Skill and Clock

                           Games of Skill and Clock

Games of Skill

A game of 100 means that a person among the contestants who scores 100 points first, is the winner. If A scores 1_.00 points, while B scores only 85 points, we say that A can give B 15 points.

E5.  A can give B 10 points in the game of 150. If B can give C 12 points in the game of 120, then how many points can A give to C in the game of 150?

Sol.                   A                    B                    C

150                 140

—                    120                108

150                 140        126

So, A can give 24 (150 — 126) points to C.

Circular Races

When two or more persons start from the same place at the same time and travel round a circle in the same direction or in opposite directions, then

They will be first together at the starting point again after an interval of time which is the LCM of the times in which each of them makes one complete round.

E6.  A and B walk around a circular path of circumference 1200 metres. A walks @ 150 m/min and B® 70 m/min. If they start from the same point and walk in the same direction, then when will they be first together again at the starting point?

Sol. Time for 1 round by A = 1200/150 = 8 minutes and for B = 1200/70 = 120/7 minutes.

= Time taken to meet again at the starting point = LCM of 8 and 120/7 = 120 minutes = 2 hrs.

They will be first together again after an interval of time which is the LCM of the times in which one of the persons gains one complete round over each of the others.

E7. Assuming the data given in previous example, when will they be together again anywhere else on the course?

Sol. Now in 1 minute, A gains over B (150 — 70) = 80 metres. To gain 1200 metres over B, A will take 1200/80 x 1 =15 min.

They will be first together again after an interval of time which is required to complete a round with the relative speed.

E8.  Assuming the data given in previous example, when will they be together supposing they walk in opposite directions?

Sol. Relative speed = 150 + 70 = 220 m/min.

So, they will be together after 1200/220 = 60/11 minutes.

We hope that the concept of circular race is clear to you. If not, please go through the following example. It will help you to understand the concept thoroughly.

E9.  Along with A and B, C is also running across the circular track at the rate of 130 m/min in the same direction. When will all the three people meet (use data of previous example)

(i)      at the starting point?

(ii)     at any other point on the track?

(iii)    At what distance (nearest) from the starting point do the three meet?

Sol. (i)       Time for 1 round by A = 8 min.

Time for 1 round by B = 120/7 min.

Time for 1 round by C = 120/13 min.

So, all the three people will meet at the starting point after the time

= The LCM of 8, 120/7 and 120/13 i.e. 120 minutes.

(ii)     A gains 80 metres per minute over B.

So they will meet every 1200/80 = 15 minutes. A gains 20 metres per minute over C.

So they will meet every 1200/20 = 60 minutes.

So, all the three people will meet after the time which is equal to the LCM of 15 and 60 i.e. after 60 minutes or 1 hour.

(iii)    They are meeting after 60 minutes from the start. Distance covered by A in 60 minutes

= 150 x 60 = 9000 m = 600m from the starting point. Distance covered by B in 60 minutes

= 70 x 60 = 4200 m = 600m from the starting point. Distance covered by C in 60 minutes

= 130 x 60 = 7800 m = 600m from the starting point. So, the three people will meet at 600 m from the starting point.

ElO.Now, if we assume that C starts running in the opposite direction, then the three will once again meet at the starting point after the same time interval. But can they meet anywhere else? If Yes, after what time interval and what minimum distance from the starting point?

              Sol. C starts running in the opposite direction

            The three will once again meet at the starting point after the same time interval i e 2 hours

Now A & C will meet after 1200/280 = 30/7 minutes from the from the starting point,  starting point. B & C will

Meet after 1200/200 = 6 minutes from the starting point.

So, in this case the three people will meet after the time which is equal to the LCM of 30/7 and 6 i.e. after 30 minutes from the starting point.

Distance covered by A in 30 minutes

= 150 x 30 = 4500 m = 900m from the starting point.

Distance covered by B in 30 minutes

= 70 x 30 = 2100 m = 900m from the starting point. Distance covered by C in 30 minutes

= 130 x 30 = 3900 m = 300m from the starting point.

The minimum distance at which they will meet = 300m from

the starting point.

Important

Throughout this chapter, units of quantities have a crucial importance. e.g. while solving a problem if we take distance in metres, we should take speed in m/sec and time in seconds. If proper units are not used and conversions not effected, then you'll find yourself in trouble.

 Some important points

Two persons starting at the same time and from the same point along a circular path will be together again for the

First  time, when the faster gains one complete round over the other. Time taken by faster person to complete one round over the other 

=     Length of race course

         Relative speed

Three persons starting at the same time and from the same point along a circular path will be together for the first time after the start at a time which is the LCM of the time taken by the fastest to gain a complete round each over the other two.

A overtakes B at 1/n th of xth round means, when A has  (x  - 1/n) rounds , B has completed  [( x – 1) – 1/n] rounds.

Also  ,  A's Speed     =      Rounds completed by A in a given time                      

           B's Speed           Rounds completed byB in the same time

 e.g., A overtakes B in the middle of the 4^th round implies, when A has completed 7/2^. rounds, B has completed 5/2  rounds. Therefore,

A's Speed  =    7/2 / 5/2  =       7 : 5.

B's Speed

                                                   

Clocks

All of us have seen a clock. The dial of a clock or watch is a circle whose circumference is divided into 60 equal parts, called  minute spaces.

In a clock, while the larger hand (minute hand) moves 60 minute spaces, the smaller hand (hour hand) moves only 5 minute spaces.

So, "The ratio between speeds of minute hand and hour hand of a watch will always = 12 : 1".

Clearly, in 60 minutes, the larger hand gains 55 minutes on the smaller hand or the minute hand gains 1 minute space on the hour hand in 12/11 minutes.

This is the fundamental principle which is used throughout in solving problems on clocks.

Some important points

1 minute division = 360⁰ / 60  = 6⁰ apart.

Each hour number is evenly and equally separated by five minute divisions (= 5 x 6°) = 30⁰ apart.

In one minute, the minute hand moves one minute division or 6°.

In one minute, the hour hand moves  ( ½ ) ⁰.     

In one minute, the minute hand gains  ( 11/2 )⁰    over the hour hand.

Every hour, both the hands coincide once.

If both the hands start moving together from the same position , both the hands will coincide after 360 x 2/11 =          65  x 5/11      minutes.                                        

When the two hands are at right angles, they are 15 minute spaces apart. This happens twice every hour.

When the hands are in opposite directions, they are 30 minute spaces apart. This happens once every hour.

The hands are in the same straight line when they are coincident or opposite to each other.

When the hands are together, they are 0° apart.

In the following table required angles are shown for various positions of the hands.

	Required Angle (A°)
Both hands to be coincident	0°
Both hands to be at right angle	90°
Both hands to be in opposite direction	180°
Both hands to be in straight line	0° or 180°

Incorrect Clock

An incorrect clock either gains or loses time, so, time interval indicated by such a clock will be different from that actually shown by a true clock.

          True time interval

    Time interval in incorrect clock

=                          1

1±hour gained /lost in1 hour by incorrect clock

(+) when incorrect clock gains time

 (-) when incorrect clock loses time

In a correct (true) clock, both hands coincide at an interval of 65 x 5/11 minutes.

But, if both hands coincide at an interval of x minutes (≠ 65 x 5/11) of correct time, then the clock is incorrect and total time gained /lost = 60T  x 65 x 5/11/x   -      x /x min (in T hours of x correct time)

                                                                    

Important

  If the above expression becomes (+) ve, time is gained and clock is too fast.

 If the above expression becomes (-) ve, then time is lost and clock is too slow.

E11.A clock is set right at 1 p.m. If it gains one minute in an hour, then what is the true time when the clock indicates 6 p.m. the same day?

Sol. Time interval indicated by incorrect clock = 6 p.m - 1 p.m = 5 hours.

Time gained by incorrect clock in one hour

         = + 1 minute   =       1

                       60 hour.

  Using the formula,              True time interval

           Time interval in incorrect clock

=  1/ 1 + hour gained in 1 hour by incorrect clock

=                 True time interval   =                1

                                                       5                                                   1+60  

           True time interval =   5 x  60 = 4 x 56/61

                                                 61         

True time = 1 p.m. + 4x 56/ 61 hrs.

= 5 p.m. + 56/61 hrs. = 5 p.m. + 56/61 x 60 min.

= 55 x 5/61 minutes past 5.

For students practice

E12. The minute hand of a clock overtakes (or coincides) the hour hand at intervals of 65 minutes of correct time. How much does the clock gain or lose in 12 hours?