The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

1) 59
2) 19
3) 21
4) 42

• 
  
  Let the 5 natural numbers in increasing order be: a, b, b, b, a+100
  Note that a < b < a+100
  
  (a+b+b+b+a+100)5=150(a+b+b+b+a+100)5=150
  2a+3b=6502a+3b=650
  
  Since a < b, let's find the point where a = b
  2a + 3a = 650
  a = 130 = b
  But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But that decreases the largest number too, so increase b by another 1.
  So a = 127, b = 132 give us the numbers as 127, 132, 132, 132, 227
  
  Since b < a+100, let's find the point where b = a+100
  2a + 3(a+100) = 650
  a = 70, b = 170
  But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But that increases the largest number too, so decrease b by another 1.
  So a = 73, b = 168 give us the numbers as 73, 168, 168, 168, 173
  
  Values of a will be: 73, 76, 79, ....127 (Difference of 3 to make b a natural number)
  This is an AP.
  Last term = First term + (n - 1)*Common difference
  127 = 73 + (n - 1)*3
  n = 19
  So the last term (a+100) will also take 19 distinct values.
  
  Answer (B)
  
• 8 years agoHelpfull: Yes(1) No(0)