150+ RRB NTPC Previous Solved Papers PDF |
Free Download Now |

RRB NTPC Free Study Material PDF |
Free Download Now |

RRB NTPC Free Mock Test |
Attempt It Now |

Quant Booster eBook – A Complete Maths Shortcut eBook |
GET Book Now |

SSC CGL,CPO,GD 2018 Study Material & Book Free PDF |
Free Download Now |

Q1 Option A

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.

Q2. Option D

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.

Q3. Option B

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

Q4. Option C

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

A=AE=√(〖AF〗^2+〖EF〗^2 )

=√(〖60〗^2+〖80〗^2 )

=√(3600+6400)

=√10000

=1000

So from starting point his father was 100 metre away.

Q5 Option D

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

AD= √(AC^2+CD^2 )

= √(4^2+3^2 )

=√25=5

So AD is 5 Km also with reference to starting point Kunal’s direction is North-East.

Q6. Option C

Now, Gaurav distance from his initial position A to E

AE = (AD+DE) = 40 + 20 = 60 metres.

Q7. Option B

Now dog is facing North

Q8. Option A

Final direction will be north-east with reference to the starting position.

Q9. Option C

The movements of Rohit are shown in figure.

Rohit’s distance from the starting point A will be

AE = AD+DE = 20 + 15 = 35 metre

And direction with reference to the starting point is east.

Q10. Option B

Distance from the P to Q is 10 metres and direction of Q with reference to to P is west.

GovernmentAdda Recommends |
Online Tyari Mock Test |

RRB Group D Speed Test Attempt Now |
Attempt Now |

Railway RRB Pshyco Complete Package |
Get It Now |

SSC CHSL, CGL Mock Test (1 free+30 Paid) |
Attempt Now |

RRB JE Free Study Material PDF Books |
Free Download Now |

FCI Free Study Material PDF Books |
Free Download Now |

SSC JE Free Study Material PDF Books |
Free Download Now |

Download Quant Power Bank (1000+ Pages) |
Free Download Now |

Reasoning & GI Booster -Power Bank Book (1200+ Pages) |
Free Download Now |