a,b,c are non-negative integers such that 20a+30b+31c=365, then a+b+c =?

• vivek i think the question will be 28a+30b+31c=365
  now in a normal(non-leap year) year we have month having 28 days 1, 30 days 4 months, 31 days 7. so a+b+c=12
• 9 years agoHelpfull: Yes(15) No(1)
• 12
  per yr 365 days
  so 12 months
  
• 9 years agoHelpfull: Yes(3) No(2)
• all brothers.you are fully wrong.the question cant be changed according to your suits in any competetive exams.so please solve it.be an an ENGINEER,not a RATTU TOTA>
  according to the question the solution is following:
  20a+30b+31c=365
  20a+30b+31c=20x3+30x5+31x5
  compaRING TWO SIDES WE GET A=3,B=5,C=5 now,a+b+c=13.
• 9 years agoHelpfull: Yes(3) No(1)
• answer is 90 based on the year concept
• 9 years agoHelpfull: Yes(1) No(2)
• its not 20a it should be 28a
  so according to our calender
  a+b+c=12
• 9 years agoHelpfull: Yes(1) No(0)
• This is tricky ques:
  here a,b,c are the months and total days are 365.
  therefore a=1,b=4,c=7.
  and sum is 12.
• 9 years agoHelpfull: Yes(0) No(0)
• 20a+30b+31=365
  a=1
  b=6
  c=5 so totaly 12 months.
  
• 9 years agoHelpfull: Yes(0) No(0)
• 12 or
• 9 years agoHelpfull: Yes(0) No(0)
• The given question is calendar based.We have to take calculation of days in a particular month
  28--Feb(a=1)
  30-apr,june,sep,nov(b=4)
  31-Jan,March,May,July,Aug,Oct,Dec(c=7)
  The question asked is a+b+c =7+4+1=12
• 9 years agoHelpfull: Yes(0) No(0)
• in order to obtain the unit digit as 5 we have to multiply the 31 with 5 anh hence 155 then remaining can be obtained 210 by taking value ans is 14
• 9 years agoHelpfull: Yes(0) No(1)
• 1+4+7=12 a month having 31 days that is 7 and a month having 30 days that is 4 and 1 is feb.
  
• 9 years agoHelpfull: Yes(0) No(0)