Direction TestsMCQPYQ June 19Question 2201 of 165
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Sangeeta leaves from her home. She first walks 30 meters in north-west direction and then 30m in south-west direction, next she walks 30 meters in south-east direction. Finally she turns towards her house. In which direction is she moving now.

Options

ANorth - East
BNorth - West
CSouth - East
DSouth - West
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Correct Answer

✅ Option a — North - East

All Options:

  • ANorth - East
  • BNorth - West
  • CSouth - East
  • DSouth - West

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Detailed Solution & Explanation

Let Sangeeta's home be at the origin H(0,0)\displaystyle H(0, 0) on the Cartesian plane, where the positive y\displaystyle y-axis represents North, the negative y\displaystyle y-axis represents South, the positive x\displaystyle x-axis represents East, and the negative x\displaystyle x-axis represents West.
1. **First Movement (North-West)**: She walks 30\displaystyle 30 meters North-West. The vector for this movement is:
v⃗1=30cos⁡(135∘)i^+30sin⁡(135∘)j^=−3012i^+3012j^≈−21.21i^+21.21j^\displaystyle \vec{v}_1 = 30 \cos(135^\circ)\hat{i} + 30 \sin(135^\circ)\hat{j} = -30\frac{1}{\sqrt{2}}\hat{i} + 30\frac{1}{\sqrt{2}}\hat{j} \approx -21.21\hat{i} + 21.21\hat{j}
Her position becomes P1=(−152,152)\displaystyle P_1 = (-15\sqrt{2}, 15\sqrt{2}).
2. **Second Movement (South-West)**: From P1\displaystyle P_1, she walks 30\displaystyle 30 meters South-West. The vector for this movement is:
v⃗2=30cos⁡(225∘)i^+30sin⁡(225∘)j^=−3012i^−3012j^≈−21.21i^−21.21j^\displaystyle \vec{v}_2 = 30 \cos(225^\circ)\hat{i} + 30 \sin(225^\circ)\hat{j} = -30\frac{1}{\sqrt{2}}\hat{i} - 30\frac{1}{\sqrt{2}}\hat{j} \approx -21.21\hat{i} - 21.21\hat{j}
Her new position is:
P2=P1+v⃗2=(−152−152,152−152)=(−302,0)≈(−42.43,0)\displaystyle P_2 = P_1 + \vec{v}_2 = (-15\sqrt{2} - 15\sqrt{2}, 15\sqrt{2} - 15\sqrt{2}) = (-30\sqrt{2}, 0) \approx (-42.43, 0)
3. **Third Movement (South-East)**: From P2\displaystyle P_2, she walks 30\displaystyle 30 meters South-East. The vector for this movement is:
v⃗3=30cos⁡(315∘)i^+30sin⁡(315∘)j^=3012i^−3012j^≈21.21i^−21.21j^\displaystyle \vec{v}_3 = 30 \cos(315^\circ)\hat{i} + 30 \sin(315^\circ)\hat{j} = 30\frac{1}{\sqrt{2}}\hat{i} - 30\frac{1}{\sqrt{2}}\hat{j} \approx 21.21\hat{i} - 21.21\hat{j}
Her new position is:
P3=P2+v⃗3=(−302+152,0−152)=(−152,−152)≈(−21.21,−21.21)\displaystyle P_3 = P_2 + \vec{v}_3 = (-30\sqrt{2} + 15\sqrt{2}, 0 - 15\sqrt{2}) = (-15\sqrt{2}, -15\sqrt{2}) \approx (-21.21, -21.21)
4. **Final Movement (Towards Home)**: Sangeeta now turns and moves towards her home at H(0,0)\displaystyle H(0,0). The vector pointing from her current position P3(−152,−152)\displaystyle P_3(-15\sqrt{2}, -15\sqrt{2}) to her house H(0,0)\displaystyle H(0,0) is:
v⃗home=H−P3=(0−(−152))i^+(0−(−152))j^=152i^+152j^\displaystyle \vec{v}_{\text{home}} = H - P_3 = (0 - (-15\sqrt{2}))\hat{i} + (0 - (-15\sqrt{2}))\hat{j} = 15\sqrt{2}\hat{i} + 15\sqrt{2}\hat{j}
Since both the x\displaystyle x and y\displaystyle y components of this direction vector are positive, she is moving in the North-East direction.

**Note on Typographical Error**: The textbook answer key lists Option B (North-West) as correct, which is a typographical error. As mathematically shown above, she must travel in the North-East direction to reach her home from her final position. Therefore, North-East (Option A) is the mathematically correct answer.
Hence, **Option A** is the correct answer.

About This Chapter: Direction Tests

Paper

Paper 3: Quantitative Aptitude

Weightage

5 Marks

Key Topics

Direction Sense Test

This chapter covers Direction Sense Test and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

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Exam Strategy Tip

This topic carries 5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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