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9 votes
9 votes

Consider a well functioning clock where the hour, minute and the seconds needles are exactly at zero. How much time later will the minutes needle be exactly one minute ahead ($1/60$ th of the circumference) of the hours needle and the seconds needle again exactly at zero?

Hint: When the desired event happens both the hour needle and the minute needle have moved an integer multiple of $1/60$ th of the circumference.

  1. $144$ minutes
  2. $66$ minutes
  3. $96$ minutes
  4. $72$ minutes
  5. $132$ minutes 
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5 Answers

Best answer
11 votes
11 votes
The minute needle should be exactly one minute ahead of hour needle.

Difference between min. needle and hr. needle is equal to one minute.

In 1 minute, distance covered by minute needle $=\dfrac{360 ^{\circ}}{60}= 6^{\circ}$

Suppose, after $x$ minutes, hour needle and minute needle are separated by $6^{\circ}$ .

In $x$ minutes, distance covered by minute needle $=\dfrac{ 360 ^{\circ} }{60} \times x = 6x ^{\circ}$

In $x$ minutes, distance covered by hour needle $=\dfrac{ 360 ^{\circ} }{12*60} \times x = \dfrac{x}{2} ^{\circ}$

$\therefore$ Difference between minute needle and hour needle in $x$ minutes $=\left(6x ^{\circ} - \dfrac{x}{2} ^{\circ}\right)$

$\therefore$ $(6x ^{\circ} - \dfrac{x}{2} ^{\circ}) = 6 ^{\circ}$

$\implies 12x - x = 12$

$\implies11x = 12$

$\implies x = \dfrac{12}{11}$

It is given in question that second hand is $0$ or minute hand has traversed an integral multiple of minutes. So, smallest possible value of $x = 12 \times 11 = 132$ minutes.

Option e.
selected by
3 votes
3 votes

Let the event takes place at $H$ hr $M$ min

Given the condition We can see $11M$ - $60H$ = $12.$

Only e satisfies this.

edited by
2 votes
2 votes

Correct option: (E)

Angular diff. every min. b/w min. hand and hour hand = 5.5 degree

Initially, angular diff. b/w hour and min. hand = 0 degree
Suppose, after x min, angular diff. = 6 degree, (1/60 th of the circumference = (1/60) * 360 degree)
Therefore, angular distance covered in x min. = 6 degree
=> 5.5degree * x = 6 degree
=> x = (12/11) min

Also, the second hand is again exactly at zero after completing a full rotation , i.e. every minute.

Therefore, the conditions given in the question are satisfied after y min. such that
y = lcm((12/11), 1) min. = lcm(12, 11) min. = 132 min. 
  

1 votes
1 votes

Option (e) is correct ..

As minute hand progresses 6∘ in one minute & Hour hand progresses 1/2 in one minute , And second Hand 360∘ in one minute ..So it is always at 0 for all options..

Answer:

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