The average of 10 numbers is 40.2. Later it is found that 2 number have been wrongly added.the first is 18 greater than the actual number and the second number added is 13 instead of 31. find the correct average:
(A)40.2 (B)40.4 (c)40.6 (D)40.8 (E)41

• total=40.2*10=402
  402-18+(31-13)=402
  avg remains same
  40.2 ans
• 13 years agoHelpfull: Yes(13) No(0)
• sum of the 10 numbers=40.2*10=402.
  According to the problem, the first number is 18 greater than the actual number.
  so, sum=402-18,
  And the second number added is 13 instead of 31.
  So, sum=402-18-13+31=402.
  Therefore average=402/10=40.2(a)
• 13 years agoHelpfull: Yes(5) No(0)
• 40.2 itself first is 18 greater then 13 instead of 31 so 18 lesser so answer is +18 -18=0 so 40.2
• 13 years agoHelpfull: Yes(4) No(3)
• ya,40.2 absolutly
• 13 years agoHelpfull: Yes(4) No(0)