A student finds the average of ten 2digit numbers.while copying numbes he writes one number with its digit interchanged.As a result his answer is 1.8 less than the correct answer.the difference of the of the digits of the numbers in which he made mistake is

• Let Enterchanged number be 10b+a
  and original be 10a+b
  
  now the average difference is 1.8
  so (10b+a-(10a+b))/10 = 1.8 (total number is 10 here )
  
  => 9b-9a = 18
  => b-a =2 answer
• 7 years agoHelpfull: Yes(3) No(0)
• Original average be x = total sum / 10 = x........... equ1
  lets say the number is ab = 10a+b
  the wrong number he took = 10b+a
  difference betwwen both numbers = 9(a-b)
  now the avg becomes = [total sum - 9(a-b)] /10= x-1.8 ....... equ 2
  solving equ. 1 and equ. 2
  9(a-b) = 18
  Required difference = 18
• 7 years agoHelpfull: Yes(0) No(2)
• elitmus me aya tha ye?
• 7 years agoHelpfull: Yes(0) No(0)
• Change in average = 1.8
  Change in weight = 1.8 × 10 = 18
  9 × (difference of digits) = 18
  
  ∴ Difference of digits = 18/9 = 2
• 5 years agoHelpfull: Yes(0) No(0)