StreamsIn water,the direction along the stream is called DownStream. And, the direction against the stream is called

UpStream. (Speed of water is sometimes mentioned as speed of current (or) Rate of Current) If the speed of a boat or man in still water is u or Ssw

km/hr, and the speed or Rate of the stream is v or Rs or Ss km/hr, then : (u>v)

DownStream Speed(a,Dss) = (u+v)=(Ssw+Rs) km/hrUpStream Speed (b,Uss) = (u-v)=(Ssw-Rs)km/hr If the DownStream Speed is a or Dss km/hr, and the

UpStream Speed is b or Uss km/hr, then Speed in Still Water(u,Ssw)

= a+b2

=Dss+Uss2

km /hr

Rate of Stream (v, Rs)

= ab2

=DssUss2

km /hr

A man can row u km/hr in still water. If in a stream

which is flowing at v km/hr, it takes him t hrs to row to a place and back, the distance between

the two places is : = t(u2v 22u )km /hr A man rows a certain distance DownStream in t1 hrs

and returns (UpStream) the same distance in t2 hrs. Ifthe stream flows at the rate of v km/hr then :Speed of

the man in still water is = = v( t1+t 2t 2t 1)km /hr If in the above case speed of man in still water is u

km/hr then :Speed of the Stream is

= u(t 2t1t 1+t 2)km /hrEx: A man can row 30 km upstream and 44 km downstream in 10 hrs. Also, he can row 40 km upstream and 55 km downstream in 13 hrs. Find the rate of the current and the speed of the man in still water.Sol: (By use of multiple cross multiplication)Arrange the given figures in the following form:Upstream - DownStream - Time

30 44 1040 55 13

Upstream Speed of man(30554044)(55104413)

=11022

=5km / hr

DownStream Speed of man(30554044)(30134010)

=11010

=11km /hr

Speed of man=(5+11)2

=8km /hr Speed of Stream=(115)2

=3km / hr

Note: How do the denominators of above two formulae differ? For upstream speed we use the figures of downstream speedand time and For downstream speed we use the figures of upstream speed and time. Numerator remain the same in both Formulae.Ex:If a man's rate with current is 12 km/hr and rate of current is 1.5 km/hr, then :Man' s rate against Current is=Man' s rate withCurrent 2Rate of Current =12 2*1.5 = 9 km/hr

Ex:A boat travels upstream from B to A and downstream from A to B in 3 hours. If the speed of boat in still water is 9 km/hr and the speed of the current is 3 km/hr, the distance between A and B is: (Consider T = Total Time)

Distance = TotalTime(Speed in Still water2Speed of Current2)

2Speed in still waterDistance= 3(9

232)(29)

=12km

Ex:The speed of a boat in still water is 6 km/hr and the speed speed of stream is 1.5 km/hr. A man rows to a place at a distance of 22.5 km and comes back to the starting point. Find the total time taken by him.

Total Time (T )= 2DistanceSpeed in still waterSpeed in Still water 2Speed of Current2

= 222.56621.52

=8hrs

Ex:A can row a certain distance down a stream in 6 hrs and return the same distance in 9 hrs. If the stream flows at the rate

of 2 14

km/hr, find how far he can row in an hour in still water.

Speed in Still Water = Rate of StreamSum of Up and Down Stream TimesDifference of Up and Down streamTime

=(94 (6+9))(96)

= 454=11 1

4km /hr

Ex:The current of the stream runs at the rate of 4 km/hr. A boat goes 6 km and back to the starting point in 2 hrs. The speed of the boat in still water is ____km/hr.Let Speed of boat in still water = x km/hrSpeed of current = 4 km/hr ; Speed of Upstream=(x-4)km/hr ; Speed of DownStream=(x+4) km/hr,

Now 6(x4)

+ 6(x+4)

=2=> (x8)(x+2)=0 We reject negative Value, So Speed of boat in still water=8km/hr

Direct Formula => 6=(2( x242))

(2x)=> x=8km / hr

NOTE : a km /hr=a( 518

)m/ sec ; a m / sec=a( 185)km/hr

KNKumar 1

Time and Distance

Speed=DistanceTime

;Time=DistanceSpeed

Distance=speedTime If the speed of a body is changed in the ratio a:b,

then the ratio of time taken changes in the ratio b:a. If a certain distance is covered at x km/hr and the

same distance is covered at y km/hr then the average speed during the whole journey is

= 2xyx+ y

km /hr

If two trains or two bodies start at the same time frompoints A and B towards (Opposite Directions) each other and after crossing they take a and b sec in reaching B and A respectively, then (A ' s Speed ) :(B ' s Speed )=b : a

If the Ratio of speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance

is 1a

: 1b=b :a

Average Speed=Total DistanceTotal TimeEx:A man covers a certain distance between his house and office on scooter. Having an average speed of 30 km/hr, he is lateby10 min. However, with a speed of 40 km/hr, he reaches his office 5 min earlier. Find the distance between his house and office.

Required Distance = Product of Two Speeds * Difference between Arrival TimesDifference of Two Speeds

= 30+4040-30

*10+560

=30km

Note:10 min late and 5 min earlier making difference of (10+5)=15 min. As the other units are in km/hr, the difference in time should also be changed into hours. If both timings are late:10 min late and 5 min late, then difference between difference in arrival timings

=10-5=5 min

If one time is late and another is on time:15 min late and on time, i.e. 0 min, then difference in arrival timings

=10-0=10 min.Ex:A boy goes to school at a speed of 3 km/hr and returns to the village at a speed of 2 km/hr. If he take 5 hrs in all, what is the distance between the village and the school?

Required Distance = Total TimeTakenProduct of TwoSpeedsAdditionof TwoSpeeds

= 5323+2

=6km

Ex:A motor car does a journey in 10 hrs, the the first half at 21 km/hr and the second half at 24 km/hr. Find the distance.

Distance= 2TimeS1S2S1+S2

=210212421+24

=224km

Where ,S1=Speed During First Half and S2=Speed During Second Half of JourneyNote: Half of the journey means Half of the Distance not the TimeEx:A man travels 600 km by train at 80 km/hr, 800 km by ship at 40 km/hr, 500 km by aeroplane at 400 km/hr and 100 km by car at 50 km/hr. What is the average speed for entire distance?

Total Distance=600+800+500+100=2000 km ; Total TimeTaken=60080

+80040

+500400

+10050

=1234

hrs.

Average Speed=2000* 4123

km /hr=65 5123

km /hr.

Ex:Two guns were fired from the same place at an interval of 12 min but a person in the train approaching the place hears the second shot 10 min after the first. The speed of the train, if speed of sound is 330 m/s, is Assume that the person were static: In that case, he would hear the sound in 12 mins, but he heard the sound in 10 minutes, and that's because during these 10 minutes the person was moving towards the sound.

Thus these 10 minutes of person moving "saved" the sound its 2 minutes of moving.Let the distance between the shooting point and person when the second shot was made was d. If the person were

static, this distance would be covered in 12 mins.So, d =12 * s (s speed of sound)= (12 * 60) * 330 [Converting 12 mins into seconds] ----- (i)

Since the person moves towards the sound, he heard the sound 10 minutes, which means that the distance d was covered in 10 minutes at their combined speed (i.e. the sound also traveled and the person also)

Their combined speed was s + p (p is the speed of person or train)So, d = 10 * (s + p) = (10 * 60) * (330 + p) [Converting 10 mins into seconds] ----- (ii)

Comparing (i) and (ii) Hence p = 66 m/s.Ex: A carriage driving in a fog passed a man who was walking at the rate of 3 km/hr in the same direction. He could see the carriage for 4 minutes and it was visible to him upto a distance of 100 meters. What was the speed of the carriage?

TheDistance travelled by man in 4minutes=31000460

=200meters

Distance travelled by the carriage in 4minutes=(200+100)=300meters

Speed of carriage=300604*1000

km /hr=4 12km / hr

KNKumar 2

Ex:The Distance between two stations, Delhi and Amritsar, is 450 km. A train starts at 4 PM from Delhi and moves towards Amritsar at an average speed of 60 km/hr. Another train starts from Amritsar at 3:20 PM and moves towards Delhi at at an average speed of 80 km/hr. How far from Delhi will the two trains meet and at what time?Note:(1) In this Problem we have to find the two trains meeting point distance is from Delhi from which train 'A' moves towards Amritsar. So, consider 'x' as the distance from Delhi to trains meeting point (the distance covered by train 'A' from Delhi to meeting point). (2) If they asked the meeting point distance from Amritsar from which train 'B' moves towards Delhi, then consider 'x' asdistance from Amritsar to trains meeting point (the distance covered by train 'B' from Amritsar to meeting point). In this we consider Note(1) Meeting Point Train A Train B Delhi Amritsarfor Note(1) x 450-x

for Note(2) 450-x x

The difference between the two trains journey

time is 40 min (i.e. 4060

hr ). Second train starts moving after

40 min of first train. Here the distance is same for both trains but the journey time depends on speed of train, so we get equation as

(we consider Note(1))(450x )

80 x

60=40

60hr => x=170km

which isDistance fromDelhiFor Note(2) we can just subtract 170 km from total distance= 450-170 = 280 km.

To find at what Time we need to calculate time taken by train 'A' to cover 170 km = 17060

=2hrs 50min

=> Train 'A' Starts at 4:00 PM= 4+2 hrs 50 min=06:50 PM.

For Note(2) To find at what Time we need to calculate time taken by train 'B' to cover 280 km = 28080

=3hr 30min

Train B starts at 3:20 PM, so meeting time is equal to 3hr 20 min + 3 hr 30 min = 06:50 PM.Formula:

Meeting Point fromDelhi (First train Starting Point)=S1(Total DistanceS2 |T1-T2| S1+S2 )Meeting Point Distance from Amritsar (2nd Train Starting Point )=S2(Total Distance+S1 |T1-T2|S1+S2 )Where S1, S2 are Speeds of First and Second Trains, and T1,T2 are Starting Times of First and Second Trains respectively.

Ex: Walking 34

of his usual speed, a person 10 min late to

his office. Find his Usual time to cover the distance.

Usual Time= Late Time

( 1given Speed 1)= 10

( 1(34)1)=30min

Ex: Running 43

of his usual speed, a person improves his

timing by 10 minutes. Find his usual time to cover the distance.

Usual Time= Improved Time

(1 1given speed )= 10

(134)=40min

Ex:A train travelling 25 km/hr leaves Delhi at 9 AM and another train travelling 35 km/hr starts at 2 PM in the same direction. How many km from Delhi will they be together?

Meeting Point ' s Distance from Starting Point=S1S 2Difference in Time Difference in Speed

Where S 1 and S2are the Speeds of the first and Second trains respectively

Required Distance=25*35*(2 PM-9 AM)(3525)

= 25*35*510

=437 12km

Ex: Two men A and B walk from P to Q, a distance of 21 km, at 3 km/hr and 4 km/hr respectively. B reaches Q, returns immediately and meets A at R. Find the distance from P to R.

In this problem A travels PR, B travels PQ+QR, i.e. both A and B together Completed 2PQ distance, R is any point between P and Q. Now the rates of A and B are 3:4 and they have walked 42 km.

Thedistance travelled by A(PR) is=3742=18km P R Q

Note:When the ratio of speeds of A and B is a:b, then in this case: Distance travelled by A=

2Distance of two points aa+b

Distance travelled by B=

2Distance of two points ba+b

KNKumar 3

Ex:Two cars A and B started moving from the same point at the same time towards opposite direction(A towards North and B is towards South). If the speed of car A is 34.5 km/h and that of car B is 41.5 km/h, after how much time will they be 684 km apart?(in hrs)

Their relative Speed = 34.5+41.5=76 ; Distance between them is 684 km , Time=68476

=9hrs ; B A

If they are moving in same direction and distance=686 km ; Relative speed =41.5-34.5=7 km/h, Time=6867

=98hrs

Ex:Two runners cover the same distance at the rate of 15 km/hr and 16 km/hr respectively. Find the distance travelled whenone takes 16 min longer than the other.

Let Distance= x km ,

Time takenby1st Runner= x15

; Time taken by 2nd Runner= x16

; Then giventhat x15

x16

=1660

=> x=64km

Formula:

Distance =Difference in Time to Cover the Distance * Multiplicationof SpeedsDifference of Speeds

= 1516(1615)

(1660)=64kmEx: A man rode out a certain distance by train at the rate of 25 km/hr and walked back at the rate of 4 km/hr. The whole journey took 5 hours 48 minutes. What distance did he ride?

Let the Distance be x km. ;Then time spent in journeyby train= x25

hrs ;Time spent in journeyby walking= x4hrs

Therefore , x25

+ x4=5hours 48minutes=> x=20 km.

Formula: Distance=Total Time * Multiplicationof two Speeds Sum of Speeds

= 5 4860

* 25*425+4

=20km.

Note: Here Total Time during both types of journey is given whereas in the previous example the Diffrence in Time between both types of journey were given.Ex: A man takes 8 hrs to walk to certain place and ride back. However, he could have gained 2 hrs, if he had covered bothways by riding. How long would he have taken to walk both ways?Walking Time+Riding Time=8 hrs (1) ; 2 Riding Time=8-2=6 hrs (2) ; From (1) & (2) => 2 Walking Time=10 hrs

BothWayWalkingTime=One waywalking Time andone way RidingTime+Gain inTime=8+2=10hrs.Ex: A man takes 12 hrs to walk to certain place and ride back. However, if he walks both ways he needs 3 hrs more. How long would he have taken to ride both ways? Both Way Riding Time = 12-3 = 9 hrs.

BothWay RidingTime = One waywalking Timeand one way RidingTimeExtra required time for walkingboth ways

Ex: A man leaves a point P at 6 AM and reaches the point Q at 10 AM. Another man leaves the point Q at 8 AM, and reaches point P at 12 noon. At what time do they meet?Let the Distance PQ=A km ; And they meet x hrs after the first man starts.

Average speed of First man= A(106)

= A4km /hr ; Average speed of Second man= A

(125)= A

4km /hr

Distance travelled by first man= Ax4

km

They meet x hrs after the first man starts. The second man, as he starts 2 hrs late, meets after (x-2) hrs from his start.

Therefore, the distance travelled by the second man = A(x-2)4

km

Now , Ax4

+A (x2)4

=A=> x = 3hrs ; They meet at 6 AM+ 3 hrs=9 AM

Formula: Since both the persons take equal time of 4 hrs to cover the distance, their meeting time will be exactly in the middle of 6 AM and 12 noon, i.e. at 9AM. But what happens when they take different times? In that case the following formula is useful

They will meet at=First ' s Starting Time+TimeTakenby First(2nd's Arrival Time - 1st's Starting Time) Sum of Time Taken by Both

=6 AM+(10:00 - 6:00)(12:00 - 6:00)(10:00 - 6:00)+(12:00 - 8:00)

=6 :00 AM+ (46)(4+4)

=9: 00 AM

Ex: A person covers a distance in 40 minutes if he run at a speed of 45 km/h on an average. Find the speed at which he must run to reduce the time of journey to 30 minutes. => 45*40= S2 *30 =>S2=60 km/hr

Speed and Time taken are inversely proportional. Therefore , S1T 1=S 2T 2=S 3T 3. ...WhereS1 ,S2 ,S3.... are the speeds andT1 ,T2 ,T3...are the time taken to travel the same Distance.

KNKumar 4

Ex:Without any stoppage a person travels a certain distance at an average speed of 80 km/h, and with stoppages he covers the same distance at an average speed of 60 km/hr. How many minutes per hour does he stop.

Timeof Rest per Hour= Difference of SpeedSpeed without Stoppage

=80-6080

=14hrs=15minutes

Ex: One aeroplane started 30 minutes later than the scheduled time from a place 1500 km away from its destination. To reach the destination at the scheduled time the pilot had to increase the speed by 250 km/hr. What was the speed of the aeroplane per hour during the journey?Let it takes x hrs in second case.

ThenSpeed=1500x

= 1500

(x+12)+250

(or)

(1500(x+12)1500x)( x(x+12))

=250

750=250x( x+12)=> x=32 ,2Neglect negative Values.

Therefore, the plane takes 32

hrs in second case,

i.e. 32+1

2=2 hrs in normal case.

Thus, normal speed = 15002

=750 km /hr

Quicker MethodWe arrange the given information in Two columns as below.

Lesser Time : Increase in Speed

12

: 250

1 : 500

32

: 750

2 : 1000

The Ratio is continued until we get the two ratios such that their cross-products give the distance between the points.Thus, we find our answer as:

With the speed of 1000 km/hr the plane takes 32

hrs and

with the speed of 750 km/hr the plane takes 2 hrs.Therefore, normal speed is 750 km/hr

Ex:An aeroplane scheduled on hour later than its scheduled time from a place 1200 km away from its destination. To reach the destination at the scheduled time the pilot had to increase the speed by 200 km/hr. What was the speed of the aeroplane per hour in normal case?Lesser Time : Increase in Speed 1 : 200 2 : 400 3 : 600

Thus, the normal time is 3 hrs and normal speed is 400 km/hr

Ex:A train was late by 6 minutes. The driver increased its speed by 4 km/hr. At the next station, 36 km away, the train reached on time. Find the original speed of train.Lesser Time : Increase in Speed

110

hr (6min) : 4 km /hr

210

hr (12min) : 8km /hr

310

hr (18min) : 12 km /hr

Lesser Time : Increase in Speed

410

hr (24min) : 16 km /hr

510

hr (30min) : 20 km /hr

610

hr (36min) : 24 km /hr

Thus the Normal speed is 36 km/hr.

Lesser Time : Increase in Speed

710

hr (42min) : 28km /hr

810

hr (48min) : 32km / hr

910

hr (54min) : 36km /hr

1hr (60 min) : 40 km/hr

Ex:When a man travels equal distance at speeds V1 and V2 km/hr, his average speed is 4 km/hr. But when he travels at these speeds for equal times his average speed is 4.5 km/hr. Find the difference of the two speeds.Suppose the equal distance=D km. Then time taken with V1

and V2 speeds are DV1hrs and D

V2hrs respectively.

Average speed=Total Distance Total Time

= 2D

( DV1+ DV2)=2*V1*V2

V1+V2=4km / hr

In second case, average speed= V1+V22

=4.5 kn/hr

That is; V1+V2=9 and V1*V2==18Now ,(V1V2)2=(V1+V2)2 4V1V2=8172=9V1V2=3km /hr andV1=3km /hr ;V2=6km /hr

Direct Formula:V1V2=(44.5(4.54))=3km / hr

KNKumar 5

Ex: A person travels for 3 hrs at the speed of 40 km/hr and for 4.5 hrs at the speed of 60 km/hr. At the end of it, he finds that

he has covered 35

of total distance. At what average speed should he travel to cover the remaining distance in 4 hrs?

Total DistanceCovered in(3+4.5)hrs=340+4.560=390km. ; Now , since 35of distance=390 km

25of distance=390(53)(25)=260km. ; Average speed for remaining Distance=2604 =65kmDirect Formula :The average speed for the remaining distance=(R1T1+R2T2)

T ( 1f 1)= (403+604.5)

4(531)=3902

42=65km /hr.

Ex: A person travelled 120 km by steamer, 450 km by train and 60 km by horse. It took 13 hrs 30 mins. If the rate of the train is 3 times that of the horse and 1.5 times that of the steamer, find the rate of the train per hour.Suppose the speed of horse= x km/hr. Then speed of the train=3x km/hr, and speed of steamer=2x km/hr

Now , 1202x

+4503x

+60x=13.5hrs => x=20 ;

Speed of train=3x=60 km /hr ,Speed of steamer=2x=40 km /hr ,

Speed of Horse=20 km /hr

Quicker method: Arrange the information like: Train Steamer Horse

Distance 450 km 120 km 60 kmSpeed 3 2 1Total Time=13.5 hrs

Speed of Train=(45021+12031+6032)(13.521)

=60km /hr

Speed of Steamer=(45021+12031+6032)(13.531)

=40km /hr

Speed of Horse=(45021+12031+6032)(13.532)

=20 km /hr

Ex: A man covers a certain distance on scooter. Had he moved 3 km/h faster, he would have taken 40 minutes less. If he hadmoved 2 km/h slower, he would have taken 40 minutes more. Find the distance (in km) and original speed.Suppose the Distance is D km and the initial speed is x km/hr.

Then Dx+3

=Dx40

60 and D

x-2=D

x+40

60 => D

x-2D

x=2

3 and 2D

x(x-2)=2

3 => 3D

x(x+3)= 2D

x(x-2)=> 3x6=2x+6

=> x=12 km /hr and D=23(1215

3)=40km.

Quicker method: In the above question when time reduced in arrival (40 minutes) is equal to the time increased in arrival (40 minutes) then

Speed= 2Increase in SpeedDecrease in SpeedDifference in increase and Decrease in Speeds

=2(32)(32)

=12km /hr.

Now , Distance=(12+3)*(12-2)(12+3)-(12-2)

Difference in Arrival Times=(15*105

)(40+4060

)=40km

Ex: A, B and C can walk at the rates of 3,4 and 5 km/hr respectively. They start from poona at 1, 2, 3 o'clock respectively. When B catches A, B sends him back with a message to C. When will C get the message?

Speed Starting TimeA 3km /hr 1o ' clockB 4km /hr 2o ' clockC 5km / hr 3o ' clock

A takes lead of 3 km from B.Relative speed of A and B=4-3=1 km/hr.

B catches A after 31

= 3 hr, i.e., at 2+3=5 o'clock.

A returns at 5 o'clock and from a distance of 3*4=12 km from poona.In the mean time C covers a distance of 5*2=10 km from poona.Thus, A and C are 12-10=2 km apart at 5 o'clock.Relative speed of A and C=3+5=8 km/hr.

Thus they meet after 28

= 14

hr=15 min.

Thus, C will get the message at 5:15 o'clockEx:A thief is spotted by a policeman from a distance of 200 meters. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief 10 km/hr, and that of the policeman 12 km/hr, how far will have the thief runbefore he is overtaken? Quicker Method:Relative Speed=12-10=2 km/hr ; The thief will caught

after= 0.22

= 110

hr ; Distance covered by the thief

before he gets caught = 10110

=1km

Thedistance covered by the thief before he gets caught

= Lead of Diatance Relative Speed

Speed of Thief

= 0.2 km2 km /hr10 km /hr=1km

KNKumar 6

TrainsNOTE :

a km /hr=a 518

m /sec ; a m /sec=a185

km /hr

Relative Speed:1. When two trains or two bodies are moving in opposite

directions with u m/s and v m/s, then their relative speed is sum of their speeds i.e. (u+v) m/s.

2. When two trains or two bodies are moving in same direction with u m/s and v m/s, where u>v, then their relative speed isdifference of their speeds i.e.(u-v) m/s.

3. If two trains or two bodies start at the same time from points A and B towards (Opposite Directions) each other and after crossing they take a and b sec in reaching B and A respectively, then(A ' s Speed ) :(B ' s Speed )=(a :b)

Distance:1. When a train of length l meters passes a platform of length b

meters, it should travel the length equal to sum of the lengthsof Train and Platform, i.e.(l+b) meters.

2. When a train of length l meters passes a pole or man it should

travel the length equal to the length of the Train i.e. l meters.

Time:1. Time taken by a train of length l meters to pass

a pole or a standing man or a signal post is equalto the time taken by the train to cover l meters.

2. Time taken by a train of length l meters to pass a stationary object of length b meters is equal to the time taken by the train to cover (l+b) meters.

3. If two trains of length a meters and b meters aremoving in opposite directions at u m/s and v m/s, then time taken by the trains to cross each

other is = (a+b)(u+v )

sec

4. If two trains of length a meters and b meters aremoving in same directions at u m/s and v m/s, then time taken by the faster train to cross

slower train is = (a+b)(uv )

sec (Such that u>v)

Ex:Two trains 121 meters and 99 meters in length respectively are running in opposite directions, one at the rate of 40 km/hr and the other at the rate of 32 km/hr. In whattime will they be completely clear of each other from the moment they meet?Realative Speed (Opposite Direction)=40+32=72 km/hr=20 m/sec., Total Distance=121+99=220 meters.

Required time= Total LengthRelative Speed

= 22020

=11 sec

Realative Speed (Opposite Direction)=40-32=8 km/hr

= 209m /sec. , Total Distance=121+99=220 meters.

Required time= Total LengthRelative Speed

= 22020

9=99 sec

Ex: A train 110 meters in length travels at 60 km/hr. In what time will it pass a man who is walking at 6 km an hour (i)against it ; (ii)in the same direction?(We have to consider length of the man or pole is zero)

(i)Relative speed=60+6=66 km/hr= 553m / sec

Required Time= 11055

3 =6 sec

(ii)Relative speed=60-6=54 km/hr=15 m/sec.

Required time= 11015

= 7 13sec

Ex: Two trains are moving in the same direction at 50 km/hr and 30 km/hr. The faster train crosses a man in the slower train in 18 seconds. Find the length of the faster train.

Relative speed=(50-30) km/hr

=(20518 )m /sec=509 m / secDistance covered in 18 sec at this speed

= 18509

=100 meters

Length of faster train= 100 meters

Ex: Two trains start at the same time from Hyderabad and Delhi and proceed towards each other at the rate of 80 km/hr and 95 km/hr respectively. When they meet, it is found that one train has travelled 180 km more than the other. Find the distancebetween Delhi and Hyderabad.

Distance=Difference in Distance travelled Sum of SpeedsDifference in Speeds

=180(17515 )=2100 kmEx:Two trains for Delhi leave Jaipur at 08:30 a.m. and 09:00 a.m. And travel at 60 and 75 km/hr respectively. How many km from Jaipur will the two trains meet?

Meeting Point ' s Distance from Starting Point=S1S 2Difference in Time Difference in Speed

WhereS 1 and S2are the Speeds of the first and Second trains respectively

Required Distance=60*75*(9:00 AM-8:30 AM)(7560)

=150 km

Ex: Without stoppage a train travels at an average speed of 75 km/hr and with stoppages it covers the same distance at an average speed of 60 km/hr. How many minutes per hour does the train stop?

Timeof rest per hour=Difference in average speed Speed without stoppage

=75-6060

=15hr=12minutes

KNKumar 7

Ex: A train passes by a stationary man standing on the platform in 7 seconds and passes by the platform completely in 28 seconds. If the length of th platform is 330 meters, what is the length of the train?(In this Problem man = Stationary Man)

Length of theTrain=Length of the Platform Difference in Time

Time taken to cross a pole or man= 33028-7

7=110meters

Ex: Two stations A and B are 110 km apart on a straightline. One train starts from A at 8 a.m. And travels towards B at 40 km/hr. Another train starts from B at 10 a.m. And travels towards A at 50 km/hr. At what time will they meet?

TrainsMeeting Time=S T1+(Total Distance+|S T1ST2 |S2)

(S1+S 2)=S T2+

(Total Distance| ST1S T2 |S 1)(S1+S 2)

Where S T1 , ST2Starting Times of First and Second Trains respectively, S 1 , S 2 Speeds of First and Second Trains respectively

= 8a.m.+110 + |10 a.m.-8 a.m.|*50 40+50

= 8a.m.+21090

=10 : 20a.m.

(21060

90 minutes=140minutes=2hr 20minutes.)

= 10 a.m.+110 - |10 a.m.-8 a.m.|*40 40+50

= 10 a.m.+3090

=10 : 20a.m.

(3090 means

306090 minutes=20minutes.)

Ex: A train leaves Delhi for Amritsar at 2:45 pm and goes at the rate of 50 km/hr. Another train leaves Amritsar for Delhi at 1:35 pa and goes at the rate of 60 km/hr. If the distance between Delhi and Amritsar is 510 km, at what distance from Delhi will the two trains meet?

Let the two rains meet at x km from Delhi. Then their meeting time = 2 :45 pm+ x50

=1 :35 pm+(510x)60

=> (2 : 45pm1 :35 pm)+( x50 +x

60 )=51060 =

172 => 1

16 hr+ x

50+6050*60 =

172

=> x ( 1103000

)=1727

6=44

6 => x=( 44

6)( 3000

110)=200km.

Direct Formula:

The meeting Point Distance from Delhi(FirstTrain ' s Starting Point)=S1(Total DistanceS 2| ST1S T2 |)

S1+S 2

The meeting Point Distance from Amritsar (2nd Train ' s Starting Point)=S 2(Total Distance+S1| ST1ST2 |)

S 1+S 2Where S T1 , ST2Starting Times of First and Second Trains respectively,

S 1 , S 2 Speeds of First and Second Trains respectively TheMeeting Point Distance from Delhi

(First Train' s Starting Point )

= 50(510 607060)

50+60=50440

110=200km

TheMeeting Point Distance from Amritsar(2ndTrain' s Starting Point )

= 60(510+507060)

50+60=603060

11060=310 km

In the above expression |T 1 T 2 | = | 2: 45 pm 1 :35 pm |= |45+25| minutes=70minutes=7060

hr.

Ex: Two trains of length 100 m and 80 m respectively run on parallel lines of rails. When running in the same direction the faster train passes the slower train in 18 sec, but when they are running in opposite directions with the same speed as earlier,they pass each other in 9 seconds. Find the speed of each train.Let the speeds of the trains be x m/s and y m/s. When they are moving in the same direction, the relative speed=(x-y) m/s

x y=100+8018

=10 Similarly , x+ y=100+809=20 Solving we get, x=15 m/s and y=5 m/s.

Formula:

Speed of FasterTrain=Average Lengthof TwoTrains( 1Opposite direction's time + 1Same Direction's time )Speed of Slower Train=Average Lengthof TwoTrains( 1Opposite direction's time 1Same Direction's time )

Speed of Faster Train=100+802 ( 118+19)=15m / s ;Speed of Slower Train=100+802 ( 11819)=5m/ s

KNKumar 8

Ex: A train overtakes two persons who are walking in the same direction as the train is moving, at the rate of 2 km/hr and 4 km/hr and passes them completely in 9 and 10 seconds respectively. Find the speed and length of train.

Speeds of twomenare 2 km /hr=59m / sand 4 km /hr=10

9m / s.; Let speed of train=x m / s ,

then , Relative speeds=(x59)m /s and (x109 )m /s. =>(x59)9=(x109 )10 => x=55

9m/ s => x=22km /hr and lengthof the train=(559 59)=50mDirect formula :Length of the Train=

|2 5184518 |90

| 109 |=50m

Formula : Length of the train= Relative SpeedTime taken to pass a man

Length of the Train = | Difference in Speeds of Two men |Multiplication of Times |Difference in time |

This formula is same for both men are in same or opposite Direction to TrainEx: A train passes a pole in 15 seconds and passes a platform 100m long in 25 seconds. Find its length.

Length of the Train= Time to pass a pole * Length of PlatformDifference in Time to cross a pole and Platform

= 15*100(2515)

=150m

Ex:Two trains are running in opposite direction with speeds of 62 km/hr and 40 km/hr respectively. If the length of one trainis 250 meters and they cross each other in 18 seconds, the length of the other train in meters.

Length of theOther train=Relative SpeedTime to cross eachotherLengthof First Train

=(102 518)18 250=260mEx:A goods train and a passenger train are running on parallel tracks in the same direction. The driver of the goods train observes that the passenger train coming from behind overtakes and crosses his train completely in 60 seconds. Whereas a passenger on the passenger train marks that he crosses the goods train in 40 seconds. If the speeds of the trains be in the ratio of 1:2, find the ratio of their lengths.Let speeds of two trains are x m/s and 2x m/s respectively. And Let lengths of two trains are A m and B m respectively.

Then ,A+B2x-x

=60 ----(1) and A2x-x

=40 ----(2)

=> A+BA

=6040

=> A:B=2 :1

Quicker Approach: The man in the passenger train crosses the goods train in 40 seconds. This implies that the man in the goods train can observe that the passenger train passes his in 60-40=20 seconds. (This is only because relative velocity for both the persons are the same.)

Therefore, we may conclude that a person takes double the time to cross the goods train than to cross the passenger train. Thus the ration of their lengths=4020=2:1.

Ex: A train after travelling 50 km meets with an accident and then proceeds at 34

of its former speed and arrives at its

destination 35 minutes late. Had the accident occurred 24 km further, it would have reached the destination only 25 minuteslate. The speed of the train is___.Quicker Approach:We may conclude that the speeds of the train upto 50 km are the same in both the cases. And also, the speeds after (50+24=)74 km are the same in both the cases. Thus the difference in time (35 min-25 min=10 min) is only due to the difference in speeds for the 24 km journey.

Now, if the speeds of the train is x km/hr then 24

(3x4 )24

x=10

60=> x=860

10=48 km /hr

Direct Formula:

Speed of train=24(134)

(34(3525)60 )=646

3=48km /hr. (134 )shows fractional change in speed.

Initially , it was1(suppose) . After accident it was reduced to 34

.The denomenator has two parts 34

and(352560).34is the changed fractional speed. 35-25

60is the difference (in hour) in arrival Times.

KNKumar 9

Ex: A train covers a distance between stations A and B in 45 minutes. If the speed is reduced by 5 km/hr, it will cover the same distance in 48 minutes. What is the distance between the two stations A and B (in km)? Also, find the speed of the train.Let the Distance is x kmand Speed of the trainis y km /hr.

We have tworelationships :

(1) xy=45

60=> x=3

4y ; (2) x

y5=48

60=> x=4

5( y5)

from (1)and (2)34y=4

5( y5)=> y=80km /hr ; x= 3

480=60km

Direct Formula:

Speed of theTrain= 484845

5=80 km /hr ;

and Distance=5 4548(4845)

160

=60 km

( 160 is for changingminutes intohours.)Ex: Two places P and Q are 162 km apart. A train leaves P for Q and at the same time another train leaves Q for P. Both the trains meet 6 hrs after they start moving. If the train travelling from P to Q travels 8 km/hr faster than other train, find the speed of the two trains.

Suppose the speeds of two trains are p km /hr and q km/hr

respectively. Thus p+q=1626

=27(1), pq=8(2)

from (1) and (2) => p=17.5 km /hr ; q=9.5km /hr

Direct Formula :

Speeds of Trains=162+6826

=17.5 km/hr

and 1626826 =9.5km /hr.

Formula:Speeds of the Trains

= Total Distance+( | Difference in meeting Time and Starting Time | * | Difference in Speeds of Train | )2 * | Difference in meeting Time and Starting Time |Ex: Two trains A and B start from Delhi and Patna towards Patna and Delhi respectively. After Passing each other they take 4 hours 48 minutes and 3 hours 20 minutes to reach Patna and Delhi respectively. If the train fro Delhi is moving at 45 km/hr then find the speed of the other train.

Formula: Speed of theOther Train=Speed of the First Train( TimeTaken First Train after MeetingTimeTakenby Second Train after Meetin )Speed of the Other Train = 45 4 453 13 =45 245 310=45 3625=54 km/hr

Ex: The speed of two trains are in the ratio x : y. They are moving in the opposite directions on parallel tracks. The first traincrosses a telegraph pole in 'a' seconds where as the second train crosses a telegraph pole in 'b' seconds. Find the time taken by the trains to cross each other completely.

Speed of two trainsare in the ratio x : y ; CASE(i) : Time takenby TwoTrains to cross a TelegraphPloe ' a ' seconds and ' b ' seconds

Suppose the Speeds are Ax m / s and Ay m / s.Then length of First Train=Axa=Axa meters and Lengthof Second Train=Ayb=Ayb meters.

Time to cross each other=Sum of Lengths Sum of Speeds

= Axa+AybAx+Ay

=ax+byx+ y

seconds

CASE(ii) : Time takenbyTwoTrains to cross aTelegraph Ploe for both Trainsis same ' a ' secondsSuppose the Speeds are Ax m / s and Ay m / s.

Then length of First Train=Axa=Axa meters and Length of Second Train=Aya=Aya meters.

Time to cross each other=Sum of Lengths Sum of Speeds

= Axa+AyaAx+Ay

=ax+ayx+ y

=a (x+ y)(x+ y )

=a seconds

NOTE :In above bothcases the Result DOES NOT depend on RATIOOF SPEEDS

Ex: The speed of two trains are in the ratio 7 : 9. They are moving in the opposite directions on parallel tracks. The firsttrain crosses a telegraph pole in 4 seconds where as the second train crosses a telegraph pole in 6 seconds. Find the time taken by the trains to cross each other completely.

Direct Formula : 74+967+9

=8216

=5 16seconds

Ex: The speeds of the two trains are in the ratio 3:4. They aregoing in opposite directions along parallel tracks. If each takes 3 seconds to cross a telegraph post, find the time taken by th trains to cross each other completely?

Direct Formula : 3(3+4)3+4

=217=3 seconds

KNKumar 10

Ex: A train 75 meters long overtook a person who was walking at the rate of 6 km/hr, and passed him in 7 12seconds.

Subsequently it overtook a second person, and passed him in 6 34seconds. At what rate was the second person

travelling?Formula:Speed of 2nd person=Relative Speed of Train with respect to1st person + speed of 1st person - Relative Speed of Trainwith respect to2nd person

=( 75152 185 )+6(75274185 )=36+640=2km /hr

Ex: Two Trains running at the rates of 45 and 36 km/hr respectively, on parallel rails in opposite directions, are observed to pass each other in 8 seconds, and when they are running in the same direction at the same rate as before, a person sitting in the faster train observes that he passes the other train in 30 seconds. Find the lengths of the trains.Relative Speed of two trains=45+36=81 km/hr(Two trains are moving Opposite Directions)

=(81( 518))=22 12 m /sLength of both the Trains= 45

28 =180 m

Now, when two trains are moving in the same direction, the

relative speed = 4536=9km /hr=9518

=52m / s

The man sitting in the faster train passes the length of the slower train in 30 seconds.

Length of the Slower Train= 52

*30=75 m.

Length of the Faster Train= 180-75=105 m.

Formula:

Length of Slower Train=30Relative Speed of TwoTrains=30(4536) 518

=75m.

Length of Faster Train=Total Lengthof BothTrainsLength of Slower Train=8(45+36) 518

75=105m.

Ex:Two trains measuring 100 and 80 meters respectively, run on parallel lines of rails. When travelling in opposite directions they are observed to pass each other in 9 seconds, but when they are running in the same direction at the same rates as before, the faster train passes the other in 18 seconds. Find the speed of the two trains in km/hr.

Relative Speed= Total DistanceTime to pass Each Other

R1=Relative Speed (Moving in Same Direction )

= 100+8018

=10m / s

R2=Relative Speed (Moving inOpposite Directions)

= 100+809

=20m / s

Speed of Faster Train=R1+R2

2

= 10+202

=15m / s=15185=54km /hr

Speed of Slower Train=R2R1

2

= 20102 =5m / s=5185 =18km /hr

Ex: A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 km/h and 4 km/hr respectively and passes them completely in 9 and 10 seconds respectively. The length of the train is in meters.

Whenall areMoving in the Same Direction ,

Length of Train= Relative Speed of Twomen Product of Times to PassThemDifference of Times toPassThem

=(42) 5

18910

109=50meters

KNKumar 11

Pipes and Cisterns Inlet: A pipe connected with a tank or reservoir or cistern,

that fills it. Outlet:A pipe connected with a tank or reservoir or

cistern, emptying it. If a pipe can fill a tank in 'x' hours, then the part filled in 1

hour = 1x

If a pipe can empty a tank in 'y' hours, then the part

emptied in 1 hour = 1y

If a pipe can fill a tank in 'x' hours and another pipe can empty the full tank in 'y' hours, then net part filled in 1 hour, when

both the pipes are opened (y>x) = (1x 1y ). Time Taken to fill the tank, when both pipes are

opened = xyyx

hrs.

If a pipe can fill a tank in 'x' hours and another pipe can empty the full tank in 'y' hours, then net part emptied in 1 hour, when both the pipes are opened (x>y) =

(1y1x ).

Time Taken to empty the tank, when both pipes

are opened = xyx y

hrs.

If a pipe can fill the tank in 'x' hours and another canfill the same tank in 'y' hours, then net part filled in

1hour, when both pipes are opened = (1x+ 1y ). Time Taken to fill the tank = xy

y+x hrs.

If a pipe fills a tank in 'x' hours and another fills the same tank in 'y' hours, but a third one empties the full tank in 'z' hours, and all of them are opened together,

the net part filled in 1 hour = (1x+ 1y1z ). Time Taken to fill the tank =

xyzyz+zx xy

hrs.

A pipe can fill a tank in 'x' hours. Due to a leak in the bottom it is filled in 'y' hours. If the tank is full, the time taken by the leak to empty the tank =

xyyx

hrs.

Ex: In what time would a cistern be filled by three pipes whose diameters are 1 cm, 1 13

cm and 2 cm running

together, when the largest alone will fill it in 61 minutes, the amount of water flowing in by each pipe being proportional to the square of its diameter?

In 1 minute the pipe of 2 cm diameter fills 161

of the cistern.

In 1 minute the pipe of 1 cm diameter fills 161

14

of the

cistern.

In 1 minute the pipe of 1 13

cm diameter fills 161

49

of the

cistern.

In 1 minute ( 161+ 1614+ 4619) = 136 of the cistern filled.

Therefore, The whole is filled in 36 minutes

Note: We are given that amount of water flowingis proportional to square of the diameter of the

pipe. Since 2 cm diameter fills 161

of the

cistern,

=> 1 cm Diameter fills 161 (12)

2

= 161

14

of

the cistern

=> 1 13=4

3cm Diameter fills

161

14(43)

2

= 161

49

of the cistern.

Ex: Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 18 minutes?Let B be closed after 'x' minutes. Then, part filled by (A+B) in x min.+Part filled by A in (18-x) min. = 1.

x( 124+ 132)+(18 x) 124=1 => x=8.So , B should beclosed after 8min

Direct Formula:

Pipe B should be closed after (11824)32=8minNote: If they asked for pipe A (should be closed after how much TIme), then Direct formula becomes

(11832)24=10min30 sec.Ex: If two pipes function simultaneously, the reservoir is filled in 12 hrs. One pipe fills the reservoir 10 hrs faster than the other. How many hours does the faster pipe take to fill the reservoir?Let the first pipe fills the tank in 'x' hrs. Then the slower pipe fills the tank in (x+10) hrs.

When both of them are opened, the reservoir will be filled in x( x+10)x+( x+10)

=12 => x=20,6.

But x can't be ve , hence the faster pipe will fill the reservoir in 20 hrs.KNKumar 12

Ex: Three pipes A,B and C can fill a cistern in 6 hrs. Afterworking together for 2 hours, C is closed and A and B fill the cistern in 8 hrs. Then find the time in which the cisterncan be filled by pipe C.

A+B+C can fill in 1 hr = 16

of cistern.

A+B+C can fill in 2 hrs = 26=1

3of cistern.

Unfilled part =(113 =)23 is filled by A+B in 8 hrs(A+B) can fill the cistern in 83

2= 12 hrs.

And we have (A+B+C) can fill the cistern in 6 hrs.C=(A+B+C)-(A+B) can fill the cistern in

126126

=12hrs.

Ex: A tank has a leak which would empty it in 8 hrs. A tapis turned on which admits 6 liters a minute into the tank, and it is now emptied in 12 hrs. How many liters does the tank hold?

The filler tap can fill thetank in128128

=24hrs.

Capacity of tank=24606=8640 liters.Ex: A tank is normally filled in 8 hours but takes 2 hours longer to fill because of a leak in its bottom. If the cistern is full, in hoe many hrs will the leak empty it?Suppose the leak can empty the tank in x hrs.

Then part of cistern filled in 1 hr = 181x= x8

8x

Cistern will be filled in 8xx8

hrs.

Now, 8xx8

=8+2=10hrs=> x=40hrs.

It is clear from the question that the filler pipe fill the tank in 8 hrs and if both the filler and the leak work together, the tank is filled in 10 hrs. Therefore, the leak will empty

the tank in 810108

=40 hrs

Ex: A pipe can fill a tank in 12 minutes and another pipe in 15 minutes, but a third pipe can empty it in 6 minutes. The first two pipes are kept open for 5 minutes in the beginning and then the third pipe is also opened. In what time is the cistern emptied?

Cistern filled in 5 minutes = 5( 112+ 115)= 34.Net work done by 3 pipes in 1 minute

=( 112+ 115)16= 160 . ve sign shows that 160 partis emptied in 1minute.

34

part is emptied in 6034=45minutes.

Ex: If three taps are opened together, a tank is filled in 12 hrs. One of the taps can fill it in 10 hrs and another in 15 hrs. How does the third tap work?We have to find the nature of the third tap whether it is afiller or a waste pipe.Let it be a filler pipe which fills in x hrs.

Then 1015x1015+10x+15x

=12

=> 150x=15012+25x12=> x=12 sign shows that the third pipe is a waste

pipewhichempties the tank in 12hrs.Ex: A, B and C are three pipes connected to a tank. A and B together fill the tank in 6 hrs. B and C together fill the

tank in 10 hrs. A and C together fill the tank in 7 12hrs.

In how much time will A, B and C fill the tank separately?A+B fill in 6 hrs. ; B+C fill in 10 hrs. ;

A+C fill in 7 12=15

2hrs.

2(A+B+C) fill in

610152

610+6152+1015

2

=6515180

=52hrs.

A+B+C fill the tank in 5 hrs

Now, A[=(A+B+C)-(B+C)] fills in 105105

=10hrs.

Similarly,

B fills in

1525

152 5

=15hrs ; andC fills in 5665=30hrs.

Ex: Two pipes can separately fill a tank in 20 hrs and 30 hrs respectively. Both the pipes are opened to fill the tank

but when the tank is 13

full a leak develops in the tank

through which 13

of the water supplied by both the

pipes leak out. What is the total time taken to fill the tank?Time taken by the two pipes to fill the tank

= 203020+30

hrs=12hrs ; 13

of tank is filled in 123=4 hrs

Now, 13

of the supplied water leaks out.

=> The filler pipes are only 113=2

3as efficient as

earlier.=> the work of (12-4=) 8 hrs will be completed now in

823=83

2=12 hrs. => Total Time=4+12=16 hrs.

Quicker Method: Since 13

of supplied water leaks out,

the leakage empties the tank in 12*3=36 hrs. Now, time taken to fill the tank by the two pipes and the

leakage = 36123612

=18hrs.

Time taken by the two pipes and the leakage to fill 23

of

the tank = 1823

=12hrs.

Total Time=4 hrs+12 hrs=16 hrs.

KNKumar 13

Ex: Two Pipes A and B can fill a cistern in 60 minutes and75 minutes respectively. There is also an outlet C. If all thethree pipes are opened together, the tank is full in 50 minutes. How much time will be taken by C to empty the full tank?

Work done byC in 1min.= ( 160+ 175 150)= 1100.C canempty the full tank in100minutes.

Ex: A tap can fill a tank in 6 hours. After half the tank is filled , three more similar taps are opened. What is the total time taken to fill the tank completely?Time taken by one tap to fill half the tank= 3hrs.

Part filled by the four taps in 1 hour = 416= 2

3.

Remaining part = 112= 1

2.

=> 23

: 12

: :1 : x => x=1213

2= 3

4hrs=45min.

Total Time=3 hrs 45 min.Ex: A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both pipes are open, how long will it take to empty or fill the tank completely?Clearly, Pipe B is faster than pipe A and so, the tank will

be emptied. Part to be emptied = 25.

Part emptied by (A+B) in 1 minute =(16 110)= 115.=> 1

15: 25

: :1 : x => x=25115=6min.

So, the tank will be emptied in 6 min.Ex: Three pipes A,B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of solution R inthe tank after 3 minutes?Part filled by (A+B+C) in 3 minutes

= 3( 130+ 120+ 110)=(31160)=1120 .Part filled by C in 3 minutes = 3

10.

Required Ratio =( 3102011)= 611 .Ex: Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both taps are open then due to leakage, it took 30 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?

Part filled by (A+B) in 1 hour =(15+ 120)=14 .So, A and B together can fill the tank in 4 hours.

Work done by the leak in 1 hour =(14 29)= 136 .leak will empty the tank in 36 hours.

By direct formula: = 44.54.54

=36hrs.

Ex: One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:Let the slower pipe alone fill the tank in x minutes.

Then, faster pipe will fill it in x3

minutes .

Then 1x+3x= 1

36 => x=144min.

Ex: A tank is filled in 5 hours by three pipes A,B and C. The pipe C is twice as fast as B and B is twice as fast as A.How much time will pipe A alone take to fill the tank?Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take x2

and x4

hours

respectively to fill the tank.

=> 1x+ 2x+4x=1

5 7

x=1

5 x=35hrs.

So pipe A in 35 hrs, Pipe B in 17.5 hrs and Pipe C in 8.75 hrs alone can fill the empty tank.Ex: A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in thesame time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x-5) and (x-9) hours respectively to fill the tank.

=> 1x+ 1x5

= 1x9

=> x=15hrs.

Neglectingve Values.Ex: Two pipes A and B can fill a tank in 12 minutes and 15 minutes respectively. If both the taps are opened simultaneously, and the tap A is closed after 3 minutes, then how much more time will it take to fill the tank by tap B?

Part fill in 3min.= 3( 112+ 115)=3 960= 920 .Remaining part= 1 9

20=11

20.

Part filled by B in 1min= 115

.

115

: 1120

::1 : x => x=(1120115)=8 14=8min15 sec.Remaining part is filled by B in8min15 seconds.

Ex: Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallon per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:Work done by the waste pipe in 1 minute

= 115

( 120+ 124)=( 115 11120)= 140[ve signmeans emptying ]

Volume of 140

part=3 gallons.

Volume of whole =(3*40) gallons=120 gallons.

KNKumar 14

Ex: Two pipes A and B can fill a tank in 15 hours and 20 hours respectively while a third pipe C can empty the full tank in 25 hours. All the three pipes are opened in the beginning. After 10 hours, C is closed. In how much time, will the tank be full?

Part filled in 10 hours =10( 115+ 120 125)=2330 .Remaining part = 1 23

20= 7

30.

(A+B)'s 1 hour work =( 115+ 120)= 760 .760

: 730

::1 : x => x= 730

1607=2hours.

The tank will be full in (10+2)hours=12 hrs.Ex: Two pipes A and B can fill a cistern in 12 minutes and15 minutes respectively. While a third pipe C can empty the full tank in 6 minutes. A and B kept open for 5 minutesin the beginning and then C is also opened. In what time isthe cistern emptied?

Part filled in 5 min. = 5( 112+ 115)=(5 960)=34 .Part emptied in 1 min. when all the pipes are opened

= 16( 112+ 115)=( 16 320)= 160 .

Now, 160

part is emptied in 1 min.

34

part will be emptied in (6034)=45min.Ex: Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be fullin:

(A+B)'s 1 hour's work =( 112+ 115)= 320 .(A+C)'s 1 hour's work =( 112+ 120)= 215 .

Part filled in 2 hrs =( 320+ 215)=1760 .Part filled in 6 hrs. =(31760 )=1720 .Remaining Part =(11720)= 320 .Now, it is the turn of A and B and 3

20part is filled by A

and B in 1 hour.Total time taken to fill the tank=(6+1) hrs=7 hrs.Ex: A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400m3 .The emptying capacty of the pump is 10m3/minutehigher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?Let the filling capacity of the pump be x m3/minute .Then, emptying capacity of the pump

= (x+10)m3/minute .

So,2400x

2400(x+10)

=8 => ( x50)(x+60)=0

=> x=50. (NeglectingveValues of x) .Ex: A leak in the bottom of a tank can empty the full tank in 8 hours. An inlet pipe fills water at the rate of 6 liters a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 12 hours. How many litersdoes the cistern hold?

Work done by the inlet in 1 hour =(18 112)= 124 .Work done by the inlet in 1 min. =( 124 160)= 11440 .Volume of 1

1440part=6 liters.

Volume of whole = (1440*6) liters = 8640 liters.

KNKumar 15

Work and WagesNote: Wages are distributed in Proportion to the work done and in indirect (or inverse) proportion to the time taken by the individual.TotalWage = One person' sOne Day ' sWage * Number of Persons * Number of Days

Ex: A can do a work in 6 days and B can do the same work in 5 days. The contract for the work is Rs 220. How much shall B get if both of them work together?Method I:

A' s1dayWork = 16

; B ' s1Day ' sWork=15

Ratio of Wages=16

: 15=5 :6 ; B ' s Share= 220

5+66=Rs120.

Method II:As wages are distributed in inverseproportion of number of days, their share should be in the ratio 5:6.

B's Share = 22011

6=Rs120.

Ex: A man can do a work in 10 days. With the help of a boy he can do the same work in 6 days. If they get Rs 50for that work, what is the share of that boy?

The boy can do the work in = 106106

=15days.

Man's Share : Boy's Share=15:10=3:2

Man's Share = 5053=Rs30.

Ex: A, B and C can do a work in 6, 8 and 12 days respectively. Doing that work together they get an amount of Ra 1350. What is the share of B in that amount?Direct Method: A's Share : B's Share : C's Share=B's time * C's time : A's time * C's time : A's time * B'stime =96:72:48=4:3:2

=> B ' s Share=13509

3=Rs.450.

Ex: A,B and C contract a work for Rs 550. Together, A

and B are supposed to do 711

of the work. How much

does C get?

(A+B)did 711

work andC did =(1 711)= 411work.(A+B)' s Share : C ' s Share= 7

11: 411

=7: 4

C's Share = 55011

4=Rs200.

Ex: Two men undertake to do a piece of work for Rs 200. One alone could do it in 6 days, the other in 8 days.With the asssistance of a boy they finish it in 3 days. How should the money be divided?

1stman ' s 3days work = 36;

2ndman' s3days work=38

The boy's 3 days work =1(36+38)=18The share will be in the ratio = 3

6: 3

8: 18=4: 3:1

1 st man's Share = 2008

4=Rs100

2 nd man's Share = 2008

3=Rs 75

The boy's Share = 2008

1=Rs 25

Ex: Wages for 45 women amount to Rs 15525 in 48 days. How many men must work 16 days to receive Rs 5750, the daily wages of a man being double those of a woman?

Wage of a women for a day = 155254548

=Rs 11516.

Thus, wage of a man for a day = 211516

=Rs 1158

Now , Number of men

= Total WageNumber of days * 1man' s1day ' sWage

= 5750816115=25men

Ex: A and B undertake to do a work for Rs 56. A can do it alone in 7 days and B in 8 days. If with the assistance of a boy they finish the work in 3 days then the boy gets Rs___.A's 3 days work+B's 3 days work+Boy's 3 days work=1

37+3

8+Boy ' s3days work=1

Boy ' s3days work=1(37+38)=1156Ratio of Shares=3

7: 38

: 1156

=24 : 21:11

Boy ' sShare= 5624+21+11

11=Rs11.

Ex: A sum of money is sufficient to pay A's wages for 21 days or B's wages for 28 days. The money is sufficient to pay the wages of both for___days.sum of money is sufficient to pay A wages for 21 days

money sufficient to pay A for 1 day = 121

sum of money is sufficient to pay B wages for 28 days

money sufficient to pay B for 1 day = 128

money sufficient to pay A & B for 1 day =1

21+ 1

28=(21+28)

2128= 49

588= 1

12thus, The same money is sufficient to pay the wages of bothfor 12 daysFormula:

Number of Days=Multiplication of Number of DaysAdditionof Days

Number of days= 212821+28

=12 days.

KNKumar 16

Ex: 3 men and 4 boys can earn Rs 756 in 7 days. 11 men and 13 boys can earn Rs 3008 in 8 days. In what time will 7 men with 9 boys earn Rs 2480?(3m+4b) in 1 day earn

Rs 7567

=Rs108(1)

(11m+13b) in 1 day earn

Rs 30088

=Rs376(2)

From (1), we see that to earn Re 1 in 1 day there

should be 3m+4b108

persons. Similarly, from

(2), to earn Re 1 in 1 day there should be11m+13b

376persons.

And also; 3m+4b108

=11m+13b376

=> mb=5

3Since both the LHS and the RHS denote the samenewline quantity: Number of persons earning Re 1 in 1 day.Now, from (1)(3m+4b) in 1 day earn Rs 108

or, 3m+435m in 1 day earn Rs 108

or, 27m5

in 1 day earn Rs 108

1m in 1 day earns Rs 108527

=Rs 20

Thus, we get that a man earns Rs 20 daily and a boy earns

Rs 2035=Rs 12daily.

(7m+9b) earn Rs (7*20+9*12)=Rs 248 in 1 day.(7m+9b) earn Rs 2480 in 10 days.Quicker method:Using Cross-Multiplication-Division Rule.Men Boys Earnings Days 3 4 756 7 11 13 3008 8

Now,

Men(330088 117567 )=Boys(137567 430088 )=>m(337611108)=b(108134376)

=> mb= 5

3.Ex: A,B and C together earn Rs 1350 in 9 days. A and C together earn Rs 470 in 5 days. B and C together earn Rs 760 in 10 days.Find the daily earning of C.

Daily earning of (A+B+C )=13509

=Rs150(1)

Daily earning of (A+C )=4705

=Rs94(2)

Daily earning of (B+C )=76010

=Rs76(3)

From (1)and (2)daily earning of B=15094=Rs56(4)From (3)and (4)daily earning of C=7656=Rs 20(5)From (2)and (5)daily earning of A=9420=Rs74.

Ex: A can do a piece of work in 10 days, B in 15 days. They work for 5 days. The rest of the work was finished by C in 2days. If they get Rs. 1500 for the whole work, the daily wages of B and C are:Part of the work done by A

=( 1105)=12Part of the work done by B

=( 1155)=13Part of the work done by C

= 1(12+13)=16

A's Share : B's Share : C's Share = 12

: 13

: 16=3: 2:1

A's Share =(361500)=Rs.750 ; B's Share =(261500)=Rs.500 ; C's Share =(161500)=Rs.250Daily Wages: A=750

5=Rs.150 B=500

5=Rs.100 C=250

2=Rs.125

Daily Wages of (B+C)=Rs. 225

Ex: If A can do 14

of a work in 3 days and B can do

16

of same work in 4 days, how much will A get if both

work together and are paid Rs. 180 in all?Whole work done by A in (3*4)=12, B in (4*6)=24 days.

A ' sWage :B ' sWage= 112

: 124

=2 :1

A ' s Share=23180=Rs.120,

B ' s Share=180120=Rs.60Ex: A,B and C together earn Rs. 300 per day, while A and C together earn Rs.188 and B and C together earn Rs. 152.The daily wager of C is:

B ' s Daily earnings=300188=Rs.112 ;A' s Dialy earnings=300152=Rs.148 ;C ' s Dialy earnings=300(112+148)=Rs.40

Ex: A,B and C are employed to do a piece of work for Rs.

529. A and B together are supposed to do 1923

of work

and B and C together 823

of work. What amount should

A be paid?

Work doneby A=(1 823)=1523=> A:(B+C)=1523 :

823=15 :8

So , A ' s Share=(1523529)=Rs.345.KNKumar 17

Time and WorkNote: If M 1 persons can do W 1 works in D1 days andM 2 persons can do W 2 works in D2 days then we have

M 1D1W 2=M 2D 2W 1. More men less days and conversely more days less men. More men More work and conversely more work more men. More days more work and conversely more work more days. If we include the working hours (say T 1 and T 2 ) for twogroups then the relation ship is

M 1D1T 1W 2=M 2 D2T 2W 1 Again, if the efficiency (say E1 and E2 ) of the persons in two groups is different then the relationship is

M 1D 1T 1 E1W 2=M 2D 2T 2 E2W 1 If A can do a piece of work in x days, and B can do it in y days then A and B working together will

do the same work in xyx+ y If A,B and C can do a work in x,y and z days respectively then all of them working together can

finish the work in xyzxy+ yz+zx If A and B together can do a piece of work in x days and A alone can do it in y days, then B alone

can do the work in xyyxEx: A can do a piece of work in 5 days. How many days will he take to complete 3 works of same type?As ' A ' is the only person to do thework in bothcases , soM 1=M 2=1,D1=5,W 1=1,D2=? ,andW 2=3=>M 1 D1W 2=M 2 D2W 1 => 53=D21=> D 2=15days.

Ex: 16 men can do a piece of work in 10 days. How many men are needed to complete the work in 40 days?Work is same in both cases , soW 1=W 2=1,D1=10,M 1=16 ,D 2=40, andM 2=?=>M 1 D1W 2=M 2 D2W 1 => 1610=M 240 =>M 2=4men.

Ex: 40 men can cut 60 trees in 8 hrs. If 8 men leave the job how many trees will be cut in 12 hours?M 1=40,D1=8(We cantakeT 1 instead of D1) ,W 1=60(cutting of trees is takenas work ) ,M 2=408=32,D2=12(Wecan takeT 2 instead of D2) ,W 2=?=>M 1 D1W 2=M 2 D2W 1 or M 1T 1W 2=M 2T 2W 1 => 408W 2=321260 =>W 2=72 trees

Ex: 5 men can prepare 10 toys in 6 days working 6 hrs a day. Then in how many days can 12 men prepare 16 toys working 8 hrs a day?

M 1 D1T 1W 2=M 2D2T 2W 1=> 56616=12D2810

=> D2=5661612810

=3days.

By Rule of Fraction:1. We have to find number of days, so write givennumber of days first.2. Number of men increases => Work will be done in less days => multiplying factor should be

less than 1, which is 512.

3. Number of toys increases => it will take more days => multiplying factor should be more than 1,

which is 1610.

4. Number of working hours increases => it will take less days => multiplying factor should be less

than 1, which is 68.

Thus , required number of days

= 6( 512)(1610)(68)=3days.Note: In rule of fraction the multiplying factor is decided such that how the known variable changes the value of unknown variable.(Increase or Decrease). Ex: more men less days, here men known, days unknown variable, number of men inversely proportional to number of days, so, the fraction is always less than 1.

Ex: A can do a piece of work in 5 days, B can do it in 6 days and Ccan do it in 12 days. (i) How long it will take if A and B work together? (ii) How long it will take if A, B and C work together?

(i) (A+B)cando thework in= 565+6

=2 811

days.

(ii) (A+B+C )can do thework in

= 561256+612+512

=2 29days.

Ex: A and B together can do a piece of work in 6 days and A alone can do it in 9 days. In how many days can B alone do it?

Balone can do thewhole work in= 6969

=18days.

Ex: A and B can do a piece of work in 12 days, B and C in 15 days, C and A in 20 days. How long would each take separately to do the same work?

(A+B)can do in 12days ,(B+C )can do in 15days ,(A+C )can do in 20days.

2(A+B+C)= 1215201215+1220+1520=5 days.

(A+B+C)=10days.(less menmoredays )

Now , Acando thework in 10151510

=30 days

[As A=(A+B+C )(B+C )]

B cando thework in 10202010=20days

[As B=(A+B+C )(A+C )]

Acando thework in 10121210

=60days

[AsC=(A+B+C )(B+A)]

KNKumar 18

Ex: 38 men, working 6 hours a day can do a piece of work in 12 days. Find the number of days in which 57 men working8 hrs a day can do twice the work. Assume that 2 men of the first group do as much work in 1 hour as 3 men of the

second group do in 1 12hrs.

Ratio of efficiency of persons in first group to the second group = E1 :E2=(31.5) :(21)=4.5: 2

Now, M 1D1T 1 E1W 2=M 2 D2T 2 E2W 1 D2=381264.52

57821=27days.

Note: Less number of persons from the 1st Group do the same work in Less Number of days, so they are more efficient.M 1,M 2 Number of Men in 1st and 2nd Groups respectively.W 1,W 2 Represents the work done by 1st and 2nd Groups respectively.D1,D2 Represents Number of days taken by 1st and 2nd Groups respectively.E1, E2 Represents the Efficiency of men's in 1st and 2nd Groups respectively.T 1,T 2 Represents Number of working Hours taken by 1st and 2nd Groups respectively.

Ex: Two women, Ganga and Saraswati, working separetely can mow a field in 8 and 12 hrs respectively. If they work in stretches of one hour alternately, Ganga beginning at 9a.m., when will the mowing be finished?

In the first hour Ganga mows 18

of the field.

In the second hour Saraswati mows 112

of the field.

In the first 2 hours (18+ 112 =) 524 of the field mown.In 8 hrs 5

244=5

6of the field is mown.(we calculated

for 4 pairs of hours only because if we calculate for 5 pairsof hours, the work done is more than 1.)

Remaining Field to be mown =(156)=16 of the field.

In the 9th hour Ganga mows 18

of the field.

Saraswati will finish the mowing of (16 18)= 124 of the field in ( 124 112)or 12 of anhour.The total time required is (8+1+12)=9 12 hours.Thus the work will be finished at

9+9 12=18 1

2=6 1

2p.m.

Ex: A and B together can do a piece of work in 12 days which B and C together can do in 16 days. After A has been working at it for 5 days, and B for 7 days, C takes up and finishes it alone in 13 days. In how many days could each do the work by himself?

(A+B)'s 1 day work= 112

; (B+C)'s 1 day work= 116

Given that (A's 5 days+B's 7 days+C's 13 days) work =1(A's 5 days+B's 5 days+B's 2 days+C's 2 days+C's 11 days) work=1

((A+B)'s 5 days+(B+C)'s 2 days+C's 11 days) work=1512

+ 216

+C ' s 11days work=1 ; C ' s 11dayswork=1( 512+ 216)=1124

C ' s 1 day work= 112411

= 124

B ' s 1 day work= 116

124

= 148

A' s 1 day work= 1121

48=1

16A,B and C can do the work in

16, 48 and 24 days respectively.Ex: To do a certain work B would take three times as long as A and C together and C twice as long as A and B together. The three men together complete the work in 10 days. How long would each take separately?3 times B's daily work=(A+C)'s daily workAdd B's daily work to both sides.

4 times B's daily work=(A+B+C)'s daily work= 110

B's dialy work= 140

2 times C's daily work=(A+B)'s daily workAdd C's daily work to both sides.

3 times C's daily work=(A+B+C)'s daily work= 110

C's daily work= 130

A's daily work = 110

( 140+ 130)= 124.A,B and C can do the work in 24,40 and 30 days respectively.

Ex: 8 men and 16 women can do a work in 8 days. 40 men and 48 women can do the same work in 2 days. How many days are required for 6 men and 12 women to do the same work?We will compare the capacity of a man and woman.8 days*(8 men+16 women)= 1 work (1) 2 days*(40 men+48 women)= 1 work (2)'x' days*(6men+12 women)=1 work (3) => x=?=> 8(8m+16w)=2(40m+48w)=1 work => 1m=2w

Now, 8m+16w=16w+16w=32w [from(1)]Now, 6m+12w=12w+12w=24w [from(3)]M 1=32w ; D 1=8 ;M 2=24w ; D2=?

M 1D1=M 2 D2 => D2=328

24=10 2

3days.

KNKumar 19

Ex: 1 man or 2 women or 3 boys can do a work in 44 days. Then in how many days will 1 man, 1 women and 1 boy do the work? Note:Required Number of Days

= 1

[ 1441+ 1442+ 1443 ] = 44123

6+3+2=24 days

1441

= Number of Men in AND-PartNumber of Days * Number of Men in OR-Part

1442

= Number of women in AND-PartNumber of Days*Number of women in OR-Part

1443=

Number of Boys in AND-PartNumber of Days * Number of Boys in OR-Part

Ex: If 12 men or 15 women can reap a field in 14 days, how long will 7 men and 5 women take to reap it?

Required Number of Days = 1

[ 71412+ 51415 ] = 2442

24+42=168

11=15 3

11days.

Ex: A certain number of men can do a work in 60 days. If here were 8 men more it could be finished in 10 days less. How many men are there?'x' men can do the work in 60 days and (x+8) men do the work in (60-10=)50 days.

M 1D1=M 2 D2 => 60x=50(x+8) => x=40men

Original Number of Workers= Number of MOREWorkers * Number of Days taken by the Second GroupNumber of LessDays

Original Number of Workers= 8(6010)10 =850

10 =40men.

Ex: A is thrice as fast as B, and is therefore able to finish awork in 60 days less than B. Find the time in which they can do it working together.If A works 3 times as fast,

he'll do the job in 1/3 the time: A=13B

BA=60 => B13B=60

Multiply through by 3:

3BB=180 =>2B=180 => B=1802

=> B=90

A=1390 => A=30

So A can do the job in 30 days, B can do the job in 90 days

Working together, they can do = 309030+90

=22.5days

Ex: A takes as much time as B and C together take to finish a job. A and B working together finish the job in 10 days. C alone can do the same job in 15 days. In how many days can B alone do the same work?

(A+B)+(C) can do in 151015+10

=6days.

Since A's days=(B+C)'s daysB+C can do in 6*2=12 days.

B[B={B+C}-C] can do in 15121512

=60days.

Ex: A is twice as good a workman as B. Together, they finish the work in 14 days. In how many days can it be done by each separately?Let B finish the work in 2x days,Then, A finish the work in x days (A is twice active as B)

(A+B) finish the work in 2x2

3x=14 => x=21

A takes 21 days and B takes 42 days.

Ex: 5 men and 2 boys working together can do four times as much work per hour as a man and a boy together. Compare the work of a man with that of a boy.Given that (5m+2b)'s 1 day's work=(1m+1b)'s 4 day's work=> (5m+2b)'s 1 day's work=(4m+4b)'s 1 day's work

=>5m+2b=4m+4b =>m=2b => mb=2

1That is, a man is twice as efficient as a boy.Ex: A and B can do a work in 45 and 40 days respectively.They began the work together, but A left after some time and B finished the remaining work in 23 days. After how many days did A leave?B works alone 23 days.

Work done by B in 23 days= 2340

work

A+B do together 12340

=1740

work

Now, A+B do 1 work in 404540+45

=404585

days.

A+B do 1740

work in 404585

1740

=9days.

Quicker Method: 404540+45(402340 )=9days.

Ex: A can do a work in 25 days and B can do the same work in 20 days. They work together for 5 days and then Agoes away. In how many days will B finish the work?A+B can do the work in 5 days

= 5[ 125+ 120 ]= 5452520= 920Rest of the work = 1 9

20=11

20

B will do the rest of the work in 201120

=11days.

KNKumar 20

Note :Efficient persontakes less time. In other wordswe may say thatEfficiency (E) is indirectly proportional to number of days (D) taken to complete a work

then , E 1D

or E= KD

,(where K is aConstant.)=> ED=Constant => E1 D1=E2 D2=E3D 3=....=EnD n

Ex: A builder decided to build a house in 40 days. He employed 100 men in the beginning and 100 more after 35days and completed the construction in stipulated time. If he had not employed the additional men, how many days behind the schedule would it have been finished?Let 100 men only complete the work in 'x' days.Work done by 100 men in 35 days+work done by 200 menin (40-35=) 5 days=1

35x+2005

100x=1 => x=45days

Therefore, if additional men were not employed, the work would have lasted 45-40=5 days behind schedule time.Quicker Approach: 200 men do the rest of the work in 40-35=5 days.100 men can do the rest of the work in

5200100

=10days.

Required number of days=10-5=5 days.Ex: A team of 30 men is supposed to do a work in 38 days. After 25 days, 5 more men were employed and the work was finished one day earlier. How many days would it have been delayed if 5 more men were not employed?35 men do the rest of the job in 12 days.[12=38-25-1]

30 men can do the rest of the job in 123530

=14days.

Thus the work would have been finished in 25+14=39 days, that is, (39-38=)1 day after scheduled time.Ex: A contractor undertakes to dig a canal 12 km long in 350 days and employs 45 men. After 200 days he finds that only 4.5 km of the canal has been completed. Find thenumber of extra men he must employ to finish the work intime.M 1 D1W 2=M 2D2W 1=> 452007.5=M 21504.5

=>M 2=452007.5

1504.5=100

Required Number of persons to beadded = 10045=55men

Ex: There is sufficient food for 400 men for 31 days. After28 days 280 men leave the place. For how many days will the rest of the food last for the rest of the men?

40031=28400+(400280)x=> 12400=11200+120x => x=10days.

Method II: The rest of the food will last for (31-28=) 3 days if nobody leaves th place.

Thus the rest of the food will last for 3( 400400280) days for the 120 men left=10 daysNote: For less persons the food will last longer, therefore

3 is multiplied by 140120

, a more than-one fraction.

Ex: A, B and C can do a work in 8, 16 and 24 days respectively. They all begin together. A continues to work till it is finished, C leaving off 2 days and B one day

before its completion. In what time is the work finished?Let the work be finished in x days.Then A's x day's work+B's (x-1) day's work+C's (x-2) day's work = 1x8+ x1

16+ x2

24=1 => 6x+3x3+2x4

48=1

=> 11x=55 => x=5days.Ex: A, B and C can do a work in 16 days, 12 4

5days

and 32 days respectively. They started the work together but after 4 days A left. B left the work 3 days before the completion of the work. In how many days was the work completed?Suppose the work is completed in 'x' days.A's 4 day's work+B's (x-3) day's work+C's x day's work=1

416

+( x3)5

64+ x

32=1

16+5x15+2x64

=1 => x=9days.

Ex: A started a work and left after working for 2 days. Then B was called and he finished the work in 9 days. HadA left the work after working for 3 days, B would have finished the remaining work in 6 days. In how many days can each of them, working alone, finish the whole work?Let A and B do the work in 'x' days and 'y' days respectively.Now, work done by A in 2 days+ work done by B in 9 days=1

2x+ 9

y=1 ; similarly 3

x+ 6

y=1

Let 1x=a ; 1

y=b.

2a+9b=1(1) and 3a+6b=1--(2)

a=15

=> x=5 ;b= 115

=> y=15days.

Quicker method:

A will finish the work in 392696

=153=5days.

For B, we should use the above result

B does 125=3

5work in 9days.

B does 1work in 953=15days.

Ex: A can copy 75 pages in 25 hrs. A and B together can copy 135 pages in 27 hrs. In what time can B copy 42 pages?

A can copy 7525

=3 pages in 1 hr.

A+B can copy 13527

=5 pages in 1 hr.

B can copy 5-3=2 pages in 1 hr.

B can copy 42 pages in 422=21hrs.

KNKumar 21

Ex: A does half as much work as B in three-fourth of the time. If together they take 18 days to complete a work, how much time shall B take to do it?Suppose B does the work in x daysThen ,

Adoes 12work in 3x

4days => Adoes1work in 3x

2days

Given that (A+B) can dothe same work in 18 days.

(A+B)dothe work inx3x

2

x+ 3x2

=18 => x=30 days.

Bdoes the whole work in 30days.Ex: If 5 men and 3 boys can reap 23 acres in 4 days, and 3men and 2 boys can reap 7 acres in 2 days, how many boys must assist 7 men in order that they may reap 45 acres in 6 days?

(5m+3b) ' s 1day ' swork=234

=> 423

(5m+3b)=1work(1)

(3m+2b)' s1day ' s work=72

=> 27(3m+2b)=1work(2)

From (1) and (2) 5m+3b3m+2b

= 2347

2=23

14 => m=4b.

5m+3b=5*4+3=23 boys [from (1)]M 1D 1W 2=M 2D 2W 1.

M 2=M 1D1W 2D2W 1

=23445623

=30boys.

3074=2boys should assist them.Note:To reap 45 acres in 7 days 30 boys are needed, already 7 men are there that is (7*4=)28 boys, so number of boys required to assist 7 men are 30-28=2 boys.Ex: A, B and C can finish a work in 10, 12 and 15 days respectively. If C stops after 2 days, how long would it take A and C to finish the remaining work?

A+B+C in 2 days, do

2( 110 + 112+ 115)work=214=12 work.Now, C with draws. A+B will do the whole work in

101210+12

=6011

days.

A+B will do 12work in 30

11=2 8

11days.

Ex: B can do a job in 6 hrs, B and C can do it in 4 hrs and

A, B and C in 2 23hrs. In how many hrs can A and B do

it?

A+B+C cando the work in 83hrs(1)

B+C cando the work in 4hrs(2)B cando the work in 6hrs (3)

From (2) and (3), C can do it in 4664

=12hrs(4)

From (1) and (4) A+Bcando it in

=

8312

1283

=32328

=247=3 3

7hrs.

Ex: A group of men decided to do a work in 10 days, but five of them became absent. If the rest of the group did thework in 12 days, find the original number of men.Suppose there were 'x' men originally. Then,

M 1D 1=M 2 D2 => 10x=12( x5)

=> x=125|2-10|

x=30men

Ex: A can do a certain job in 12 days. B is 60% more efficient than A. How many days does B alone take to do the same job?Ratio of Times taken by A and B=160:100=8:5Suppose B alone takes 'x' days to do the job.

8:5 : :12 : x => 8x=512 => x=7 12days.

Ex: A takes twice as much time as B or thrice as much time to finish a piece of work by C. Working together, they can finish the work in 2 days. B can do the work alone in:

Let A, B and C take x , x2

and x3hours respectively to

finish the work.

Then (1x+ 2x+3x )=12 => 6x=12 => x=12So, B takes 6 hrs to finish the work.

Ex: X can do 14

of a work in 10 days, Y can do 40% of

the work in 40 days and Z can do 13

of the work in 13

days. Who will complete the work first?Whole work will be done by X in (10*4)=40 days.

Wholework will be doneby Y in=(4010040 )=100 daysWhole work will be done by Z in (13*3)=39 days.

Z will complete the work First.Ex: P,Q and R are three typists who working simultaneously can type 216 pages in 4 hours. In one hour,R can type as many pages more than Q as Q can type morethan P. During a period of five hours, R can type as many as P can during 7 hours.How many pages does each of them type per hour?Let the number of pages typed in one hour by P,Q and R be x,y and z respectively. Then,

x+y+z= 2144

=> x+y+z=54 (1) newline

z-y=y-x => 2y=x+z (2) newline

5z=7x => x= 57z (3) newline

solving (1),(2) and (3) we get x=15,y=18 and z=21

KNKumar 22

Ex: Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much timewill they take, working together on two different computers to type an assaignment of 110 pages?

Number of pages typed by Ronald in 1hour = 326=16

3

Number of pages typed by Elan = 405=8.

Number of pages typed by both in 1hour = 163+8=49

3Time takenby both to type110 pages

=(110 340)hrs=8 14 hrs=8hr 15min.Ex: A and B together can complete a work in 12 days. A alone can complete it in 20 days. If B does the work only for half a day daily, then in how many days A and B together will complete the work?

B ' s 1daywork=( 1121over 20)= 130 .Now, (A+B)'s 1 day's work =( 120+ 160)= 115[B works for half a day only so, Number of working days doubles]So, A and B together will complete the work in 15 days.Ex: A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes the remaining work in 42 days. In how much time will A and B, working together, finish the work?

Work doneby A in 10days= 180

=18

Remaining work

(work done in 42days by B)=118=78

Wholework doneby B in 4287=48 days.

(A+B)can finish thework in= 804880+48=30days.

Ex: Two workers A and B are engaged to do a work. A working alone takes 8 hours more to complete the job thanif both worked together. If B worked alone, he would need

4 12hours more to complete the job than they both

working together. What time would they take to work together?Let A and B together take x hours to complete the work. Then A alone takes (x+8) hrs and B alone takes

(x+92)hrs to complete the work. Then1

( x+8)+ 1

(x+92)=1x

=> 1x+8

+ 22x+9

=1x

=> x (4x+25)=(x+8)(2x+9) => x=6 hrsEx: A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 a.m. While machine P is closed at 11 a.m. And the remaining two machines complete the work. Approximately at what time will the work be finished?

(P+Q+R) ' s 1hour work=(18+ 110+ 112)= 37120.Work done by P ,Q and R in 2hours=( 371202)= 3760.

Remaining work=(13760)= 2360.(Q+R) ' s1hour ' s work=( 110+ 112)=1160

1160

work is doneby Q andR in 1hour.

2360

work will doneby Q and R in

(60112360)=2311 hours=2hours.So ,the work will be finished approximately

2hours after 11a.m. , i.e.around 1 p.m.Ex: A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave.How many days more will A take to finish the work?

2(A+B+C ) ' s 10days work

=( 130+ 124+ 120)10days= 11610=58RemainingWork=1 5

8=3

8

A' s1day work= 116

124

= 148

.

Now , 148

work is doneby A in 1day.

So , 38work will be done by Ain(4838)=18 days.

Ex: A, B and C together can complete a piece of work in 10 days. All the three started working at it together and after 4 days A left. Then B and C together completed the work in 10 more days. A alone could complete the work in:

Work done by A , B andC in 4days=( 1104)=25 . ; Remaining work=125=35 .35 work doneby (B+C) in 10days. ; Wholework doneby (B+C) in 10

53=

503 days.

(A+B+C )' s1day ' s work= 110,

; (B+C )' s1day ' s work= 350

. ; A' s1day ' s work= 110

350

= 125

.

A alone could complete thework in 25days.KNKumar 23

Ex: 12 men can complete a piece of work in 4 days, while 15 women can complete the same work in 4 days. 6 men start working on the job and after working for 2 days, all of them stopped working. How many women should be put on the job to complete the remaining work, if it is to be completed in 3 days?

1man' s 1day ' swork= 1124

= 148

1woman ' s1day ' s work= 1154

= 160

6men' s2 day ' s work= 648

2=14

Remaining work=(114)=34Now , 1

60work is doneby1woman in1day ,

Then 34 work will be donebyM 1D1W 2=M 2 D2W 1

=> 1134=M 23

160

=>M 2=15women.

Ex: Tweleve children take sixteen days to complete a work which can be completed by eight adults in tweleve days. Sixteen adults started working and afterthree days ten adults left and four children joined them. How many days will they take to complete the remaining work?

1Child ' s1day ' s work= 11612

= 1192

1 Adult ' s1day ' swork= 1812

= 196

work done in 3days by Adults=( 196163)=12Remaining work=112=

12 .

(6 Adults+4Childrens) ' s1 daywork= 696

+ 4192

= 112

W 2 D1=W 1 D2 =>121=

112D2 => D2=6days.

Ex: Sixteen men can complete a work in tweleve days. Twenty-four children can complete the same work in eighteen days. Tweleve men and eight children started working and after eight days three more children joined them. How many days will they now take to complete?

1man' s1day ' s work= 1192

1children ' s1day ' s work= 1432

(12men+8children) ' s work done

in 8days = 8( 12192+ 8432)=3554Remaining work=13554=

1954

(12men+11 children)' s

1day ' s work =