# Clock puzzles Questions

A) 20 | B) 22 |

C) 24 | D) 48 |

Explanation:

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clcok only).

So, in a day, the hands point in the opposite directions 22 times.

A) 145 | B) 150 |

C) 155 | D) 160 |

Explanation:

Angle traced by hour hand in 12 hrs = 360º.

Angle traced by hour hand in 5 hrs 10 min. *i.e., 31/6 hrs *= ${\left(\frac{360}{12}*\frac{31}{6}\right)}^{\xb0}$= 155º

A) 2 p.m. on Tuesday | B) 2 p.m. on Wednesday |

C) 3 p.m. on Thursday | D) 1 p.m. on Friday |

Explanation:

Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours.

The Watch gains $\left(2+4\frac{4}{5}\right)$ min. or $\frac{34}{5}$ min.in 170 hrs.

Now, $\frac{34}{5}$min.are gained in 170 hrs.

2 min.are gained in$\left(170*\frac{5}{34}*2\right)$ hrs=50 hrs

Watch is correct 2 days 2 hrs. after 12 p.m. on Monday *i.e.,* it will be correct at 2 p.m. on Wednesday

A) 120 degrees | B) 125 degrees |

C) 130 degrees | D) 135 degrees |

Explanation:

Angle traced by hour hand in 12 hrs. = 360º.

Angle traced by it in 11/3 hrs= ${\left(\frac{360}{12}*\frac{11}{3}\right)}^{\xb0}$=${110}^{\xb0}$

Angle traced by minute hand in 60 min. = 360º.

Angle traced by it in 40 min. =(360/60*40)${\xb0}^{}$=240${\xb0}^{}$

Required angle (240 - 110)º = 130º.

A) 5 degrees | B) 10 degrees |

C) 15 degrees | D) 25 degrees |

Explanation:

Since at 4 : 20 the minute hand will be at 4 and the angle between them will be same as the distance covered in degree by the hour hand in 20 minutes.

Required angle = distance of hour hand = speed × time = $\frac{1}{2}$× 20 =10 degrees.

A) 180/14 min past 10 | B) 180/11 min past 9 |

C) 148/7 min past 10 | D) 154/11 min past 9 |

Explanation:

At 9 o’clock, the hour hand is at 9 and the minutes hand is at 12, i.e., the two hands are 15 min. spaces apart.

So, the minute hand should gain = (30 - 15) minutes = 15 minutes

55 minutes will be gained in 60 min.

15 minutes spaces will be gained in ((60/55) x 15) min. = 180/11 min.

The hands will be in the same straight line but not together i.e.,in 180 degrees at 180/11 min. past 9.

A) 1 10/11 minutes past 2 | B) 1 10/11 minutes past 3 |

C) 1 11/10 minutes past 3 | D) 11 10/11 minutes past 2 |

Explanation:

Since, in one hour, two hands of a clock coincide only once, so, there will be value.

Required time $T=\frac{2}{11}\left(H\times 30+{A}^{o}\right)$ minutes past H.

Here H - initial position of hour hand = 2 (since 2 O'clock)

A° = Required angle = 0° (Since it coincides)

$T=\frac{2}{11}\left(2\times 30+{0}^{o}\right)$ minutes past 2

=> $\mathbf{1}\frac{\mathbf{10}}{\mathbf{11}}$ minutes past 2