Clocks problems
Posted by
Ravi Kumar at Thursday, March 24, 2011
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Clocks problems
Clocks problems
Simple Problems:
Type1:
Find the angle between the hour hand and the minute hand
of a clock when the time is 3.25
solution : In this type of problems the formulae is as follows
30*[hrs(min/5)]+(min/2)
In the above problem the given data is time is 3.25. that is
applied in the
formulae
30*[3(25/5)]+(25/2)30*(1525)/5+25/2
= 30*(10/5)+25/2
= 300/5+25/2
= 600+(25/2)=475/10=47.5
i.e 47 1/20
therefore the required angle is 47 1/20.
Note:The sign must be neglected.
Another shortcut for type1 is :
The formulae is
6*x(hrs*60+X)/2
Here x is the given minutes,
so in the given problem the minutes is 25 minutes,
that is applied in the given formulae
6*25(3*60+25)/2
150205/2
(300205)/2=95/2
=47 1/20.
therefore the required angle is 47 1/20.
Type2:
At what time between 2 and 3 o' clock will be the hands of a
clock be together?
Solution : In this type of problems the formulae is
5*x*(12/11)
Here x is replaced by the first interval of given time.
here i.e 2. In the above problem the given data is between
2 and 3 o' clock
5*2*12/11 =10*12/11=120/11=10 10/11min.
Therefore the hands will coincide at 10 10/11 min.past2.
Another shortcut for type2 is:
Here the clocks be together but not opposite
to each other so the angle is 0 degrees. so the formulae is
6*x(2*60+x)/2=06*x(120+x)/2=012x120x=0
11x=120
x=120/11=10 10/11
therefore the hands will be coincide at 10 10/11 min.past2.
Medium Problems
Type3:
At what time between 4 and 5 o'clock will the hands of a clock
be at rightangle?
Solution : In this type of problems the formulae is
(5*x + or 15)*(12/11)
Here x is replaced by the first interval of given time
here i.e 4
Case 1 : (5*x + 15)*(12/11)
(5*4 +15)*(12/11)
(20+15)*(12/11)
35*12/11=420/11=38 2/11 min.
Therefore they are right angles at 38 2/11 min .past4
Case 2 : (5*x15)*(12/11)
(5*415)*(12/11)
(2015)*(12/11)
5*12/11=60/11 min=5 5/11min
Therefore they are right angles at 5 5/11 min.past4.
Another shortcut for type 3 is:
Here the given angle is right angle i.e 900.
Case 1 : The formulae is 6*x(hrs*60+x)/2=Given angle
6*x(4*60+x)/2=90
6*x(240+x)/2=90
12x240x=180
11x=180+240
11x=420
x=420/11= 38 2/11 min
Therefore they are at right angles at 38 2/11 min. past4.
Case 2 : The formulae is (hrs*60+x)/2(6*x)=Given angle
(4*60+x)/2(6*x)=90
(240+x)/2(6*x)=90
240+x12x=180
11x+240=180
240180=11x
x=60/11= 5 5/11 min
Therefore they art right angles at 5 5/11 min past4.
Type 4:
Find at what time between 8 and 9 o'clock will the hands of a
clock be in the same straight line but not together ?
Solution : In this type of problems the formulae is
(5*x30)*12/11
x is replaced by the first interval of given time Here i.e 8
(5*830)*12/11
(4030)*12/11
10*12/11=120/11 min=10 10/11 min.
Therefore the hands will be in the same straight line but not
together at 10 10/11 min.past 8.
Another shortcut for type 4 is:
Here the hands of a clock be in the same
straight line but not together the angle is 180 degrees.
The formulae is (hrs*60+x)/2(6*x)=Given angle
(8*60+x)/26*x=180
(480+x)/2(6*x)=180
480+x12*x=360
11x=480360
x=120/11=10 10/11 min.
therefore the hands will be in the same straight line but not
together at 10 10/11 min. past8.
Type 5:
At what time between 5 and 6 oâ€™ clock are the hands of a
3 minutes apart ?
Solution : In this type of problems the formuae is
(5*x+ or  t)*12/11
Here x is replaced by the first interval of given time here xis 5.
t is spaces apart
Case 1 : (5*x+t)*12/11
(5*5+3)*12/11
28*12/11 = 336/11=31 5/11 min
therefore the hands will be 3 min .apart at 31 5/11 min.past5.
Case 2 : (5*xt)*12/11
(5*53)*12/11
(253)*12/11=24 min
therefore the hands wi be 3 in apart at 24 min past 5.
Typicalproblems
problems:
A watch which gains uniformly ,is 5 min,slow at 8 o'clock in
the morning on sunday and it is 5 min.48 sec.fast at 8 p.m on
following sunday. when was it correct?
Solution :
Time from 8 am on sunday to 8 p.m on following sunday = 7 days
12 hours = 180 hours
the watch gains (5+(5 4/5))min .or 54/5 min. in 180 hours
Now 54/5 minare gained in 180 hours.
Therefore 5 minutes are gained in(180*5/54*5)hours=83 hours20 min.
=3 days11hrs20min.
therefore watch is correct at 3 days 11 hours 20 minutes after 8 a.m
of sunday
therefore it wil be correct at 20 min.past 7 p.m on wednesday