Test: Distance And Direction- 1 - Bank Exams MCQ

As per the given information;

Point Q is in South East of Point P.

Distance between P and Q (PQ) = √(PO² + OQ²)

PQ = √(40² + 30²)

PQ = √(1600 + 900)

PQ = √2500

PQ = 50 m

Hence, "50 metres, Southeast" is the correct answer.

Sonu started from his house and walked 3 km Eastward, then he turned right and walked 2 km, then again he turned left and walked 1 km. Again he turned left and walked 3 km, then turned right and walked 2 km. He turned right again and walked 1 km and reached his school.

Shortest distance between Sonu's home and school is, Home to point A + AD + point D to school = 3 + 2 + 1 = 6 km.
Hence, the distance between Sonu's home and the school is 6 km.
Thus, this is the correct answer.

• If South-east becomes North and North East becomes West, therefore, the whole figure moves through 135º. Hence, West will be South-East.
• See, Actual figure is rotating 135º anticlockwise.
• So, When West will be rotated by same degree anticlockwise. It will hold the place of south-East.

Final distance from original position = AE = AD + DE = 60 metres

To solve this, let's break down Hemant's movements step by step and visualize his path:

1. Hemant starts from his house and walks 3 km eastward.
2. He turns right and walks 2 km south.
3. He turns left and walks 1 km east.
4. He turns right and walks 2 km south.
5. He turns right and walks 1 km west and reaches his school.

Now, let's consider the overall east-west and north-south movement:

• East-West: Hemant initially walked 3 km east, then 1 km east, and finally 1 km west. So, the net eastward distance is 3 + 1 − 1 = 3 km.
• North-South: Hemant walked 2 km south, then another 2 km south, so he walked a total of 2 + 2 = 4 km south.

The minimum distance from Hemant’s house to the school can be found using the Pythagorean theorem (since he moved in perpendicular directions):

Thus, the minimum distance between Hemant’s house and his school is: D: 5 km

To solve this question, let's analyze Sushant's path step-by-step based on the directions provided:

1. Sushant starts at his house and walks 5 km north.
2. He then turns left and walks 3 km west.
3. Next, he turns right and walks 5 km north again.
4. Finally, he turns right and walks 3 km east.

Now let’s assess Sushant's final position relative to his starting point:

• North-South movement: He walked 5 km north, then another 5 km north, totaling 10 km north.
• East-West movement: He walked 3 km west and then 3 km back east, which brings him back to the original east-west position he started from.

Given this analysis, Sushant's final destination is directly north of his house without any deviation east or west. Therefore, the correct answer is:

D: 10 km North

Rohan is walking in south now

The man starts riding in south direction then he rides 2 km in the east (for him his left is the east direction)and then it rides 4km in the south (for him his right is in south direction).

Let's break down Namratha's movements step by step:

1. First Movement: She walks 14 meters west.
2. Second Movement: She turns right and walks 14 meters north.
3. Third Movement: She turns left and walks 10 meters west.
4. Fourth Movement: She turns left again and walks 14 meters south.

Now, let's analyze her position.

• After the first movement, she's 14 meters west of the starting point.
• After the second movement, she's 14 meters west and 14 meters north of the starting point.
• After the third movement, she's 24 meters west and 14 meters north of the starting point.
• After the fourth movement, she moves 14 meters south, so she's now 24 meters west and at the same latitude as her starting point (because the south movement cancels out the north movement).

To find the shortest distance between her starting point and her present position, we just need to calculate the straight-line distance (which is the horizontal distance since her vertical displacement is zero).

Since she is 24 meters west of her starting point, the shortest distance is:

Distance = 24 meters

So the correct answer is 24 meters (Option 3).

To find out how far Jenny is from her original point, we will break down her journey step by step:

1. North: Jenny walks 2.5 km north.
2. West: She turns and walks 2 km west.
3. South: She turns again and walks 1.5 km south.
4. East: Finally, she turns and walks 2 km east.

Now, let's analyze her final position relative to the start:

• North-South movement: Jenny initially walked 2.5 km north, then 1.5 km south, resulting in a net movement of 2.5 − 1.5 = 1.0 km north.
• East-West movement: She walked 2 km west and then 2 km east. This means she ended up exactly where she started in terms of east-west movement, as these cancel each other out.

So, Jenny's final position is 1.0 km north of her starting point. Thus, she is: D: 1 km

A diagram is made according to the directions given in the question. Clearly A will be South-West of C.

From the figure given, clearly the direction is OC which is South-East.

We can solve this problem using the Pythagorean theorem.
First, let's break down Ayush's movements:
1. He walks 100m south.
2. He turns left (which means he is now facing east) and walks 200m.
3. He turns right (which means he is now facing south again) and walks 500m.
Now, let's determine his horizontal (east-west) and vertical (north-south) displacements.
Horizontal displacement:
Ayush only moved horizontally during step 2, where he walked 200m east. Therefore, his total horizontal displacement is 200m.
Vertical displacement:
Ayush moved vertically during steps 1 and 3. In step 1, he walked 100m south, and in step 3, he walked 500m south. Therefore, his total vertical displacement is 100m + 500m = 600m.
Now we can use the Pythagorean theorem to calculate the distance between his starting point (his house) and his final position:
Distance = √(horizontal displacement² + vertical displacement²)
Distance = √(200m² + 600m²)
Distance = √(40,000m² + 360,000m²)
Distance = √(400,000m²)
Distance = 200√10m
So the correct answer is 3 (200√10m).

Step 1: Tanuj starts from point P and walks 40 metres south.
Step 2: He takes a left turn (facing east) and walks 30 metres to reach point Q.
Applying the Pythagorean Theorem:
The path forms a right-angled triangle with:
One side = 40 metres (southward movement)
Another side = 30 metres (eastward movement)
Using the Pythagorean theorem to find the distance from P to Q:
PQ = √(40² + 30²) = √(1600 + 900) = √2500 = 50 metres
Direction of Q from P:
Since Tanuj moved south first and then east, Q is located South-East of P.
Answer:
c) 50 metres, South-East

Required distance= 20 +12
= 32 m in south direction

The last position of the man is P and OEP is a Right angled triangle in which
EP = 3 km and OE = 4 km
Thus,
OP = √(3² +4²)
OP = √25
OP = 5 km.

Let's analyze the path taken by A and the distances involved:

1. A moves towards B's house: A starts moving north towards B's house which is 4 km away.
2. A covers three-fourth of the distance: He covers 4 × 3/4 = 3 km towards the north.
3. A takes a left turn: At this point, A is 1 km away from B's house (since he has covered 3 km out of 4 km). He then takes a left turn, which would be towards the west and runs 2 km.

Now, A is 1 km south of B's house and 2 km west of a direct line north from B’s house. To find the distance from point P (A's new location) to B's house, we use the Pythagorean theorem:

Thus, A would need to cover a distance of: B: √5km  

1. A walks three-fourths of the distance towards B's house, which is:
(3/4) × 4 km = 3 km (north).
After walking 3 km north, A takes a left turn and runs 2 km west.

2. Now, we have a right-angled triangle, where:
- One leg (north) = 3 km,
- Another leg (west) = 2 km.

3. The distance from A’s current position (P) to his house is the hypotenuse of this right-angled triangle:
Distance = √(3² + 2²) = √(9 + 4) = √13 km.

Thus, the distance from A’s current position to his own house is √13 km.

P is in South-West of Y.