Clocks

General Concepts: The face or dial of a watch is a circle whose circumference is divided into 60 equal parts, called minute spaces. A clock has two hands, the Smaller one is called the hour hand or short hand while the larger one is called the minute hand or long hand. Important points: a) In every 60 minutes, the minute hand gains 55 minutes on the hour hand b)In every hour, both the hands coincide once ,i.e 0 degrees. c)the hands are in the same straight line when they are coincident or opposite to each other. i.e 0 degrees or 180 degrees. d)when the two hands are at right angles, they are 15 minute spaces apart,i.e 90 degrees. e)when the hands are in the opposite directions,they are 30 minute spaces apart,i.e 180 degrees. f)Angle traced by hour hand in 12hrs = 360 degrees. g)Angle traced by minute hand in 60 min = 360 degrees. If a watch or a clock indicated 8.15,when the correct time is 8, it is said to be 15 minutes too fast. On the other hand, if it indicates 7.45, when the correct time is 8,it is said to be 15 minutes slow. h)60 min --> 360 degrees 1 min --> 60

i)the hands of a clock coincide in a day or 24 hours is 22 times, in 12hours 11minutes. j)the hands of clock are straight in a day is 44 times . k)the hands of a clock at right angle in a day is 44 times . l)the hands of a clock in straight line but opposite in direction is 22 times per day 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*(15-25)/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)/2 150-205/2 (300-205)/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=012x-120-x=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*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. Complex Problems 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. Typicalproblems

A watch which gains uniformly ,is 5 min,slow at 8 o'clock in problems:

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