Six persons are arranged in descending order of their weights from left to right. Only T is lighter than Q, whose weight is $17 ext{ kg}$. S is twice as heavy as Q and $14 ext{ kg}$ lighter than N. There are two persons between M and S. R is not the second heaviest among all.
How many persons are lighter than S?

Question

Six persons are arranged in descending order of their weights from left to right. Only T is lighter than Q, whose weight is $17 	ext{ kg}$. S is twice as heavy as Q and $14 	ext{ kg}$ lighter than N. There are two persons between M and S. R is not the second heaviest among all.
How many persons are lighter than S?

Accepted Answer

**Solution:**

1. Given that Q weighs $17 	ext{ kg}$ and only T is lighter than Q, so T weighs less than $17 	ext{ kg}$.

2. S is twice as heavy as Q, so:
   $$	ext{Weight of S} = 2 	imes 17 = 34 	ext{ kg}$$

3. S is $14 	ext{ kg}$ lighter than N, so:
   $$	ext{Weight of N} = 34 + 14 = 48 	ext{ kg}$$

4. There are two persons between M and S. Since the order is descending by weight from left to right, and S is at weight $34 	ext{ kg}$, M must be heavier than S with two persons in between.

5. R is not the second heaviest, so M or N must be the second heaviest.

6. Arranging in descending order by weight:
   - M (heaviest)
   - N ($48 	ext{ kg}$)
   - R
   - S ($34 	ext{ kg}$)
   - Q ($17 	ext{ kg}$)
   - T (lightest)

7. Persons lighter than S are Q and T, which makes **2 persons**.

**Answer:** Two

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