SOLUTION: four young children are seated next to each other on a bench. The child on the left cannot be older than the child on the right. The ages of 1st and 3rd children differ by 8 years

Question 1188708: four young children are seated next to each other on a bench. The child on the left cannot be older than the child on the right. The ages of 1st and 3rd children differ by 8 years. No child is below 2 years old, neither any of them is older than 10 years old. Then 2nd child is 7 years younger than the 4th child .What are the ages of the four children?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

I will assume the "child on the left" is the first child and the one on the right is the fourth....

Let's call them, left to right, A, B, C, and D:

(A, B, C, D)

Given:
(1) A is not older than D
(2) The age difference between A and C is 8 years
(3) The minimum and maximum ages are 2 and 10
(4) B is 7 years younger than D

From (2) and (3), the ages of A and C are either 2 and 10, or 10 and 2:

(5) (2, B, 10, D) or (10, B, 2, D)

From (4), the ages of B and D are either 2 and 9, or 3 and 10:

(6) {A, 2, C, 9) or (A, 3, C, 10)

Combining (5) and (6), we get four possibilities:
(2, 2, 10, 9)
(2, 3, 10, 10)
(10, 2, 2, 9)
(10, 3, 2, 10)

Those four possibilities all satisfy given conditions (2), (3), and (4); we need to see which one(s) satisfy condition (1).

A is older than D (so condition (1) is not satisfied) in only one of the four possibilities; so we see that three of the four possibilities we have satisfy (1):

(2, 2, 10, 9)
(2, 3, 10, 10)
(10, 3, 2, 10)

All three of these sets of ages satisfy all the conditions of the problem. Since the statement of the problem suggests that there is a single solution, the problem is faulty.

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