Tag: time

The two hands of a clock make $\displaystyle { 180 }^{ o }$ when they face each other in a straight line. 
This happens 11 times in 12 h. 
At 6 am and 6 pm, two hands form $\displaystyle { 180 }^{ o }$
Number of times they form $\displaystyle { 180 }^{ o }$ = 11 

Since the watch is  5 minutes slow at 12 noon on a Sunday and 5 minutes fast on the subsequent Wednesday at 6:00 p.m, it will show the right time at a time exactly mid way of theses 2 times.
hours passed=24+24+24+6=78hrs
mid way will be 39hrs after Sunday noon, which will be 3:00 a.m. tuesdsay.

When the hands are moving anticlockwise, we can assume that the time shown is the mirror image of actual time.
Assume that there is a line drawn between 12 and 6 on the clock, this can be a line of symmetry.
When we take the mirror image of this clock, the hands that are on the right side of this line will appear equidistant on the left side of the line and vice versa.
The time indicated is 8:25, the hour mid way after 8 and the minute hand on 5.
Its mirror image will become, the hour mid way before 4 and the minute hand on 7, i.e, 3:35.

In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.

To be together again, the minute hand must gain 60 minutes over the hour hand.

55 minutes are gained in 60 min.

60 min. are gained in [(60/55) * 60] min = $65\frac { 5 }{ 11 } $min.

Therefore, loss in 60 minutes =  $65\frac { 5 }{ 11 } -60=5\frac { 5 }{ 11 } $ min.

Loss in 24 hours = $5\frac { 5 }{ 11 } *\frac { 24*60 }{ 60 } $ = $130\frac { 10 }{ 11 } $min.

Therefore, the clock loses $130\frac { 10 }{ 11 } $ minutes in 24 hours.

Let the hours passed be $x$,
time shown on $clock 1=12:00 a.m$ saturday $(24*6+12hrs)$
$3600x-5x=156*60*60$,
$x=156.21hrs=156hrs$ and $13mins$
actual time $=12:13 a.m.$ saturday.

In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.

To be together again, the minute hand must gain 60 minutes over the hour hand.

55 minutes are gained in 60 min.

60 min. are gained in [(60/55) * 60] min = $65\dfrac { 5 }{ 11 } $min.

But they are together after 65 min.

Therefore, gain in 65 minutes =  $65\dfrac { 5 }{ 11 } -60=\dfrac { 5 }{ 11 } $ min.

Gain in 24 hours = $\dfrac { 5 }{ 11 } * \dfrac { 24*60 }{ 65 } $ = 1440/143 min.

Therefore, the clock gains $(10 + 10/143 )$ minutes in $24$ hours.

Let original speed be v and distance to breakdown be x. The equations derived from the two scenarios (30km total) lead to v = 10 km/hr and x = 10 km.

Let original speed be v and distance to breakdown be x. The equations derived from the two scenarios (30km total) lead to v = 10 km/hr and x = 10 km.

Ramu =3 kmph Somu =4 kmph Both starts at 7 AM Distance =2km

Time taken by Ramu ⇒S=v×t

2=3×t

t=23hr=23×60

=40min

Time taken by Somu ⇒2=4×t

t=12hr=30 min.

∴ Somu reaches school 10 mins earlier than Ramu.