**CSIR-NET General Aptitude:** **CSIR NET** General Aptitude plays a crucial role to qualify for the exam as it contains 30 marks out of 200. So, to prepare well for Part A, aspirants need to follow some quick and smart preparation strategies. We at BYJU'S Exam Prep have always come up with different and quick study notes to help you practice the concepts during your last-minute preparation for the exam.

Logical reasoning is one of the scoring sections of the **CSIR-NET** exam. Due to the pandemic, the exam dates have been rescheduled till July month, as the exam is near and candidates should go through the **CSIR-NET Syllabus** to ensure which topics are left to prepare for the exam. We all are aware of the famous quote *i.e.,**Practice makes the man perfect* and to ensure you get enough questions for practise, we are sharing the **Concept of Direction Sense Test for CSIR-NET 2021**. We will be providing the **answer and solution pdf** to cross-check the answers. Scroll down the article below and practice these questions to ace **CSIR-NET 2021**!

## Concept of Direction Sense Test

As the name of the topic clearly explains that we deal with finding distance or direction in the questions of this topic. But for dealing with the questions from this topic, we must have a very clear idea about two things.

- Basic Directions
- Pythagoras theorem

**1. Basic Directions: **We have 8 basic directions which should be crystal clear to us for attempting distance and direction questions.

One key point that should be kept in mind is that If not mentioned we always assume that the person is facing north.

**2. Pythagoras theorem: **According to this theorem,” The square of Hypotenuse is always equal to the sum of the squares of the other two sides of the right angle triangle”.

Suppose we have a triangle having base p, height q and hypotenuse r. Then according to this theorem:

**p ^{2}+q^{2}=r^{2}**

Now you have the basics required to attempt Distance & Direction questions. So let us try to look at a few questions on the same so that you will get to know the proper approach to solve these questions.

**Some other basics:**

1.B is to the east of A.

2. B is to the west of A.

3. B is to the north of A.

4. B is to the south of A.

5. B is to the North East of A.

6. B is to the North West of A.

7. B is to the South East of A.

8. B is to the South West of A.

**Direction: Ashok started walking towards South. After walking 50 meters he took a right turn and walked 30 meters. He then took a right turn and walked 100 meters. He then took a left turn and walked 30 meters and stopped. How far and in which direction was he from the starting point?**

**Solution:**

Ashok started walking towards the south.

After walking 50 meters...

..he took a right turn…

Some people have doubt in deciding left or right in direction questions; they can replace **right by clockwise **and **left by anticlockwise**. So now moving right (clockwise) from the tip of the arrow.

…and walked 30 meters.

He took a right turn…

…and walked 100 meters.

He took a left turn…

…and walked 30 meters.

Now for finding how far he has moved, we will check two things:

- Horizontal displacement
- Vertical displacement

Horizontal displacement = 30+30 = 60m

Vertical displacement = 100-50 = 50m

Final displacement = √(60^{2}+50^{2}) = √(3600+2500) = √6100 = 10√61 m

Now for finding the direction with reference to initial position, we will draw a line joining two points which will give us the direction.

We can clearly see that the direction in which Ashok moved is northwest.

So the final answer for this question will be “Ashok moved 10√61 meters in northwest direction”.

**Directions: Jay starts his van from point X and covers a distance of 10 km towards west, then he turns north and covers a distance of 7 km. Again, he takes a right turn and covers 25 km. Now he covers 6 km, after taking a left turn. At last, he takes a left turn and covers 15 km and stops at point Z.**

**Solution:**

Jay starts his van from point X and covers a distance of 10 km towards west

then he turns north

and covers a distance of 7 km

Again, he takes a right turn

…and covers 25 km

Now he covers 6 km, after taking a left turn.

At last, he takes a left turn…

…and covers 15 km and stops at point Z.

**Q. Towards which direction was the van running before stopping at point Z?**

**North****East****West****South****None of these**

**Solution:** We can clearly see that van was running towards the west before stopping at point Z.

So the correct answer is C.

**Q. How far is Jay from point X?**

**23 km****25 km****17 km****50 km****None of these**

**Soution:**

Horizontal movement = 10-25+15 = 0 kms

Vertical movement = 7+6 = 13 km

So final movement = = √(0^{2}+13^{2}) = √169 = 13 km

Direction of movement is North.

**Question - Point B is 40√2m to the East of point A. Point C is 50m in the direction, which is 225 degree anticlockwise north of point B. Point D is 30m to the North-East of point C. Rahul started from point A, after crossing point B reaches to point E in**the east

**of point A. He took a right turn and after walking some distance he reaches to point D. find the total distance travelled by him?**

**Solution -**Let us first draw some lines and take points, F and G such that Line CF is perpendicular to BE and also line DG is the extension of line CD, which meets at G when BE line is extended in east direction of point E.

For BE and ED:-

In triangle BCG, Angle C is 90 degree, angle B and G are of 45 degrees, and the total length of line CG must be 50m (symmetry).

BG ²

_{=}BC

^{2}+ CG

^{2}(PT theorem)

BG

^{2}= 50

^{2}+ 50

^{2}= 2*50

^{2}

BG = 50 √2

Length of DG = CG – CD = 50 – 30 = 20m.

In triangle DEG

DE/DG = Sin45

DE/20 = 1/ √2, DE = 20/ √2 = 10√2

EG/DG = Cos45

EG/20 = 1/√2, EG = 20/√2 = 10√2

The length BE = BG – EG = 50√2 - 10√2 = 40√2

Total Distance covered by him = AB + BE + ED = 40 √2 + 40 √2 + 10 √2 = 90 √2

## Shadow Based Concepts of Direction Sense

**In Morning (During Sunrise)**

The sun rises in East direction. At the time of the sunrise/ morning, if a man is standing, the shadow of a man always falls on West.

**In Evening (During Sunset )**

The sun rises in the West direction. At the time of the sunset/ evening, if a man is standing, the shadow of a man always falls on East.

**At 12'o clock **

There is no shadow at this time. At this time, the sun is exactly above our head, so no shadow is formed at 12'o clock.

**Question: One morning, Sonu and Akhil were standing to face each other. Sonu’s shadow fell towards his right. Which direction was Akhil facing?**

**Solution**- In the morning the sun is in the East so the shadow will form in the West. Thus, Sonu’s right is west i.e Sonu is facing south. Hence, Akhil is facing north.

**Key points related to Distance & Direction**

- Always remember the basic directions.
- Pythagoras theorem is valid only for a right angle triangle.
- The direction of the right turn is always in the clockwise direction.
- The direction of the left turn is always in an anti-clockwise direction.
- The direction of the North is to the upwards side.
- The Direction of East is to the right.
- The direction of the West is to left.
- Always approach the question step by step.
- In the End, join the initial and final point to get to know the distance and relative direction.
- If in any of the questions, the relative direction is given. i.e. P is to the north of Q, then you can use the basic directions to get the location of P and Q.

## Download PDF for Direction Sense Test

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