Two parallel streets run north–south. A person walks north along the first street and wants to reach the second street, which lies somewhere to his right (east). At one point, he turns 150° to the right and walks for 15 minutes at 20 km/h. Then he turns 60° to the left and walks for 20 minutes at 30 km/h, finally reaching the second street. What is the distance (in km) between the two parallel streets?

Introduction:
This is a challenging direction and distance problem involving vectors and angles. A person changes direction twice and eventually reaches a parallel street. The key is to resolve each leg of the journey into east–west (horizontal) and north–south (vertical) components and then use these to find the separation between the two parallel streets.

Given Data / Assumptions:

• The two streets are straight, parallel, and oriented north–south.
• Initial direction: the person walks north along the first street.
• First turn: 150° to the right from north, then walks 15 minutes at 20 km/h.
• Second turn: 60° to the left from his new direction, then walks 20 minutes at 30 km/h.
• He ends up exactly on the second street.

Concept / Approach:
We choose a coordinate system: north as positive y-axis and east as positive x-axis. The streets are vertical (north–south) lines, so the separation between them is the net east–west (x) displacement from the starting point to the final position. We compute the displacement vectors for each leg using speed × time for distance and trigonometry for direction.

Step-by-Step Solution:
Step 1: Distance of first leg: 20 km/h * (15/60) h = 5 km.Step 2: From north, a right turn of 150° gives a bearing 150° clockwise from north, equivalent to a direction 60° below east (i.e., 60° south of east).Step 3: Components of first leg: x₁ = 5 * cos 60° = 5 * 0.5 = 2.5 km east; y₁ = 5 * (−sin 60°) = 5 * (−√3/2) ≈ −4.33 km (southward).Step 4: Distance of second leg: 30 km/h * (20/60) h = 10 km.Step 5: A left turn of 60° from this heading brings him to due east (90° bearing), so the second leg is purely eastward.Step 6: Components of second leg: x₂ = 10 km east; y₂ = 0 km.Step 7: Total eastward displacement: x = x₁ + x₂ = 2.5 + 10 = 12.5 km.Step 8: Total north–south displacement is irrelevant for street separation; only east–west displacement matters.

Verification / Alternative check:
Because both streets are parallel and oriented north–south, any point on each street has the same x-coordinate along that street. The person starts on the first street (x = 0) and ends on the second street (x = 12.5 km), so the horizontal separation must be 12.5 km. This confirms that 12.5 km is the correct distance between the streets.

Why Other Options Are Wrong:
Distances like 7.5 km, 10.5 km, or 15 km arise if you incorrectly project only one leg or take total path length. The problem asks for perpendicular separation between streets, not total distance walked. 9.5 km does not correspond to any correct component calculation based on the given speeds, times, and angles.

Common Pitfalls:
Typical mistakes involve misinterpreting the 150° right turn (confusing it with 30°), or forgetting that only the east–west displacement gives the distance between parallel north–south streets. Careful diagramming and consistent coordinate axes are essential for such vector-based problems.

Final Answer:
The distance between the two streets is 12.5 km.