. Hence 38%9=2
• 9 years agoHelpfull: Yes(2) No(1)
• total=38+210
  so,238/9=26 4/9
• 9 years agoHelpfull: Yes(1) No(2)
• ans is 2 try to solve..i can't
• 9 years agoHelpfull: Yes(1) No(0)
• 38 chocolates 14*2 10 left
  38 chocs 15*2 8 left
  210+38 = 248 / 9
  remainder = 5
• 9 years agoHelpfull: Yes(1) No(2)
• correct ans iz 2
• 9 years agoHelpfull: Yes(1) No(2)
• y=14*x+10
  y=15*x+8
  solve
  x=2;
  y=38
  (210+38)%9=5
• 9 years agoHelpfull: Yes(0) No(1)
• we get x=38 =min feasible chocolates 38/9 rem 2
• 9 years agoHelpfull: Yes(0) No(0)