shuvodip04 rencsee

Imagine a number line (if you are familiar with VeritasKarishma posts on weighted average)... think about it like we are dealing with absolute value i.e. the 'distance' from the Average.

___1x<------->2x___
15 ------ Avg ------- 39

Adding 15 gives Avg - 1, the Avg is 'pulled' to the left because 15 is less than the Avg
Adding 39 gives Avg + 2, Avg is 'pulled' to the right because 39 is more than the Avg

Now, subtract the 15,

___1x<------->2x___
0 ------ Avg ------- 24

The 'pull' from 0 to the Avg is still 1, and the 'pull' from the Avg to 24 is still 2. This is because the distance covered stays the same in a ratio of 1x:2x.
Thus, the total from 0 to 24 would be 3x = 24 --> x = 8, hence n+1 = 8 and originally n = 7, A.

VeritasKarishma, Bunuel
Hi, Need to practice more questions like this.. Can you guys help?

VERITAS PREP OFFICIAL SOLUTION:

What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average – 1). If instead, the person aged 39 were added to the group, there would be 39 – 15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8 – 1 = 7.