a, b, c are non negative integers such that 28a+30b+31c=365. Then a+b+c is?
a)13 b)> 13 c)

• ans is always 12.
  here a=1 i.e february.
  b=4 (april,june,sept,nov)
  c=7 (jan,march,may,july,aug,oct,dec)
• 11 years agoHelpfull: Yes(128) No(2)
• 12 is the Answer.
• 11 years agoHelpfull: Yes(12) No(3)
• Here, a + b + c = 12
  Explanation :
  a = 1, because in a month of twelve only Feb has 28 days
  b = 4, there are four months namely April,June,Sep and Nov having 30 days
  c = 7, there are 7 months namely Jan,Mar,May,Jul,Aug,Oct and Dec having 31 days
  So, a + b + c = 1 + 4 + 7 = 12.
• 8 years agoHelpfull: Yes(8) No(1)
• This is how I solved it;;
  28a+30b+31c=365
  28a+(28+2)b+(28+3)c=365
  28(a+b+c)+(2b+3c)=365
  if suppose 2b+3c is zero (suppose)
  then 365/28 is 13.something
  then 13*28=364
  
  365-364=1
  
  we cannot get 1 with 2a+3b
  
  so lets try 12
  12*28=336
  
  365-336=29
  
  we can get 29 with 2a+3b in this form
  2*4+3*7
  
  so b=4 c=7 and we assumed a+b+c=12 so a = 1
  
  put these values in the equation and it satisfies
  
  so the value of a+b+c=12
  
  :)
• 6 years agoHelpfull: Yes(2) No(1)
• please explain question
• 8 years agoHelpfull: Yes(1) No(0)
• a=1,b=7,c=4 according to monthly days
• 8 years agoHelpfull: Yes(0) No(0)
• yes bro..12 is the correct answer..
• 8 years agoHelpfull: Yes(0) No(3)
• put a=0 , b=7, c=5. it makes 0+210+155=365. so correct answer is 12
• 7 years agoHelpfull: Yes(0) No(1)
• simplify the given equation as 30(a+b+c)+2a+c=365
  and then check for (a+b+c)=14,13,12,11
  and finally you got a valid case for 12.
  so,answer is 12
  
• 7 years agoHelpfull: Yes(0) No(0)
• 1+4+7 as 12 months in year make 365 days
• 7 years agoHelpfull: Yes(0) No(0)
• In a calender,
  Number of months having 28 days = 1
  Number of months having 30 days = 4
  Number of months having 31 days = 7
  28 x 1 + 30 x 4 + 31 x 7 = 365
  Here, a = 1, b = 4, c = 7.
  
  a+b+c = 12
• 5 years agoHelpfull: Yes(0) No(0)