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*x-15)*(12/11)
(5*4-15)*(12/11)
(20-15)*(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
12x-240-x=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+x-12x=180
-11x+240=180
240-180=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*x-30)*12/11
x is replaced by the first interval of given time Here i.e 8
(5*8-30)*12/11
(40-30)*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)/2-6*x=180
(480+x)/2-(6*x)=180
480+x-12*x=360
11x=480-360
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 o’ 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*x-t)*12/11
(5*5-3)*12/11
(25-3)*12/11=24 min
therefore the hands wi be 3 in apart at 24 min past 5.
Typical Question:
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.
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*x-15)*(12/11)
(5*4-15)*(12/11)
(20-15)*(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
12x-240-x=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+x-12x=180
-11x+240=180
240-180=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*x-30)*12/11
x is replaced by the first interval of given time Here i.e 8
(5*8-30)*12/11
(40-30)*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)/2-6*x=180
(480+x)/2-(6*x)=180
480+x-12*x=360
11x=480-360
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 o’ 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*x-t)*12/11
(5*5-3)*12/11
(25-3)*12/11=24 min
therefore the hands wi be 3 in apart at 24 min past 5.
Typical Question:
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.
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