Question

A man is 5m to the South of a certain point. Another is 3m to the west of the same
point. The first man then moves 8 m towards North and Second 7m towards East.
The distance between them now is
?
a) 6 m
b) 8 m
c) 4 m
d) 5 m​

Answer 1

GIVEN:−⤵️

A man is 5m to the south of a certain point.

Another man is 3m to the west of the same point.

The first man then moves 8 m towards North and the second man moves 7 m towards East.

TO FIND:−⤵️

If the distance between the 2 men is :

a) 6 m

b) 8 m

c) 4 m

d) 5 m

SOLUTION:−⤵️

Let the certain point be A.

The first man :

Position = 5 m to the south of A

Distance covered = 8 m towards the north of A from its initial position.

The man displaces 8 m in the opposite direction.

So, distance from A = 8 - 5 = 3 m to the north of A

____________________________

The second man :

Position = 3 m to the west of A

Distance covered = 7 m towards the east of A from its initial position.

The man displaces 7 m in the opposite direction.

So, distance from A = 7 - 3 = 4 m to the east of A

____________________________

Distance between the 2 men (diagram) = BC

Applying Pythagoras Theorem,

BC² = AB² + AC²

→ BC² = 3² + 4²

→ BC² = 9 + 16

→ BC = √25

→ BC = 5 m

Therefore, the distance between the 2 men is 5 m.

The answer is d) 5 m.

Answer 2

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Let the certain point be A.

The first man Position = 5 m to the south of A

Distance covered = 8 m towards the

north of A from its initial position.

The man displaces 8 m in the opposite

direction.

So, distance from A 8 - 5 = 3 m to thenorth of A

The second man:

Position = 3 m to the west of A

Distance covered = 7 m towards the east

of A from its initial position.

The man displaces 7 m in the opposite

So, distance from A = 7 -3 = 4 m to theeast of A

Distance between the 2 men (diagram)

BC

Applying Pythagoras Theorem,

BC^2 = AB^2 + AC^2

BC ^2 =3^2 + 4^2

BC^2 = 9 + 16

BC = √25

βC= 5m

Therefore, the distance between the 2 men is 5 m.