Fundamentals of Aptitude(Discount)

CONCEPT


Let rate=R%per annum, Time= T years

1.Present worth (PW) = (100*Amount)/(100+(R*T))

   = (100* TrueDiscount)/(R*T)

2.TrueDiscount (TD) = (P.W*R*T)/100

= (Amount*R*T)/(100+(R*T))

3.Sum =(SimpeInterest*TrueDiscount)/(SimpleInterest-TrueDiscount)

4.SimpleInterest-TrueDiscount=SimpeInterest on TrueDiscount

5.When the sum is put at CompoundInterest,then

PresentWorth=Amount/(1+(R/100))^T

GeneralConcept: Suppose a man has to pay Rs.156 after 4 years 
and the rate of interest is 14%per annum 
Clearly ,Rs.100 at 14% will amount to Rs156 in 4 years 
So,the payment of Rs.100 now wil cear off the debt
of Rs156 due 4 years hence
We say that 
Sum due = Rs156 due 4 years hence
Present Worth =Rs100
True Discount=Rs.156-Rs100=Rs56
=Sumdue-PW
We define 
TrueDiscount= Interest on Present Worth(PW)
Amount = PresentWorth+TrueDiscount





SIMPLE PROBLEMS



1.Find the present worth of Rs.930 due 3 years hence at 8% per

annum.Aso find the discount?

Sol: Amount=RS.930,Time=3years,Rate=8%

TrueDiscount = (Presentworth*Time*Rate)/100

Presentworth = (Amount*100) /(100+(R*T))

Presentworth = (930*100)/(100+(8*3)

=Rs.750

TrueDiscount = (930*3*8)/(100+(8*3)) =Rs.180 (or)

TrueDiscount = Amount-Presentworth

=Rs.930-Rs.750

=Rs.180

2.The truediscount on a bill due 9 months hence at 12% per

annum is Rs540.Find the amount of the bill and its presentworth?

Sol: Time=9months=9/12years=3/4years

Rate=12%

TrueDiscount=Rs540

TrueDiscount=(Amount*100)/(100+(R*T))

=>Amount=(TrueDiscount*(100+(R*T))) /100

=(540*(100+12*(3/4))/100

=Rs.6540

Presentworth=Amount-TrueDiscount

=Rs.6540-Rs540

=Rs.6000

3.The TrueDiscount on a certain sum of money due 3 years hence is Rs.250

and SimpeInterest on the same sum for same time and same rate is Rs375

find sum and rate%?

Sol: Time=3 years

Truediscount=Rs.250

SimpeInterest=Rs375

Sum = (SimpeInterest*TrueDiscount)/(SimpeInterest-TrueDiscount)

= (375*250)/(375-250)
=Rs.750
SimpeInterest=(Principle*Time*Rate)/100
375 = ( 750*3R)/100
R=50/3%
=> Rate=162/3%


4.The difference between SimpeInterest and TrueDiscount on a certain sum
of money for 6 months at 121/2% per annum is Rs25.find the sum.

Sol: Let amount be Rs.x
SimpleInterest=(Amount*T*R)/100
TrueDiscount=(Amount*T*R)/(100+(R*T))
SI-T.D=Rs25
=>((x*6/12*25/2)/100)-((x*6/12*25/2)/(100+(6/12*25/2))=25
=>x=Rs.6800


5.The Present worth of Rs.2310 due 21/2yrs henceThe interest being 15%
per annum is
sol: Amount=Rs2310,Time=21/2yrs,Rate=15%

Presentworth = (Amount*100) /(100+(R*T))

=2310*100/(100+15*5/2)

=Rs1680.





MEDIUM PROBLEMS

1.A bill falls due in 1 year.The creditor agrees to accept
immediate payment of the half and to defer the payment of
the other half for 2 years .By this arrangement he gains
Rs.40.what is the amount of bill,if the money be worth 121/2%?

sol: Let the amount be Rs.x
If the bill falls due for 1 yr,
Truediscount=(Amount*100)/(100+R*T)
=(x*100)/(100+(25/2*1)
If Accepting immediate payment of the half and to
defer the payment of the other half for 2 years,
Truediscount=(Amount*100)/(100+R*T)
=[x/2+((x/2*100)/(100+(25/2*2))]
x/2 for immediate payment,
((x/2*100)/(100+(25/2*2)) for paying after 2 yrs
He gains Rs40
=>[x/2+((x/2*100)/(100+(25/2*2))] – (x*100)/(100+(25/2*1) = Rs40
=> x/2+2x/5-8x/9=40
=>x=Rs.3600
Amount of the bill=Rs.3600


2.If the truediscount on a sum due 2yrs hence at 14% per annum
be Rs168.The sum due is?
Sol: Time=2yrs,
Rate=14%
Truediscount=Rs168,Amount=x
TrueDiscount=(Amount*R*T)/(100+R*T)
168 =(x*14*2)/(100+(14*2))
=>x=Rs.768


3.The truediscount on Rs.2562 due 4 months hence is Rs.122.The rate % is?
Sol: Amount=Rs.2562
Time=4/12yrs
TrueDiscount=Rs122
Rate=X%

TrueDiscount=(Amount*R*T)/(100+R*T)
=>122=(2562*x*(4/12))/(100+(x*(4/12))
=>36600+122x=2562x
=>Rate=15%


4.The Truediscount on Rs1760 due after a certain time at
12% per annum is Rs160The time after which it is due is:
Sol: Amount=Rs.1760
Rate=12%
Truediscount=Rs160
Time=x
TrueDiscount=(Amount*R*T)/(100+R*T)
=>x*11*12=100+12x
=>x=5/6yr
Time=10months


5.The interest on Rs.760 for 2yrs is the same as the
Truediscount on Rs960 due 2yrs hence .If the rate of
interest is same in both cases,it is
sol:
Principal=Rs.760
Time=2yrs
amount=Rs960
P*T*R/100=(Amount*R*T)/(100+R*T)
=>(750*2*r)/100=(960*r*2)/(100+2r)
=>150r=2100
rate=14%


6.The SimpleInterest & TrueDiscount on a certain sum of
money for a given time & at a given rate are Rs85
& Rs.80.The sum is.............
sol:
S.I= Rs.85
TrueDiscount=Rs80
Sum=(S.I.*TD)/(SI-T.D)
=Rs.1360

7.A trader owes a maerchant Rs10028 due 1yr hence The trader
wants to settle the account after 3 months .If the rate of
interest is 12%per annum How much cash should he pay?
Sol:
Time=1yr but he settles account after 3months
so time = 9months
Cash to pay=(Amount*R*T)/(100+R*T)
=(10028*100)/(100+12*(9/12))
=Rs.9200


8.A man buys a watch for Rs1950 in cash and sells it
for Rs2200 at a credit of 1yrThe rate of interest is 10%
per annum Then he gains / loose---------------amount?
Sol:
If he sells it for Rs2200 at a credit of 1yr then the
present worth of that amount is
Presentworth = (Amount*100) /(100+(R*T))
=(2200*100)/(100+10*1)
=2000/-
So he gains 2000-1950=Rs.50


9.A owes Rs.1573 payable 11/2yrs hence ,Aso B owes A Rs.144450
payabe 6months hence If they want to settle the accunt forth
with ,keeping 14% as the rate of interest ,then who should pay
& how much?
Sol:
B->A-----------Amount=Rs.1573
Rate=14%
Time=3/2
Presentworth = (Amount*100) /(100+(R*T))
=(1573*100)/(100+(3/2)*14)
=Rs1300
A->B------------Amount=Rs.1444.50
Rate=14%
Time=1/2
Presentworth = (Amount*100) /(100+(R*T))
=(1444.50*100)/(100+(1/2)*14)
=Rs1350
so B should pay Rs.1350


10.If Rs20 is the TrueDiscount on Rs260due after a certain time.
What wil be trueDiscount on same sum due after ½ of former time,
the rate of interest beng the same?
Sol:

Simple Interest on (260-20) for a gven time=Rs20
Simple Interest on (260-20) for a half time=Rs20*(1/2)
=Rs10
True Discount on Rs.250=Rs10
True Discount on Rs.260=Rs10*260/250
=Rs10.40


11.A has to pay Rs.220 to B after 1yr.B asks A to pay Rs.110
in cash and defer the payment of Rs,.110 for 2 yrs.A agrees
to it.If the rate of interest be 10% per annum in this mode
of payment:...
............ sol:
A has to pay =Present worth of Rs220 due 1yr hence
=Rs.(220*100)/(100+(10*1))
=Rs.200 A actually pays=Rs.110+Presentworth of Rs110 due 2 yrs hence
=[110+ (Amount*100) /(100+(R*T))]
=[110+((110*100)/(100+(10*2)))]
=Rs192.66
So A gains Rs(200-192.66)=Rs.7.34 





COMPLEX PROBLEMS

1.If Rs.10 be allowed as truedscount on a bill of Rs.110 at the
end of a certaimn time ,then discount allowed on the same sum
due at the end of double the time..
sol:
Amount=Rs110
TrueDiscount=Rs10
Present worth=Amount-TrueDiscount
=Rs110-10
=Rs.100
SI on Rs.100 for a certain time =Rs.10
SI on Rs.100 for doube the time =Rs.20
TrueDiscount on (100+20)=120-100
=Rs20
TrueDiscount on Rs.110 =(110*20)/120
=Rs18.33


2.A man wants to se hs scooter .There are two offers one at
Rs12000 cash and other at acredit of Rs12880 to be paid
after 8 months ,money being at 18% per annum which is better offer?
Sol:
offer1=Rs12000
offer2 Present worth= (Amount*100) /(100+(R*T))
=Rs.12880*100/(100+(18*(8/12))
=Rs11,500
The first offer is better as if he gains 500/-
if he sells at Rs.12000 is better


3.Goods were bought for Rs.600 and sold the same day for
Rs.688.80 at a credit of 9 months and thus gaining 2%
rate of interest per annum is
sol:
MethodI:
Amount=Rs.688.50
Time=9.12
Gaining 2%----------->Presentworth=102%of 600=(102*600)/100
=Rs.612
Presentworth=(Amount*100)/(100+(T*R))
612=(688.80*100)/(100+((9/2)*R))
=>R=162/3%
MethodII:
TrueDiscount=688.50-612
=Rs.76.50
Rate=TrueDiscount*100/(P.W*T)
=(76.50*100)/(612*(9/12))
=162/3%


4.The present worth of Rs.1404 due in 2 equal halfyearly instalments
at 8% per annu8m S.I is
SOl:
Presentworth=(Amount*100)/(100+(T*R))
PresentWorth=Presentworth of Rs.702 6 months hence +
Presentworth of Rs.702 1yr hence
= (702*100)/(100+(1/2*8)) + (702t*100)/(100+(1*8))
=Rs.1325

Fundamentals of Aptitude(RACES AND GAMES OF SKILL)

RACES AND GAMES OF SKILL

RACES AND GAMES OF SKILL

Races :- A contest of speed in running ,riding,driving,sailing or 
         rowing is called a Race



Race Course :-The ground or path on which contests are made is called
              a race course



STARTING POINT :-The point from which a race begins is called starting point.



Winning point or goal:-The point set to bound a race is called a winning point.



Dead Heat Race:-If all the persons contesting a race reach the goal exactly 
                 at the same time,then the race is called a dead heat race.



Start:-suppose A and B are two contestants in a race .If before the start of
         the race,A is at the satrtint point and B is ahead of A by 12 metres.
         Then we say that "A gives B a start 12 metres.



->To cover a race of 100metres in this case,A will have to cover 100m while B 
  will have to cover 88m=(100-12) 



->In a100m race 'A can give B 12m' or 'A can give B a start of 12m' or 'A beats 
  B by 12m'means that while A runs 100m B runs 88m.


GAMES:- A game of 100m,means that the person among the contestants who 
          scores 100 points first is the winner.



If A scores 100 points while B scores only 80 points then we say that 'A can give 
B 20 points'.

RACES AND GAMES OF SKILL


PROBLEMS :- 1) In a 1 km race,A beats B by 28 m or 7sec.Find A's time over the course? Sol: B covers 28 m in 7sec so,B's time over the course = 7/28 *1000 =250 sec A's time over the course =250-7 =243 sec = 4min ,3 sec. 

2) A runs 1 3/4 times as fast as B.If A gives B a start of 84 m,how far must be winning post be so that A and B might reach it at the same time ? Sol: Ratio of rates of A:B =7/4 :1 =7 :4 In a game of 7 m A gains 3m over B. 3m are gained by A in a race of 7m 84 m are gained by A in a race of 7*84/3 =196 m Winning post must be 196m away from the starting point. 

3)A can run 1km in 3 min ,10sec and B can cover same distance in 3 min 20 sec. By what distance A beat B? Sol: clearly A beats B by 10 sec. Distance covered by B in 10 sec =1000/200 *10 =50 m A beats B by 50 metres. 

4) In a 100m race,A runs at 8km per hour.If A gives b a start of 4 m and still beats him by 15 sec,what is the speed of B? Sol: 8000 m -------60*60 sec 100m ------- 60*60*1000/8000 =45 sec. Time taken by A to cover 100m =45 sec. B covers 100-4 m =96 min 45 sec =60 sec B's speed =96 *60*60/60*1000 =5.76 km/hr 

5) A and B take part in 100m race .A runs at 5km per hour. A gives B a start of 8 m and still beats him by 8 sec. What is the speed of B? Sol : A'speed = 5km/hr =5*5/18 =25/18 m/s Time taken by A to cover 100m =100*18/25 =7.2 sec Time taken by A to cover 92m = 72+8 =80 sec B's speed =92*18/80*5 = 4.14 kmph. 

6) A runs 1 2/3 times as fast as B.If A gives B a start of 80 m, how far must the winning post be so that A and B might reach if at the same time? Sol: Ratio of the speed of A and B =5/3 :1 Thus in a race of 5m ,A gains 2m over B 2m are gained by in a race of 5m 80 m will be gained by A in a race of 5/2* 80 =200 m 

7) A,B and C are three contestans in a kmrace .If A can give B a start of 40 m and A can give C a start of 64 m,how amny metres start can b give C? Sol: A covers 100m,B covers (1000-40) =960 m C covers 1000-64 m or 936m when B covers 960 m,C covers 936 m when B covers 1000 m,C covers 936*1000/960 m =975 m B can give C a start of 1000-975 or 25 m. 

8) In a 100m race,A covers the distance in 36 sec and B in 45 sec.In this race A beats B by? Sol: Distance covered by B in 9secs =100*9/45 =20 m A beats B by 20m 

9) In a 200 m race A beats B by 35m or 7 sec.What is the A's time over the course? Sol: B runs 35 m in 7sec. B covers 200m in =7*200/35 = 40 sec. B's time over the course =40 sec A's time over the course =40-7 =33 sec.

10) In a 300 m race A beats B by 22.5 m or 6 sec.What is the B's time over the course? Sol: B runs 22.5 m in 6sec. B runs 300m in =6*300*2/45 =80 sec. B's time over the course =80 sec.

11) A can run 22.5 m while B runs 25 m.In a kilometre race B beats A by? Sol: B runs 25 m ,A runs 45/2 m B runs 1000 A runs = 1000*45/2*25 =900m B beats A by 100m

12)In a 500 m race, the ratio of the speeds of two contestants A and B is 3:4.A has a start of 140 m.Then ,A win by B? Sol: The speeds of A and B =3:4 To reach the winning post A will have to cover a distance of 500-140 m i.e 360 m while A covers 3m ,b covers 4m A covers 360m B covers 4/3*360=480m Thus when A reaches the winning post, b covers 480m and therefore remains 20m behind. A wins by 20m.

13) In a 100m race,A can beat B by 25 m and B can beat c by 4m.In the same race A can beat C by/ Sol: If A:B =100 :75 B :C=100 :96 then A :C =A/B*B/C=100/75* 100/96 = 100/72 A beats C by 100 -72 m=28 m

14) In a 100 race,A can give B 10 mand C 28 m.In the same race B can give C? Sol: A:B =100 :90 A :C=100 :72 B:C =B/A *A/C =90/100*100/72 =90/72 When B runs 90 C runs 72 when B runs 100 C runs =72*100/90 B beats C by 20m

15) In a 100m race ,A beats B by 10 m and C by 13 m.In the race of 180m. B will beat c by? Sol :A : B =100 :90 A/B =100/90 A/C =100/87 B/C =B/A *A/C =90 /87 When B runs 90m C runs 87 When B runs 180 m then C runs =87 *180/90 B beats C by 180-174= 6m

16) In a race of 200 m, A can beat B by 31 m and C by 18m.In arace of 350 m, C will beat B by? Sol : In a race of 200 m A :B =200:169 A :C =200 :182 C/B =C/A*A/B =182/200*200/169 When C runs 182 m B runs 169 when C runs 350 m B runs =350*169/182 =325m

17) In agame of 100 points A can give B 20 points and C 28 points then B can give C? In a game of 100 points. A :B =100 :80 A :C =100 :72 B/C=B/A*A/C =80/100*100/72 = 80/72 when B runs 80m C runs 72 when B runs 100m C runs =100*72/80 =90 B can give C 10 points in agame of 100.

18)At a game of billiards,A can give B 15 points in 60 and A can give C 20 points in60.How amny points can B give C in a game of 90? Sol: A:B =60:45 A:C =60:40 B/C =B/A*A/C =45/60*60/40 =90/80 B can give C 10 points in agame of 90.

19) in agame of 80 points,A can give B 5 points and C 15 points.then how many points B can give C in agame of 60? Sol: A :B =80 :75 A :C =80:65 B/C=B/A*A/C = 75/80*80/65 =15/13 B:C =60:52

Fundamentals of Aptitude(Compound Interest)

Compound Interest:Sometimes it so happens that the borrower and
the lender agree to fix up a certain unit of time ,say yearly or half-yearly
or quarterly to settle the previous account.
In such cases ,the amount after the first unit of time becomes the principal
for the 2nd unit ,the amount after second unit becomes the principal for the
3rd unit and so on.
After a specified period ,the difference between the amount and the money
borrowed is called Compound Interest for that period.

Formulae:


Let principal=p,Rate=R% per annum Time=nyears

1.When interest is compounded Annually,
Amount=P[1+(R/100)]n
2.When interest is compounded Halfyearly,
Amount=P[1+((R/2)100)]2n
3.When interest is compounded Quaterly,
Amount=P[1+((R/4)100)]4n
4.When interest is compounded Annually,but time in fractions say 3 2/5 yrs
Amount=P[1+(R/100)]3[1+((2R/5)/100)]
5.When rates are different for different years R1%,R2%,R3% for 1st ,2nd ,
3rd yrs respectively
Amount=P[1+(R1/100)][1+(R2/100)][1+(R3/100)]
6.Present Worth of Rs.X due n years hence is given by
Present Worth=X/[1+(R/100)]n







SIMPLE PROBLEMS

1.Find CI on Rs.6250 at 16% per annum for 2yrs ,compounded annually.
Sol:
Rate R=16,n=2,Principle=Rs.6250
Method1:
Amount=P[1+(R/100)]n
=6250[1+(16/100)]2
=Rs.8410
C.I=Amount-P
=8410-6250
=Rs.2160
Method2:
Iyear------------------6250+1000
\\Interest for 1st yr on 6250
II yr---------------6250+1000+160
\\Interest for I1yr on 1000
C.I.=1000+1000+160
=Rs.2160


2.Find C.I on Rs.16000 at 20% per annum for 9 months compounded quaterly
Sol:
MethodI:
R=20%
12months------------------------20%
=> 3 months------------------------5%
For 9 months,there are '3' 3months
--------16000+800
--------16000+800+40
--------16000+800+40+10+2
=>Rs.2522
MethodII:
Amount=P[1+(R/100)]n
=16000[1+(5/100)]3
=Rs.18522
C.I=18522-16000
=Rs.2522





MEDIUM PROBLEMS

1.The difference between C.I and S.I. on a certain sum at 10% per annum
for 2 yrs is Rs.631.find the sum
Sol:
MethodI:
NOTE:
a) For 2 yrs -------->sum=(1002D/R2)
b) For 3 yrs -------->sum=(1003D/R2(300+R))
Sum=1002*631/102
=Rs.63100
MethodII:
Let the sum be Rs.X,Then
C.I.=X[1+(10/100)]2-X
S.I=(X*10*2)/100
=X/5
C.I-S.I.=21X/100-X/5
=X/100
X/100=631
X=Rs.63100


2.If C.I on a certain sum for 2 yrs at 12% per annum is Rs.1590.
What would be S.I?
sol:
C.I.=Amount-Principle
Let P be X
C.I=X[1+(12/100)]2-X
=>784X/625-X=1590
=>X=Rs.6250
S.I=(6250*12*2)/100=Rs.1500


3.A sum of money amounts to Rs.6690 after 3 yrs and to Rs.10035b
after 6 yrs on C.I .find the sum
sol:
For 3 yrs,
Amount=P[1+(R/100)]3=6690-----------------------(1)
For 6 yrs,
Amount=P[1+(R/100)]6=10035----------------------(2)
(1)/(2)------------[1+(R/100)]3=10035/6690
=3/2
[1+(R/100)]3=3/2-----------------(3)
substitue (3) in (1)
p*(3/2)=6690
=>p=Rs.4460
sum=Rs.4460


4.A sum of money doubles itself at C.I in 15yrs.In how many yrs will it
become 8 times?
sol:
Compound Interest for 15yrs p[1+(R/100)]15
p[1+(R/100)]15=2P
=>p[1+(R/100)]n=8P
=>[1+(R/100)]n=8
=>[1+(R/100)]n=23
=>[1+(R/100)]n=[1+(R/100)]15*3
since [1+(R/100)] =2
n=45yrs


5.The amount of Rs.7500 at C.I at 4% per annum for 2yrs is
sol:
Iyear------------------7500+300(300------Interest on 7500)
IIyear ----------------7500+300+12(12------------4% interest on 300)
Amount=7500+300+300+12
=Rs.8112


6.The difference between C.I and S.I on a sum of money for 2 yrs at
121/2% per annum is Rs.150.the sum is
sol:
Sum=1002D/R2=( 1002*150) /(25/2)2=Rs.9600


7.If the S.I on sum of money at 15% per annum for 3yrs is Rs.1200,
the C.I on the same sum for the same period at same rate is---------
sol:
S.I=1200
P*T*R/100=1200
P*3*5/100=1200
=>P=Rs.8000
C.I for Rs.8000 at 5% for 3 yrs is-------------8000+400
-------------8000+400+20
-------------8000+400+20+20+1
C.I =400+400+20+400+20+20+1
=Rs.1261





COMPLEX PROBLEMS

1.A certain sum amounts to Rs.7350 in 2 yrs and to Rs.8575
in 3 yrs.Find the sum and rate%?
sol:
S.I. on Rs.7350 for 1yr =Rs.(8575-7350)
=Rs.1225
S.I. on Rs.7350 for 2yrs=Rs.2*1225
=Rs.2450
PTR/100=2450
=>P*2*R/100=2450
Since S.I. on Rs.7350 for 1yr =Rs.(8575-7350)
=Rs.1225
Rate R=100*1225/(7350*10
=16 2/3%
Since it is C.I ,Let sum be Rs.X
Then X[1+(R/100)]2=7350
=>X[1+(50/100)]2=7350
=>X=7350*(36/49)
Sum=Rs.5400


2.If the difference between C.I compounded halfyearly and S.I on
a sum at 10% per annum for one yr is Rs.25 the sum is
sol:
p[1+((R/2)/100)]2n-PTR/100=25
P[1+((10/2)/100)]2n-P*1*10/100=25
=>P=Rs.400


3.A man borrowed Rs.800 at 10 % per annum S.I and immediately lent
the whole sum at 10% per annum C.I What does he gain at the end of 2yrs?
sol:
C.I.=Rs.[800[1+(10/100)2]-800]=Rs.168
S.I=Rs.[800*10*2/100]=Rs.160
Gain=C.I-S.I=Rs(168-160)
=Rs.8


4.On what sum of money will be S.I for 3 yrs at 8% per annum be half
of C.I on Rs.400 for 2 yrs at 10% per annum?
sol:
C.I on Rs.400 for 2yrs at 10%=Rs.[400*[1+(10/100)]2-400]
=Rs.84
Required S.I =1/2*84=42/-
New S.I=Rs.42,Time=3yrs Rate=8%
Sum=Rs.[100*42/(3*8)]
=Rs.175


5.A sum of money placed at C.I doubles itself in 5yrs .It will amount to
8 times itself in-------------
sol:
p[1+(R/100)]5=2P
=>[1+(R/100)]5=2
To become 8 times =>8P
p[1+(R/100)]5=2^3P
=[1+(R/100)]^(5*3)
=[1+(R/100)]^15
n=15years 

Fundamentals of Aptitude(ALLIGATION OR MIXTURES)

ALLIGATION OR MIXTURES



Important Facts and Formula:
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.




ALLIGATION OR MIXTURES

SOLVED PROBLEMS



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.

Fundamentals of Aptitude(BOATS AND STREAMS)

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

Formulas:

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







BOATS AND STREAMS

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

Fundamentals of Aptitude(Trains)

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

Fundamentals of Aptitude(Time and Distance)

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
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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
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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