Answer 1: Option A
Explanation:
The man firstly faces the direction OA. On moving 45 degree clockwise[Please check carefully always if clockwise or anticlockwise], he faces the direction OB.
Now again he moved 180 degree clockwise, now he will be facing OC. From here he moved 270 degree anticlockwise, Finally he is facing OD, which is South west.
Now again he moved 180 degree clockwise, now he will be facing OC. From here he moved 270 degree anticlockwise, Finally he is facing OD, which is South west.
Answer 2: Option D
Explanation:
The man firstly faces the direction OA. On moving 45 degree clockwise, he faces the direction OB.
Now again he moved 180 degree clockwise, now he will be facing OC. From here he moved 45 degree anticlockwise, Finally he is facing OD, which is South direction.
Now again he moved 180 degree clockwise, now he will be facing OC. From here he moved 45 degree anticlockwise, Finally he is facing OD, which is South direction.
Answer 3: Option B
Explanation:
Raviraj starts from home at A, moves 20 Km in south upto B. Then he turns right and moves 10 Km upto C, then he turns right and moves 20 Km upto D, then he turns lefts and moves 20 Km upto E.
So from image it is clear that, if he moves straight then he will have to move AD+DE, AD = BC = 10 Km
So, he will have to move 10 + 20 = 30 Km
So from image it is clear that, if he moves straight then he will have to move AD+DE, AD = BC = 10 Km
So, he will have to move 10 + 20 = 30 Km
Answer 4: Option C
Explanation:
Clearly, the child moves from A to B 90 metres eastwards upto B, then turns right and moves 20 metre upto C, then turns right and moves upto 30 metre upto D. Finally he turns right and moves upto 100 metre upto E.
So AB = 90 metre, BF = CD = 30 metre,
So, AF = AB - BF = 60 metre
Also DE = 100 metre, DF = BC = 20 metre
So, EF = DE - DF = 80 metre
as we can see in image that triangle AFE is a right angled triangle and we are having two sides, need to calculate third one, so we can apply Pythagoras theorem here
\begin{aligned}
A = AE = \sqrt{AF^2 + EF^2} \\
= \sqrt{(60)^2+(80)^2} \\
= \sqrt{3600+6400} \\
= \sqrt{10000} = 100
\end{aligned}
So from starting point his father was 100 metre away.
So AB = 90 metre, BF = CD = 30 metre,
So, AF = AB - BF = 60 metre
Also DE = 100 metre, DF = BC = 20 metre
So, EF = DE - DF = 80 metre
as we can see in image that triangle AFE is a right angled triangle and we are having two sides, need to calculate third one, so we can apply Pythagoras theorem here
\begin{aligned}
A = AE = \sqrt{AF^2 + EF^2} \\
= \sqrt{(60)^2+(80)^2} \\
= \sqrt{3600+6400} \\
= \sqrt{10000} = 100
\end{aligned}
So from starting point his father was 100 metre away.
Answer 5: Option D
Explanation:
Clearly, Kunal moves from A 10 Km northwards upto B, then moves 6 Km southwards upto C, turns towards east and moves 3 km upto D.
Then AC = (AB-BC) = 4 Km
So Kunal distance from starting point A
\begin{aligned}
AD = \sqrt{AC^2 + CD^2} \\
= \sqrt{4^2+3^2} \\
= \sqrt{25} = 5
\end{aligned}
So AD is 5 Km also with reference to starting point Kunal's direction is North-East.
Then AC = (AB-BC) = 4 Km
So Kunal distance from starting point A
\begin{aligned}
AD = \sqrt{AC^2 + CD^2} \\
= \sqrt{4^2+3^2} \\
= \sqrt{25} = 5
\end{aligned}
So AD is 5 Km also with reference to starting point Kunal's direction is North-East.
Answer 6: Option C
Explanation:
Please check the movements of Gaurav in the figure.
Now, Gaurav distance from his initial position A to E
AE = (AD+DE) = 40 + 20 = 60 metres.
Now, Gaurav distance from his initial position A to E
AE = (AD+DE) = 40 + 20 = 60 metres.
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