Umesh went directly from P to Q, a distance of 9 feet. He then turned right…
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Umesh went directly from P to Q, a distance of 9 feet. He then turned right and walked 4 feet. After this, he turned right again and walked a distance equal to that from P to Q. Finally, he turned right once more and walked 3 feet. How far is he now from P?
- A.
6 feet
- B.
5 feet
- C.
1 foot
- D.
0 feet
Show answer & explanation
Correct answer: C
Concept: In a walking/turning problem, split the path into two perpendicular axes -- the direction of the very first leg, and the direction faced after the first right turn. A right turn rotates the walking direction by 90 degrees; two legs along the same axis add when they point the same way and cancel when they point opposite ways (fully cancelling when they are equal in length).
Application:
Leg 1 (P to Q, 9 ft): this sets the first axis, call it axis A. Movement so far: 9 ft along axis A, 0 ft along axis B (the perpendicular axis).
Leg 2 (turn right, 4 ft): facing shifts onto axis B. Movement: 4 ft along axis B.
Leg 3 (turn right, 9 ft): facing returns to axis A, pointing opposite to Leg 1. Since this leg's length (9 ft) equals Leg 1's length, it exactly cancels Leg 1's displacement -- axis A movement returns to 0.
Leg 4 (turn right, 3 ft): facing returns to axis B, opposite to Leg 2. This leg subtracts from the axis B offset built by Leg 2: 4 minus 3 = 1 ft remains along axis B.
Net position: 0 ft along axis A, 1 ft along axis B, so the straight-line distance from P is 1 ft.
Cross-check:
Placing P at the origin and facing north for Leg 1: Q = (0, 9). Turning right (facing east) and walking 4 ft reaches (4, 9). Turning right (facing south) and walking 9 ft reaches (4, 0). Turning right (facing west) and walking 3 ft reaches (1, 0). The distance from P (0, 0) to (1, 0) is 1 ft, confirming the axis method.
Required distance from P = 1 foot.