Six boys \( A, B, C, D, E \) and \( F \) are marching in a line. They are arranged according to their heights, the tallest being at the back and the shortest in the front. \( F \) is between B and \( A \) . \( E \) is shorter than D but taller than \( C \) who is taller than \( A \) . \( E \) and \( F \) have two boys between them A is not the shortest among them. If we start counting from the shortest, which boy is fourth in the line?
To solve this puzzle, we need to combine all the clues to create a single height ranking of the six boys from shortest to tallest.
- The key clue provides the base of the ranking: "E is shorter than D but taller than C who is taller than A."
- This clue breaks down into a single, clear chain of heights: D > E > C > A (where > means "is taller than").
- Now we need to place F and B using the clue: "F is between B and A".
- In our current ranking (D > E > C > A), A is the shortest person so far. For F to be between A and B, both F and B must be shorter than A.
- This gives us the order: A > F > B.
- We can now combine our two chains to get the complete order from tallest to shortest:
D > E > C > A > F > B
- Let's check the final clue: "E and F have two boys between them".
- In our final ranking `D > E > C > A > F > B`, the boys positioned between E and F are indeed C and A. Our order is correct.
The question asks who is fourth in the line, counting from the shortest. The line is arranged with the shortest boy at the front.
Final Answer: The fourth boy from the shortest is C.
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