Average properties

CAT Exam
Say, you have the following set of consecutive integers: 2, 3, 4, 5, 6, 7, 8 What is the average of this set? There are 7 consecutive integers here and the average is 5, the middle number. Say the set is changed to: 2, 3, 4, 5, 6, 7, 8, 9 (another consecutive number is added to the extreme right). Now what is the average? It is the average of the two middle numbers (5+6)/2 = 5.5. Let’s edit the set one more time: 1, 2, 3, 4, 5, 6, 7, 8, 9 (another consecutive number is added to the extreme left). The average now is 5 again. Whenever you add a number on either side of a set of consecutive integers, the average changes by 0.5. This is obvious because odd number of consecutive integers have the middle number as the average and an even number of consecutive integers have the average of two middle numbers as the average. Since every time you add an integer, the number of integers changes from odd to even or from even to odd, the average changes by 0.5. By the same logic, what happens when you remove an integer from either extreme? Given a set 3, 4, 5, 6, 7, 8, 9, how will its average change if you remove 3? The average of 3, 4, 5, 6, 7, 8, 9 is 6, and the average of 4, 5, 6, 7, 8, 9 is 6.5 — the average increases to 6.5 because you removed a small number. Now how will the average change if you remove 9 instead of 3? The average of 3, 4, 5, 6, 7, 8, 9 is 6, and the average of 3, 4, 5, 6, 7, 8 is 5.5 — here, the average decreases to 5.5 because you removed a large number. So, every time you add or remove a number from one of the extremes, the average will move by 0.5. What happens if you remove a number from somewhere in the middle? The average changes but by how much? When you remove the greatest or the least number, the average changes by 0.5. So when you remove some other number, the average will change by something less than 0.5. For example, from the set 3, 4, 5, 6, 7, 8, 9, if you remove 8, the average changes from 6 to 5.667. If instead, you remove 7, the average changes to 5.833. A few takeaways:
  1. When you remove an integer very close to the average, the average changes by very little. If you remove the average, the average doesn’t change (changes by 0). When you remove a number close to the extreme, the average changes by a larger number (up to a maximum of 0.5).
  2. When you remove a number less than the average, the average increases. When you remove a number more than the average, the average decreases.
  3. When you remove the smallest number, the average increases by 0.5. When you remove the greatest number, the average decreases by 0.5.
Now, a question based on this concept: In a class, the teacher wrote a set of consecutive integers beginning with 1 on the blackboard. A student erased one number. The average of the remaining numbers was 29(14/19). What was the number that the student erased? (A) 13 (B) 16 (C) 28 (D) 36 (E) 50 Solution: The numbers on the board: 1, 2, 3, 4, … The new average is 29(14/19). Since the average changes by not more than 0.5 when you remove an integer from a set of consecutive integers, the original average was either 29.5 or 30. So originally there were either 58 numbers (average 29.5) or 59 numbers (average 30). When you remove a number, you are left with either 57 numbers or with 58 numbers. Now, the new average will tell you whether you are left with 57 numbers or 58 numbers. The denominator is 19 in the fraction, so when you divide the sum of all remaining integers by the number of integers, the number of integers (denominator) is 19 or a multiple of 19 — 57 is a multiple of 19, 58 is not. So you must have been left with 57 integers and the original number of integers must be 58. This means the original average must have been 29.5. The original average of 29(1/2) increases to 29(14/19), i.e. an increase of 14/19 – 1/2 = 9/38. When an integer was removed, the average increased by 9/38 so the integer must be less than the original average. Now use the concept of average that we have learned. One integer was bringing the rest of the numbers down by 9/38 each so the integer must have been (9/38)*57 = 13.5, which is less than the original average of 29.5. This means the integer that was removed must have been (29.5 – 13.5) = 16, so the answer is B.

Category :

CAT Exam

Share This :

Join us MBA CET 2025