TIFR CSE 2013 | Part A | Question: 20

Question

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.

• A. $144$ minutes
• B. $66$ minutes
• C. $96$ minutes
• D. $72$ minutes
• E. $132$ minutes 

Answer 1

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.

Comments

Thank you ma'am

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Let x be the number of minutes.

$6^{\circ}x - \tfrac{1}{2}^{\circ}x = 360^{\circ}k + 6^{\circ}$

We will have to find x and k such that x is a whole number (i;e without any decimal part).

Put k=1 then x=66.54$^{\circ}$ (we want whole numbers only hence we don't take this as answer)

Put k=2 then x=132$^{\circ}$. (This is the required answer.)

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please explain how that equation is formed?

Answer 2

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

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

Only e satisfies this.

Comments

Can you please explain how this equation arised?

I am not able to follow.

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@Manoja Rajalakshmi A Minute hand angle resets after an hour (360 degrees). If we have total minutes expressed in hours and minutes (H:M), we should consider the minutes part for angle traversed for minute hand, whereas for hour hand we need to consider full minutes (60*H+M).

So, 6*M - (60*H+M)*0.5 = 6 => 11M - 60H = 12 (the equation in @sudipta roy's derivation)

132 minutes will be 2 hours and 12 minutes, minute hand traverses 12*6 = 72 degrees and hour hand traverses 132/2 = 66 degrees, that satisfies above equation.

Hope this helps.

Answer 3

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. 
  

Answer 4

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..