APTITUDE

Numbers
H.C.F and L.C.M
Decimal Fractions
Simplification
Square and Cube roots
Average
Problems on Numbers
Problems on Ages
Surds and Indices
Percentage
Profit and Loss
Ratio And Proportions
Partnership
Chain Rule
Time and Work
Pipes and Cisterns
Time and Distance
Trains
Boats and Streams
Alligation or Mixture
Simple Interest
Compound Interest
Logorithms
Areas
Volume and Surface area
Races and Games of Skill
Calendar
Clocks
Stocks ans Shares
True Discount
Bankers Discount
Oddmanout and Series
Data Interpretation
probability
Permutations and Combinations
Puzzles
BACK

CLOCKS



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.