Clock Medium Problems
Type 3 : At what time between 4 and 5 o'clock will the hands of a clock be at
right angle?
Solution : In this type of problems the formula 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 5/11 min
Therefore they are right angles at 5 5/11 min past 4.
Another shortcut for type 3 is :Here the given angle is right angle i.e 900.
Case 1 : The formula 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 formula 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 formula 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 formula 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 formula 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.
