December 2012

PROBLEMS ON 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
Top
(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
Top

(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

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PROBLEMS ON TRAINS

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PROBLEMS ON TRAINS

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PROBLEMS ON 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

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PROBLEMS ON Square and Cube Roots

Formula:

The Product of two same numbers in easiest way as follow.
Example:let us calculate the product of 96*96
Solution: Here every number must be compare with the 100.
See here the given number 96 which is 4 difference with the 100.
so subtract 4 from the 96 we get 92 ,then the square of the
number 4 it is 16 place the 16 beside the 92 we get answer
as 9216.

9 6
- 4
--------------
9 2
--------------
4*4=16
9 2 1 6

therefore square of the two numbers 96*96=9216.

Example: Calculate product for 98*98
Solution: Here the number 98 is having 2 difference when compare
to 100 subtract 2 from the number then we get 96 square the
number 2 it is 4 now place beside the 96 as 9604

9 8
- 2
-------------
9 6
-------------
2*2=4
9 6 0 4.
so, we get the product of 98*98=9604.

Example: Calculate product for 88*88
Solution: Here the number 88 is having 12 difference when compare
to 100 subtract 12 from the 88 then we get 76 the square of the
number 12 is 144 (which is three digit number but should place
only two digit beside the 76) therefore in such case add one to
6 then it becomes 77 now place 44 beside the number 77 we will get
7744.
88
-12
------------
76
-----------
12*12=144

76
+ 144
--------------------
7744
--------------------

Example: Find the product of the numbers 46 *46?
Solution:consider the number 50=100/2. Now again go comparision with
the number which gets when division with 100.here consider the number
50 which is nearer to the number given. 46 when compared with the
number 50 we get the difference of 4. Now subtract the number 4 from
the 46, we get 42. As 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
42/2=21 now square the the number 4 i.e, 4*4=16
just place the number 16 beside the number 21
we get 2116.
4 6
4
----------------
4 2 as 50 = 100/2

42/2=21
now place 4*4=16 beside 21
2 1 1 6

Example: Find the product of the numbers 37*37
Solution:
consider the number 50=100/2
now again go comparision with the number which gets when
division with 100.
here consider the number 50 which is nearer to the number given.
37 when compared with the number 50 we get the difference of 13.
now subtract the number 13 from the 37, we get 24.
as 50 got when 100 get divided by 2.
so, divided the number by 2 after subtraction.
24/2=12
now square the the number 13 i.e, 13*13=169
just place the number 169 beside the number 21
now as 169 is three digit number then add 1 to 2 we get
1t as 13 then place 69 beside the 13
we get 1369.

3 7
1 3
-----------------
2 4 as 50 = 100/2

24/2=12
square 13* 13=169

1 2
+ 1 6 9
-----------------------
1 3 6 9
-------------------------
Top
Example: Find the product of 106*106
Solution: now compare it with 100 ,
The given number is more then 100
then add the extra number to the given number.
That is 106+6=112
then square the number 6 that is 6*6=36
just place beside the number 36 beside the 112,then
we get 11236.
1 0 6
+ 6
---------------------
1 1 2
--------------------
now 6* 6=36 place this beside the number 112, we get
1 1 2 3 6

Square root: If x2=y ,we say that the square root of y
is x and we write ,âˆšy=x.

Cube root: The cube root of a given number x is the number
whose cube is x. we denote the cube root of x by x1/3 .

Examples:

1.Evaluate 60841/2 by factorization method.

Solution: Express the given number as the product of prime
factors. Now, take the product of these prime factors choosing
one out of every pair of the same primes. This product gives the
square root of the given number.

Thus resolving 6084 in the prime factors ,we get 6084
2 6024
2 3042
3 1521
3 507
13 169
13
6084=21/2 *31/2 *131/2
60841/2=2*3*13=78.
Answer is 78.

2.what will come in place of question mark in each of the following
questions?

i)(32.4/?)1/2 = 2
ii)86.491/2 + (5+?1/2)2 =12.3

Solution: 1) (32.4/x)1/2=2
Squaring on both sides we get
32.4/x=4
=>4x=32.4
=>x=8.1

Answer is 8.1

ii)86.491/2 + âˆš(5+x2)=12.3

solutin:86.491/2 + (5+x1/2 )=12.3
9.3+ âˆš(5+x1/2 )=12.3
=> âˆš(5+x1/2 ) =12.3-9.3
=> âˆš(5+x1/2 )=3
Squaring on both sides we get
(5+x1/2 )=9
x1/2 =9-5
x1/2 =4
x=2.
Answer is 2.

3.âˆš 0.00004761 equals:

Solution: âˆš (4761/108)
âˆš4761/âˆš 108
. 69/10000
0.0069.
Answer is 0.0069

4.If âˆš18225=135,then the value of
âˆš182.25 + âˆš1.8225 + âˆš 0.018225 + âˆš0.00018225.

Solution: âˆš(18225/100) +âˆš(18225/10000) +
âˆš(18225/1000000) +âˆš(18225/100000000)
=âˆš(18225)/10 + (18225)1/2/100 +
âˆš(18225)/1000 + âˆš(18225)/10000
=135/10 + 135/100 + 135/1000 + 135/10000
=13.5+1.35+0.135+0.0135=14.9985.
Answer is 14.9985.

5.what should come in place of both the question
marks in the equation (?/ 1281/2= (162)1/2/?) ?

Solution: x/ 1281/2= (162)1/2/x
=>x1/2= (128*162)1/2
=> x1/2= (64*2*18*9)1/2
=>x2= (82*62*32)
=>x2=8*6*3
=>x2=144
=>x=12.

6.If 0.13 / p1/2=13 then p equals

Solution: 0.13/p2=13
=>p2=0.13/13
=1/100
p2=âˆš(1/100)
=>p=1/10
therefore p=0.1
Answer is 0.1
Top
7.If 13691/2+(0.0615+x)1/2=37.25 then x is equals to:

Solution
37+(0.0615+x)1/2=37.25(since 37*37=1369)
=>(0.0615+x)1/2=0.25
Squaring on both sides
(0.0615+x)=0.0625
x=0.001
x=10-3.
Answer is 10-3.

8.If âˆš(x-1)(y+2)=7 x& y being positive whole numbers then
values of x& y are?

Solution: âˆš(x-1)(y+2)=7
Squaring on both sides we get
(x-1)(y+2)=72
x-1=7 and y+2=7
therefore x=8 , y=5.
Answer x=8 ,y=5.

9.If 3*51/2+1251/2=17.88.then what will be the
value of 801/2+6*51/2?

Solution: 3*51/2+1251/2=17.88
3*51/2+(25*5)1/2=17.88
3*51/2+5*51/2=17.88
8*51/2=17.88
51/2=2.235
therefore 801/2+6 51/2=(16*51/2)+6*1/25
=4 51/2+6 51/2
=10*2.235
=22.35
Answer is 22.35

10.If 3a=4b=6c and a+b+c=27*âˆš29 then Find c value is:

Solution: 4b=6c
=>b=3/2*c
3a=4b
=>a=4/3b
=>a=4/3(3/2c)=2c
therefore a+b+c=27*291/2
2c+3/2c+c=27*291/2
=>4c+3c+2c/2=27*291/2
=>9/2c=27*291/2
c=27*291/2*2/9
c=6*291/2

11.If 2*3=131/2 and 3*4=5 then value of 5*12 is

Solution:
Here a*b=(a2+b2)1/2
therefore 5*12=(52+122)1/2
=(25+144)1/2
=1691/2
=13
Answer is 13.

12.The smallest number added to 680621 to make
the sum a perfect square is

Solution: Find the square root number which
is nearest to this number
8 680621 824
64
162 406
324
1644 8221
6576
1645
therefore 824 is the number ,to get the nearest
square root number take (825*825)-680621
therefore 680625-680621=4
hence 4 is the number added to 680621 to make it
perfect square.

13.The greatest four digit perfect square number is

Solution: The greatest four digit number is 9999.
now find the square root of 9999.
9 9999 99
81
189 1819
1701
198
therefore 9999-198=9801 which is required number.
Answer is 9801.

14.A man plants 15376 apples trees in his garden and arranges
them so, that there are as many rows as there are apples trees
in each row .The number of rows is.

Solution: Here find the square root of 15376.
1 15376 124
1
22 53
44
244 976
976
0
therefore the number of rows are 124.

15.A group of students decided to collect as many paise from
each member of the group as is the number of members. If the
total collection amounts to Rs 59.29.The number of members
in the group is:

Solution: Here convert Money into paise.
59.29*100=5929 paise.
To know the number of member ,calculate the square root of 5929.
7 5929 77
49
147 1029
1029
0
Therefore number of members are 77.

16.A general wishes to draw up his 36581 soldiers in the form
of a solid square ,after arranging them ,he found that some of
them are left over .How many are left?
Solution: Here he asked about the left man ,So find the
square root of given number the remainder will be the left man
1 36581 191
1
29 265
261
381 481
381
100(since remaining)
Therefore the left men are 100.

17.By what least number 4320 be multiplied to obtain number
which is a perfect cube?

Solution: find l.c.m for 4320.
2 4320
2 2160
2 1080
2 540
2 270
3 135
3 45
3 15
5
4320=25 * 33 * 5
=23 * 33 * 22 *5
so make it a perfect cube ,it should be multiplied by 2*5*5=50
Answer is 50.

18.3(4*12/125)1/2=?

Solution: 3(512/125)1/2
3(8*8*8)1/2/(5*5*5)
3(83)1/2/(53)
((83)/(53))1/3
=>8/5 or 1 3/5.

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PROBLEMS ON Simple Intrest

Simple Intrest

Important Facts and Formulae:

Principal or Sum:- The money borrowed or lent out for a
certain period is called Principal or the Sum.

Interest:- Extra money paid for using others money is
called Interest.

Simple Interest:- If the interest on a sum borrowed for
a certain period is reckoned uniformly,then it is called
Simple Interest.

Formulae:
Principal = P
Rate = R% per annum
Time = T years. Then,

(i)Simple Interest(S.I)= (P*T*R)/100

(ii) Principal(P) = (100*S.I)/(R*T)
Rate(R) = (100*S.I)/(P*T)
Time(T) = (100*S.I)/(P*R)

Simple Problems

1.Find S.I on Rs68000 at 16 2/3% per annum for 9months.

Sol:- P=68000
R=50/3% p.a
T=9/12 years=4/3 years
S.I=(P*R*T)/100
=(68000*(50/3)*(3/4)*(1/100))
=Rs 8500

Note:If months are given we have to converted into
years by dividing 12 ie., no.of months/12=years

2.Find S.I on Rs3000 at 18% per annum for the period from
4th Feb to 18th April 1995

Sol:- Time=(24+31+18)days
=73 days
=73/365=1/5 years
P= Rs 3000
R= 18% p.a
S.I = (P*R*T)/100
=(3000*18*1/5*1/100)
=Rs 108
Remark:- The day on which money is deposited is not
counted while the day on which money is withdrawn is
counted.

3. In how many years will a sum of money becomes triple
at 10% per annum.

Sol:- Let principal =P
S.I = 2P
S.I = (P*T*R)/100
2P = (P*T*10)/100
T = 20 years
Note:
(1) Total amount = Principal + S.I
(2) If sum of money becomes double means Total amount
or Sum
= Principal + S.I
= P + P = 2P

Medium Problems

1.A sum at Simple interest at 13 1/2% per annum amounts
to Rs 2502.50 after 4 years.Find the sum.

Sol:- Let Sum be x. then,
S.I = (P*T*R)/100
= ((x*4*27)/(100*2))
= 27x/100
Amount = (x+(27x)/100)
= 77x/50
77x/50 = 2502.50
x = (2502.50*50)/77
= 1625
Sum = 1625

2. A some of money becomes double of itself in 4 years
in 12 years it will become how many times at the same
rate.

Sol:- 4 yrs - - - - - - - - - P
12 yrs - - - - - - - - - ?
(12/4)* P =3P
Amount or Sum = P+3P = 4 times

3. A Sum was put at S.I at a certain rate for 3 years.
Had it been put at 2% higher rate ,it would have
fetched Rs 360 more .Find the Sum.

Sol:- Let Sum =P
original rate = R
T = 3 years
If 2% is more than the original rate ,it would have
fetched 360 more ie., R+2
(P*(R+2)*3/100) - (P*R*3)/100 = 360
3PR+ 6P-3PR = 36000
6P = 36000
P = 6000
Sum = 6000.

4.Rs 800 amounts to Rs 920 in 3yrs at S.I.If the interest
rate is increased by 3%, it would amount to how much?

Sol:- S.I = 920 - 800 = 120
Rate = (100*120)/(800*3) = 5%
New Rate = 5 + 3 = 8%
Principal = 800
Time = 3 yrs
S.I = (800*8*3)/100 = 192
New Amount = 800 + 192
= 992

5. Prabhat took a certain amount as a loan from bank at
the rate of 8% p.a S.I and gave the same amount to Ashish
as a loan at the rate of 12% p.a . If at the end of 12 yrs,
he made a profit of Rs. 320 in the deal,What was the
original amount?

Sol:- Let the original amount be Rs x.
T = 12
R1 = 8%
R2 = 12%
Profit = 320
P = x
(P*T*R2)/100 - (P*T*R1)/100 =320
(x*12*12)/100 - (x*8*12)/100 = 320
x = 2000/3
x = Rs.666.67

6. Simple Interest on a certail sum at a certain rate is
9/16 of the sum . if the number representing rate percent
and time in years be equal ,then the rate is.

Sol:- Let Sum = x .Then,
S.I = 9x/16
Let time = n years & rate = n%
n = 100 * 9x/16 * 1/x * 1/n
n * n = 900/16
n = 30/4 = 7 1/2%

Complex Problems

1. A certain sum of money amounts t 1680 in 3yrs & it
becomes 1920 in 7 yrs .What is the sum.

Sol:- 3 yrs - - - - - - - - - - - - - 1680
7 yrs - - - - - - - - - - - - - 1920
then, 4 yrs - - - - - - - - - - - - - 240
1 yr - - - - - - - - - - - - - ?
(1/4) * 240 = 60
S.I in 3 yrs = 3*60 = 18012
Sum = Amount - S.I
= 1680 - 180
= 1500
we get the same amount if we take S.I in 7 yrs
I.e., 7*60 =420
Sum = Amount - S.I
= 1920 - 420
= 1500

2. A Person takes a loan of Rs 200 at 5% simple Interest.
He returns Rs.100 at the end of 1 yr. In order to clear
his dues at the end of 2yrs ,he would pay:

Sol:- Amount to be paid
= Rs(100 + (200*5*1)/100 + (100*5*1)/100)
= Rs 115

3. A Man borrowed Rs 24000 from two money lenders.For one
loan, he paid 15% per annum and for other 18% per annum.
At the end of one year,he paid Rs.4050.How much did he
borrowed at each rate?

Sol:- Let the Sum at 15% be Rs.x
& then at 18% be Rs (24000-x)
P1 = x R1 = 15
P2 = (24000-x) R2 = 18
At the end of ine year T = 1
(P1*T*R1)/100 + (P2*T*R2)/100 = 4050
(x*1*15)/100 + ((24000-x)*1*18)/100 = 4050
15x + 432000 - 18x = 405000
x = 9000
Money borrowed at 15% = 9000
Money borrowed at 18% = (24000 - 9000)
= 15000

4.What annual instalment will discharge a debt of Rs. 1092
due in 3 years at 12% Simple Interest ?

Sol:- Let each instalment be Rs x
(x + (x * 12 * 1)/100) + (x + (x * 12 * 2)/100) + x = 1092
28x/25 + 31x/25 + x =1092
(28x +31x + 25x) = (1092 * 25)
84x = 1092 * 25
x = (1092*25)/84 = 325
Each instalement = 325

5.If x,y,z are three sums of money such that y is the simple
interest on x,z is the simple interest on y for the same
time and at the same rate of interest ,then we have:

Sol:- y is simple interest on x, means
y = (x*R*T)/100
RT = 100y/x
z is simple interest on y,
z = (y*R*T)/100
RT = 100z/y
100y/x = 100z/y
y * y = xz

6.A Sum of Rs.1550 was lent partly at 5% and partly at 5%
and partly at 8% p.a Simple interest .The total interest
received after 3 years was Rs.300.The ratio of the money
lent at 5% to that lent at 8% is:

Sol:- Let the Sum at 5% be Rs x
at 8% be Rs(1550-x)
(x*5*3)/100 + ((1500-x)*8*3)/100 = 300
15x + 1500 * 24 - 24x = 30000
x = 800
Money at 5%/ Money at 8% = 800/(1550 - 800)
= 800/750 = 16/15

7. A Man invests a certain sum of money at 6% p.a Simple
interest and another sum at 7% p.a Simple interest. His
income from interest after 2 years was Rs 354 .one
fourth of the first sum is equal to one fifth of the
second sum.The total sum invested was:

Sol:- Let the sums be x & y
R1 = 6 R2 = 7
T = 2
(P1*R1*T)/100 + (P2*R2*T)/100 = 354
(x * 6 * 2)/100 + (y * 7 * 2)/100 = 354
6x + 7y = 17700 â€”â€”â€”(1)
also one fourth of the first sum is equal to one
fifth of the second sum
x/4 = y/5 => 5x - 4y = 0 â€”â€” (2)
By solving 1 & 2 we get,
x = 1200 y = 1500
Total sum = 1200 +1500
= 2700

8. Rs 2189 are divided into three parts such that their
amounts after 1,2& 3 years respectively may be equal,
the rate of S.I being 4% p.a in all cases. The Smallest
part is:

Sol:- Let these parts be x,y and[2189-(x+y)] then,
(x*1*4)/100 = (y*2*4)/100 = (2189-(x+y))*3*4/100
4x/100 = 8y/100
x = 2y
By substituting values
(2y*1*4)/100 = (2189-3y)*3*4/100
44y = 2189 *12
y = 597
Smallest Part = 597

9. A man invested 3/3 of his capital at 7% , 1/4 at 8% and
the remainder at 10%.If his annual income is Rs.561. The
capital is:

Sol:- Let the capital be Rs.x
Then, (x/3 * 7/100 * 1) + ( x/4 * 8/100 * 1)
+ (5x/12 * 10/100 * 1) = 561
7x/300 + x/50 + x/24 = 561
51x = 561 * 600
x = 6600

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Permutations and Combinations

Formulae:

Factorial Notation:
Let n be positive integer.Then ,factorial n dentoed by n!
is defined as n! = n(n-1)(n-2). . . . . . . .3.2.1
eg:- 5! = (5 * 4* 3 * 2 * 1)
= 120
0! = 1
Permutations:
The different arrangements of a given number of things by
taking some or all at a time,are called permutations.
eg:- All permutations( or arrangements)made with the letters
a,b,c by taking two at a time are (ab,ba,ac,ca,bc,cb)

Numbers of permutations:
Number of all permutations of n things, taken r at a time is
given by nPr = n(n-1)(n-2). . .. . . (n-r+1)
= n! / (n-r)!

An Important Result:
If there are n objects of which p1 are alike of one kind;
p2 are alike of another kind ; p3 are alike of third kind and
so on and pr are alike of rth kind, such that
(p1+p2+. . . . . . . . pr) = n
Then,number of permutations of these n objects is:
n! / (p1!).(p2!). . . . .(pr!)

Combinations:
Each of different groups or selections which can be formed by
taking some or all of a number of objects,is called a combination.
eg:- Suppose we want to select two out of three boys A,B,C .
then ,possible selection are AB,BC & CA.
Note that AB and BA represent the same selection.

Number of Combination:
The number of all combination of n things taken r at atime is:
nCr = n! / (r!)(n-r)!
= n(n-1)(n-2). . . . . . . tor factors / r!
Note: nCn = 1 and nC0 =1

An Important Result:
nCr = nC(n-r)
Top
Problems

1.Evaluate 30!/28!

Sol:- 30!/28! = 30 * 29 * (28!) / (28!)
= 30 * 29 =870

2.Find the value of 60P3

Sol:- 60P3 = 60! / (60 â€“ 3)! = 60! / 57!
= (60 * 59 *58 * (57!) )/ 57!
= 60 * 59 *58
= 205320

3. Find the value of 100C98,50C 50

Sol:- 100C98 = 100C100-98)
= 100 * 99 / 2 *1
= 4950
50C50 = 1

4.How many words can be formed by using all the letters of the
word â€œDAUGHTRâ€ so that vowels always come together &
vowels are never together?

Sol:-
(i) Given word contains 8 different letters
When the vowels AUE are always together we may suppose
them to form an entity ,treated as one letter
then the letter to be arranged are DAHTR(AUE)
these 6 letters can be arranged in 6p6 = 6!
= 720 ways
The vowels in the group (AUE) may be arranged in 3! = 6 ways
Required number of words = 760 * 6 =4320

(ii)Total number of words formed by using all the letters of
the given words

8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= 40320
Number of words each having vowels together
= 760 * 6
= 4320
Number of words each having vowels never together
= 40320 â€“ 4320
= 36000

5.In how many ways can a cricket eleven be chosen out of a batch
of 15 players.

Sol:- Required number of ways
= 15C 11 = 15C (15-11)
= 15 C 4
15C4 = 15 * 14 * 13 * 12 / 4 * 3 * 2 *1
= 1365

6.In how many a committee of 5 members can be selected from 6men
5 ladies consisting of 3 men and 2 ladies

Sol:- (3 men out of 6) and (2 ladies out of 5) are to be chosen
Required number of ways
=(6C3 * 5C2)
= 200

7.How many 4-letter word with or without meaning can be formed out
of the letters of the word 'LOGARITHMS' if repetition of letters is
not allowed

Sol:- 'LOGARITHMS' contains 10 different letters
Required number of words
= Number of arrangements of 100 letters taking
4 at a time
= 10P4
= 10 * 9 * 8 * 7
= 5040

8.In how many ways can the letter of word 'LEADER' be arranged

Sol:- The word 'LEADER' contains 6 letters namely
1L,2E,1A,1D and 1R
Required number of ways
= 6! / (1!)(2!)(1!)(1!)(1!)
= 6 * 5 * 4 * 3 * 2 *1 / 2 * 1
=360
Top
9.How many arrangements can be made out of the letters of the word
'MATHEMATICS' be arranged so that the vowels always come
together

Sol:- In the word ' MATHEMATICS' we treat vowels
AEAI as one letter thus we have MTHMTCS(AEAI)
now we have to arrange 8 letters out of which M occurs
twice ,T occurs twice & the rest are different
Number of ways of arranging these letters
= 8! / (2!)(2!)
= 10080
now AEAI has 4 letters in which A occurs 2 times and the rest
are different
Number of ways of arranging these letters
= 4! / 2! = 12
Required number of words = (10080 * 12)
= 120960

10.In how many different ways can the letter of the word 'DETAIL' be
arranged in such a way that the vowels occupy only the odd positions

Sol:- These are 6 letters in the given word , out of which
there are 3 vowels and 3 consonants
Let us mark these positions as under

(1)(2) (3) (4)(5)(6)
now 3 vowels can be placed at any of the three places out of 4
marked 1,3,5
Number of ways of arranging the vowels = 3P3 = 3! =6
Also,the 3 consonants can be arranged at the remaining 3 positions
Number of arrangements = 3P3 = 6
Total number of ways = (6 * 6) =36

11.How many 3 digit numbers can be formed from the digits 2,3,5,6,7
and 9 which are divisible by 5 and none of the digits is repeated?

Sol:- Since each desired number is divisible by 5,
so we much have 5 at the unit place. The hundreds place
can now be filled by any of the remaining 4 digits .so, there
4 ways of filling it.
Required number of numbers = (1 * 5 * 4)
= 20
12.In how many ways can 21 books on English and 19 books on Hindi
be placed in a row on a self so that two books on Hindi may not
be together?

Sol:- In order that two books on Hindi are never together,
we must place all these books as under:
X E X E X . . . . . . . . . . X E X
Where E denotes the position of an English and X that of
a Hindi book.
Since there are 21 books on English,the number of places
marked X are therefore 22.
Now, 19 places out of 22 can be chosen in
22 C 19 = 22 C 3 =22 * 21 * 20 / 3 * 2 *1
Hence the required number of ways = 1540

13.Out of 7 constants and 4 vowels how many words of 3 consonants
and 2 vowels can be formed?

Sol:- Number of ways of selecting (3 consonants out of 7) and
(2 vowels out of 4)
= 7C3 * 4C2
= 210
Number of groups each having 3 consonants and 2 vowels = 210
Each group contains 5 letters
Number of ways of arranging 5 letters among themselves
= 5! = (5 * 4 * 3 * 2 * 1)
= 210
Required number of words = (210 * 210)
= 25200

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PROBLEMS BASED ON 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.

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PROBLEMS BASED ON PERCENTAGES

Simple problems:

1 . Express the following as a fraction.

a) 56%
SOLUTION:
56/100=14/25

b) 4%
SOLUTION:
4/100=1/25

c) 0.6%
SOLUTION:
0.6/100=6/1000=3/500

d) 0.08%
SOLUTION:
0.08/100=8/10000=1/1250

2. Express the following as decimals

a) 6%
SOLUTION:
6% = 6/100=0.06

b) 0.04%
SOLUTION:
0.04% = 0.04/100=0.0004

3 . Express the following as rate percent.
i).23/36

SOLUTION:
= (23/36*100) %
= 63 8/9%
ii).6 Â¾

SOLUTION:
6 Â¾ =27/4
(27/4 *100) % =675 %

4. Evaluate the following:
28% of 450 + 45% of 280 ?

SOLUTION:
=(28/100) *450 + (45/100) *280
= 28 * 45 / 5
= 252

5. 2 is what percent of 50?

SOLUTION:
Formula : (IS / OF ) *100 %
= 2/50 *100
= 4%

6. Â½ is what percent of 1/3?

SOLUTION:
=( Â½) / (1/3) *100 %
= 3/2 *100 %
= 150 %

7. What percent of 2 Metric tonnes is 40 Quintals?

SOLUTION:
1 metric tonne =10 Quintals
So required percentage=(40/(2*10))*100%
= 200%

8. Find the missing figure .
i) ? % of 25 = 2.125

SOLUTION :
Let x% of 25 = 2.125.then
(x/100) *25 =2.125
x = 2.125 * 4
= 8.5

ii) 9% of ? =6.3

SOLUTION:
Let 9 % of x = 6.3.
Then 9/100 of x= 6.3
so x = 6.3 *100/7
= 70.

9. Which is the greatest in 16 2/3 %, 2/15,0.17?

SOLUTION:
16 2/3 % = 50/3 %
=50/3 * 1/100
=1/6
= 0.166
2 / 15 =0.133
So 0.17 is greatest number in the given series.

10.If the sales tax be reduced from 3 Â½ % to 3 1/3 % ,
then what difference does it make to a person who
purchases an article with marked price of RS 8400?

SOLUTION:

Required difference = 3 Â½ % of 8400 â€“ 3 1/3 % of 8400
=(7/2-10/3)% of 8400
=1/6 % of 8400
= 1/6* 1/100* 8400
= Rs 14.

11. A rejects 0.08% of the meters as defective .How many
will he examine to reject 2?

SOLUTION:
Let the number of meters to be examined be x.
Then 0.08% of x=2.
0.08/100*x= 2
x= 2 * 100/0.08
=2 * 100 * 100/8
= 2500

12.65 % of a number is 21 less than 4/5 of that number.
What is the number?

SOLUTION: Let the number be x.
4/5 x- (65% of x) = 21
4/5x â€“ 65/100 x=21
15x=2100
x=140

13. Difference of two numbers is 1660.If 7.5 % of one number
is 12.5% of the other number. Find two numbers?

SOLUTION:
Let the two numbers be x and y.
7.5% of x=12.5% of y'
So 75x=125 y
3x=5y
x=5/3y.
Now x-y=1660
5/3y-y=1660
2/3y=1660
y=2490
So x= 2490+1660
=4150.
So the numbers are 4150 , 1660.

14. In expressing a length 81.472 KM as nearly as possible
with 3 significant digits ,Find the % error?

SOLUTION:
Error= 81.5-81.472=0.028
So the required percentage = 0.028/81.472*100%
= 0.034%
Top

15. In an election between two persons ,75% of the voters
cast their votes out of which 2% are invalid. A got
9261 which 75% of the total valid votes. Find total
number of votes?

SOLUTION:
Let x be the total votes.
valid votes are 98% of 75% of x.
So 75%(98%(75% of x))) = 9261
==> 75/100 *98 /100 * 75 100 *x = 9261
x= 1029 * 4 *100 *4 / 9
= 16800
So total no of votes = 16800

16 . A's maths test had 75 problems i.e 10 arithmetic, 30
algebra and 35 geometry problems.Although he answered
70% of arithmetic ,40% of algebra and 60 % of geometry
problems correctly he didn't pass the test because he
got less than 60% of the problems right. How many more
questions he would have needed to answer correctly to
get a 60% passing grade.

SOLUTION:
70% of 10 =70/100 * 10
=7
40% of 30 = 40 / 100 * 30
= 12
60 % of 35 = 60 / 100 *35
= 21
So correctly attempted questions = 7 + 12 + 21
=40.
Questions to be answered correctly for 60% grade
=60% of 75
= 60/100 *75
=45.
So required questions=45-40 = 5

17 . If 50% of (x â€“ y) = 30% of (x + y) then what percent
of x is y ?

SOLUTION:
50/100(x-y) =30/100(x+y)
Â½ (x-y)= 3/10(x+y)
5x-5y=3x+3y
x=4y
So Required percentage =y/x*100 %
=y/4y *100 %
= 25%.

18 .If the price of tea is increased by 20% ,find how much
percent must a householder reduce her consumption of tea
so as not to increase the expenditure?

SOLUTION:
Reduction in consumption= R/(100+R) *100%
=20/120 *100
= 16 2/3 %

19.The population of a town is 176400 . If it increases
at the rate of 5% per annum ,what will be the
population 2 years hence? What was it 2 years ago?

SOLUTION:
Population After 2 years = 176400[1+5/100]2
=176400 * 21/20 *21/20
=194481
Population 2 years ago = 176400/(1+5/100)2
= 176400 * 20/21 *20/ 21
=160000

20.1 liter of water is add to 5 liters of a 20 % solution
of alcohol in water . Find the strength of alcohol in
new solution?

SOLUTION:
Alcohol in 5 liters = 20% of 5
=1 liter
Alcohol in 6 liters of new mixture = 1liter
So % of alcohol is =1/6 *100=16 2/3%

21.If A earns 33 1/3 more than B .Then B earns less
than A by what percent?

SOLUTION:
33 1/3 =100 /
Required Percentage = (100/3)/(100 + (100/3)) *100 %
= 100/400 *100 = 25 %

22.A school has only three classes which contain
40,50,60 students respectively.The pass percent of
these classes are 10, 20 and 10 respectively . Then
find the pass percent in the school.

SOLUTION:
Number of passed candidates =
10/100*40+20/100 *50+10/100 * 60
=4+10+6
=20
Total students in school = 40+50+60 =150
So required percentage = 20/150 *100
= 40 /3
=13 1/3 %

23. There are 600 boys in a hostel . Each plays either
hockey or football or both .If 75% play hockey and
45 % play football ,Find how many play both?

SOLUTION:
n(A)=75/100 *600
=450
n(B) = 45/100 *600
= 270
n(A^B)=n(A) + n(B) â€“ n(AUB)
=450 + 270 -600
=120
So 120 boys play both the games.

24.A bag contains 600 coins of 25p denomination and
1200 coins of 50p denomination. If 12% of 25p coins
and 24 % of 50p coins are removed, Find the percentage
of money removed from the bag ?

SOLUTION:
Total money = (600 * 25/100 +1200 *50/100)
=Rs 750
25p coins removed = 12/100 *600
=72
50p coins removed = 24/100 *1200
=288
So money removed =72 *1/4 +288 *1/2
= Rs 162
So required percentage=162/750 *100
=21 .6%

25. P is six times as large as Q.Find the percent that
Q is less than P?

SOLUTION:
Given that P= 6Q
So Q is less than P by 5Q.
Required percentage= 5Q/P*100 %
=5/6 * 100 %
=83 1/3%

26.For a sphere of radius 10 cm ,the numerical value of
surface area is what percent of the numerical value
of its volume?

SOLUTION:
Surface area = 4 *22/7 *r2
= 3/r(4/3 * 22/7 * r3)
=3/r * VOLUME
Where r = 10 cm
So we have S= 3/10 V
=3/10 *100 % of V
= 30 % of V
So surface area is 30 % of Volume.

27. A reduction of 21 % in the price of wheat enables
a person to buy 10 .5 kg more for Rs 100.What is
the reduced price per kg.

SOLUTION:
Let the original price = Rs x/kg
Reduced price =79/100x /kg
==> 100/(79x/100)-100/x =10.5
==> 10000/79x-100/x=10.5
==> 10000-7900=10.5 * 79 x
==> x= 2100/10.5 *79
So required price = Rs (79/100 *2100/10.5 *79) /kg
= Rs 2 per kg.

28.The length of a rectangle is increased by 60 % .
By what percent would the width have to be decreased
to maintain the same area?

SOLUTION:
Let the length =l,Breadth= b.
Let the required decrease in breadth be x %
then 160/100 l *(100-x)/100 b=lb
160(100-x)=100 *100
or 100-x =10000/160
=125/2
so x = 100-125/2
=75/2=37.5

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General knowledge

PROBLEMS ON TRAINS

PROBLEMS ON TRAINS

PROBLEMS ON TRAINS

PROBLEMS ON Time and Work

PROBLEMS ON Square and Cube Roots

PROBLEMS ON Simple Intrest

Permutations and Combinations

PROBLEMS BASED ON Pipes and Cisterns

PROBLEMS BASED ON PERCENTAGES

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