Kenneth ChengMathematical modeling
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A person walks n km east, turns right 150°, then walks 3 km. The final distance from the start is √3 km. Find n.

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TOPIC 01

2022 JM01

The turning angle is 150°; the triangle’s interior angle is its supplement 30°.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

A person walks n km east, turns right 150°, then walks 3 km. The final distance from the start is √3 km. Find n.

Official paper · jm01-2022 · I.15 · PDF 3

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option A333\sqrt3
  2. Option B3/3\sqrt3/3
  3. Option C1/31/3
  4. Option D33
  5. √3 or 2√3

Working and explanation

BUILD THE REASONING

Hint 1
The second displacement has components 3cos150° and −3sin150°.
Hint 2
Square the final distance.
Worked solution

  1. Form the distance equation.

    (n−33/2)2+(−3/2)2=3(n-3\sqrt3/2)^2+(-3/2)^2=3

  2. Factor the resulting quadratic.

    n2−33n+6=(n−3)(n−23)=0n^2-3\sqrt3n+6=(n-\sqrt3)(n-2\sqrt3)=0

E: n=√3 or 2√3.

Checks and common pitfalls: The turning angle is 150°; the triangle’s interior angle is its supplement 30°.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Form the distance equation.
    (n−33/2)2+(−3/2)2=3(n-3\sqrt3/2)^2+(-3/2)^2=3
  • Factor the resulting quadratic.
    n2−33n+6=(n−3)(n−23)=0n^2-3\sqrt3n+6=(n-\sqrt3)(n-2\sqrt3)=0

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