If a number consists of each digit of each 1 to 9 then the difference of its sums of alternate digits.
a)0 b)11 c)22 d)33

• As per this question 1 more condition was there that the number is divisible by 11.
  
  Ans : option b that is 11.
  
  explanation :
  
  sum of the digits from 1 to 9 is 45.
  and as we know if a number is divisible by 11 then the sum of digits at odd and even places would be zero or multiple of 11.
  
  now as per the question we have x + y = 45
  take x as sum of digits at odd or even place of the number.
  take y as sum of digits at odd or even place of the number.
  
  so we get x = 45 - y.
  as per the given options we have :
  x - y = 0 or x- y = 11 or x -y = 22 or x -y = 33 we can get to know which is satisfying the equation.
  
  so as a whole we get 45 - y - y = 0 as we put the value of x from the equation x + y = 45.
  
  so out of 4 only 2 options are giving the integer value. ie 45-2y=11 and 45-2y=33.
  
  =>45-2y=33
  =>2y=78
  =>y=39.
  now x+y =45 therefore x =6.
  as x = 6. we cant get 6 as the sum of the digits at even or odd places. therefore it cannot be possible .
  
  next is 45 -2y= 11.
  =>2y = 56
  =>y=28.
  therefore x = 45-28
  => x= 17.
  Now we can get 17 as a sum of digits = 17 at odd places and 28 at even places. see the arrangement as 2+9+3+8+5+7+6+1 .
  
  and the difference of 28 and 17 is 11 that is divisible by 11 hence the answer is 11.
• 10 years agoHelpfull: Yes(16) No(3)
• Ans:- a) 0
  
  Solution:-
  
  1+2+3+4+5+6+7+8+9 = 45
  sum of alternate digit
  9+8+7+6+5+4+3+2+1 = 45
  
  then difference
  =45-45
  =0
• 10 years agoHelpfull: Yes(5) No(7)
• there is 1 more condition that number is divisible by 11
• 10 years agoHelpfull: Yes(3) No(0)