# Aptitude - Clock - Discussion

Discussion Forum : Clock - General Questions (Q.No. 15)

15.

At what time between 9 and 10 o'clock will the hands of a watch be together?

Answer: Option

Explanation:

To be together between 9 and 10 o'clock, the minute hand has to gain 45 min. spaces.

55 min. spaces gained in 60 min.

45 min. spaces are gained in | 60 | x 45 | min or 49 | 1 | min. | |

55 | 11 |

The hands are together at 49 | 1 | min. past 9. |

11 |

Discussion:

47 comments Page 1 of 5.
Sank said:
9 years ago

We can solve this question as a speed - time question.

Initially at 9'o clock hour hand will be at 9 and minute hand will be at 12.

In 12 hour a hour hand gains angle = 360 degree.

So speed of hour hand will be 0.5 deg/min.

In 60 minutes a minute hand gains = 360 degree.

So speed of hour hand will be 6 deg/min.

Let after hour hand covers angle x both the hands coincide.

So angle covered by hour hand = x and by minute hand = 270+x.

x/0.5= (270+x)/6 (as time taken by both will be same).

x = 270/11.

As 1 min = 6 degrees, so we can convert above angle(x) in degree to minute.

So in minute it will be (270/11)/6 = 45/11.

So the time when both of the clock hands will coincide is 49 1/11.

Initially at 9'o clock hour hand will be at 9 and minute hand will be at 12.

In 12 hour a hour hand gains angle = 360 degree.

So speed of hour hand will be 0.5 deg/min.

In 60 minutes a minute hand gains = 360 degree.

So speed of hour hand will be 6 deg/min.

Let after hour hand covers angle x both the hands coincide.

So angle covered by hour hand = x and by minute hand = 270+x.

x/0.5= (270+x)/6 (as time taken by both will be same).

x = 270/11.

As 1 min = 6 degrees, so we can convert above angle(x) in degree to minute.

So in minute it will be (270/11)/6 = 45/11.

So the time when both of the clock hands will coincide is 49 1/11.

Akhilesh said:
8 years ago

My solution is:

Both the hands will meet somewhere between 9 and 10 i.e after the minute hand has reached 9 ( after 45 minutes). In these 45 mins the hour hand must also have traveled forward by some angle.

The angles covered by the hour hand in 45 mins will be 22.5. (how? in one min it covers 1/2 that's why!).

Now, let's suppose it takes another x mins for both of them to reach the point where they will coincide.

The angle traveled by hour hand will be x/2 and by the minute hand will be 6x.

Now in order to coincide with the hour hand, the minute hand has to cover the total angle the hour hand has moved forward.

So, 6x = x/2 + 22.5.

Both the hands will meet somewhere between 9 and 10 i.e after the minute hand has reached 9 ( after 45 minutes). In these 45 mins the hour hand must also have traveled forward by some angle.

The angles covered by the hour hand in 45 mins will be 22.5. (how? in one min it covers 1/2 that's why!).

Now, let's suppose it takes another x mins for both of them to reach the point where they will coincide.

The angle traveled by hour hand will be x/2 and by the minute hand will be 6x.

Now in order to coincide with the hour hand, the minute hand has to cover the total angle the hour hand has moved forward.

So, 6x = x/2 + 22.5.

(3)

Kapil said:
1 decade ago

It's very simple..

in 12 hour a hour hand gains angle= 360 degree

so in one hour it will gain = 360/12 = 30 degree

similarly,

in 60 minutes a minute hand gains = 360 degree

so in 1 minute it will gain = 360/60= 6 degree

so angle gain by hour hand at any time = 30*[y+(x/60)] degree

& angle gain by minute hand=6*x degree

where y=no. of hour (in this question it is '9')

&

x=minutes after the y O' clock..

angle gained by hour hand should be Equuleus to minute hand..

=> 30*[9+(x/60)]=x*6

=> 270+ x/2=6x

=> 540+ x=12x

=> 11x= 540

=> x= 49 1/11

in 12 hour a hour hand gains angle= 360 degree

so in one hour it will gain = 360/12 = 30 degree

similarly,

in 60 minutes a minute hand gains = 360 degree

so in 1 minute it will gain = 360/60= 6 degree

so angle gain by hour hand at any time = 30*[y+(x/60)] degree

& angle gain by minute hand=6*x degree

where y=no. of hour (in this question it is '9')

&

x=minutes after the y O' clock..

angle gained by hour hand should be Equuleus to minute hand..

=> 30*[9+(x/60)]=x*6

=> 270+ x/2=6x

=> 540+ x=12x

=> 11x= 540

=> x= 49 1/11

Gaurav Mittal said:
1 decade ago

Friends.

We know that the hour and the minute hand will coincide somewhere between 9 and 10. At 9:45, the angle between the hour & the minute hand is 22.5 degrees (the hour hand gains 0.5 degree per minute, i.e. 45/2 degrees in 45 mins).

Now, assume that the minute and the hour hand coincide at x mins after 9:45.

So, in x mins, the hour hand moves x/2 degrees while the minute hand moves 6x degrees.

So, 6x - x/2 = 22.5 (i.e. from the original position to the new position)

So, x = 45/11 = 4 1/11.

So, the time at their meeting is 49 1/11.

We know that the hour and the minute hand will coincide somewhere between 9 and 10. At 9:45, the angle between the hour & the minute hand is 22.5 degrees (the hour hand gains 0.5 degree per minute, i.e. 45/2 degrees in 45 mins).

Now, assume that the minute and the hour hand coincide at x mins after 9:45.

So, in x mins, the hour hand moves x/2 degrees while the minute hand moves 6x degrees.

So, 6x - x/2 = 22.5 (i.e. from the original position to the new position)

So, x = 45/11 = 4 1/11.

So, the time at their meeting is 49 1/11.

Rohan said:
9 years ago

Let QH and QM be angular speed of hour and minute hand respectively.

QH = 360/720 = 0.5 (degree/minute) QM = 360/60 = 6 (degree/minute).

Relative angular speed (QM-QH) = 11/2 = 5.5.

Now at 9:45 angle between minute and hour hand will be = 22.5 deg.

Time taken for minute hand to coincide hour hand = 22.5/5.5 = 45/11.

= 45/11 = 4+1/11.

Hence time at which hour and minute hand coincide is:

Hour = 9 minute = 45+4+1/11 = 49+1/11 or 45 1/11 past 9.

QH = 360/720 = 0.5 (degree/minute) QM = 360/60 = 6 (degree/minute).

Relative angular speed (QM-QH) = 11/2 = 5.5.

Now at 9:45 angle between minute and hour hand will be = 22.5 deg.

Time taken for minute hand to coincide hour hand = 22.5/5.5 = 45/11.

= 45/11 = 4+1/11.

Hence time at which hour and minute hand coincide is:

Hour = 9 minute = 45+4+1/11 = 49+1/11 or 45 1/11 past 9.

Laxmi said:
9 years ago

Hello guys.

In previous question one member said that for angle formula.

Less 6 clock use formula theta = 11/2*M-(30*H).

And above 6 clock use formula theta = (30*H)-11/2*M.

This question this above 6.

But @Pradeep use formula is less 6.

I totally confused which one is correct formula in this question.

And when we use angle both the formulas. Any one please explain.

In previous question one member said that for angle formula.

Less 6 clock use formula theta = 11/2*M-(30*H).

And above 6 clock use formula theta = (30*H)-11/2*M.

This question this above 6.

But @Pradeep use formula is less 6.

I totally confused which one is correct formula in this question.

And when we use angle both the formulas. Any one please explain.

Shree Hema said:
1 decade ago

@manish

better use this formula

theta=11(m)/2-30(h)

if both hands coincide degree ll be 0

if both hands r in opposite direction degree will be 180

if both hands r in right angle degree will be 90

here it coincides so 0 degree

we have to find min only so keep that parameter(m) as it is, use minimum hour for hour parameter(h=9)

0=11(m)/2-30(9)

m=49 1/11

Got It???

better use this formula

theta=11(m)/2-30(h)

if both hands coincide degree ll be 0

if both hands r in opposite direction degree will be 180

if both hands r in right angle degree will be 90

here it coincides so 0 degree

we have to find min only so keep that parameter(m) as it is, use minimum hour for hour parameter(h=9)

0=11(m)/2-30(9)

m=49 1/11

Got It???

Niraj kumar said:
1 decade ago

When the hands of watch together at that time it is obvious that angle made by hour's hand = angle made by minute's hand.

Let us assume both hand meet at time (9+x/60), where x = minute.

Now, (9+x/60)*360/12=x*360/60 (angle made by both hand).

After solving this equation, we find x=540/11. So ans=9hr, 540/11min. i.e = Option C.

Let us assume both hand meet at time (9+x/60), where x = minute.

Now, (9+x/60)*360/12=x*360/60 (angle made by both hand).

After solving this equation, we find x=540/11. So ans=9hr, 540/11min. i.e = Option C.

(2)

Bhavesh Kirange said:
9 years ago

I don't understand the concept here you are assuming. In question from 9 to 10 o'clock the hands should be meet each other on 9.50 o'clock in real situation by watch.

But how could you say here they are meeting on 9.45 o'clock means we are suppose to assume hour hand at stable position?

But how could you say here they are meeting on 9.45 o'clock means we are suppose to assume hour hand at stable position?

Goku said:
1 decade ago

Keep it simple friends! Just use this formula.

12/11 (minutes to be gained). See in the question between 9 to 10'o clock we know that at 9:45 the hands will coincide so 45 minutes is to be gained from 9 o'clock.

So 12/11* (45) =49 1/9 (Answer).

12/11 (minutes to be gained). See in the question between 9 to 10'o clock we know that at 9:45 the hands will coincide so 45 minutes is to be gained from 9 o'clock.

So 12/11* (45) =49 1/9 (Answer).

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