Important Facts and Formulae:
1.Allegation:It is the rule that enables us to find the
ratio in which two of more ingredients at the given price
must be mixed to produce a mixture of a desired price.
2.Mean Price:The cost price of a unit quantity of the
mixture is called the mean price.
3.Rule of Allegation:If two ingredients are mixed then
Quantity of Cheaper / Quantity of Dearer =
(C.P of Dearer – Mean Price) /(Mean Price–C.P of Cheaper).
C.P of a unit quantity of cheaper(c)
C.P of unit quantity of dearer(d)
Mean Price(m)
(d-m) (m-c)
Cheaper quantity:Dearer quantity = (d-m):(m-c)
4.Suppose a container contains x units of liquid from which
y units are taken out and replaced by water. After n
operations the quantity of pure liquid = x (1 – y/x)n units.
Simple Problems
1.In what ratio must rice at Rs 9.30 per Kg be mixed with rice
at Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg?
Solution:
C.P of 1 Kg rice of 1st kind 930 p
C.P of 1 Kg rice of 2n d kind 1080p
Mean Price 1000p
80 70
Required ratio=80:70 = 8:7
2.How much water must be added to 60 liters of milk at
11/2 liters for Rs 20 so as to have a mixture worth
Rs 10 2/3 a liter?
Solution: C.P of 1 lit of milk = 20*2/3 = 40/3
C.P of 1 lit of water 0 C.P of 1 lit of milk 40/3
Mean Price 32/3
8/3 32/3
Ratio of water and milk =8/3 : 32/3 = 1:4
Quantity of water to be added to 60 lit of milk
=1/4*60=15 liters.
3.In what ratio must water to be mixed with milk to gain
20% by selling the mixture at cost price?
Solution:Let the C.P of milk be Re 1 per liter
Then S.P of 1 liter of mixture = Re.1
Gain obtained =20%.
Therefore C.P of 1 liter mixture = Rs(100/120*1) =5/6
C.P of 1 liter of water 0 C.P of 1 liter of milk1
Mean Price 5/6
1/6 5/6
Ratio of water and milk =1/6 : 5/6 = 1:5.
4.In what ratio must a grocer mix two varieties of pulses
costing Rs 15 and Rs 20 per Kg respectively so as to get
a mixture worth Rs 16.50 per Kg?
Solution:
Cost of 1 Kg pulses of 1 kind 15 Cost of 1 Kg pulses of
2nd kind 20
Mean Price Rs 16.50
3.50 1.50
Required ratio =3.50 : 1.50 = 35:15 = 7:3.
5. 4Kg s of rice at Rs 5 per Kg is mixed with 8 Kg of rice
at Rs 6 per Kg .Find the average price of the mixture?
Solution:
rice of 5 Rs per Kg rice of 6 Rs per Kg
Average price Aw
6-Aw Aw-5
(6-Aw)/(Aw-5) = 4/8 =1/2
12-2Aw = Aw-5
3Aw = 17
Aw = 5.66 per Kg.
6.5Kg of rice at Rs 6 per Kg is mixed with 4 Kg of rice to
get a mixture costing Rs 7 per Kg. Find the price
of the costlier rice?
Solution: Using the cross method:
rice at Rs 6 per Kg rice at Rs x per Kg
Mean price Rs 7 per Kg
5 4
x-7:1=5:4
4x-28 = 5
4x=33=>x=Rs 8.25.
Therefore price of costlier rice is Rs 8.25 per Kg
Medium Problems
1.A butler stole wine from a butt of sherry which contained
40% of spirit and he replaced,what he had stolen by wine
containing only 16% spirit. The butt was then of 24%
strength only. How much of the butt did he steal?
Solution:
Wine containing 40%spirit Wine containing 16% spirit
Wine containing 24% spirit
8 16
They must be mixed in the ratio of =1:2.
Thus 1/3 of the butt of sherry was left and hence the
butler drew out 2/3 of the butt.
2.The average weekly salary per head of the entire staff
of a factory consisting of supervisors and the laborers
is Rs 60.The average salary per head of the supervisors
is Rs 400 and that of the laborers is Rs 56.Given that
the number of supervisors is 12.Find the number of
laborers in the factory.
Solution:
Average salary of laborer Rs 56 Average salary of
supervisors Rs 400
Average salary of entire staff Rs 60
340 4
Number of laborer / Number of Supervisors = 340 / 4=85/1
Thus,if the number of supervisors is 1,number of
laborers =85.
Therefore if the number of supervisors is 12 number of
laborers 85*12=1020.
3.The cost of type 1 rice is Rs 15 per Kg and type 2 rice
is Rs 20 per Kg. If both type1 and type 2 are mixed in the
ratio of 2:3,then the price per Kg of the mixed variety
of rice is?
Solution:Let the price of the mixed variety be Rs x per Kg.
Cost of 1 Kg of type 1 rice Rs 15 Cost of 1 Kg of type 2
rice Rs 20
Mean Price Rs x
20-x x-15
(20-x) /( x-15) = 2/3
=> 60-3x = 2x-30
5x = 90=>x=18.
4.In what ratio must a grocer mix two varieties of tea worth
Rs 60 a Kg and Rs 65 a Kg so that by selling the mixture
at Rs 68.20 a Kg he may gain 10%?
Solution:S.P of 1 Kg of the mixture = Rs 68.20,gain =10%
S.P of 1 Kg of the mixture = Rs (100/110*68.20)=Rs 62.
Cost of 1 Kg tea of 1st kind 60 Cost of 1 Kg tea of 2nd
kind 65
Mean Price Rs 62
3 2
Required ratio =3:2.
5.A dishonest milkman professes to sell his milk at cost price
but he mixes t with water and there by gains 25% .The
percentage of water in the mixture is?
Solution:Let C. P of 1 liter milk be Re 1.
Then S.P of 1 liter mixture=Re 1. Gain=25%
C.P of 1 liter mixture =Re(100/125*1) = Re 4/5.
C.P of 1 liter milk Re 1 C.P of 1 liter of water 0
Mean Price 4/5
4/5 1/5
Ratio of milk to water =4/5 : 1/5 = 4:1
Hence percentage of water n the mixture=1/5*100=20%.
12.A merchant has 1000Kg of sugar,part of which he sells
at 8% profit and the rest at 18% profit. He gains 14% on the
whole. The quantity sold at 18% profit is?
Solution:
Profit on 1st part 8% Profit on 2nd part 18%
Mean Profit 14%
4 6
Ratio of 1st and 2nd parts =4:6 =2:3.
Quantity of 2nd ind =3/5*1000Kg =600 Kg.
6.A jar full of whiskey contains 40% alcohol. A part of
this whiskey is replaced by another containing 19% alcohol
and now the percentage of alcohol was found to be 26%.
The quantity of whiskey replaced is?
Solution:
Strength of first jar 40% Strength of 2nd jar 19%
Mean Strength 26%
7 14
So,ratio of 1st and 2nd quantities =7:14 =1:2
Therefore required quantity replaced =2/3.
7.A container contains 40lit of milk. From this container
4 lit of milk was taken out and replaced by water.
This process was repeated further two times.
How much milk is now contained by the container?
Solution:Amount of milk left after 3 operations = 40(1-4/40)3lit
=(40*9/10*9/10*9/10)
= 29.16 lit
Complex Problems
1.Tea worth Rs 126 per Kg are mixed with a third variety in
the ratio 1:1:2. If the mixture is worth Rs 153 per Kg ,
the price of the third variety per Kg will be?
Solution:
Since First and second varieties are mixed in equal proportions
so their average price =Rs (126+135)/2 = 130.50.
So the mixture is formed by mixing two varieties ,one at
Rs 130.50 per Kg and the other at say Rs x per Kg in the
ratio 2:2 i e,1:1 we have to find x.
Costof 1Kg tea of 1st kind RS 130.50 Costof 1Kg tea of 2n d
kind Rs x.
Mean Price Rs 153
x-153 22.50
(x=153)/22.5 = 1 =>x-153 = 22.5
x = 175.50.
Price of the third variety =Rs 175.50 per Kg.
2.The milk and water in two vessels A and B are in the ratio 4:3
and 2:3 respectively. In what ratio the liquids in both the
vessels be mixed to obtain a new mixture in vessel c
consisting half milk and half water?
Solution:Let the C.P of milk be Re 1 per liter.
Milk in 1 liter mixture of A = 4/7 liter.
Milk in 1 liter mixture of B = 2/5 liter.
Milk in 1 liter mixture of C = 1/2 liter.
C.P of 1 liter mixture in A=Re 4/7
C.P of 1 liter mixture in B=Re 2/5.
Mean Price = Re 1/2.
By rule of allegation we have:
C.P of 1 liter mixture in A C.P of 1 liter mixture in B
4/7 2/5
Mean Price ½
1/10 1/14
Required ratio = 1/10 : 1/14 = 7:5.
3.How many Kg s of wheat costing him Rs 1.20,Rs 1.44
and Rs 1.74 per Kg so that the mixture may be worth
Rs 1.41 per Kg?
Solution:
Step1:Mix wheat of first and third kind to get a mixture
worth Rs 1.41 per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of
3rd kind 174p
Mean Price 141p
33 21
They must be mixed in the ratio =33:21 = 11:7
Step2:Mix wheats of 1st and 2n d kind to obtain a mixture
worth of 1.41.per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of 2n d
kind 144p
Mean Price 141p
3 21
They must be mixed in the ratio = 3:21=1:7.
Thus,Quantity of 2n d kind of wheat / Quantity of
3rd kind of wheat = 7/1*11/7= 11/1
Quantities of wheat of 1st :2n d:3rd = 11:77:7.
4.Two vessels A and B contain spirit and water mixed in
the ratio 5:2 and 7:6 respectively. Find the ratio n which
these mixture be mixed to obtain a new mixture in vessel
c containing spirit and water in the ratio 8:5?
Solution:Let the C.P of spirit be Re 1 per liter.
Spirit in 1 liter mix of A = 5/7 liter.
C.P of 1 liter mix in A =5/7.
Spirit in 1 liter mix of B = 7/13 liter.
C.P of 1 liter mix in B =7/13.
Spirit in 1 liter mix of C = 8/13 liter.
C.P of 1 liter mix in C =8/13.
C.P of 1 liter mixture in A 5/7 C.P of 1 liter mixture
in B 7/13
Mean Price 8/13
1/13 9/91
Therefore required ratio = 1/13 : 9/91 = 7:9.
5.A milk vendor has 2 cans of milk .The first contains 5% water
and the rest milk. The second contains 50% water. How much milk
should he mix from each of the container so as to get 12 liters
of milk such that the ratio of water to milk is 3:5?
Solution:Let cost of 1 liter milk be Re 1.
Milk in 1 liter mixture in 1st can = 3/4 lit.
C.P of 1 liter mixture in 1st can =Re 3/4
Milk in 1 liter mixture in 2n d can = 1/2 lit.
C.P of 1 liter mixture in 2n d can =Re 1/2
Milk in 1 liter final mixture = 5/8 lit.
Mean Price = Re 5/8.
C.P of 1 lt mix in 1st Re3/4 C.P of 1 lt mix in 2nd Re1/2
Mean Price 5/8
1/8 1/8
There ratio of two mixtures =1/8 :1/8 = 1:1.
So,quantity of mixture taken from each can=1/2*12
= 6 liters.
6.One quantity of wheat at Rs 9.30 p
er Kg are mixed
with another quality at a certain rate in the ratio 8:7.
If the mixture so formed be worth Rs 10 per Kg ,what is
the rate per Kg of the second quality of wheat?
Solution:Let the rate of second quality be Rs x per Kg.
C.P of 1Kg wheat of 1st 980p C.P of 1 Kg wheat of 2nd 100x p
Mean Price 1000p
100x-1000p 70 p
(100x-1000) / 70 = 8/7
700x -7000 = 560
700x = 7560 =>x = Rs 10.80.
Therefore the rate of second quality is Rs10.80
7.8lit are drawn from a wine and is then filled with water.
This operation is performed three more times.The ratio of
the quantity of wine now left in cask to that of the water
is 16:81. How much wine did the cask hold originally?
Solution:
Let the quantity of the wine in the cask originally be
x liters.
Then quantity of wine left in cask after
4 operations = x(1- 8/x)4lit.
Therefore x((1-(8/x))4)/x = 16/81.
(1- 8/x)4=(2/3) 4
(x- 8)/x=2/3
3x-24 =2x
x=24.
8.A can contains a mixture of two liquids A and B in the
ratio 7:5 when 9 liters of mixture are drawn off and the
can is filled with B,the ratio of A and B becomes 7:9.
How many liters of liquid A was contained by the can initially?
Solution:
Suppose the can initially contains 7x and 5x liters
of mixtures A and B respectively .
Quantity of A in mixture left = (7x- (7/12)*9 )lit
= 7x - (21/4) liters.
Quantity of B in mixture left = 5x - 5/12*9
= 5x - (15/4) liters
Therefore (7x – 21/4)/ (5x – 15/4+9)=7/9
(28x-21)/(20x +21)= 7/9
(252x -189)= 140x +147
112x = 336
=> x=3.
So the can contains 21 liters of A.
9.A vessel is filled with liquid,3 parts of which are water
and 5 parts syrup. How much of the mixture must be drawn off
and replaced with water so that the mixture may be
half water and half syrup?
Solution:
Suppose the vessal initially contains 8 liters of liquid.
Let x liters of this liquid be replaced with water
then quantity of water in new mixture
= 3-(3x/8)+x liters.
Quantity of syrup in new mixture = 5 - 5x/8 liters.
Therefore 3 - 3x/8 +x = 5 - 5x/8
5x+24 = 40-5x
10x = 16.
x= 8/5.
So part of the mixture replaced = 8/5*1/8 =1/5.
Tuesday, 12 November 2013
Alligation or Mixtures
Boats and Streams
Important facts:
1)In water, the direction along the stream is called down stream.
2)Direction against the stream is called upstream.
3)The speed of boat in still water is U km/hr and the speed of
stream is V km/hr then
speed down stream =U + V km/hr
speed up stream = U – V km/hr
Formulae:
If the speed down stream is A km/hr and the speed up stream is
B km/hr then speed in still water = ½(A+B) km/hr
rate of stream =1/2(A-B) km/hr
Problems:
1. In one hour a boat goes 11 km long the stream and 5 km
against the stream. The speed of the boat in still water is?
Sol:
Speed in still water = ½ ( 11+5) km/hr= 8 kmph
2.A man can row 18 kmph in still water. It takes him thrice
as long as row up as to row down the river. find the rate
of stream.
Sol:
Let man's rate up stream be xkmph
then, in still water =1/2[3x+x]=2x kmph
so, 2x= 18, x=9
rate upstream =9kmph
rate downstream =27 kmph
rate of stream = ½ [27-9]
= 9kmph
3.A man can row 71/2kmph in still watre . if in a river
running at 1.5 km an hour, if takes him 50 min to row to
place and back. how far off is the place?
Sol: speed down stream =7.5+1.5=9kmph
speed upstream =7.5-1.5=6kmph
let the required distence x km. then ,
x/9+x/6=50/60 = 2x+3x= 5/6*18
5x=15, x=3
Hence, the required distence is 3 km
4.A man can row 3 quarters of a km aganist the stream is
111/4 min. the speed of the man in still water is ?
Sol: rate upstream = 750/625 m/sec =10/9 m/sec
rate downstream =750/450 m/sec = 5/3 m/sec
rate in still water =1/2[10/9+5/3] = 25/18 m/sec
= 25/18*18/5=5 kmph
5.A boat can travel with a speed of 13 kmph in still water.
if the speed of stream is 4 kmph,find the time taken by
the boat to go 68 km downstream?
Sol: Speed down stream = 13+4= 17 kmph
time taken to travel 68km downstream =68/17 hrs
= 4 hrs
6.A boat takes 90 min less to travel 36 miles downstream then
to travel the same distence upstream. if the speed of the
boat in still water is 10 mph . The speed of the stream is :
Sol: Let the speed of the stream be x mph .
then, speed downstream = [10+x]mph
speed upstream =[10-x] mph
36/[10+x] - 36/[10-x] = 90/60 =72x*60= 90[100-x2]
(x+50)(x-2) =0
x=2 kmph
7.At his usual rowing rate, Rahul 12 miles down stream in a
certain river in 6 hrs less than it takes him to travel the
same distence upstream. but if he could double his usual
rowing rate for his 24 miles roundthe down stream 12 miles
would then take only one hour less than the up stream 12 miles.
what is the speed of the current in miles per hours?
Sol: Let the speed in still water be x mph and the speed of
the curren be y mph.
then, speed upstream = (x-y)
speed downstream =(x+y)
12/(x-y) - 12/(x+y) = 6
6(x2 – y2) m= 2xy => x2 – y2 =4y -(1)
and 12/(2x-y) - 12/(2x+y) =1 => 4x2 – y2 = 24y
x2= ( 24y + y2)/4 -->(2)
from 1 and 2 we have
4y+ y2 =( 24y+y2)/4
y=8/3 mph
y= 22/3 mph
8.There is a road beside a river. two friends started from
a place A, moved to a temple situated at another place B
and then returned to A again. one of them moves on a cycle
at a speed of 12 kmph, while the other sails on a boat at a
speed of 10 kmph . if the river flows at the speedof 4 kmph,
which of the two friends will return to place A first ?
Sol: Clearly, The cyclist moves both ways at a speed of 12 kmph
so, average speed of the cyclist = 12 kmph
the boat sailor moves downstream = (10+4) = 14 kmph
upstream =(10-4) = 6 kmph
So, average speed of the boat sailor =[ 2*14*6]/[14+6] kmph
=42/5 kmph =8.4 kmph
Since, the average speed of the cyclist is greater, he will
return to A first.
9.A boat takes 19 hrs for travelling downstream from point A to
point B. and coming back to a point C midway between A and B.
if the velocity of the sream is 4 kmph. and the speed of the
boat in still water is 14 kmph. what is the distence between
A and B?
Sol:
speed downstream =14+4 =18 kmph
speed upstream = 14 -4 = 10 kmph
let the distence between A and B be x km. then,
x/18 + (x/2)/10 = 19
x/18 + x/20 =19
19x/180 =19 =>x = 180km
Hence, the distence between A and B bw 180 km
Trains
General Concept:
(1) Time taken by a train x mt long in passing a signal post
or a pole or a standing man = time taken by the train to cover x mt
(2) Time taken by a train x mt long in passing a stationary
object of length y mt = time taken by the train to cover x+y mt
(3) Suppose two trains or two bodies are moving in the same
direction at u kmph and v kmph such that u > v then their
relative speed is u-v kmph
(4)If two trains of length x km and y km are moving in opposite
diredtions at u kmph and vmph,then time taken by the train to
cross each other = (x+y)/(u+v) hr
(5) Suppose two trains or two bdies are moving in opposite direction
at u kmph and v kmph then,their relative speed = (u+v) kmph
(6)If two train start at the same time from 2 points A & B towards
each other and after crossing they take a & b hours in reaching B & A
respectively then A's speed : B's speed = (b^1/2 : a^1/2 )
Problems
(1)Find the time taken by a train 180m long,running at 72kmph in
crossing an electric pole
Solution:
Speed of the train =72*5/18m/s =20 m/s
Distance move din passing the pole = 180m
Requiredtime = 180/20 = 9 seconds
(2)A train 140 m long running at 60kmph.In how much time will it
pass a platform 260m long.
Solution:
Distance travelled =140 + 260 m =400 m,
speed = 60 * 5/18 = 50//3 m
time=400*3 / 50 = 24 Seconds
(3)A man is standing on a railway bridge which is 180 m.He finds
that a train crosses the bridge in 20 seconds but himself in
8 sec. Find the length of the train and its sppeed
Solution:
i)D=180+x
T = 20 seconds
S= 180+x / 20 ------------ 1
ii)D=x
T=8 seconds
D=ST
x=8S ------------- 2
Substitute 2 in 1
S=180 + 8 S / 20
S=15 m/s
Length of the train,x is 8 *15 = 120 m
(4)A train 150m long is running with a speed of 68 mphIn wht
time will it pass a man who is running at a speed of 8kmph in
the same direction in which the train is going
Solution:
Relative Speed = 68-8=60kmph*5/18 = 50/3 m/s
time= 150 * 3 /50 =9sec
5)A train 220m long is running with a speed of 59 k mph /..In
what time will it pass a man who is running at 7 kmph in the
direction opposite to that in which train is going.
Solution:
Relative Speed = 59+7=66kmph*5/18 = 55/3 m/s
time= 220/55 * 3 =12sec
(6)Two trains 137m and 163m in length are running towards each
other on parallel lines,one at the rate of 42kmph & another at
48 mph.In wht time will they be clear of each other from the
moment they meet.
Solution:
Relative speed =42+48 = 90 *5/18 = 25m/s
time taken by the train to pass each other = time taken to cover
(137+163)m at 25 m/s
= 300 /25 s =12 s
(7)A train running at 54 kmph takes 20 sec to pass a platform.
Next it takes 12 sec to pass a man walking at 6kmph in the same
direction in which the train is going.Find length of the train
and length of platform
Solution:
Relative speed w.r.t man = 54-6=48kmph
the length of the train is 48 * 5/18 * 12 =160m
time taken to pass platform =20 sec
Speed of the train = 54 * 5/18 =15m/s
160+x =20 *15
x=140m
length of the platform is 140m
(8)A man sitting in a train which is travelling at 50mph observes
that a goods train travelling in opposite irection takes 9 sec
to pass him .If the goos train is 150m long fin its speed
Solution:
Relative speed =150/9 m/s =60 mph
speed of the train = 60-50 =10kmph
(9)Two trains are moving in the sam e direction at 65kmph and
45kmph. The faster train crosses a man in slower train in18sec.the
length of the faster train is
Solution:
Relative speed =65-45 kmph = 50/9 m/s
Distancce covered in18 s =50/9 * 18 = 100m
the length of the train is 100m
(10)Atrain overtakes two persons who are walking in the same
direction in which the train is going at the rate of 2kmph an
4kmph and passes them completely in 9 sec an 10 sec respectively.
The length of train is
Solution:
2kmph = 5/9 m/s
4 mph =10/9 m/s
Let the length of the trainbe x meters and its speed is y m/s
then x / (y- 5/9) = 9 and x / (y- 10/9) = 10
9y-5 =x and 10(9y-10)=9x
9y-x=5 and 90y-9x=100
on solving we get x=50,lenght of trains
(11) Two stations A & B are 110 km apart on a straight line.
One train starts from A at 7am and travels towards B at 20kmph.
Another train starts from B at 8am an travels toward A at a speed
of 25kmph.At what time will they meet
Solution:
Suppose the train meet x hr after 7am
Distance covered by A in x hr=20x km
20x+25(x-1) = 110
45x=135
x=3
So they meet at 10 am
(12)A traintravelling at 48kmph completely crosses another train
having half its length an travelling inopposite direction at 42kmph
in12 sec.It also passes a railway platform in 45sec.the length of
platform is
Solution:
Let the length of the first train be x mt
then,the length of second train is x/2 mt
relative speed = 48+42 kmph =90 * 5/18 m/s = 25m/s
(x+ x/2)/25 =12
x=200
Length of the train is 200m
Let the length of the platform be y mt
speed f the first train = 48*5/18 m/s = 40/3 m/s
200+y * 3/40 = 45
y=400m
(13)The length of a running trsain in 30% more than the length of
another train B runnng in the opposite direction.To find out the
speed of trtain B,which of the following information given in the
statements P & Q is sufficient
P : The speed of train A is 80 kmph
Q : They too 90 sec to cross each other
(a) Either P & Q is sufficient
(b)Both P & Q are not sufficient
(c)only Q is sufficient
(d)Both P & Q are neeed
Ans: B
Solution:
Let the length of th e train A be x mt
Length of the train B = 130/100 x mt =13x/10 mt
Let the speed of B be y mph,speed of the train A=80mph
relative speed= y+80 * 5/18 m/s
time taken by the trains t cross each other is gven by
90 = (x + 13x/10)/ (5y+400 / 18)
to find y,clearly xis also needed
so,both P & Q are not sufficient
(14)The speed of a train A,100m long is 40% more than then the speed
of another train B,180m long running in opposite direction.To fin out
the speed of B,which of the information given in statements P & Q is
sufficient
P :The two trains crossed each other in 6 seconds
Q : The difference between the spee of the trains is 26kmph
(a)Only P is sufficient
(b)Only Q is sufficient
(c)Both P & Q are needed
(d)Both P & Q are not sufficient
Ans : A
Solution:
Let speed of B be x kmph
then,speed of A =140x/100 kmph =7x/5 mph
relative speed = x + 7x/5 =2x/3 m/s
time taken to cross each other = (100+180)*3/2x s =420/x s
now,420/x = 6
x=70 mph
thus,only P is sufficient
(15)The train running at certain speed crosses astationary enginein
20 seconds.to find out the sped of the train,which of the following
information is necessary
(a)Only the length of the train
(b)only the length of the engine
(c)Either the length of the train or length of engine
(d)Both the length of the train or length of engine
Ans : D
Solution:
Since the sum of lengths of the tran and the engine is needed,
so both the length must be known
Time and Distance
Formulae:
I)Speed = Distance/Time
II)Time = Distance/speed
III) Distance = speed*time
IV) 1km/hr = 5/18 m/s
V)1 m/s = 18/5 Km/hr
VI)If the ratio of the speed of A and B is a:b,then the ratio of
the time taken by them to cover the same distance is 1/a : 1/b
or b:a
VII) suppose a man covers a distance at x kmph and an equal
distance at y kmph.then the average speed during the whole
journey is (2xy/x+y)kmph
Problems
1)A person covers a certain distance at 7kmph .How many meters
does he cover in 2 minutes.
Solution::
speed=72kmph=72*5/18 = 20m/s
distance covered in 2min =20*2*60 = 2400m
2)If a man runs at 3m/s. How many km does he run in 1hr 40min
Solution::
speed of the man = 3*18/5 kmph
= 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km
3)Walking at the rate of 4knph a man covers certain distance
in 2hr 45 min. Running at a speed of 16.5 kmph the man will
cover the same distance in.
Solution::
Distance=Speed* time
4*11/4=11km
New speed =16.5kmph
therefore Time=D/S=11/16.5 = 40min
Complex Problems
1)A train covers a distance in 50 min ,if it runs at a speed
of 48kmph on an average.The speed at which the train must run
to reduce the time of journey to 40min will be.
Solution::
Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=40/60hr=2/3hr
New speed = 40* 3/2 kmph= 60kmph
2)Vikas can cover a distance in 1hr 24min by covering 2/3 of
the distance at 4 kmph and the rest at 5kmph.the total
distance is?
Solution::
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km
3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr.
the usual time is.
Solution::
Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½
4)A man covers a distance on scooter .had he moved 3kmph
faster he would have taken 40 min less. If he had moved
2kmph slower he would have taken 40min more.the distance is.
Solution::
Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x --------------1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x -----------------2
divide 1 & 2 equations
by solving we get x = 40
5)Excluding stoppages,the speed of the bus is 54kmph and
including stoppages,it is 45kmph.for how many min does the bus
stop per hr.
Solution::
Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min
6)Two boys starting from the same place walk at a rate of
5kmph and 5.5kmph respectively.wht time will they take to be
8.5km apart, if they walk in the same direction
Solution::
The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs
7)2 trains starting at the same time from 2 stations 200km
apart and going in opposite direction cross each other ata
distance of 110km from one of the stations.what is the ratio of
their speeds.
Solution::
In same time ,they cover 110km & 90 km respectively
so ratio of their speed =110:90 = 11:9
8)Two trains start from A & B and travel towards each other at
speed of 50kmph and 60kmph resp. At the time of the meeting the
second train has traveled 120km more than the first.the distance
between them.
Solution::
Let the distance traveled by the first train be x km
then distance covered by the second train is x + 120km
x/50 = x+120 / 60
x= 600
so the distance between A & B is x + x + 120 = 1320 km
9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the
theft is discovered at 3pm and the owner sets off in another car
at 75kmph when will he overtake the thief
Solution::
Let the thief is overtaken x hrs after 2.30pm
distance covered by the thief in x hrs = distance covered by
the owner in x-1/2 hr
60x = 75 ( x- ½)
x= 5/2 hr
thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm
10)In covering distance,the speed of A & B are in the ratio
of 3:4.A takes 30min more than B to reach the destion.The time
taken by A to reach the destinstion is.
Solution::
Ratio of speed = 3:4
Ratio of time = 4:3
let A takes 4x hrs,B takes 3x hrs
then 4x-3x = 30/60 hr
x = ½ hr
Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr
11)A motorist covers a distance of 39km in 45min by moving at a
speed of xkmph for the first 15min.then moving at double the
speed for the next 20 min and then again moving at his original
speed for the rest of the journey .then x=?
Solution::
Total distance = 39 km
Total time = 45 min
D = S*T
x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km
x = 36 kmph
12)A & B are two towns.Mr.Fara covers the distance from A t0 B
on cycle at 17kmph and returns to A by a tonga running at a
uniform speed of 8kmph.his average speed during the whole
journey is.
Solution::
When same distance is covered with different speeds,then the
average speed = 2xy / x+y
=10.88kmph
13)A car covers 4 successive 3km stretches at speed of
10kmph,20kmph,30kmph&:60kmph resp. Its average speed is.
Solution::
Average speed = total distance / total time
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph
14)A person walks at 5kmph for 6hr and at 4kmph for 12hr.
The average speed is.
Solution::
avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph
15)A bullock cart has to cover a distance of 80km in 10hrs.
If it covers half of the journeyin 3/5th time.wht should be its
speed to cover the remaining distance in the time left.
Solution::
Time left = 10 - 3/5*10
= 4 hr
speed =40 km /4 hr
=10 kmph
16)The ratio between the speeds of the A& B is 2:3 an
therefore A takes 10 min more than the time taken by B to reach
the destination.If A had walked at double the speed ,he would
have covered the distance in ?
Solution::
Ratio of speed = 2:3
Ratio of time = 3:2
A takes 10 min more
3x-2x = 10 min
A's time=30 min
--->A covers the distance in 30 min ,if its speed is x
-> He will cover the same distance in 15 min,if its speed
doubles (i.e 2x)
17)A is twice as fast as B and B is thrice as fast as C is.
The journey covered by B in?
Solution::
speed's ratio
a : b = 2: 1
b : c = 3:1
Time's ratio
b : c = 1:3
b : c = 18:54
(if c covers in 54 min i..e twice to 18 min )
18)A man performed 3/5 of the total journey by ratio 17/20 by
bus and the remaining 65km on foot.wht is his total journey.
Solution::
Let total distance is x
x-(3/5x + 17/20 x) =6.5
x- 19x/20 = 6.5
x=20 * 6.5
=130 km
19)A train M leaves Meerat at 5 am and reaches Delhi at 9am .
Another train N leaves Delhi at 7am and reaches Meerut at 1030am
At what time do the 2 trains cross one another
Solution::
Let the distance between Meerut & Delhi be x
they meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e 7/2 hr
speed of M =x/4
speed of N = 2x/7
Distance covered by M in y+2 hr + Distance covered by N in
y hr is x
x/4 (y+2) +2x/7(y)=x
y=14/15hr or 56 min
20)A man takes 5hr 45min in walking to certain place and riding
back. He would have gained 2hrs by riding both ways.The time he
would take to walk both ways is?
Solution::
Let x be the speed of walked
Let y be the speed of ride
Let D be the distance
Then D/x + D/y = 23/4 hr -------1
D/y + D/y = 23/4 – 2 hr
D/y = 15/8 --------2
substitute 2 in 1
D/x + 15/8 = 23/4
D/x = 23/4 -15/8 =46-15/8 =31/8
Time taken for walk one way is 31/8 hr
time taken to walk to and fro is 2*31/8 = 31/4 hr
=7 hr 45 min
Pipes and Cisterns
Important Facts:
1.INLET:A pipe connected with a tank or cistern or a reservoir,
that fills it, it is known as Inlet.
OUTLET:A pipe connected with a tank or a cistern or a reservoir,
emptying it, is known as Outlet.
2. i) If a pipe can fill a tank in x hours, then :
part filled in 1 hour=1/x.
ii)If a pipe can empty a tank in y hours, then :
part emptied in 1 hour=1/y.
iii)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where y>x), then on
opening both the pipes, the net part filled in
1 hour=(1/x -1/y).
iv)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where x>y), then on opening
both the pipes, the net part filled in 1 hour=(1/y -1/x).
v) If two pipes A and B can fill a tank in x hours and y hours
respectively. If both the pipes are opened simultaneously, part
filled by A+B in 1 hour= 1/x +1/y.
Simple Problems
1)Two pipes A& B can fill a tank in 36 hours and 45 hours respectively.
If both the pipes are opened simultaneously, how much time will be
taken to fill the tank?
Sol: Part filled by A in 1 hour=1/36
Part filled by B in 1 hour= 1/45;
Part filled by (A+B)'s in 1 hour=1/36 +1/45= 9/180 =1/20
Hence, both the pipes together will fill the tank in 20 hours.
2)Two pipes can fill a tank in 10 hours & 12 hours respectively. While
3rd pipe empties the full tank n 20 hours. If all the three pipes
operate simultaneously,in how much time will the tank be filled?
Sol: Net part filled in 1 hour=1/10 +1/12 -1/20
=8/60=2/15
The tank be filled in 15/2hours= 7 hrs 30 min
3)A cistern can be filled by a tap in 4 hours while it can be emptied
by another tap in 9 hours. If both the taps are opened simultaneously,
then after how much time will the cistern get filled?
Sol: Net part filled in 1 hour= 1/4 -1/9= 5/36
Therefore the cistern will be filled in 36/5 hours or 7.2 hours.
4)If two pipes function simultaneously, the reservoir will be filled in
12 days.One pipe fills the reservoir 10 hours faster than the other.
How many hours does it take the second pipe to fill the reservoir.
Sol: Let the reservoir be filled by the 1st pipe in x hours.
The second pipe will fill it in (x+10) hours
1/x + (1/(x+10))= 1/12
=> (2x+10)/((x)*(x+10))= 1/12
=> x=20
So, the second pipe will take 30 hours to fill the reservoir.
5)A cistern has two taps which fill it in 12 min and 15 min respectively.
There is also a waste pipe in the cistern. When all the three are opened,
the empty cistern is full in 20 min. How long will the waste pipe take to
empty the full cistern?
Sol: Work done by a waste pipe in 1 min
=1/20 -(1/12+1/15)= -1/10 (-ve means emptying)
6)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?
Sol: Time taken by one tap to fill the half of the tank =3 hours
Part filled by the four taps in 1 hour=4/6=2/3
Remaining part=1 -1/2=1/2
Therefore, 2/3:1/2::1:x
or x=(1/2)*1*(3/2)=3/4 hours.
i.e 45 min
So, total time taken= 3hrs 45min.
7)A water tank is two-fifth full. Pipe A can fill a tank in 10 min. And B
can empty it in 6 min. If both pipes are open, how long will it take to
empty or fill the tank completely ?
Sol: Clearly, pipe B is faster than A and So, the tank will be emptied.
Part to be emptied=2/5.
Part emptied by (A+B) in 1 min= 1/6 -1/10=1/15
Therefore, 1/15:2/5::1:x or x=((2/5)*1*15)=6 min.
So, the tank be emptied in 6 min.
8)Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for
Bucket P to fill the empty drum. How many turns it will take for both the
buckets P&Q, having each turn together to fill the empty drum?
Sol: Let the capacity of P be x lit.
Then capacity of Q=x/3 lit
Capacity of the drum=60x lit
Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45
Complex Problems
1)Two pipes can fill a cistern in 14 hours and 16 hours respectively. The
pipes are opened simultaneously and it is found that due to leakage in the
bottom it took 32min more to fill the cistern. When the cistern is full,
in what time will the leak empty it?
Sol: Work done by the two pipes in 1 hour= 1/14+1/16=15/112
Time taken by these two pipes to fill the tank=112/15 hrs.
Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours
Therefore, work done by (two pipes + leak) in 1 hr= 1/8
work done by leak n 1 hour=15/112 -1/8=1/112
Leak will empty full cistern n 112 hours.
2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After
7 min, C also opened. In how much time, the tank s full.
Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20
Remaining part= 1-7/20=13/20
Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60
Now, 1/60 part is filled in 1 min.
13/20 part is filled n (60*13/20)=39 min
Total time taken to fill the tank=39+7=46 min
3)Two pipes A&B can fill a tank in 24 min and 32 min respectively. If
both the pipes are opened simultaneously, after how much time B should
be closed so that the tank is full in 18 min.
Sol: Let B be closed after x min, then part filled by (A+B) in x min+
part filled by A in (18-x) min=1
x(1/24+1/32) +(18-x)1/24 =1
=> x=8
Hence B must be closed after 8 min.
4)Two pipes A& B together can fill a cistern in 4 hours. Had they been
opened separately, then B would have taken 6 hours more than A to fill
the cistern. How much time will be taken by A to fill the cistern
separately?
Sol: Let the cistern be filled by pipe A alone in x hours.
Pipe B will fill it in x+6 hours
1/x + 1/x+6=1/4
Solving this we get x=6.
Hence, A takes 6 hours to fill the cistern separately.
5)A tank is filled by 3 pipes with uniform flow. The first two pipes
operating simultaneously fill the tan in the same time during which
the tank is filled by the third pipe alone. The 2nd pipe fills the tank
5 hours faster than first pipe and 4 hours slower than third pipe. The
time required by first pipe is :
Sol: Suppose, first pipe take x hours to fill the tank then
B & C will take (x-5) and (x-9) hours respectively.
Therefore, 1/x +1/(x-5) =1/(x-9)
On solving, x=15
Hence, time required by first pipe is 15 hours.
6)A large tanker can be filled by two pipes A& B in 60min and 40 min
respectively. How many minutes will it take to fill the tanker from
empty state if B is used for half the time & A and B fill it together for
the other half?
Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24
Suppose the tank is filled in x minutes
Then, x/2(1/24+1/40)=1
=> (x/2)*(1/15)=1
=> x=30 min.
7)Two pipes A and B can fill a tank in 6 hours and 4 hours respectively.
If they are opened on alternate hours and if pipe A s opened first, in
how many hours, the tank shall be full.
Sol: (A+B)'s 2 hours work when opened alternatively =1/6+1/4 =5/12
(A+B)'s 4 hours work when opened alternatively=10/12=5/6
Remaining part=1 -5/6=1/6.
Now, it is A's turn and 1/6 part is filled by A in 1 hour.
So, total time taken to fill the tank=(4+1)= 5 hours.
8)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
alternatively, the tank will be full in.
Sol: (A+B)'s 1 hour's work=1/12+1/15=9/60=3/20
(A+C)'s 1 hour's work=1/20+1/12=8/60=2/15
Part filled in 2 hours=3/20+2/15=17/60
Part filled in 2 hours=3/20+2/15= 17/60
Part filled in 6 hours=3*17/60 =17/20
Remaining part=1 -17/20=3/20
Now, it is the turn of A & B and 3/20 part is filled by A& B in 1 hour.
Therefore, total time taken to fill the tank=6+1=7 hours
9)A Booster pump can be used for filling as well as for emptying a tank.
The capacity of the tank is 2400 m3. The emptying capacity of the tank is
10 m3 per minute higher 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?
Sol: Let, the filling capacity of the pump be x m3/min
Then, emptying capacity of the pump=(x+10) m3/min.
So,2400/x – 2400/(x+10) = 8
on solving x=50.
10)A leak in the bottom of a tank can empty the full tan in 8 hr. An inlet
pipe fills water at the rate of 6 lits a minute. When the tank is full,
the inlet is opened and due to the leak, the tank is empty in 12 hrs.
How many liters does the cistern hold?
Sol: Work done by the inlet in 1 hr= 1/8 -1/12=1/24
Work done by the inlet in i min= (1/24)*(1/60)=1/1440
Therefore, Volume of 1/1440 part=6 lit
Volume of whole=(1440*6) lit=8640 lit.
11)Two pipes A and B can fill a cistern in 37 ½ min and 45 minutes
respectively. Both the pipes are opened. The cistern will be filled in
just half an hour, if the pipe B is turned off after:
sol: Let B be turned off after x min. Then,
Part filled by (A+B) in x min+ part filled by A in (30-x)min=1
Therefore, x(2/75+1/45)+(30-x)(2/75)=1
11x/225 + (60-2x)/75=1
11x+ 180-6x=225
x=9.
So, B must be turned off after 9 minutes.
Time and Work
Important Facts:
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.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 variables
follow some definite rule. It said that the two variables vary
directly or inversely.Its notation is X/Y=k, where k is called
constant. This variation is called direct variation. XY=k. This
variation 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.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)
Problems
1)If 9 men working 6 hours a day can do a work in 88 days. Then 6 men
working 8 hours a day can do it in how many days?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so (9*6*88/1)=(6*8*d/1)
on solving, d=99 days.
2)If 34 men completed 2/5th of a work in 8 days working 9 hours a day.
How many more man should be engaged to finish the rest of the work in
6 days working 9 hours a day?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so, (34*8*9/(2/5))=(x*6*9/(3/5))
so x=136 men
number of men to be added to finish the work=136-34=102 men
3)If 5 women or 8 girls can do a work in 84 days. In how many days can
10 women and 5 girls can do the same work?
Sol: Given that 5 women is equal to 8 girls to complete a work
so, 10 women=16 girls.
Therefore 10women +5girls=16girls+5girls=21girls.
8 girls can do a work in 84 days
then 21 girls ---------------?
answer= (8*84/21)=32days.
Therefore 10 women and 5 girls can a work in 32days
4)Worker A takes 8 hours to do a job. Worker B takes 10hours to do the
same job. How long it take both A & B, working together but independently,
to do the same job?
Sol: A's one hour work=1/8.
B's one hour work=1/10
(A+B)'s one hour work=1/8+1/10 =9/40
Both A & B can finish the work in 40/9 days
5)A can finish a work in 18 days and B can do the same work in half the
time taken by A. Then, working together, what part of the same work they
can finish in a day?
Sol: Given that B alone can complete the same work in days=half the time
taken by A=9days
A's one day work=1/18
B's one day work=1/9
(A+B)'s one day work=1/18+1/9=1/6
6)A is twice as good a workman as B and together they finish a piece of
work in 18 days.In how many days will A alone finish the work.
Sol: if A takes x days to do a work then
B takes 2x days to do the same work
=>1/x+1/2x=1/18
=>3/2x=1/18
=>x=27 days.
Hence, A alone can finish the work in 27 days.
7)A can do a certain work in 12 days. B is 60% more efficient than A. How
many days does B alone take to do the same job?
Sol: Ratio of time taken by A&B=160:100 =8:5
Suppose B alone takes x days to do the job.
Then, 8:5::12:x
=> 8x=5*12
=> x=15/2 days.
8)A can do a piece of work n 7 days of 9 hours each and B alone can do it
in 6 days of 7 hours each. How long will they take to do it working together
8 2/5 hours a day?
Sol: A can complete the work in (7*9)=63 days
B can complete the work in (6*7)=42 days
=> A's one hour's work=1/63 and
B's one hour work=1/42
(A+B)'s one hour work=1/63+1/42=5/126
Therefore, Both can finish the work in 126/5 hours.
Number of days of 8 2/5 hours each=(126*5/(5*42))=3days
9)A takes twice as much time as B or thrice as much time to finish a piece
of work. Working together they can finish the work in 2 days. B can do the
work alone in ?
Sol: Suppose A,B and C take x,x/2 and x/3 hours respectively finish the
work then 1/x+2/x+3/x=1/2
=> 6/x=1/2
=>x=12
So, B takes 6 hours to finish the work.
10)X can do ¼ of a work in 10 days, Y can do 40% of work in 40 days and Z
can do 1/3 of work in 13 days. Who will complete the work first?
Sol: Whole work will be done by X in 10*4=40 days.
Whole work will be done by Y in (40*100/40)=100 days.
Whole work will be done by Z in (13*3)=39 days
Therefore,Z will complete the work first.
Complex Problems
1)A and B undertake to do a piece of workfor Rs 600.A alone can do it in
6 days while B alone can do it in 8 days. With the help of C, they can finish
it in 3 days, Find the share of each?
Sol: C's one day's work=(1/3)-(1/6+1/8)=1/24
Therefore, A:B:C= Ratio of their one day’s work=1/6:1/8:1/24=4:3:1
A's share=Rs (600*4/8)=300
B's share= Rs (600*3/8)=225
C's share=Rs[600-(300+225)]=Rs 75
2)A can do a piece of work in 80 days. He works at it for 10 days & then B alone
finishes the remaining work in 42 days. In how much time will A and B, working
together, finish the work?
Sol: Work done by A in 10 days=10/80=1/8
Remaining work=(1-(1/8))=7/8
Now, work will be done by B in 42 days.
Whole work will be done by B in (42*8/7)=48 days
Therefore, A's one day's work=1/80
B’s one day's work=1/48
(A+B)'s one day's work=1/80+1/48=8/240=1/30
Hence, both will finish the work in 30 days.
3)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 more
than P. During a period of five hours, R can type as many pages as P can
during seven hours. How many pages does each of them type per hour?
Sol:Let the number of pages typed in one hour by P, Q and R be x,y and z
respectively
Then x+y+z=216/4=54 ---------------1
z-y=y-x => 2y=x+z -----------2
5z=7x => x=5x/7 ---------------3
Solving 1,2 and 3 we get x=15,y=18, and z=21
4)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 time will they take, working together on two different computers
to type an assignment of 110 pages?
Sol: Number of pages typed by Ronald in one hour=32/6=16/3
Number of pages typed by Elan in one hour=40/5=8
Number of pages typed by both in one hour=((16/3)+8)=40/3
Time taken by both to type 110 pages=110*3/40=8 hours.
5)Two workers A and B are engaged to do a work. A working alone takes 8 hours
more to complete the job than if both working together. If B worked alone,
he would need 4 1/2 hours more to compete the job than they both working
together. What time would they take to do the work together.
Sol: (1/(x+8))+(1/(x+(9/2)))=1/x
=>(1/(x+8))+(2/(2x+9))=1/x
=> x(4x+25)=(x+8)(2x+9)
=> 2x2 =72
=> x2 = 36
=> x=6
Therefore, A and B together can do the work in 6 days.
6)A and B can do a work in12 days, B and C in 15 days, C and A in 20 days.
If A,B and C work together, they will complete the work in how many days?
Sol: (A+B)'s one day's work=1/12;
(B+C)'s one day's work=1/15;
(A+C)'s one day's work=1/20;
Adding we get 2(A+B+C)'s one day's work=1/12+1/15+1/20=12/60=1/5
(A+B+C)'s one day work=1/10
So, A,B,and C together can complete the work in 10 days.
7)A and B can do a work in 8 days, B and C can do the same wor in 12 days.
A,B and C together can finish it in 6 days. A and C together will do it in
how many days?
Sol: (A+B+C)'s one day's work=1/6;
(A+B)'s one day's work=1/8;
(B+C)'s one day's work=1/12;
(A+C)'s one day's work=2(A+B+C)'s one day's work-((A+B)'s one day
work+(B+C)'s one day work)
= (2/6)-(1/8+1/12)
=(1/3)- (5/24)
=3/24
=1/8
So, A and C together will do the work in 8 days.
8)A can do a certain work in the same time in which B and C together can do it.
If A and B together could do it in 10 days and C alone in 50 days, then B alone
could do it in how many days?
Sol: (A+B)'s one day's work=1/10;
C's one day's work=1/50
(A+B+C)'s one day's work=(1/10+1/50)=6/50=3/25
Also, A's one day's work=(B+C)’s one day's work
From i and ii ,we get :2*(A's one day's work)=3/25
=> A's one day's work=3/50
B's one day’s work=(1/10-3/50)
=2/50
=1/25
B alone could complete the work in 25 days.
9) A is thrice as good a workman as B and therefore is able to finish a job
in 60 days less than B. Working together, they can do it in:
Sol: Ratio of times taken by A and B=1:3.
If difference of time is 2 days , B takes 3 days
If difference of time is 60 days, B takes (3*60/2)=90 days
So, A takes 30 days to do the work=1/90
A's one day's work=1/30;
B's one day's work=1/90;
(A+B)'s one day's work=1/30+1/90=4/90=2/45
Therefore, A&B together can do the work in 45/2days
10) 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&B,
working together, finish the work?
Sol: Work Done by A n 10 days =10/80=1/8
Remaining work =1-1/8=7/8
Now 7/8 work is done by B in 42 days
Whole work will be done by B in 42*8/7= 48 days
=> A's one day's work =1/80 and
B's one day's work =1/48
(A+B)'s one day's work = 1/80+1/48 = 8/240 = 1/30
Hence both will finish the work in 30 days.
11) 45 men can complete a work in 16 days. Six days after they started working,
so more men joined them. How many days will they now take to complete the
remaining work?
Sol: M1*D1/W1=M2*D2/W2
=>45*6/(6/16)=75*x/(1-(6/16))
=> x=6 days
12)A is 50% as efficient as B. C does half the work done by A&B together. If
C alone does the work n 40 days, then A,B and C together can do the work in:
Sol: A's one day's work:B's one days work=150:100 =3:2
Let A's &B's one day's work be 3x and 2x days respectively.
Then C's one day's work=5x/2
=> 5x/2=1/40
=> x=((1/40)*(2/5))=1/100
A's one day's work=3/100
B's one day's work=1/50
C's one day's work=1/40
So, A,B and C can do the work in 13 1/3 days.
13)A can finish a work in 18 days and B can do the same work in 15 days. B
worked for 10 days and left the job. In how many days A alone can finish the
remaining work?
Sol: B's 10 day's work=10/15=2/3
Remaining work=(1-(2/3))=1/3
Now, 1/18 work is done by A in 1 day.
Therefore 1/3 work is done by A in 18*(1/3)=6 days.
14)A can finish a work in 24 days, B n 9 days and C in 12 days. B&C start the
work but are forced to leave after 3 days. The remaining work done by A in:
Sol: (B+C)'s one day's work=1/9+1/12=7/36
Work done by B & C in 3 days=3*7/36=7/12
Remaining work=1-(7/12)=5/12
Now , 1/24 work is done by A in 1 day.
So, 5/12 work is done by A in 24*5/12=10 days
15)X and Y can do a piece of work n 20 days and 12 days respectively. X started
the work alone and then after 4 days Y joined him till the completion of work.
How long did the work last?
Sol: work done by X in 4 days =4/20 =1/5
Remaining work= 1-1/5 =4/5
(X+Y)'s one day's work =1/20+1/12 =8/60=2/15
Now, 2/15 work is done by X and Y in one day.
So, 4/5 work will be done by X and Y in 15/2*4/5=6 days
Hence Total time taken =(6+4) days = 10 days
16)A does 4/5 of work in 20 days. He then calls in B and they together finish
the remaining work in 3 days. How long B alone would take to do the whole work?
Sol: Whole work is done by A in 20*5/4=25 days
Now, (1-(4/5)) i.e 1/5 work is done by A& B in days.
Whole work will be done by A& B in 3*5=15 days
=>B's one day's work= 1/15-1/25=4/150=2/75
So, B alone would do the work in 75/2= 37 ½ days.
17) A and B can do a piece of work in 45 days and 40 days respectively. They
began to do the work together but A leaves after some days and then B completed
the remaining work n 23 days. The number of days after which A left the work was
Sol: (A+B)'s one day's work=1/45+1/40=17/360
Work done by B in 23 days=23/40
Remaining work=1-(23/40)=17/40
Now, 17/360 work was done by (A+B) in 1 day.
17/40 work was done by (A+B) in (1*(360/17)*(17/40))= 9 days
So, A left after 9 days.
18)A can do a piece of work in 10 days, B in 15 days. They work for 5 days.
The rest of work finished by C in 2 days. If they get Rs 1500 for the whole
work, the daily wages of B and C are
Sol: Part of work done by A= 5/10=1/2
Part of work done by B=1/3
Part of work done by C=(1-(1/2+1/3))=1/6
A's share: B's share: C's share=1/2:1/3:1/6= 3:2:1
A's share=(3/6)*1500=750
B's share=(2/6)*1500=500
C's share=(1/6)*1500=250
A's daily wages=750/5=150/-
B's daily wages=500/5=100/-
C's daily wages=250/2=125/-
Daily wages of B&C = 100+125=225/-
19)A alone can complete a work in 16 days and B alone can complete the same
in 12 days. Starting with A, they work on alternate days. The total work will
be completed in how many days?
(a) 12 days (b) 13 days (c) 13 5/7 days (d)13 ¾ days
Sol: (A+B)'s 2 days work = 1/16 + 1/12 =7/48
work done in 6 pairs of days =(7/48) * 6 = 7/8
remaining work = 1- 7/8 = 1/8
work done by A on 13th day = 1/16
remaining work = 1/8 – 1/16 = 1/16
on 14th day, it is B’s turn
1/12 work is done by B in 1 day.
1/16 work is done by B in ¾ day.
Total time taken= 13 ¾ days.
So, Answer is: D
20)A,B and C can do a piece of work in 20,30 and 60 days respectively. In how
many days can A do the work if he is assisted by B and C on every third day?
Sol: A's two day's work=2/20=1/10
(A+B+C)'s one day's work=1/20+1/30+1/60=6/60=1/10
Work done in 3 days=(1/10+1/10)=1/5
Now, 1/5 work is done in 3 days
Therefore, Whole work will be done in (3*5)=15 days.
21)Seven men can complete a work in 12 days. They started the work and after
5 days, two men left. In how many days will the work be completed by the
remaining men?
(A) 5 (B) 6 (C ) 7 (D) 8 (E) none
Sol: 7*12 men complete the work in 1 day.
Therefore, 1 man's 1 day's work=1/84
7 men's 5 days work = 5/12
=>remaining work = 1-5/12 = 7/12
5 men's 1 day's work = 5/84
5/84 work is don by them in 1 day
7/12 work is done by them in ((84/5) * (7/12)) = 49/5 days = 9 4/5 days.
Ans: E
22).12 men complete a work in 9 days. After they have worked for 6 days, 6 more
men joined them. How many days will they take to complete the remaining work?
(a) 2 days (b) 3 days (c) 4 days (d) 5days
Sol : 1 man's 1 day work = 1/108
12 men's 6 days work = 6/9 = 2/3
remaining work = 1 – 2/3 = 1/3
18 men's 1 days work = 18/108 = 1/6
1/6 work is done by them in 1 day
therefore, 1/3 work is done by them in 6/3 = 2 days.
Ans : A
23).A man, a woman and a boy can complete a job in 3,4 and 12 days respectively.
How many boys must assist 1 man and 1 woman to complete the job in ¼ of a day?
(a). 1 (b). 4 (c). 19 (d). 41
Sol : (1 man + 1 woman)'s 1 days work = 1/3+1/4=7/12
Work done by 1 man and 1 women n 1/4 day=((7/12)*(1/4))=7/48
Remaining work= 1- 7/48= 41/48
Work done by 1 boy in ¼ day= ((1/12)*(1/4)) =1/48
Therefore, Number of boys required= ((41/48)*48)= 41 days
So,Answer: D
24)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.
(A) 15 (B) 18 (C) 22 (D) data inadequate
Sol: one man's one day's work= 1/48
one woman's one day's work=1/60
6 men's 2 day's work=((6/48)*2)= ¼
Remaining work=3/4
Now, 1/60 work s done in 1 day by 1 woman.
So, ¾ work will be done in 3 days by (60*(3/4)*(1/3))= 15 woman.
So, Answer: A
25)Twelve children take sixteen days to complete a work which can be completed
by 8 adults in 12 days. Sixteen adults left and four children joined them. How
many days will they take to complete the remaining work?
(A) 3 (B) 4 ( C) 6 (D) 8
Sol: one child's one day work= 1/192;
one adult's one day's work= 1/96;
work done in 3 days=((1/96)*16*3)= 1/2
Remaining work= 1 – ½=1/2
(6 adults+ 4 children)'s 1 day's work= 6/96+4/192= 1/12
1/12 work is done by them in 1 day.
½ work is done by them 12*(1/2)= 6 days
So, Answer= C
26)Sixteen men can complete a work in twelve days. Twenty four children can
complete the same work in 18 days. 12 men and 8 children started working and
after eight days three more children joined them. How many days will they now
take to complete the remaining work?
(A) 2 days (B) 4 days ( C) 6 days (D) 8 days
ol: one man's one day's work= 1/192
one child's one day's work= 1/432
Work done in 8 days=8*(12/192+ 8/432)=8*(1/16+1/54) =35/54
Remaining work= 1 -35/54= 19/54
(12 men+11 children)'s 1 day's work= 12/192 + 11/432 = 19/216
Now, 19/216 work is done by them in 1 day.
Therefore, 19/54 work will be done by them in ((216/19)*(19/54))= 4 days
So,Answer: B
27)Twenty-four men can complete a work in 16 days. Thirty- two women can
complete the same work in twenty-four days. Sixteen men and sixteen women
started working and worked for 12 days. How many more men are to be added to
complete the remaining work in 2 days?
(A) 16 men (B) 24 men ( C) 36 men (D) 48 men
Sol: one man's one day's work= 1/384
one woman's one day's work=1/768
Work done in 12 days= 12*( 16/384 + 16/768) = 12*(3/48)=3/4
Remaining work=1 – ¾=1/4
(16 men+16 women)'s two day's work =12*( 16/384+16/768)=2/16=1/8
Remaining work = 1/4-1/8 =1/8
1/384 work is done n 1 day by 1 man.
Therefore, 1/8 work will be done in 2 days in 384*(1/8)*(1/2)=24men
28)4 men and 6 women can complete a work in 8 days, while 3 men and 7 women
can complete it in 10 days. In how many days will 10 women complete it?
(A) 35 days (B) 40 days ( C) 45 days (D) 50 days
Sol: Let 1 man's 1 day's work =x days and
1 woman's 1 day's work=y
Then, 4x+6y=1/8 and 3x+7y=1/10.
Solving these two equations, we get: x=11/400 and y= 1/400
Therefore, 1 woman's 1 day's work=1/400
=> 10 women will complete the work in 40 days.
Answer: B
29)One man,3 women and 4 boys can do a piece of work in 96hrs, 2 men and 8 boys
can do it in 80 hrs, 2 men & 3 women can do it in 120hr. 5Men & 12 boys can do
it in?
(A) 39 1/11 hrs (B) 42 7/11 hrs ( C) 43 7/11 days (D) 44hrs
Sol: Let 1 man's 1 hour's work=x
1 woman's 1 hour's work=y
1 boy's 1 hour's work=z
Then, x+3y+4z=1/96 -----------(1)
2x+8z= 1/80 ----------(2)
adding (2) & (3) and subtracting (1)
3x+4z=1/96 ---------(4)
From (2) and (4), we get x=1/480
Substituting, we get : y=1/720 and z= 1/960
(5 men+ 12 boy)'s 1 hour's work=5/480+12/960 =1/96 + 1/80=11/480
Therefore, 5 men and 12 boys can do the work in 480/11 or 43 7/11hours.
So,Answer: C
Chain Rule
Important Facts:
Direct Proportion: Two Quantities are said to be directly
proportional, if on the increase (or decrease) of th one, the
other increases(or decreases) to the same extent.
Ex:(i) Cost is directly proportional to the number of articles.
(More articles, More cost).
(ii) Work done is directly proportional to the number of men
working on it. (More men, more work).
Indirect Proportion: Two Quantities are said to be
indirectly proportional,if on the increase of the one , the other
decreases to the same extent and vice-versa.
Ex:(i) The time taken by a car covering a certain distance is
inversely proportional to th speed of the car.(More speed,
less is the time taken to cover the distance).
(ii) Time taken to finish a work is inversely proportional to
the number of persons working at it.
(More persons, less is the time taken to finish a job).
Nte: In solving Questions by chain rule, we compare every
item with the term to be found out.
Problems
1)If 15 toys cost Rs.234, what do 35 toys cost ?
Sol: Let the required cost be Rs. x then
more toys more cost(direct proportion)
15:35:: 234:x
(15*x)=(234*35)
x=(234*35) /(15)= 546 Rs
2)If 36 men can do a piece of work in 25hours, in how many hours
will 15men do it?
Sol: Let the required number of hours be x.
less men more hours(Indirect proportion).
15:36::25:x
(15*x)=(36*25)
x=(36*25) /15
x=60
For 15 men it takes 60 hours.
3)If 9 engines consume 24metric tonnes of coal, when each is working
8 hours a day, how much coal will be required for 8 engines, each
running 13 hours a day, it being given that 3 engines of former type
consume as much as 4 engines of latter type?
Sol: Let 3 engines of former type consume 1 unit in 1 hour.
4 engines of latter type consume 1 unit in 1 hour.
1 engine of former type consumes 1/3 unit in 1 hour.
1 engine of latter type consumes ¼ unit in 1 hour.
Let required consumption of coal be x units.
Less engines, less coal consumed.(direct)
More working hours, more coal consumed(direct)
Less rate of consumption, less coal consumed (direct)
9:8
8:13 :: 24:x
1/3:1/4
(9*8*(1/3)*x)=(8*13*(1/4)*24)
24x=624
x=26 metric tonnes.
Complex Problems
1)A contract is to be completed in 46 days and 117 men were set to work,
each working 8 hours a day. After 33 days, 4/7 of the work is completed.
How many additional men may be employed so that the work may be
completed in time, each man now working 9 hours a day?
Sol: 4/7 of work is completed .
Remaining work=1- 4/7
=3/7
Remaining period= 46-33
=13 days
Less work, less men(direct proportion)
less days, more men(Indirect proportion)
More hours/day, less men(Indirect proportion)
work 4/7:3/7
Days 13:33 :: 117:x
hrs/day 9:8
(4/7)*13*9*x=(3/7)*33*8*117
x=(3*33*8*117) / (4*13*9)
x=198 men
So, additional men to be employed=198 -117=81
2)A garrison had provisions for a certain number of days. After 10 days,
1/5 of the men desert and it is found that the provisions will now last
just as long as before. How long was that?
Sol: Let initially there be x men having food for y days.
After, 10 days x men had food for ( y-10)days
Also, (x -x/5) men had food for y days.
x(y-10)=(4x/5)*y
=> (x*y) -50x=(4(x*y)/5)
5(x*y)-4(x*y)=50x
x*y=50x
y=50
3)A contractor undertook to do a certain piece of work in 40 days. He
engages 100 men at the beginning and 100 more after 35 days and completes
the work in stipulated time. If he had not engaged the additional men,
how many days behind schedule would it be finished?
Sol: 40 days- 35 days=5 days
=>(100*35)+(200*5) men can finish the work in 1 day.
4500 men can finish it in 4500/100= 45 days
This s 5 days behind the schedule.
4)12 men and 18 boys,working 7 ½ hors a day, can do a piece if work in
60 days. If a man works equal to 2 boys, then how many boys will be
required to help 21 men to do twice the work in 50 days, working
9 hours a day?
Sol: 1man =2 boys
12men+18boys=>(12*2+18)boys=42 boys
let the required number of boys=x
21 men+x boys
=>((21*2)+x) boys
=>(42+x) boys
less days, more boys(Indirect proportion)
more hours per day, less boys(Indirect proportion)
days 50:60
hrs/day 9:15/2 :: 42:(42+x)
work 1:2
(50*9*1*(42+x))=60*(15/2)*2*42
(42+x)= (60*15*42)/(50*9)= 84
x=84-42= 42
=42
42 days behind the schedule it will be finished.
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