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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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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.
by Boss
selected by
+1

ma'am, I am not able to understand this line "smallest possible value of x=12×11=132" ...

+4
see, value of $x$ should be multiples of $11$ i.e. $11, 22, 33, ...$

Here,  only $66$ and $132$ are the multiples of $11$.

Now, in $66$ minutes, min. needle covers $6^{\circ} * 66 = 396^{\circ}$

$= 360^{\circ} + 36^{\circ}$

$= 36^{\circ}$

hr. needle covers $\dfrac{66}{2}^{\circ} = 33^{\circ}$

distance covered by (min. needle - hr. needle) = $36^{\circ} - 33^{\circ}$ = $3^{\circ}$

so, $66$ minutes can be omitted.

Now, in $132$ minutes, min. needle covers $6^{\circ} * 132 = 792^{\circ}$

$= (360*2) ^{\circ}+ 72^{\circ}$

$= 72^{\circ}$

hr. needle covers $\dfrac{132}{2} ^{\circ}= 66 ^{\circ}$

distance covered by (min. needle - hr. needle) = $72^{\circ} - 66^{\circ}$ = $6^{\circ}$

So, $132$ minutes will be the correct answer.
0
Thank you ma'am
0
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.)

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

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

Only e satisfies this.

edited
0
Can you please explain how this equation arised?

I am not able to follow.
0

@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.

+1 vote

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..

by Boss
+1 vote

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.

by Junior
Please correct me if I am wrong. I think answer is (b) 66 min.

Reason :

Here it is given current time is 12:00 pm. After 66 min , time will be 1:06 p.m . So , hour's hand will be at 1 and minute's hand will be exactly 1 place ahead.
by Active
0
At 1:06 hours handle will be slightly over 1.
0
Option (e) is correct