Time and Work - Examples

Q
If A can do a work in 10 days, B can do it in 20 days and C in 30 days in how many days will the three together do it?

Soln:
The efficiencies are A = 1/10, B = 1/20 and C = 1/30
So work done per day by the three = 1/10 + 1/20 + 1/30 = 11/60 => No of days = 60/11 = 5.45 days.


Q
If A and B can do a work in 10 days , B and C can do it in 20 days and C and A can do it in 40 days in what time all the three can do it?

Soln:

A+B = 1/10
B+C = 1/20
C+A = 1/40
Adding all the three we get 2(A+B+C) = 7/40 => A+B+C = 7/80 => No of days = 80/7 days.

If A can do a work in 12 days, B can do it in 18 days and C in 24 days. All the three started the work. A left after two days and C left three days before the completion of the work. How many days are required to complete the work?

Soln:

Let the total no of days be x.

A worked only for 2 days, B worked for x days and C worked for x-3 days.

So, mA + nB + oC = 1
ð      2(1/12) + x(1/18) + (x-3)(1/24) = 1
ð      12 + 4x + 3(x-3) = 72
ð      x = 69 / 7 days.

Note:

The ratio of dividing wages = ratio of efficiencies = ratio of parts of work done

Q:

A can do a work in 10 days and B can do it in 30 days and C in 60 days. If the total wages for the work is Rs. 1800 what is the share of A?

Soln:

Ratio of wages = 1/10 : 1/30 : 1/60 = 6 : 2 : 1  (Multiplying each term by LCM 60)

So total 9 equal parts in Rs. 1800 => each part = Rs. 200 => share of A = 6 parts = Rs. 1200.
Applying the same logics to pipes and cistern

Q:

A pipe can fill a tank in 5 hrs but because of a leak a the bottom it takes 1 hr extra. In what time can the leak alone empty the tank?

Soln:

Let the filling pipe be A.
A = 1 / 5.

But with the leak L,  A – L = 1 / 6   ( A-L because leak is outlet)

So, 1/L = 1 / 5 – 1/ 6 = 1/30 => Leak can empty the tank in 30 hrs.

Q:

A pipe A can fill the tank in 10 hrs, B can fill it in 20 hrs and C can empty in 40 hrs. All are opened at the same time. After how many hours shall the pipe B be closed such that the tank can be filled in 10 hrs?

Soln:

Let the pipe B be closed after x hrs.

Then A worked for 10 hrs, B worked for x hrs and C worked for 10 hrs.

mA + nB – oC = 1    (since C is outlet)

10(1/10) + x(1/20) – 10(1/40) = 1

x = 5 hrs.

Linear Races and Circular races

Linear Races

Linear Races :- when we have straight tracks for races.


Terms used to define linear races & their actual meanings:-
A gives B a start of 10 meters : B starts a race 10 meters ahead of A.

A gives B a start of 10 seconds : B starts 10 seconds before A.

A beats B by 10 meters : When A reaches finishing line, B is 10 meters behind.

A beats B by 10 seconds : B takes 10 seconds more than A to finish race.

A beats B by 10 meters or 10 seconds: B takes 10 seconds to cover 10 meters. Speed of B is 1 m/s.

Beat time : difference between time take by loser & winner.

Winner's distance: Length of the race track.

Time take by winner =  time taken by loser - beat time.

Dead Heat : Tie


Q1. X beats Y by 60 meters or by 12 seconds in a 2 Km race.Find Speed of X, Speed of Y, Time taken by X & time taken by Y.

Solution:-

Speed of Y = 60 / 12 = 5 m/s

Time taken by Y = 2000 / 5 = 400 minutes

Time taken by X = 400 - 60 = 340 minutes
Speed of X = 2000 / 340 = 5.88 m/s


Q2. In a 1000 m race, Neeta beats Geeta by 50 m & Seeta by 100 m. By what distance will Geeta beat Seeta ?
(1) 48.36 m (2) 50 m(3) 52.64 m (4) 51.28 m (5) 52.36 m


Solution :-
Distance covered by Neeta : Geeta in winning time by Neeta is = 1000 : 950 = 20:19.

Since time taken is same, ratio of their speed is also 20:19
Distance covered by Neeta : Seeta in winning time by Neeta is = 1000 : 900 = 20:18.
Since time taken is same, ratio of their speed is also 20:18
Hence Ratio of speeds of Geeta to Seeta would be 19:18.

Distance covered by Seeta when Geeta covers 1000 m = 1000 * 18/19 = 947.36
So Geeta beats Seeta by 1000 - 947.36 = 52.64 m. Hence answer option 3.


Q3. M beats N by 30 m or 5 seconds. Which of the following statements is/are true ?
I. Speed of N can not be found.
II. Speed of M can be found.
III. Distance covered by N can be found.
IV. N takes 6 more seconds to meet M.

1. Statement I is true.
2. Statement I, II & III are true.
3. Statement I & III are true.
4. All statements are true.
5. All statements are false.

Solution:-
By given data only speed of N can be found. To find other parameters we need length of track. Hence no statement is true. So answer option 5. 

Q4. A gives B a start of 15 seconds. A can run at speed of 80 mps & B can run at speed of 40 mps. In how much time will A  meet B on straight track after B starts race?
(1) 10 s (2) 12 s(3) 15 s  (4) 20 s (5) 30 s

Solution:-
Distance covered by B in early 15 seconds = 40 * 15 = 600 m
Relative speed of A & B = 80 - 40 = 40 mps
Time taken after A begin = 600 / 40 = 15 seconds.
Total time after B begins race = 15 + 15 = 30 seconds.Hence answer option 5. 

Q5. In above question, what would have been speed of A so that it would meet B after 2 mins after B begins his race ? (All other data remains same)
(1) 42.52 m/s(2) 45 m/s (3) 45.15 m/s (4) 45.71 m/s (5) 50 m/s

Solution:-
Relative Distance = 600 meters.
Time = 120 - 15 = 105 seconds
Relative Speed = 600 / 105 = 5.71 m/s
Speed of A = 40 + 5.71 = 45.71 m/s

Circular races

Circular Races : Circular races are on circular tracks where one can meet other person more than once.

When two persons A & B starts from same point at same time on a circular track then we can find

I. after how much time they meet for first time :- they meet for first time when one covers one more lap than other person. Relative distance would be length of track & using relative speed, time taken can be found.


II. After how much time they will meet for first time at starting point : this can be find out by taking LCM of time taken  by individual to cover one lap.


Q1. Two person X & Y start from the same point and move along a circular track of 60 m. Speed of X is 5 m/s & speed of Y is 7 m/s. After how much time will they meet for the first time ?

(1) 30 s (2) 15 s (3) 12 s (4) 16 s (5) Can not be determined


Solution:-

Since we don't know the directions of X & Y we can not determined answer. It is possible that they are running in same direction or they might be running in opposite direction. Hence answer option 5.



Q2. Two friends Raj & Rahul start a race on circular track of 240 m from same point in same direction at same time. Speed of the Raj is 20 m/s & that of Rahul is 25 m/s. After how much time will they meet for first time ?

(1) 10 s (2) 12 s (3) 24 s (4) 36 s (5) 48 s




Solution:-

Since same direction is same their related speed is : 25 - 20 = 5 m/s

Related distance to meet for first time is one lap of track = 240 m

Time taken : 240 / 5 = 48 seconds.




Q3. In above question, what would be time taken if they are running in opposite direction ?

(1) 3.33 s (2) 5.33 s (3) 8.33 s (4) 10 s (5) 12 s





Solution:-

Since opposite direction is same their related speed is : 25 + 20 = 45 m/s

Related distance to meet for first time is one lap of track = 240 m

Time taken : 240 / 45 = 5.33 seconds.



Q4. Two friends Raj & Rahul start a race on circular track of 500 m from same point in same direction at same time. Speed of the Raj is 20 m/s & that of Rahul is 25 m/s. After how much time will they meet for first time at starting point?

(1) 20 s (2) 25 s (3) 100 s (4) 200 s (5) 500 s


Solution:- 

Time taken to meet at starting point would be when both complete laps at same time. That is LCM of their time taken to complete track.

Time taken by Raj to complete track = 500/20 = 25 s

Time taken by Rahul to complete track = 500/25 = 20 s

Time taken to meet for first time at starting point = LCM(20,25) = 100 s.


Q5. In above question, what would be time taken to meet for first time if they are moving along circular track in opposite direction ?

(1) 20 s (2) 25 s (3) 100 s (4) 200 s (5) 500 s


Solution:- 

Time taken to meet at starting point would be when both complete laps at same time. That is LCM of their time taken to complete track.

Time taken by Raj to complete track = 500/20 = 25 s

Time taken by Rahul to complete track = 500/25 = 20 s

Time taken to meet for first time at starting point = LCM(20,25) = 100 s.


Meeting at starting point in circular races is independent of direction.



when more than 2 people are running in circular track. For e.g. 3 persons X, Y & Z.

I. after how much time they meet for first time :- It can be found by determining the time taken between two people & then between three.
II. After how much time they will meet for first time at starting point : LCM of time taken by X, Y & Z.

Q6. If X, Y & Z are starts their race by moving along a circular track of length 120 m from same point at same time in same direction. Find the time taken for them to meet for first time if speed of X is 2m/s, Y is 3m/s & that of Z is 5 m/s.
(1) 40s (2) 60s (3) 100 s (4) 120s (5) 240s

Solution:-
All three will meet only when X meets Y.
relative distance = 120 m
relative speed of X & Y = 3 - 2 = 1 m/s
time taken for them to meet for first time = 120 s


All three will meet only when X meets Z.
relative distance = 120 m
relative speed of X & Z = 5 - 2 = 3 m/s
time taken for them to meet for first time = 120 /3 =  40 s

All three will meet for first time when X meets Y & Z together for first time : LCM (120,40)= 120 s


Q7. in above question, what time they will meet for first time at starting point ?
(1) 40s (2) 60s (3) 100 s (4) 120s (5) 240s


Solution:- 

Time taken to meet at starting point would be when all complete laps at same time. That is LCM of their time taken to complete track.

Time taken by X to complete track = 120/2 = 60 s

Time taken by Y to complete track = 120/3 = 40 s
Time taken by Z to complete track = 120/5 = 24 s

Time taken to meet for first time at starting point = LCM(60,40,24) = 120 s.
It is independent of directions. Let them run in any direction. You don't worry whenever we are finding their first meet at starting point.


Q8. If X, Y & Z are starts their race by moving along a circular track of length 120 m from same point at same time. Y & Z are running in same direction while X is running in opposite direction. Find the time taken for them to meet for first time if speed of X is 2m/s, Y is 3m/s & that of Z is 5 m/s.
(1) 40s (2) 60s (3) 100 s (4) 120s (5) 240s

Solution:-
All three will meet only when X meets Y.
relative distance = 120 m
relative speed of X & Y = 3 + 2 = 5 m/s
time taken for them to meet for first time = 120/5 = 24 s


All three will meet only when Y meets Z.
relative distance = 120 m
relative speed of X & Z = 5 - 3 = 2 m/s
time taken for them to meet for first time = 120 /2 =  60 s

All three will meet for first time when Y meets X & Z together for first time : LCM (24,60)= 120 s.

Q9. If A overtakes B for the first time in the middle of 6th lap. Find ratio of speed of A to B. We know they started their race from same point at same time.
(1) 6:5 (2) 11:9 (3) 5:6 (4) 9:11 (5) Can not be determined.

Solution:-
Overtakes means same direction.
A overtakes B for first time when he covers 5.5 laps. Same time A would cover 4.5 laps.
Ratio of speeds = ratio of distance covered = 5.5 : 4.5 = 55:45 = 11:9

Clocks funda for cat

I. 1 minute space = 6° degrees.

II. 60 minute space = 360°

III. In one hour, minute hand moves 60 minutes spaces where hour hand moves 5. Hence In an hour, minute hand gain 55 minute space over hour hand.  

IV. In 1 minute, hour hand moves 0.5° and minute hand moves 6°. Hence in 1 minute, the minute hand gains 5.5° or 11/2 degrees over the hour hand.

V. The hour & the minute hand coincide every 720/11 minutes. 

Incorrect Clock :-

I. It is said to gain time (fast) if hour hand & minute hand coincide in less than 720/11 minutes.

II. It is said to lose time (slow) if hour hand & minute hand coincide in more than 720/11 minutes.

Q1. After 5 O' clock, at what time will the hour & minute hand of the clock coincide for the first time ? (Time in HH:MM:SS format)

(1) 05:22:16 (2) 05:24:27 (3) 05:27:16 (4) 05:27:27 (5) 05:28:27

Solution :- 

At five of clock distance between hour and & minute hand is 150°
Speed of hour hand is 0.5° per minute.
Speed of minute hand is 6° per minute. 
Relative speed of minute hand & hour hand - 6°- 0.5° = 5.5° per minute
Time taken to meet = 150 / 5.5 = 27.27 minutes = 27 minutes 16 seconds

Q2. What is smallest angle between the minute hand & hour hand at 10.30 ?
(1) 225° (2) 120° (3) 132° (4) 315° (5) 135°

Solution:-
At 10 O' clock distance will be 300°.
In next 30 minutes, minute hand will move 180°(6° per minute) & hour hand will move 15°(0.5° per minute) in clockwise direction. Angular distance between them will be 135° or 225°. Hence answer is 135°. Answer option 5.

Q3. Minute hand & hour hand coincided after 60 minutes. Which of the following options is true ?
(1) Clock is gaining time.
(2) Clock is losing time.
(3) Clock is neither losing nor gaining time.
(4) It will again coincide after 61 minutes.


Solution :- We know that clock is said to gain time (fast) if hour hand & minute hand coincide in less than 720/11 minutes. Hence option 1.


Q4. A watch was running 8 minutes behind time on Sunday noon on date X. It was running 13 minutes ahead of time on next Sunday noon. After how much time after noon on X the clock was showing correct time ?
(1) 24 hours (2)  32 hours (3)  64 hours (4)  76 hours (5)  108 hours

Solution:-
First Sunday Noon : - 8 minutes
Second Sunday Noon : + 13 minutes
Time to cover in week time(168 Hours) = 21 minutes.
Time to gain from first Sunday noon to show correct time = 8 minutes.
21 minutes gain takes 168 hours.
8 minute gain will take = 168 * 8 / 21 = 64 hours from first Sunday noon. Hence answer option 3.

Time and Work

1.If A can do a piece of work in n days, then A's 1 day work=1/n

2.If A's 1 day's work=1/n, then A can finish the work in n days.

Ex: If A can do a piece of work in 4 days,then A's 1 day's work=1/4.
If A's 1 day’s work=1/5, then A can finish the work in 5 days

3.Efficiency - The amount of work done be a person in 1 day is called his efficiency.
If A is thrice as good workman as B,then: Ratio of work done by A and B =3:1. Ratio of time taken by A and B to finish a work=1:3

4.Definition of Variation: The change in two different variablesfollow some definite rule. It said that the two variables varydirectly or inversely.Its notation is
X/Y=k, where k is called constant. This variation is called direct variation. 

XY=k. Thisvariation is called inverse variation.

5.Some Pairs of Variables:

i)Number of workers and their wages. If the number of workers increases, their total wages increase. If the number of days reduced, there will be less work. If the number of days is increased, there will be more work. Therefore, here we have direct proportion or direct variation.

ii)Number workers and days required to do a certain work is an example of inverse variation. If more men are employed, they will require fewer days and if there are less number of workers, more days are required.

iii)There is an inverse proportion between the daily hours of a work and the days required. If the number of hours is increased,  less number of days are required and if the number of hours is reduced, more days are required.

6.Some Important Tips:

More Men -Less Days and Conversely More Day-Less Men.
More Men -More Work and Conversely More Work-More Men.
More Days-More Work and Conversely More Work-More Days.

Number of days required to complete the given work=Total work/One day’s work.

Since the total work is assumed to be one(unit), the number of days required to complete the given work would be the reciprocal of one day’s work.
Sometimes, the problems on time and work can be solved using the proportional rule ((man*days*hours)/work) in another situation.

7.Proportion rule - If men is fixed,work is proportional to time. If work is fixed, then time is inversely proportional to men therefore,

(M1*T1/W1)=(M2*T2/W2)

8. If all the people do not work for all the time then the principle below can be used:

mA + nB + oC = 1.     (1 is the total work)

Here, m=no of days A worked
n=no of days B worked
o=no of days C worked
A,B,C = efficiencies

Note : Pipes and cisterns
When pipes are used filling the tank they are treated similar to the men working but some outlet pipes emptying the tank are present whose work will be considered negative.