If all 6 are replaced by 9,the algebric sum of all numbers from 1 to 100 (both inclusive)varies by,

a) 980 b)330 c)660 d)990

• first take the numbers which have 6 in their unit place - 06,16,26,36,46,56,66,76,86,96.
  if all 6 in unit plce is replaced with 9 then in each case the increment would be 3.
  so for all ten numbers total increment is 10*3=30.
  
  then take all numbers which have 6 in their tenth place - 60,61,62,63,64,65,66,67,68,69
  in each case if we replace 6 with 9 -90,91,92,93,94,95,96,97,98,99
  then there is increment of 30 in each case .
  so total increment for ten numbers is 10*30==300.
  so the total increment including unit and tenth place is 30+300=330.
• 8 years agoHelpfull: Yes(48) No(6)
• 100(9-6)+10(9-6)=330(b)
• 8 years agoHelpfull: Yes(8) No(1)
• you are mistaken, 99-66 will be 33 rather than 30.
  so ans is 330
• 8 years agoHelpfull: Yes(6) No(0)
• 6,16,26,36,46,56,76,86,96 replace by 9 then 3*9=27
  60,61,62,63,64,65,66,67,68,69 then increase by 30 i.e, 30*10=300
  It mst v 327
• 8 years agoHelpfull: Yes(5) No(10)
• 6,16,26,36,46,56,76,86,96 replace by 9 then 3*9=27
  60,61,62,63,64,65,66,67,68,69 then increase by 30 i.e, 30*10=300+3
  99-66=33-30
  It will be 330
• 8 years agoHelpfull: Yes(3) No(0)
• 60 - 90 , 61 - 91, ........ 66 - 99........ 69 - 99 its clear that
  9 numbers except 66 increased by 30 while 66 - 99 there is increase of 33
  so 9 * 30 = 270 and 1 * 33 = 33
  270 + 33 = 303
  
  also 6 - 9 , 16 - 19 , 26 - 36 ......... 3 is the increment in this 9 numbers leaving 66 as we counted 66 above
  
  so final total = 303 + 27 = 330
  so 9 * 3 = 27
  total = 300+30 = 330
  answer is 330
  
  now those who think it should be 327 it cant be because
  66 changing to
• 7 years agoHelpfull: Yes(2) No(0)
• it should b 327
  
• 8 years agoHelpfull: Yes(0) No(8)
• the soln is 330
• 7 years agoHelpfull: Yes(0) No(0)