17. Its is given that there are a number of chocolates in a bag. If they were to be 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?

• 14n+10=15n+8
  n=2
  210+38=248
  248%9=5
  remainder is 5
• 11 years agoHelpfull: Yes(52) No(12)
• Ans : 2
  let number = N
  N/14 = remainder 10.
  N/15 = remainder 8.
  
  largest no is 15 . so according to condition N/15 = remainder 8
  N = 15k + 8 (k=0,1,2...)
  again this number must satisfy the condition N/14 = remainder 10
  so it is possible if (15k+8)-10 is divisible by 14
  ie, (15k-2)is divisible by 14 , here we can find the minimum value of k
  putting k = 1 , (15k-2)=13 (not divisible by 14)
  putting k =2 , (15k-2)=13 yes divisible by 14
  
  so now putting the value of k in the N = (15k + 8 ) , we get N=38
  so 38/9 give 2 as a remainder .
  
• 11 years agoHelpfull: Yes(48) No(17)
• Answer-2
  This is because
  14n+10=15n+8
  n=2
  Total chocolates=38
• 11 years agoHelpfull: Yes(17) No(14)
• 14Q+10=n and 15Q + 8 = n
  So, n = 38,
  now (38+210)/9...remainder=5 Ans.
• 11 years agoHelpfull: Yes(13) No(5)
• plz dont confuse yaar,,,,,plz tellthe correct answere.....those whu are saying 2,,,,,then why the 210 is given in question yaar,,,,,
• 11 years agoHelpfull: Yes(9) No(3)
• Let number of chocolates be 38.
  210+38=248
  248%9=5
  So answer is 5
• 11 years agoHelpfull: Yes(4) No(2)
• Answer=2
  This is because
  14n+10=15n+8
  n=2
  Total chocolates=38
  and 38%9=2
• 11 years agoHelpfull: Yes(3) No(4)
• 14n+10=15n+8
  n=2
  total chocolate = 38
  (multiple of 210) no. of chocolate added with 38 chocolates
  LCM(210,38)=3990 i.e the minimum no. of chocolate which can satisfy 2 conditions and also after addition of (210*18).
  so rem(3990/9) = 1 will be the ans
  
• 11 years agoHelpfull: Yes(3) No(12)
• 14n+10=15n+8
  n=2
  now,14*2+10(rem)=38
  and15*2+8(rem)=38
  so,x=38 and divide by 9..get rem=2
  
• 11 years agoHelpfull: Yes(3) No(1)
• ANS:2
  14n+10=15n+8
  n=2
  so it consist of a total of 38.
  this condition is true when 210 choclates are added..this is the first part of the qn..
  then
  second part when there are minimum feasible choclates..that is (38).what is remainder divided by 9...38/9 rem:2
• 11 years agoHelpfull: Yes(2) No(2)
• answer will be 2.
• 11 years agoHelpfull: Yes(1) No(3)
• answer is 2
• 11 years agoHelpfull: Yes(1) No(3)
• can u explain how the answer is 2 ?
  
• 11 years agoHelpfull: Yes(1) No(0)
• 2 is correct answer
• 11 years agoHelpfull: Yes(1) No(5)
• Ans: 6
  My explanation... Taking the minimum case. If the no. of chocolates are divided among 14 people then the minimum chocolate that should be there so that every1 should get it is 1. and 10 remains So (14*1)+10= 24 is the minimum no. of chclts. Same is in the case of 15 children.. (15*1)+8=24. Now 0 is the least multiple of 210 [since according to many 0 is a multiple of all integers]
  So considering the least multiple of 210 , 0 chclts are added. So 24 remains in the bag..
  Now 24/9 gives 6 as the remainder.
  GUYS PLEASE COMMENT IF YOU THINK I'M WRONG. THIS IS JUST WHAT I THOUGHT.
• 11 years agoHelpfull: Yes(1) No(26)