Study the following information carefully and answer the questions given below:

Six persons are arranged in descending order according to their weight from left to right. Only T is lighter than Q whose weight is 17kg. S is twice heavier than Q and 14 kg lighter than N. Two persons between M and S. R is not the second heaviest among all.

How many persons are lighter than S?

The correct answer is Two.

Let the six persons be arranged from left to right in descending order of their weight:
1 (Heaviest) > 2 > 3 > 4 > 5 > 6 (Lightest)
From the given information, only T is lighter than Q. This means Q is at the 5th position and T is at the 6th position (the lightest). The weight of Q is given as 17 kg.

S is twice as heavy as Q. Thus, the weight of S is:
$S=2\times 17=34\text{kg}$
S is 14 kg lighter than N. Thus, the weight of N is:
$N=34+14=48\text{kg}$

Since N (48 kg) is heavier than S (34 kg), N must be positioned to the left of S. There are two persons between M and S, which gives two possible configurations for M and S:
- If M is at the 1st position, S must be at the 4th position.
- If S is at the 1st position, M must be at the 4th position.
Since N is heavier than S, N must be to the left of S. S cannot be at the 1st position because there would be no space to the left of S for N. Therefore, S must be at the 4th position, and M must be at the 1st position.

This gives the partial arrangement:
1: M > 2: [ ] > 3: [ ] > 4: S (34 kg) > 5: Q (17 kg) > 6: T
We have N and R left to place in the 2nd and 3rd positions. Since R is not the second heaviest among all, R cannot be in the 2nd position. Therefore, N is in the 2nd position, and R is in the 3rd position.
The final weight-wise descending order is:
M > N (48 kg) > R > S (34 kg) > Q (17 kg) > T

From this arrangement, the persons lighter than S (to the right of S) are Q and T. Thus, there are exactly two persons lighter than S.