A child is looking for his father. He went 90 meters in the east before turning to his right. He went 20 meters before turning to his right again to look for his father at his uncle's place 30 meters from this point. His father was not there. From there, he went 100 meters to his north before meeting his father in a street. How far did the son meet his father from starting point ?

Option c: 100 m. Solution: Let starting point be (0,0) . - Walks 90 m East (90, 0) . - Turns right (South) and walks 20 m (90, -20) . - Turns right (West) and walks 30 m (60, -20) . - Walks 100 m North (60, 80) . The distance from starting point (0,0) to the final position (60, 80) is: Distance = sqrt(602 + 802) = sqrt(3600 + 6400) = sqrt(10000) = 100 m Hence, Option C is the correct answer.

A child is looking for his father. He went $90$ meters in the east before turning to his right. He went $20$ meters before turning to his right again to look for his father at his uncle's place $30$ meters from this point. His father was not there. From there, he went $100$ meters to his north before meeting his father in a street. How far did the son meet his father from starting point ?

The correct answer is Option c: $100$ m. Let starting point be $(0,0)$. - Walks 90 m East $\implies (90, 0)$. - Turns right (South) and walks 20 m $\implies (90, -20)$. - Turns right (West) and walks 30 m $\implies (60, -20)$. - Walks 100 m North $\implies (60, 80)$. The distance from starting point $(0,0)$ to the final position $(60, 80)$ is: $$\text{Distance} = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100\text{ m}$$ Hence, **Option C** is the correct answer.

Direction TestsMCQMTP June 24 Series IQuestion 2326 of 165

All Questions

A child is looking for his father. He went $\displaystyle 90$ meters in the east before turning to his right. He went $\displaystyle 20$ meters before turning to his right again to look for his father at his uncle's place $\displaystyle 30$ meters from this point. His father was not there. From there, he went $\displaystyle 100$ meters to his north before meeting his father in a street. How far did the son meet his father from starting point ?

Options

\displaystyle 80

\displaystyle 90

\displaystyle 100

\displaystyle 110

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option c — $\displaystyle 100$ m

All Options:

A $\displaystyle 80$ m
B $\displaystyle 90$ m
C $\displaystyle 100$ m
D $\displaystyle 110$ m

Detailed Solution & Explanation

Let starting point be

\displaystyle (0,0)

.
- Walks 90 m East

\displaystyle \implies (90, 0)

.
- Turns right (South) and walks 20 m

\displaystyle \implies (90, -20)

.
- Turns right (West) and walks 30 m

\displaystyle \implies (60, -20)

.
- Walks 100 m North

\displaystyle \implies (60, 80)

.
The distance from starting point

\displaystyle (0,0)

to the final position

\displaystyle (60, 80)

is:

\text{Distance} = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100\text{ m}

Hence, **Option C** 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.

View Official ICAI Syllabus

Exam Strategy Tip

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

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