Weight, Height and Age of 6 boys

Question

Six Boys – S, R, V, A, H and K – are ranked on the basis of their weight, height and age such that the heaviest person is ranked 1 and the lightest is ranked 6, the tallest is ranked 1 and the shortest is ranked 6, and the oldest is ranked 1 and the youngest is ranked 6. Rank 1 is the highest rank and rank 6 is the lowest rank. Furthermore, it is known that:

1. Only for S, the ranks on the three parameters are consecutive integers.
2. S is heavier than both V and K. The rank of S in age is lower than that of R in height.
3. R is neither the oldest nor the youngest, and he is also neither the heaviest and nor the lightest.
4. There are no pair of persons out of the six in which one is ranked higher than the other on all the parameters.
5. K is the only person whose sum of ranks on the three parameters is a square of a natural number.
6. R’s highest rank is on age and his lowest rank is on weight.
7. On the age parameter, A, whose rank is not the lowest on any of the three parameters, is the only person ranked between R and H.

QUESTIONS :

Q1: What is the sum of weight-wise ranks of two boys for whom R is the only boy ranked between them on the parameter of height?

Q2: Who is the second tallest boy?

Q3: What is the sum of ranks of H on the three parameters?

Q4: How many boys are ranked lower than A on weight but are younger than him?

Please tell the approach after solving

Source : TIME

Answer 1

Let aX = rank in age of X (where X is substituted for a boy's name). Similarly, wX = rank in weight of X and hX = rank in height of X.

First, let's consider R.

From fact 3 and 6, all of aR, wR and hR is not 1 or 6. From fact 1, R's ranks cannot be 2-3-4 or 3-4-5, so it has to be 2-(3/4)-5. From fact 6 we know that aR = 2, hR = 3/4 and wR = 5.

S's ranks cannot be

1-2-3, because there can only be 3 boys with any higher rank than him (rank 1 to S's 2 and ranks 1&2 to S's 3), which leaves at least 2 boys ranked lower than him in all attributes (contradicting fact 4).

Likewise,

4-5-6 is also similarly impossible. 2-3-4 adds up to 9, which contradicts fact 5 because 9 is a square. Hence S's ranks must be 3-4-5, in any order.

Now let's look at K:

Sum of ranks can only be in 3 to 18 inclusive. Squares in those range are 4,9 and 16. 4 and 16 is impossible to achieve without contradicting fact 4 (similar to the 1-2-3 case above), so K's sum is 9. Ruling out 2-3-4 (fact 1) we have 1-2-6 and 1-3-5 as the only possible ranks for K.

We know that

aR = 2. From fact 7, aH must be 4 and aA must be 3. Which makes aS = 5. From fact 2, S is heavier than V and K so wS must be 3, wK must be 6 and wV must be 4. Also, aK= 1, aV= 6, hK= 2, hS = 4 and hR = 3.

hV must be

1, because 4-5-6 and 4-6-6 is impossible. Since A isn't ranked lowest in any attributes, hA must be 5 and hH must be 6. H isn't ranked lower than A in every attribute so wH= 1 and wA= 2.

Knowing all the ranks, the answers are straightforward:

9, K, 11 and 2 respectively

Answer 2

I will have to make a short response for now, as I'm headed out the door in a moment. Please advise if this is the correct solution, and I will elaborate how I came to this result.

10, K, 10 and 2

EDIT. Stupid me. I need my money for the math-classes back.

And another edit, as I misread The first questions.

9, K, 11 and 2