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APTITUDE
Numbers
Introduction:
Natural Numbers:
All positive integers are natural numbers.Ex
1,2,3,4,8,......
There are infinite natural numbers and number 1 is the least
naturalnumber.Based on divisibilit there !ould be t!o tpes of
natural numbers. Theare
Prime and composite.
Prime Numbers:
A natural number larger than unit is a prime number if itdoes
not have other divisors ex"ept for itself and unit.#ote$%&nit i
e,1 is not a prime number.
Properties Of Prime Numbers:
%'The lo!est prime number is 2.%'2 is also the onl even prime
number.%'The lo!est odd prime number is 3.%'The remainder !hen a
prime number p'() s divided b * is 1 or).+o!ever,if a number on
being divided b * gives a remainder 1 or ) need not beprime.%'The
remainder of division of the suare of a prime number p'() divideb24
is 1.%'-or prime numbers p'3, p%1 is divided b 24.%'/f a and b are
an 2 odd primes then a%b is "omposite. Also a0bis "omposite.%'The
remainder of the division of the suare of a prime number
p'()divided b 12 is 1.
Process to Check A Number s Prime or not:
Tae the suare root of the number.ound of the suare root to the
next highest integer "all this number as.he" for divisibilit of the
number # b all prime numbers belo! . /fthere is no numbers belo!
the value of !hi"h divides # then the number!ill be prime.
Example 235 is prime or not67235 lies bet!een 1) or 1*.+en"e tae
the value of (1*.rime numbers less than 1* are 2,3,),9,11 and
13.
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235 is not divisible b an of these. +en"e !e "an "on"lude that
235is a prime number.
Composite Numbers:
The numbers !hi"h are not prime are no!n as "omposite
numbers.
Co-Primes:
T!o numbers a an b are said to be "o%primes,if their +..- is
1.Example :2,3;,:4,);,:9,5;,:8,11;.....la"e value or !hi"h is
divisible b 3.
Example 2.8*4325 is not divisible b 3,sin"e sum of its
digits80*04030205(32 !hi"h is not divisible b 3.
?ivisibilit b 4$%A number is divisible b 4,if the number formed
blastt!o digits is divisible b 4.
Example 1.852*48 is divisible b 4,sin"e the number formed b the
lastt!o digits is 48 divisible b 4.
Example 2.But 945282 is not divisible b 4,sin"e the number
formed bthe last t!o digits is 82 is not divisible b 4.
?ivisibilit b )$%A number divisible b ),if its unit@s digit is
either> or ).
Example 2>82>,)>34)?ivisibilit b *$%/f the number is
divisible b both 2 and 3.
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example 3)2)* is "learl divisible b 2sum of digits
(30)020)021,!hi"h is divisible b 3Thus the given number is
divisible b *.
?ivisibilit b 8$%A number is divisible b 8 if the last 3
digitsof the number are divisible b 8.
?ivisibilit b 11$%/f the differen"e of the sum of the digits in
theodd pla"es and the sum of the digitsin the even pla"es is ero
ordivisibleb 11.
Example 4832918:8090304; % :10208;(11 !hi"h is divisible b
11.
?ivisibilit b 12$%All numbers divisible b 3 and 4 are divisible
b12.
?ivisibilit b 9,11,13$%The differen"e of the number of its
thousandsand the remainder of its division b 1>>> is
divisible b 9,11,13.
!A"IC #O$%&'A(:
%':a0b;(a0b02ab%':a%b;(a0b%2ab%':a0b;%:a%b;(4ab%':a0b;0:a%b;(2:a0b;%'a%b(:a0b;:a%b;%':a%0b0";(a0b0"02:ab0b
"0"a;%'a0b(:a0b;:a0b%ab;%'a%b(:a%b;:a0b0ab;%'a0b0"%3a b
"(:a0b0";:a0b0"%ab%b "%"a;%'/f a0b0"(> then a0b0"(3a b "
DI)I"ION A'*O$IT+%
/f !e divide a number b another number ,then?ividend ( :?ivisor
= uotient; 0 emainder
%&'TIP'ICATION !, "+O$T C&T %(T+OD"
1.Cultipli"ation b distributive la!$
a;a=:b0";(a=b0a="b;a=:b%";(a=b%a="
Example
a;)*95)8=55555()*95)8=:1>>>>>%1;)*95)8=1>>>>>%)*95)8=1)*95)8>>>>>%)*95)8)*95)232>42
b;598=1840598=81*(598=:184081*;598=1>>>(598>>>
2.Cultipli"ation of a number b )n$%ut n eros to the right of
the
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multipli"and and divide the number so formed b 2n
Example
59)43*=*2)(59)43*=)4(59)43*>>>>D1*(*>5*49)>>.
P$O*$(""ION:
A su""ession of numbers formed and arranged in a definite
ordera""ordingto "ertain definite rule is "alled a progression.
1.Arithmeti" rogression$%/f ea"h term of a progression differs
from itspre"eding term b a "onstant.This "onstant differen"e is
"alled the "ommon differen"e of the A..The n th term of this A. is
Tn(a:n%1;0d.The sum of n terms of A. n(nD2F2a0:n%1;dG.
x/mportant esults$
a.10203040)......................(n:n01;D2.
b.120220320420)2......................(n:n01;:2n01;D*.".130230330430)3......................(n2:n01;2D4
2.Heometri" rogression$%A progression of numbers in !hi"h
everterm bears a "onstant ratio !ith ts pre"eding term.i.e a,a r,a
r2,a r3.............../n H. Tn(a r n%1um of n terms n(a:1%r
n;D1%r
Problems
1.implifa.8888088808808b.11552%9823%4)*
olution$
a.88888888885892b.11552%9823%4)*(11552%:982304)*;(11552%8295(3913
2.Ihat "ould be the maximum value of J in the follo!ing
euation6)J03902J8(1114
olution$ ) J
3 9 2 J 8 11 1 4200J0(11Caximum value of J
(11%2(5 :(>,(>;
3.implif$ a.)9534>)=5555 b.835498=*2)
olution$a. )9534>)=5555()9534>)=:1>>>>%1;
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)9534>)>>>>%)9534>)()95282)*)5)b.
835498=*2)(835498=)4(835498>>>>D1*()24*939)>.
4.Evaluate 313=3130289=289
olution$a0b(1D2::a0b;0:a%b;;1D2:3130289;
0:313%289;(1D2:*>> 02*
;K:3*>>>>0*9*;(18>338 ).Ihi"h of the follo!ing
is a prime number6 a.241 b.339 ".351
olution$a.2411*'7241.+en"e tae the value of (1*.rime numbers
less than 1* are 2,3,),9,11 and 13.241 is not divisible b an of
these. +en"e !e "an"on"lude that 241 is a prime number.b.
33915'7339.+en"e tae the value of (15.
rime numbers less than 1* are 2,3,),9,11,13 and 19.339 is not
divisible b an of these. +en"e !e "an "on"ludethat 339 is a prime
number.". 3512>'7351.+en"e tae the value of (2>.rime numbers
less than 1* are 2,3,),9,11,13,19 and 15.351 is divisible b 19.
+en"e !e "an "on"ludethat 351 is not a prime number.
*.-ind the unit@s digit n the produ"t 24*9 1)3 = 341926
olution$ &nit@s digit in the given produ"t(&nit@s digit
in 9 1)3 = 192#o! 9 4 gives unit digit 1
9 1)2 gives unit digit 19 1)3 gives 1=9(9.Also 192 gives 1+en"e
unit@s digit in the produ"t (9=1(9.
9.-ind the total number of prime fa"tors in 411 =9 ) =112
6 olution$ 411 9 ) 112( :2=2; 11 =9 ) =112( 222 =9 ) =112Total
number of prime fa"tors(220)02(25
8.Ihi"h of the follo!ing numbers s divisible b
36a.)4132*b.)5*9>13
olution$ a. um of digits in )4132*()040103020*(21 divisible b
3.b. um of digits in )5*9>13()050*090>0103(31 not divisible b
3.
5.Ihat least value must be assigned to = so that th number
159=)4*2 isdivisible b 56 olution$
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The digit in pla"e of x must be 2.
1>.Ihat least number must be added to 3>>> to obtain
a number exa"tldivisible b 156
olution$Ln dividing 3>>> b 15 !e get 19 as
remainderTherefore number to be added ( 15%19(2.
11.-ind the smallest number of * digits !hi"h is exa"tl
divisible b1116
olution$mallest number of * digits is 1>>>>>Ln
dividing 1>>>> b 111 !e get 1>> as
remainder#umber to be added (111%1>>(11.+en"e,reuired number
(1>>11.
12.Ln dividing 1)5*8 b a "ertain number the uotient is 85 and
theremainderis 39.-ind the divisor6
olution$?ivisor ( :?ividend%emainder;DJuotient(:1)5*8%39; D
85(195.
13.A number !hen divided b 342 gives a remainder 49.Ihen the
samenumberis divided b 15 !hat !ould be the remainder6
olution$#umber(342 M 0 49 ( 15 = 18 M 0 15 = 2 0 5(15 : 18M 0 2;
0 5.The given number !hen divided b 15 gives 18 M 0 2 as uotient
and 5 asremainder. 14.A number being su""essivel divided b
3,),8 leaves remainders 1,4,9
respe"tivel. -ind the respe"tive remainders if the order
ofdivisors are reversed6
olution$
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1*.+o! man numbers bet!een 11 and 5> are divisible b 96
olution$The reuired numbers are 14,21,28,...........,84This is
an A. !ith a(14,d(9.>>#"learl #(1.
21.The sum of all possible t!o digit numbers formed from
threedifferent one digit natural numbers !hen divided b the sum of
theoriginal three numbers is eual to6a.18 b.22 ".3* d. none
olution$x0;0:1>x0;0:1>0x;0:1>0;0:1>0x;0:1>0;(
22:x00;Therefore sum of all possible t!o digit numbers !hen divided
b sum ofone digit numbers gives 22.
22.The sum of three prime numbers is 1>>./f one of them
ex"eeds anotherb3* then one of the numbers is6
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a.9 b.25 ".41 d*9.
olution$x0:x03*;0(1>>2x0(*4Therefore must be even prime
!hi"h is 22x02(*4('x(31.Third prime number (x03*(3103*(*9.
23.A number !hen divided b the sum of ))) and 44) gives t!o
timestheir differen"e as uotient and 3> as remainder .The number
is6a.122> b.12)> ".22>3> d.22>>3>.
olution$#umber(:)))044);=:)))%44);=203>(:)))044);=2=11>03>(22>>>>03>(22>>3>.
24.The differen"e bet!een t!o numbers s 13*).Ihen the larger
number isdivided b the smaller one the uotient is * and the
remainder is 1).The smaller number is6a.24> b.29> ".25)
d.3*>
olution$ (' x(29>euired number is 29>.
2)./n doing a division of a uestion !ith ero remainder,a
"andidatetoo 12 as divisor instead of 21.The uotient obtained b him
!as 3).The "orre"t uotient is6a.> b.12 ".13 d.2>
olution$?ividend(12=3)(42>.#o! dividend (42> and divisor
(21.Therefore "orre"t uotient (42>D21(2>.
H.C.F AND L.C.M
#acts And #ormulae:
.+ihest Common #actor:/+.C.#0 or *reatest Common
%eaure/*.C.%0 : The +.C.# of t1o or more than t1o
numbers
is the reatest
number that divides each of them e2actl.
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There are t1o methods :
.#actori3ation method: (2press each one
of the iven numbers as the product of
prime factors.
The product of least po1ers of common primefactors ives
+C#.
(2ample : #ind +C# of 45 6 746869 ;
44 678684 6 95 ;
4684 694
"ol: The prime numbers iven common numbers are4;8;9
Therefore +C# is 44 6 8 694 .
4.Division %ethod : Divide the larer number b
smaller one. No1 divide the divisor b remainder.
$epeat the process
of dividin precedin number last obtained till 3ero is
obtained as
number. The last divisor is +C#.
(2ample: #ind +C# of 87; 7; 48
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"ol:
70 48/ 7
----------
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Then 'C% is product of hihest po1ers
of all factors.
4.Common division method.
Problems:
.The +C# of 4 numbers is and 'C% is [email protected]
of numbers is 99.find other.
"ol: Other number 6 5@7B99@@.
4.#ind larest number of diits divisible b4;8;
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7.#ind least number 1hich 1hen divided b 4=;48;78;=
leaves remainders ;@;4@;7.
"ol:
4=E5 48-@5 78-4@5 =-75
Therefore number 'C% of /4=;48;78;=0 - 57@
.484 can be e2pressed as prime as :
4 484 4 45
7 57 7 4
9
prime factor is 4 64 6 7 6 7 69
8.=@8B5< 1hen e2pressed in simple form is
=@805
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97 --------- 758 758
--------- =
----------"o; +C# is 97
Therefore
=@8B5< =@8B97B5
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9.Three numbers are in ratio :4:7 and +C# is
4.#ind numbers.
"ol:
'et the numbers be 2.
Three numbers are 2;42;72
Therefore
+C# is42072/
42 -----
2042/4 42
-------- =
-------------
+C# is 2 so; 2 is 4
Therefore numbers are 4;4;75.
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Co-primes of < are /;90 and /7;80
numbers/49 6 0; /49 6 90 49;
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=.+C# and 'C% of t1o numbers are
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4.The ma2imum number of students amon them == pens
and @= pencils can be distributed in such a 1a
that
each student ets same number of pens and pencils isG
"ol:
+C# of == and @=
@=0==/ @= ------------
@0@=/= @=
-------- =
--------- Therefore +C#@
7.The least number 1hich should be added to 4@9 sothat sum
is
divisible b 8;5;;7 G
"ol: 'C% of 8;5;;7 is 5=.
On dividin 4@9 b 5= 1e et 79 as remainder.
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Therefore number to added is 5= E 79 47.
Ans1er is 47.
.The least number 1hich is a perfect sHuare and isdivisible
b
each of numbers 5;4=;4 is G
"ol: 'C% of 5;4=;4 is 4=.
4 / 4646467684=
To mAke it a perfecd sHuare multipl b 7 6 8
Therefore 4= 6 7 6 875==
Ans1er i" 75=.
DECIMAL FRACTIOFS
.Decimal fractions: #ractionin 1hich denominataons re po1ers mf
= re decimalfracpions. B= =.; B == =.=
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4.Convertion of Decimal into fractiol:-e: =.48 J 48B== =B7Ka0 If
numeratmp and defominator contain same numLeroK dechma' places;then
1e remve decieal sin. Thus;
.
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4
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F ==0 76=.===8=F7F=.5F==F=.===77.5==7
8B48 F 48B8? >B8F [email protected] =.4=7
=.The leaHt4aeon the follo1in :-a. =.4 b;B=
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4 c. =.4 d. =
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44=B4 8 =.4444 =.= =.= J =.4 J =.44;-------Jﾧ"hnce =.= is least
/=.404 is least.
.'et ! =.-4F=.98?F>-7F=.9/?-8F>=.98F=.9
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the sum of first44= terms kf series ir B65FB569FB96
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ere B "tands for Bra"et, L for Lf, ? for ?ivis/on, C
forCultipli"atiLn, A for Addition and for ubtra"tion.
-irst of all the ba"ets must be removed
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stri"tl in theordEr :; , NO , FG.
After removilg the bra"ets, !e !ant use t:e fohlo!ine
operPtions$1.Lf 2. ?ivision 3. Cultipli"ation 4. Addition ).
ubtra"tion
%odulus of a real number:Codulus nf a real number is a
defined asQaQ ( a, if a'> or %a, if a R >
Problems:
1.:)>>4 D135; S *( 6
ol$ Expession ( )>>4D 135 S * ( 3* S * ( 3>
2.Uhat mathemati"al opapations should "ome at the pla"e of 6 in
theeuation $ :2 6 * S 12 D 4 0 2 ( 11; 6
ol$ 2 6 * ( 11 0 12 D 4 S 2( 11 0 3 S 2( 122 = * ( 12
3.: 8 D 88; = 8888>88 ( 6
ol $ :1D11; = 8888>88 ( 8>8>>8
4.+o! man 1D8@s are there in2*391D2 6
ol$ :391D2; D:1D8;( :9)D2; D:1D8; ( 3>>
)
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-ind the values of 102=3 01D3=4 01D4=)0
.................01D5=1> 6
o( FK %1D3G 0F 1D3 S VG 0 FV% 1D)G 0...........
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...0F1D5%1D1>G( F K S 1D1>G(284D1) ( 2D)
*.The value of 555 of 55)D555= 555 is$
ol$ F1>>>% 4D1>>>G=555 ( 555>>>%4(
55855*
9.Along a ard 22)m lLng, 2* tree3 are planted at eual distan"e,
onetree being at ea"h eld Df the ard. !hat is the distan"e bet!een
t!o"one"ut;ve trees 6
ol$ 2* trees have 2) gaps bet!een them.+en"e , reuireP distan"e
( 22)D 2) m( 5m
8./n a garden , the2e are 1> ro!s and 12 "olumns of iango
trees. thedistan"a bet!een the t!o trees is 2 m and a distan"e of
one meter is
left from all sides of the boundar of the lengtP of the garden
is $
ol$ Ea"h ro! "ontains 12 plants.,eaving 2 "orner plants, 1>
plants in bet!een have 1> = 2 meters and1 metdr on ea"h side is
left.length ( :2> 0 2; m ( 22m
5.Eight people are plannin@ to share euall the "ost of a rental
"ar,if one pWrson !ith dra!s from the arrAngement and the others
shareeuall t+e entire "ost of the "ar, thef the share of ea"h of
thereiaining perLns in"reased b6
ol$ Lriginal share of one person ( 1D8nes share28of o#e person (
1D9in"reasW ( 1%9 S 1D8 ( 1D)*reuired fra"tions ( 1D)*;D:1D8; (
1D9
1>.A pie"e of "loth "ost s 3). if the length of 4he pie"e
!ouldhave been 4m longer Pnd ea"h meter "ost e 1 less , the
"ost!ould have remained un"hangedX ho! long is the pie"e6
ol$
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ol$> ( s 8>>
12.A man spends 2D) of his salar on house rent, 3D1> of his
salaron food, and 1D8 of his salar on "onvean"e. if he has s
14>> left!ith him, find his expenditure on food and
"onvean"e6
ol$ art of the salar left ( 1%F2D) 03D1>01D5G( 1%
33D4>(9D4> of x ( 14>>
x( F14>>=4>GD9x( 8>>>Expenditure on food (
3D1>=8>>> (s 24>>Expenditure on "onvean"e(
1D8=8>>> (s 1>>>
Averages
#ormula:
1.Average(um of uantitiesD#umber of uantities.
2.uppose a man "overs a "ertain distan"e at x mphand an eual
distan"e at mph ,then the average speedduring the !hole Yourne is
:2xDx0; mph.
(2amples:1.-ind the averagE of all these
numbers.142,149,1)3,1*),1)9.
olution$142 149 1)3 1*) 1)9+Ere "onsider the least number i.e,
142"omparing !ith othars,142 149 1)3 1*) 1 ﾧ 90) 011 0231 ﾧ#o! add
)01Z02301) ( )2D) ( 1>.8#o! add 1>.8 to 142 !e get
1)2.8:Average of All these numbers;.Anr!er is 1)2.8
2.-ind tPe25average of all these numbers.4,1>,1*,22,28
olution$
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4,1>,1*,22,28As the difberen"e of number is *Than t+e average
of thEse numbars is "entral one i.e, 1*. Ans!er is1*.
3.-ind the aterage of all these numbers.4,1>,1*,22,28,34.
olution$+ere also differen"e is *.Then middle numbers 1*,22 tae
average of theset!o numbers 1*022D2(15Therefore the average of
these numbers is 15.Ans!er is 15.
4.The average mars of a mars of a student in 4 Examinationis
4>./f he got 8> mars in )th Exam then !hat ishis ne!
average.
olution$4=4>08>(24>
Then average means 24>D)(48.Ans!er is 48.
)./n a group the average in"ome of * men is )>> and thatof
) !omen is 28>, then !hat is average in"ome of the group.
olution$*=)>>0)=28>(44>>then average is
44>>D11(4>>.Another Cethod$ here "onsider for * men*
men [\] ea"h )>>.so )th !omen is 28>.then
)>>%28>(22>.then 22>=*D11(12>.therefore
12>028>(4>>.Ans!er is 4>>.
*.The average !eight of a "lass of 3> students is 4> gs if
thetea"her !eight is in"luded then average in"reases b 2 gs
thenfind the !eight of the tea"her6
olution$3> students average !eight is 4> gs.o,!hen tea"her
!eight is added it in"reases b 2 gsso total 31 persons ,therefore
31=2(*2.#o! add the average !eight of all student to it!e get
tea"hers !eight i.e, *204>(1>2 gs.
Ans!er is 1>2 gs.
9.The average age of Cr and Crs harma 4 ears ago is 28 ears ./f
the present average age of Cr and Crs harma and their sonis 22
ears. Ihat is the age of their son.
olution$4 ears ago their average age is 28 ears.o their present
average age is 32 ears.32 ears for Cr and Crs harma then 32=2(*4
ears.
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Then present age in"luding their son is 22 ears.o 22=3 (**
ears.Therefore son age !ill be **%*4 ( 2 ears.Ans!er is 2 ears.
8.The average pri"e of 1> boos is in"reased b 19 upees
!henone of them !hose value is s.4>> is repla"ed b a ne!
boo.Ihat is the pri"e of ne! boo6
olution$1> boos Average in"reases b 19 upeesso 1>=19(
19>.so the ne! boo "ost is more and b adding its "ost
averagein"rease,therefore the "ost of ne! boo is
4>>019>()9>s.Ans!er is )9> s.
5.The average mars of girls in a "lass is *2.). The average
marsof 4 girls among them is *>.The average mars of remaining
girlsis *3,then !hat is the number of girls in the "lass6
olution$Total number of girls be x04.Average mars of 4 girls is
*>.therefore *2.)%*>(2.)then 4=2.) (1>.the average of
remaining girls is *3here >.) differen"e therefore
>.)=x(1>:sin"e !e got from 4 girls;:this is taen be"o both
should be eual;x(1>D>.)x(2>.This "lear sas that remaining
are 2> girlstherefore total is x04(2>04(24 girlsAns!er is 24
girls.
1>.-ind the average of first )> natural numbers.
olution$um of the #atural #umbers is n:n01;D2therefore for )>
#atural numbers )>=)1D2(99).the average is 99)D)>(1).)Ans!er
is 1).) .
11.The average of the first nine prime number is6
olution$rime numbers are 2,3,),9,11,13,19,15,23therefore
2030)09011013019015023(1>>
then the average 1>>D5( 11 1D5.Ans!er is 11 1D5.
12.The average of 2,9,* and x is ) and the average of and
theaverage of 18,1,*,x and is 1> .!hat is the value of 6
olution$2090*0xD4()('1)0x(2>
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('x().18010*0x0D)(1>('2)0)0()>('(2>.
13.The average of a non%ero number and its suare is ) times
thenumber.The number is
olution$The number be xthen
x0x2D2()x('x2%5x(>('x:x%5;(>therefore x(> or x(5.The
number is 5.
14.#ine persons !ent to a hotel for taing their meals . Eight
ofthem spent s.12 ea"h on their meals and the ninth spent s.8
thenthe average expenditure of all the nine. Ihat !as the total
monespent b them6
olution$The average expenditure be x.then
8=120:x08;(5x('5*0x08(5x.('8x(1>4('x(13Total mone spent
(5x('5=13(119Ans!er is s.119
1).The average !eight of A.B. is 4) Mgs./f the average !eight
ofA and B be 4> Mgs and that of Band be 43 Mgs. -ind the !eight
of B6
olution$The !eight of A,B,are 4)=3(13) Mgs.The !eight of A,B are
4>=2(8> Mgs.The !eight of B, are 43=2(8* Mgs.To get the
Ieight of B.:A0B;0:B0;%:A0B0;(8>08*%13)B(31 gs.Ans!er is 31
Mgs.
1*.The sum of three "onse"utive odd number is 48 more than the
averageof these number .Ihat is the first of these numbers6
olution$
let the three "onse"utive odd numbers are x, x02, x04.B adding
them !e get x0x020x04(3x0*.Then
3x0*%:3x0*;D3(38:given;('2:3x0*;(38=3.('*x012(114('*x(1>2('x(19.Ans!er
is 19.
19.A famil "onsists of grandparents,parents and three
grand"hildren.
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The average age of the grandparents is *9 ears,that of parents
is 3)ears and that of the grand "hildren is * ears . Ihat is the
averageage of the famil6
olution$grandparents age is *9=2(134parents age is
3)=2(9>grand"hildren age is *=3(18therefore age of famil is
13409>018(222average is 222D9(31 )D9 ears.Ans!er is 31 )D9
ears. 18.A librar has an average of )1> visitors on undas
and 24> onother das .The average number of visitors per da in a
month 3>das beginning !ith a unda is6
olution$+ere spe"ified that month starts !ith undaso, in a month
there are ) undas.Therefore remaining das !ill be 2) das.
)1>=)024>=2)(2))>0*>>>(8))> visitors.The
average visitors are 8))>D3>(28).Ans!er is 28).
15.The average age of a "lass of 35 students is 1) ears ./f the
age of the tea"her be in"luded ,then averagein"reases b 3 months.
-ind the age of the tea"her.
olution$ Total age for 35 persons is 35=1)()8) ears.#o! 4>
persons is 4>= *1D4(*1> ears:sin"e 1) ears 3 months(1)
3D12(*1D4;Age of the tea"her (*1>%)8) ears
('2) ears.Ans!er is 2) ears.
2>.The average !eight of a 1> oarsmen in a boat is
in"reasesb 1.8 Mgs .Ihen one of the "re! ,!ho !eighs )3 Mgs
isrepla"ed b ne! man. -ind the !eight of the ne! man.
olution$ Ieight of 1> oars men is in"reases b 1.8 Mgsso,
1>=1.8(18 Mgstherefore )3018(91 Mgs !ill be the !eight of the
man.Ans!er is 91 Mgs.
21.A bats man maes a s"ore of 89 runs in the 19th inningand thus
in"reases his average b 3. -ind the average
after 19th inning.
olution$ Average after 19 th inning (xthen for 1*th inning is
x%3.Therefore 1*:x%3;089 (19x('x(89%48('x(35.Ans!er is 35.
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22.The average age of a "lass is 1).8 ears .The average ageof
bos in the "lass is 1*.4 ears !hile that of the girlsis 1).4 ears
.Ihat is the ratio of bos to girls in the "lass.
olution$ atio be $1 then=1*.4 0 1=1).4 (
:01;=1).8(':1*.4%1).8;(1).8%1).4('(>.4D>.*('(2D3therefore
2D3$1('2$3Ans!er is 2$3
23./n a "ri"et eleven ,the average of eleven plaers is28 ears
.Lut of these ,the average ages of three groupsof plaers ea"h are
2) ears,28 ears, and 3> earsrespe"tivel. /f in these groups ,the
"aptain and theoungest plaer are not in"luded and the "aptain
iseleven ears older than the oungest plaers ,!hat is the age of the
"aptain6
olution$ let the age of oungest plaer be xthen ,age of the
"aptain (:x011;therefore 3=2) 0 3=28 0 3=3> 0 x 0
x011(11=28('9)08405>02x011(3>8('2x(48('x(24.Therefore age of
the "aptain (:x011;( 24011( 3) ears.Ans!er is 3) ears. 24.The
average age of the bos in the "lass is t!i"ethe number of girls in
the "lass ./f the ratio ofbos and girls in the "lass of 3* be )$1,
!hat isthe total of the age :in ears; of the bos in the "lass6
olution$ #umber of bos(3*=)D*(3>#umber of girls (*Average age
of bos (2=*(12 earsTotal age of the bos(3>=12(3*> earsAns!er
is 3*> ears.
2).-ive ears ago, the average age of and J !as1) ears ,average
age of ,J, and toda is2> ears,ho! old !ill be after 1>
ears6
olution$ Age of and J are 1)=2(3> earsresent age of and J is
3>0)=2(4> ears.Age of J and is 2>=3( *> ears.
,present age is *>%4>(2> earsAfter 1> ears
(2>01>(3> ears.Ans!er is 3> ears.
2*.The average !eight of 3 men A,B and is 84 Mgs.Another man ?
Yoins the group and the average no!be"omes 8> Mgs./f another man
E !hose !eight is3 Mgs more than that of ? ,repla"es A then
theaverage !eight B,,? and E be"omes 95 Mgs.
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The !eight of A is.
olution$Total !eight of A, B and is 84 = 3 R2>2 Mfs.
Total!eight Df A,B, and ?is 8>=4(32> MgsTherefore
?(32>%2)2(*8 Mgs.E !eight *803;(91 gsTotal !eight o- B,,? and E
( 95=4(31* Mgs:A0B00?;%:B00?0E;(22>%31* (4MgsA%E(4MgsA%9(4
gsA(9) Mgs ns!er is 9) gs
29.A team of 8 persofs Yoin in a shooting "ompetition.The best
mArsman s"ored 8) points./f he had s"ored52 points ,the average
s"ore for the3)team !ouldhave3)been 84.The team "obed !as.
olution$ +ere "onside2 the to4al s"ore be x.therefo2e
x052%8)D8(84
('x09(X92('x(*).Ans!er is **)
28.A man !hose bo!li.g a*eraee is 12.4,ta0es ) !i"etsfnr 2* runs
and there b de"rease his average b >.4.ThE number of !i"ets,taen
b3)him before his la4 mZt"h is$
Dlution2 #umber of !i"ets taen before last mat"h be x.therefore
12.4x2*Dx0)(12:sin"W average de"rease b >.4therefore
12.4%>.4(12;('12.4x02*12x02>('>.4x(34('x(34>D4('x().Ans!er
as 8).
25.The mean temperature of Conda to Iednesda !as 39 de@reesand
of Tuesda to Thursda !as 34 degrees ./d thetemperature on Thursda
!as >D)th that of CDnda.The tempera4ure on Thursda !as$
olution$The total temperaturE re"orded on Conda,IednesdZ !as
39=3(111.The total temperaturW re"orde? on Tuesda,Iedlesda,Thursda
!as3)34=3(1>2.
and also3)given that
Th(4@)C('C()D4Th:2(5C%Th(5)D4Th%Th(5Th:1D4;(5'Th(3* degrees.
3>. 1* "hiddpe# are to be divided inTo t!o gro&ps A and
Bof 1> and * "hildren. The averagW per"ent mars Dbtai#ed
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b the "hil^re. of group A is 9) an? the averaHe per"entmars of
all the 1* _hildren is 9*. Ihat is the ave2Zgeper"ent mars of
"hildren of groups B6
olutimn$ +ere given average of group A Znd !hole groups
.o,:9*=1*;%:9)=1>;D*('121*%9)>D*('4**D*(233D3(99 2D3Ans!er is
99 2D3.
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31.Lf t:e three numbers The firt is t!ibe the se"o.d andthe
se"ond is t!i"e the thir? .ThE average of the re"ipro"alof the
numbers is 9D92,the number are.
olution$
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olution$
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Anup age ( )%2 ( 3 ears. :x%*; D 18 ( 3 x%* (
)4
x(*> ears
5.C brother is 3 ears elder to me. C father !as 28 ears of
age !hen m sister !as born !hile m father !as 2* ears of age
!hen i !as born. /f m sister !as 4 ears of age !hen m brother
!as born,then !hat !as the age m father and mother
respe"tivel !hen m brother !as born.
a; 32 rs, 23rs b;32 rs, 25rs ";3) rs,25rs d;3)rs,33
rs
olution$ C brother !as born 3 ears before / !as born X
4ears after m sister !as born.
-ather@s age !hen brother !as born ( 2804 ( 32
ears. Cother@s age !hen brother !as born ( 2*%3 ( 23
ears.
S#RDS AND INDICES
"imple problems:
1.
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:5; :1>>>;9 D1>18 ( 6
:a; 1> :b; 1>> :" ; 1>>> :d;
1>>>>
olution$ :1>>>;9 D 1>18 :1>3;9 D :1>;18
( ' :1>;21 D :1>;18
(' :1>;21%18 (' :1>;3 (' 1>>>Ans $: " ;
:1>; The value of :8%2)%8%2*; is
:a; 9= 8%2) :b; 9=8%2* :" ; 8= 8%2* :d; #one
olution$ : 8%2) % 8%2* ; (' 8%2* :8%1 ; (' 9=
8%2*
Ans$ :b;
:11; 1 D :10 an%m ; 01D :10am%n; ( 6:a; > :b; 1D2 :" ; 1 :d;
an0m
olution$ 1D :10 anDam; 0 1D : 10 amDan; (' am D :am0 an ;
0 an D:am 0an ;
(' :am 0an ; D:am 0 an;(' 1
Ans$ : ";
:12; 1D:10xb%a0x"%a;01D:10xa%b0x"%b;01D:10xb%"0xa%";(6
:a; > :b; 1 : " ; xa%b%" :d; #one of the above
olution$ 1D :10xbDxa0x"Dxa; 0 1D:10xaDxb 0x"Dxb;
01D:10xbDx" 0xaDx";
(' xa D:xa 0xb0x"; 0 xbD:xa 0xb0x"; 0x"D:xa 0xb0x";
(':xa 0xb0x"; D:xa 0xb0x"; ('1 Ans$ :b;
:13; /f x(302 [c2 then the value of :[cx [\] 1D [cx; is F
[c(rootG
:a; 1 :b; 2 :" ; 2[c2 : d; 3[c3
olution$ :[cx%1D[cx;2 ( x0 1Dx%2
(' 302[c2 0 :1D302[c2 ;%2(' 302[c2 0 3%2[c2 %2
(' *%2 ( 4 :[cx%1D[cx;2 ( 4
(':[cx%1D[cx;2 ( 22:[cx%1D[cx; ( 2.
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Ans $ :b;
:14; :xbDx";b0"%a :x"Dxa;"0a%b :xaDxb;b0a%" ( 6
:a; xab" :b; 1 : "; xab0b"0"a :d; xa0b0"
olution$ Fxb%"Gb0"%a Fx"%aG"0a%b Fxa%bGa0b%"
('x:b%";:b0"%a; x:"%a;:"0a%b; x:a%b;:a0b%";('x:b2%"2%ab%a";
x:"2%a2%b"%ab; x:a2%b2%a"%b";
('x:b2%"2%ab%a"0"2%a2%b"%ab0a2%b2%a"%b"; ('
x>
('1 Ans$ :b; :1); /f 3x% ( 29 and 3x0 ( 243 then x is
eual to
:a; > :b; 2 :" ; 4 :d; *
olution$ 3x% ( 29 (' 3x% ( 33x%( 3
3x0 ( 243 (' 3x0 ( 3)x0 ( )
-rom above t!o euations x ( 4 , (1 Ans$ :" ;
:1*; /f ax ( b ( " and b2 ( a" then [\[\`euals
:a;xDx0 :b;xD2:x%; :";xD2:%x; :d;2xDx0
olution$
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:18; /f 2x ( 4 (8 and :1D2x 01D4 01D*; (24D9 then the
value of @@ is
:a; 9D1* :b; 9 D 32 :" ; 9D48 :d; 9D*4
olution$ 2x ( 4(8 2x ( 22 ( 23
x( 2 ( 3 Cultipl above euation !ith [\ 2[\` 2x ( 4(
* > :1D2x01D401D*; ( 24D9 (':1D*01D*01D*; (
24D9 (' 3 D * ( 24D9 Z ( ( 9D48 ^ Ans$ : ";
ERC$NTA%ES
EXAMPLE PROBLEMS:
(RAP'( P$OS'(%":
. (2rress the follo1in as a fraction.a0 85
L "O'&TION: / 85B==B48
b0 "O'&TION:
Q = B==B48c0 =.5
"O'&TION3 =.5B==5B===7B8==
d0 =.=
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5 UV 49B / /49B 6==0 598
.(valuate the follo1in:/ 4
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]ဠ 䀠 ] =7B5 ဥ of
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of alebra and 5= of eometr problems correctl he didnt
passthe
test because he ot less than 5= of the problems riht. +o1
man more Huestions he 1ould have needed to ans1er correctl to
et a
5= passin rade.
"O'&TION: 9= of = 9=B== 6 =
9 = of 7= = B == 6 7=
= Q 4 5= of 78 5= B == 678
L L 4 Q Q = "o correctl attempted Huestions 9 F4LF 4- Q Q
/ /
=. Yuestions to be ans1ered correctl for
5= radeL Q
5= of 98
5=B== 698
8. "o reHuired Huestions8-= 8
9 . If 8= of /2 E 0 7= of /2 F 0 then 1hat percent of 2 is
G "O'&TION:
8=B==/2-0 7=B==/2F0 UW /2-0 7B=/2F0
82-872F7 2
"o $eHuired percentae B26==
B 6==
48.
< . If the price of tea is increased b 4= ;find ho1 much
percent musta
householder reduce her consumption of tea so as not to
increasethee2penditureG
"O'&TION: $eduction in consumption $B
/==F$0 6==
4=B4= 6==
5 4B7
@.The population of a to1n is 95== . If it increases at the rate
of8
per annum ;1hat 1ill be the population 4 ears henceG Xhat
1as it 4ears aoG
"O'&TION: Population After 4 ears
95==>F8B==?4
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95== 6 4B4= 64B4=
@
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denomination. If 4 of 48p coins and 4 of 8=p coins are
removed;#ind the percentae of mone removed from the ba G
"O'&TION: Total mone /5== 6 48B== F4==
68=B==0
$s 98= 48p coins removed 4B== 65==
94 8=p coins removed 4B== 64==
4
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5=/==-20== 6== or ==-2 ====B5= 48B4 so
2 ==-48B4
98B4 79.8
ro&it an' Loss
Important #acts:
Cost Price: The pri"e at !hi"h an arti"le is pur"hased,is
"alled its "ost pri"e,abbreviated as ..
"ellin Price: The pri"e at !hi"h an arti"le is sold,is
"alled its selling pri"e,abbreviated as ..
Profit or *ain: /f .. /s greater than .. The selleris said
to have a profit or gain.
'oss:if .. /s less than .., the seller is said tohave in"urred a
loss.
#ormulae 1.Hain(:.%.;2. and sells it for s.28.*>-ind the
gain per"ent.
ol$ .(s 29.)> .(s 28.*> then
Hain(.%.(28.*>%29.)>(s 1.1>
HainW(:gain=1>>;D.W
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(:1.1>=1>>;D29.)>W(4W
2./f a radio is pur"hased for s 45> and sold for s
4*).)>-ind the lossW6
ol$ .(s 45> .(s 4*).)> %4*).)>(s 24.)>
>;D.W (:24.)>=1>>;D45>W()W
3.-ind . !hen .(s )*.2) and Hain(2>W
ol$ .(F:1>>0gainW;D1>>G=.
.(F:1>>02>;D1>>G)*.2)(s *9.)>
4.-ind . !hen .(s 8>.4>,loss()W
ol$ .(F:1>>%lossW;D1>>G=.
.(F:1>>%);D1>>G=8>.4>(s *8.34
).-ind . !hen .(s 4>.*>,gain(1*W6
ol$ .(:1>>=.;D:1>>0gainW;
.(:1>>=4>.*>;D:1>>01*;(s 3)
*.-ind . !hen .(s )1.9> ,loss(12W6
ol$ .(:1>>=.;D:1>>%lossW;
.(:1>>=)1.9>;D:1>>%12;(s )8.9)
9.A person in"urs )W loss b selling a !at"h for s 114>
. At!hat pri"e should the !at"h be sold to earn )W profit6
ol$ >%lossW;$:1st .;(:1>>0gainW;$:2nd .;
:1>>%);D114>(:1>>0);Dx x(:1>)=114>;D5)(s
12*>
8./f the "ost pri"e is 5*W of the selling pri"e,then !hat isthe
profit per"ent6
ol$ let .(s 1>> then .(s 5*profit(.%.(1>>%5*(s 4
profitW(:profit=.;D1>>W
(:4=5*;D1>>(4.19W
5.A dis"ount dealer professes to sell his goods at "ost
pri"e
but uses a !eight of 5*> gms for a Mg !eight .-ind his gain
W6
ol$ HainW(F:error=1>>;D:true value%error;GW
(F:4>=1>>;D1>>>%4>;GW(2)D*W
1>.A man sold t!o flats for s *9),5)8 ea"h .Ln one he
gains1*W !hile on the other he losses 1*W.+o! mu"h does he gain
orlose in the !hole transa"tion6
ol$ lossW(F"ommon loss or gainWD1>G2(:1*D1>;2(2.)*W
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+e sold the suit"ase for s 288> !ith 2>W profit on the
labelledpri"e .At !hat pri"e did he bu the suit"ase6
ol$ let the labelled pri"e be s xthen 12>W of x(288>
x(:288>=1>>;D12>(s 24>> .(8)W of
the 24>> :8)=24>>;D1>>(s 2>4>
).A tradesman gives 4W dis"ount on the mared pri"e and
givesarti"le free for buing ever 1) arti"les and thus gains 3)W.
Themared pri"e is above the "ost pri"e b
ol$ let the . of ea"h arti"le be s 1>> then . of 1*
arti"les(s :1>>=1*;(s 1*>> . of 1) arti"les
(1*>>=:13)D1>>;(s 21*> . of ea"h arti"le
(21*>D1)(s 144 /f . is s 5*, mared pri"e (s
1>> /f . is s 144,mared pri"e (:1>>D5*;=144(s
1)>>>
therefore mared pri"e()>W above .
*.B selling 33m of "loth ,one gains the selling pri"e of
11m.-indthe gain per"ent6
ol$ gain(. of 33m%. of 33m (11m of . . of 22m(. of
33m let . of ea"h meter be s 1,then . of 22m(s 22 . of
22m(s 33
gain(.%.(33%22(s 11 gainW(:gainD.;=1>>
(:11D22;=1>>()>W
9.The pri"e of a Ye!el, passing through three hands, rises on
the!hole b *)W.if the first and se"ond sellers earned 2>W and
2)Wprofit respe"tivel,find the per"entage profit earned b thethird
seller6
ol$ let the original pri"e of the Ye!el be s andlet the profit
earned b the third seller be xW
then :1>>0x;W of 12)W of (1*)W of
F:1>>0x;D1>>G=:12)D1>>;=:12>D1>>;=(:1*)D1>>;=
1>>0x(:1*)=1>>=1>>;D:12)=12>;
1>>0x(11> x(1>W
8.Ihen a produ"er allo!s 3*W "ommission on the retail pri"e of
hisprodu"t ,he earns a profit of 8.8W.!hat !ould be his
profitper"ent if the "ommission is redu"ed b 24W
ol$ let retail pri"e (s 1>> "ommission(s 3*
.(retail pri"e%"ommission(1>>%3*(s *4 But
profit(8.8W
.(:1>>=.;D:gain01>>;(:1>>=*4;D:1>>08.8;(s
1>>>D19
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ne! "ommission(s 12 ne! .(1>>%12(s 88
gain(88%:1>>>D19;(s 45*D19
gainW(gain=1>>D.
(:45*=19=1>>;:19=1>>>;
gainW(45.*W
5.Uias bought paper sheets for s 92>> and spent s
2>> ontransport. aing s *>>,he had 33> boxes
made,!hi"h he soldat s 28 ea"h. +is profit per"entage is
ol$ total investments(92>>02>>0*>>(s
8>>> total re"eipt(33>=28(s 524>
gain(.%. (total re"eipt%total investments
gain(524>%8>>>(s 124> gainW
(gain=1>>D.(124>=1>>D8>>>(1).)W
1>.A person earns 1)W on investment but loses 1>W on
anotherinvestment ./f the ratio of the t!o investments be 3$) ,!hat
is the
gain or loss on the t!o investments taen together6
ol$ let the investments be 3x and )x then total
investment(8x total re"eipt(11)W of 3x05>W of )x
(11)=3xD1>>05>=)xD1>>(9.5)x
loss(.%.(8x%9.5)x(>.>)x
lossW(.>)x=1>>D8x(>.*2)W
11.The profit earned b selling an arti"le for s 5>> is
double theloss in"urred !hen the same arti"le is sold for s 45>
.At !hatpri"e should the arti"le be sold to mae 2)W profit6
ol$ let . be s x 5>>%x(2:x%4)>;
3x(18>> x(s *>> .(s *>> , gain
reuired(2)W .(:1>>0gainW;=.D1>>
.(:1>>02);=*>>D1>>(s 9)>
12./f an arti"le is sold at )W gain instead of )W loss,the
sellergets s *.92 more. The . of the arti"le is6
ol$ let . be s x 1>)W of x%5)W of x(*.92
:1>)D1>>;=x%:5)D1>>;=x(*.92
xD1>(*.92 x(s *9.21.
$ATIO AND P$OPO$TION
Important #acts:
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.$atio : The ratio of t1o Hualities a and b in thesame units; is
the fraction aBb and 1e 1rite it as
a:b. In the ratio; a:b; 1e call a as the firstterm of antecedent
and b; the second term conseHuent.(2: The ratio 8:@ represents 8B@
1ith antecedent8
;conseHuent@
7$ule: The multiplication or division of each term of@ ratio b
the same non-3ero number does not affect
the ratio.
.Proportion: The eHualit of t1o ratios is
called proportion. If a:bc:d; 1e 1rite a:b::c:d and 1e
sathat a;b;c and d are in proportion. +ere a and b arecalled
e2tremes; 1hile b and c are called mean terms.
Product of meansproduct of e2tremes Thus; a:b::c:d
/b6c0/a6d0
8.#ourth proportional: If a:b::c:d; then d is called the
fourth proportional to a;b and c.
5.Third proportional: If a:b::b:c; then c is called third
proportional to a and b.
9.%ean proportional: %ean proportional bet1een a and
b is "Y$T/a6b0.
CO%PA$I"ION O# $ATIO":
Xe sa that /a:b0/c:d0 /aBb0/cBd0
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7.If aBbcBd; then /aFb0B/a-b0/cFd0B/c-d0/componend o and
dividend o0
)A$IATION:
.1e sa that 2 is directl proportional to ; if2k for some
constant k and 1e 1rite.
8.Xe sa that 2 is inversel proportional to ; if2k for some
constant and 1e 1rite.
5. R is inversel proportional to .If aBbcBdeBfBhk then
k/aFcFeF0B/bFdFfFh0 If aBb;a4Bb4;a7Bb7..............anBbn
are uneHual
fractions then the
ratio.aFa4Fa7F..........anB/bFb4Fb7F...............bn0lies bet1een
the lo1est K the hihest of the three
fractions.
"I%P'( P$O!'(%"
.If a:b 8:@ and b:c:9 #ind a:b:cG"ol: a:b8:@ and
b:c:96@B:@6B@@:57B@ a:b:c8:@:57B@4=:75:57
4.#ind the fourth proportion to ;@;4"ol: d is the fourth
proportion to a;b;c
a:bc:d :@4:2
2@64249
7.#ind third proportion to 5;75"ol: if a:bb:c then c is the
third proportion toa;b
5:7575:2 5275675 2
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.#ind mean proportion bet1een =.=< and =.<"ol: mean
proportion bet1een a and bsHuare root
of ab mean proportion sHuare root of
=.=
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.Three containers A;! and C are havin mi2tures of milk and
1ater in the ratio of :8 and 7:8 and 8:9respectivel. If the
capacities of the containers arein the ratio of all the three
containers are in theratio 8::8; find the ratio of milk to 1ater;
if the
mi2tures of all the three containers are
mi2ed toether.
"ol: Assume that there are 8==;== and 8== litersrespectivel in
the 7 containers.
Then ;1e have;
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let the incomes of A and ! be 2 and 72respectivel
let the e2pense of A and ! be 4 and respectivel
Amount of mone saved b A/income-
e2penditure02-4$s === Amount of mone saved b !72-$s
===
this can be even 1ritten as 52-4$s 4=== no1 solve and 7 to
et
2$s 8==therefore income of A268==$s 4===
income of !72768==$s 8==
.A sum of $s 54 is divided amon A;! and C. "uchthat times As
share share is eHual to 8 times !s
share and 9 times Cs share . Xhat is the share of CG"ol: times
of As share 8 times of !s share9
times of Cs share therefore ; the ratio of their share
B:B8:B9 'C% of ;8;9=
therefore; U:B8:B978:4
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artners(ip
Important #acts:
Partnership:Ihen t!o or more than t!o persons run a
businessYointl, the are "alled partners and the deal is no!n as
partnership.
$atio of Division of *ains:
1.Ihen the investments of all the partners are do the same time,
thegain or loss is distributed among the partners in the ratio of
theirinvestments.
uppose A and B invest s x and s respe"tivel for a ear in
abusiness, then at the end of the ear$
:A@s share of profit;$:B@s share of profit;(x$
2.Ihen investments are for different time periods, then
euivalent
"apitals are "al"ulated for a unit of time b
taing*>:"apital=numberof unats of time;. #o! gain or loss is
divided in the ratio of these"apitals.
uppose A invests s x for p months and B invests s for
months,then :A@s share of profit;$:B@s share of profit;(xp$
3,IoriYg afd sleeping partners$A partner !ho manages the busines
isno!n as !oring partner and the one !ho simpl i#vests the mone isa
sleeping partner.
#ormulae
1.Ihen investments od A and B are s x and s for*>a ear in
abusiness ,then*>at the end*>of the ear
:A@s share of profit;$:B@s share of profi4;(x$
2.Ihen A invests s x for p months aYd B invests s for J
months,then A@s s:are profit$Bs share of profit(xp$
"hort cuts:
1./n*>"ase of 3 A,B, investments then individual share is to
bE foundthen A(1*>>> , B(32,>>> ,
%>>,>>>
ol$ A$B$(1*$32$4> (2$4$)P
then individual share "an be easil logn.
2./f business mans A "gntributes -or ) months and B ontributes
for 5months then s+are of B in the total profit Lf s
2X,8>>> ,A ( s*>1)>>>,B (s 12>>>ol$
1)>>>=) 2 12>>>=5 *> 2) $ 3* for
3* arts(2*8>>>=:3*.*1; (s 1).1*Ba"
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DifficulT problems:
1. and J started a b&3iness investing s 8),>>> and
3 1),>>>respe"tivel. /n !hat ratio thW profit earned after
2 ears be dividedbet!een and J rWspe"tivel6
ol$ 8),>>>=2*1$ ),>>>=2 19=2 $
3=2(34$*
2.A,B and started a business b inve3ti.g s
1,2>,>>>,*1s 1,3),>>and 1,)>,>>>.-/nd
the share of ea"h ,out of an annual profit ofs )*,9>>6
ol2 atio of shares of A,B and (atio of their invetments
12>,>>>$13),>>>$1),>>>
(8$5$1> A@s share(s )*,9>>=:8D29;(s
1*,8>> B@s share (s )*,9>>=:5D2;(s
18,5>> @s share (s )*,9>>=:1>D29;(s
21,>>>
3.3 milman A,B, rented a pasture A graed his 4) "o!s for 12 dasB
graed his 3* "o!s for 1) das and " *> "o!s for 1> das./f
b@sshare of rent !as s )4> Ihat is the total rent6
ol$ 4)=12$3*=1)$*>=1> (5$5$1> 5 parts is
eual to s )4> then one part is eual to s *> total
rent(*>=28(s 1*8>
4.amu and Mrishna entered into a partnership !ith s
)>,>>> ands *>,>>>, after 4 months amu
invested s 2),>>> more !hile Mrishna!ithdra! s
2>,>>> . -ind the share of amu in the annual profit ofs
285,>>>.
ol$ amu $
Mrishna()>,>>>=409),>>>=8$*>,>>>=404>,>>>=8
(1>$9 amu@s annual profit(285>>>=:1>D19;(s
19>,>>>
).A,B, enter into partnership .A invests 3 times as mu"h as B
investsand B invests t!o third of !hat invests. At the end of the
ear ,theprofit earned is s **>>. !hat is the share of B6
ol$ let @s "apital (s x B@s "apital(s :2D3;=x
A@s "apital (3=:2D3;=x(s 2x ratio of their
"apitals(2x$:2D3;=x$x
(*x$2x3xB@s share (s **>>:2D11;(s 12>>
*.A,B and enter into a partnership b investing in the ratio of
3$2$4.After one ear ,B invests another s 2,9>,>>> and
,at the end of 2ears, also invests s 2,9>,>>>.At the
end of 3 ears ,profit are sharedin the ratio of 3$4$).-ind the
initial investment of ea"h6
ol$ /nitial investments of A,B," be s 3x, s 2x, s 4x then for
3ears
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:3x=3*;$F:2x=12;0:2x029>>>>;=24G$F:4x=24;0:4x029>>>>;=12G(3$4$)
1>8x$:92x0*4>,>>>;$:144x0324>>>>;(3$4$)
*2 1>8x$92x0*48>>>$1 ﾵ
4x0324>>>>(3$4$)
:>8x;D:92x*48>>>>;(3D4
43x(21*x01544>>>> 21*x(Z544>>>>
*2 (s 5>>> A@s initial
investmEnt(3x(3=5>,>>>(s 2,9>,>>> B_s
initial investment(2x(2=5>,>>>(s
1,8>,>>> @s initial
investment(4x(4=5>,>>>(s 1,*>,>>>
CHAIN R#LE1.?/ET LLT/L#$ T!o Juantities are said to be
dire"tlproportiLnal,iX on phe in"rease :or de"rease; of th one, the
other in"reases:orde"reases;to the same extent.
Ex$:i; ost is dire"tl proportional to the number of
arti"les.:Core
arti"les, Core "ost;.
:ii;Ior done is dire"tl proportional to the number of men
!oring onit. :Core men, more !or;.
2./#?/ET LLT/L#$ T!o Juantities are said to be
indire"tlproportional,if on the in"rease of the one , the other
de"reases to thesameextent and vi"e%versa.
Ex$:i; The time taen b a "ar "overing a "ertain distan"e is
inverselproportional to th speed of the "ar.:Core speed, less is
the time
taen to "over the distan"e;.
:ii;Time taen to finish a !or is inversel proportional to
thenumber
of persons !oring at it. :Core persons, less is the time
taen to finish a Yob;.
#LTE$ /n solving Juestions b "hain rule, !e "ompare ever item
!ith theterm to be found out.
SIMLE RO)LEMS1;/f 1) tos "ost s.234, !hat do 3) tos "ost 6
ol$
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ol$
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1 enfine of latter tpe "Dnsumes V unit in 1
hour.
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the !or in stipulated time. /f he had not engaged the additional
men,ho! man das behind s"hedule !ould it be finished6
ol$ 4> das% 3) das() das (':1>>=3);0:2>>=);
men "an finish the !or in 1 da. 4)>> men "an finish it
in 4)>>D1>>( 4) das This s ) das behind the
s"hedule.
4;12 men and 18 bos,!oring 9 K hors a da, "an do a pie"e if !or
in*> das. /f a man !ors eual to 2 bos, then ho! man bos !ill
bereuired to help 21 men to do t!i"e the !or in )> das, !oring5
hours a da6
ol$ 1man (2 bos 12men018bos(':12=2018;bos(42
bos let the reuired number of bos(x 21 men0x bos
('::21=2;0x; bos (':420x; bos less das, more
bos:/ndire"t proportion;
more hours per da, less bos:/ndire"t proportion;
das )>$*> hrsDda 5$1)D2 $$ 42$:420x;
!or 1$2 :)>=5=1=:420x;;(*>=:1)D2;=2=42 :420x;(
:*>=1)=42;D:)>=5;( 84
x(84%42( 42(42
42 das behind the s"hedule it !ill be finished.
Time an' +or,
Important #acts:
1./f A "an do a pie"e of !or in n das, then A@s 1 da !or(1Dn
2./f A@s 1 da@s !or(1Dn, then A "an finish the !or in n das.
Ex$ /f A "an do a pie"e of !or in 4 das,then A@s 1 da@s
!or(1D4. /f A@s 1 das !or(1D), then A "an finish the !or in )
das
3./f A is thri"e as good !orman as B,then$ atio of !or done bA
and B (3$1. atio of time taen b A and B to finish a !or(1$3
4.?efinition of Uariation$ The "hange in t!o different
variablesfollo! some definite rule. /t said that the t!o variables
vardire"tl or inversel./ts notation is D(, !here is "alled"onstant.
This variation is "alled dire"t variation. (. Thisvariation is
"alled inverse variation.
).ome airs of Uariables$
i;#umber of !orers and their !ages. /f the number of
!orers
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in"reases, their total !ages in"rease. /f the number of
das redu"ed, there !ill be less !or. /f the number of das
is
in"reased, there !ill be more !or. Therefore, here !e havedire"t
proportion or dire"t variation.
ii;#umber !orers and das reuired to do a "ertain !or is
anexample of inverse variation. /f more men are emploed, the!ill
reuire fe!er das and if there are less number of !orers,
more das are reuired.
iii;There is an inverse proportion bet!een the dail hours
of a!or and the das reuired. /f the number of hours is
in"reased,
less number of das are reuired and if the number of hours
isredu"ed, more das are reuired.
*.ome /mportant Tips$
Core Cen %
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8 girls "an do a !or in 84 das then 21 girls
%%%%%%%%%%%%%%%6 ans!er( :8=84D21;(32das. Therefore
1> !omen and ) girls "an a !or in 32das
4;Iorer A taes 8 hours to do a Yob. Iorer B taes 1>hours to
do thesame Yob. +o! long it tae both A X B, !oring together
butindependentl,to do the same Yob6
ol$ A@s one hour !or(1D8. B@s one hour !or(1D1>
:A0B;@s one hour !or(1D801D1> (5D4> Both A X B "an
finish the !or in 4>D5 das
);A "an finish a !or in 18 das and B "an do the same !or in half
thetime taen b A. Then, !oring together, !hat part of the same !or
the"an finish in a da6
ol$ Hiven that B alone "an "omplete the same !or in das(half
the
time taen b A(5das A@s one da !or(1D18 B@s one da
!or(1D5 :A0B;@s one da !or(1D1801D5(1D*
*;A is t!i"e as good a !orman as B and together the finish a
pie"e of!or in 18 das./n ho! man das !ill A alone finish the
!or.
ol$ if A taes x das to do a !or then B taes 2x das to do
the same !or ('1Dx01D2x(1D18 ('3D2x(1D18 ('x(29
das. +en"e, A alone "an finish the !or in 29 das.
9;A "an do a "ertain !or in 12 das. B is *>W more effi"ient
than A.+o!man das does B alone tae to do the same Yob6
ol$ atio of time taen b AXB(1*>$1>> (8$) uppose B
alone taes x das to do the Yob. Then, 8$)$$12$x ('
8x()=12 (' x(1)D2 das.
8;A "an do a pie"e of !or n 9 das of 5 hours ea"h and B alone
"an do
itin * das of 9 hours ea"h. +o! long !ill the tae to do it
!oringtogether8 2D) hours a da6
ol$ A "an "omplete the !or in :9=5;(*3 das B "an "omplete
the !or in :*=9;(42 das (' A@s one hour@s !or(1D*3 and
B@s one hour !or(1D42 :A0B;@s one hour
!or(1D*301D42()D12*
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Therefore, Both "an finish the !or in 12*D) hours.
#umber of das of 8 2D) hours ea"h(:12*=)D:)=42;;(3das
5;A taes t!i"e as mu"h time as B or thri"e as mu"h time to
finish apie"eof !or. Ioring together the "an finish the !or in 2
das. B "an dothe!or alone in 6 ol$ uppose A,B and tae x,xD2
and xD3 hours respe"tivel finishthe !or then
1Dx02Dx03Dx(1D2 (' *Dx(1D2 ('x(12 o, B taes *
hours to finish the !or.
1>; "an do V of a !or in 1> das, "an do 4>W of !or in
4> das and"an do 1D3 of !or in 13 das. Iho !ill "omplete the !or
first6
ol$ Ihole !or !ill be done b in 1>=4(4> das.Ihole !or !ill
be done b in :4>=1>>D4>;(1>> das. Ihole !or
!ill be done b in :13=3;(35 das Therefore, !ill "omplete the
!or first.
Comple2 Problems
1;A and B undertae to do a pie"e of !orfor s *>>.A alone
"an do it in* das !hile B alone "an do it in 8 das. Iith the help
of , the "anfinishit in 3 das, -ind the share of ea"h6
ol$ @s one da@s !or(:1D3;%:1D*01D8;(1D24 Therefore, A$B$(
atio of their one das!or(1D*$1D8$1D24(4$3$1 A@s share(s
:*>>=4D8;(3>>
B@s share( s :*>>=3D8;(22)@s
share(sF*>>%:3>>022);G(s 9)
2;A "an do a pie"e of !or in 8> das. +e !ors at it for 1>
das X thenB alonefinishes the remaining !or in 42 das. /n ho! mu"h
time !ill A and B,!oringtogether, finish the !or6
ol$ Ior done b A in 1> das(1>D8>(1D8 emaining
!or(:1%:1D8;;(9D8 #o!, !or !ill be done b B in 42 das.
Ihole !or !ill be done b B in :42=8D9;(48 das
Therefore, A@s one da@s !or(1D8>Bs one da@s !or(1D48:A0B;@s
one da@s !or(1D8>01D48(8D24>(1D3>
+en"e, both !ill finish the !or in 3> das.
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3;,J and are three tpists !ho !oring simultaneousl "an tpe
21*pagesin 4 hours /n one hour , "an tpe as man >ages more than
J as J "antpe morethan . ?r/ng a period of five hours, "an tpe as
man pagEs as "anduring seben hou2s. +o! man pages does ea"h of them
tpe per hour6ol$
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!hile E,a. taes ) hours to tpe ﾵ > pages.+o! mu"h time
!il< the tae, !oring togetPer on t!o diffdrent"omputdrsto tpe an
assignment of 11> pages6
ol$ #uCber o- pages tped b onald in one hour(32D*(1D3
#umber of pages tped b Elan in one hour(4>D)(8
9> #umber of pages tped b both in one
hour(::1*D3;08;(4>D3Time taen b both to tpe 11>
pages(11>=3D4>(8 hous=
);T!o !orYers A9>and B are9>engaged to Po a !or. A
!oring9>alone taes8 hoursmore to "omplete the Yob thAn if both
!oring together. /f B !oredalone,he !o)ld need 4 1D2 hours more to
"oCpete the Yob than the both !oringtogether. Ihat time !ould the
tae to do the !or toget+er.
ol$ 1D:x08;;01 ﾤ:x0:5D2;;;(1Dx(':1D:x08;;0:2D:2x05;;(1Dx
(' x:4x02);(:x08;:2x05;
' 2x2 (929> (' x2 ( 2*
(' x(*Thereford, A and B together "an do the !or in * das.
*;A and B "Al do a !or i#12 das, B and in 1) das, and A in
2>das./f A,B and !or together, the !ilh "omplete the !or i# ho!
mandas6
Dl$ :A0B;@s one da@s !orM(1D12 :B0;@s one da@s
!or"(1D1)
:A0;@s one9>da@s !or(1D2>
Adding !e get 2:A0B0;@s one da@s !or(1D1201ﾤ
1)01D2>(12D*>(1D)9> :A0B0;Ws one da9>!or(1D1>
9>o, A,B,and together "an "omplete the !or ;n 1 das.
9;A a.d B "an do a !or in 89>das, B and "an do t:e saee !L in
12das.A,B and together "an -inish it i# * das. A and together !ill
do itinho! man das'
ol$ :A0B0;@s one da@s !or(1D* :A0B;@s one da@s
!Dr(1D8 :B0;@s one da@s !gr(1D2
:A0;@s one da@s !or(2:A0B0;@s one da@s !or%::A0B;@sone
da 9> !mr0:B0;@r one da !or;
( :2D*;%:1D801D12;9> 9> (:1D3;% :)D24;
(3D24 9> 9> (1D8
L, A and together !ill do the !or in 8 das.
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8 A "a# do a "ertain !or in the same time in !hi"h B and
together "An do91it./f A and B togethWr "ould do it in 1> dAs
and a das, thenB alone"ould do it in ho! man daxs6
ol$ :A0B;@s one da@s !or(1D1> @s one dP@s
!or(1D)> :AB0;@s one da@s
!or(:1D1>01D)>;(*D)>(3D2)
Also, A@s one da@s !or(:B0;s one da@s !or -rom i and ii
,!e get $2=:A@s one da@s !or;(3D2)
(' A@s one da@s !or(3D)>B@s one das !or(:1D1>%3D)>;
(2D)> (1D2) B alone "ould "omplete the !or
in 2) das.
5; A is thri"e as good a !orman as B and therefore is able to
finish aYob
in *> das less than B. Ioring together, the "an do it in$
ol$ atio of times taen b A and B(1$3. /f differen"e of
time is 2 das , B taes 3 das /f differen"e of time is *>
das, B taes :3=*>D2;(5> das
o, A taes 3> das to do the !or(1D5> A@s one da@s
!or(1D3> B@s one da@s !or(1D5> :A0B;@s one da@s
!or(1D3>01D5>(4D5>(2D4)
Therefore, AXB together "an do the !or in 4)D2das
1>; A "an do a pie"e of !or in 8> das. +e !ors at it for
1> das and
thenB alone finishes the remaining !or in 42 das. /n ho! mu"h
time !illAXB,!oring together, finish the !or6
ol$ Ior ?one b A n 1> das (1>D8>(1D8 emaining !or
(1%1D8(9D8 #o! 9D8 !or is done b B in 42 das Ihole !or
!ill be done b B in 42=8D9( 48 das (' A@s one da@s !or
(1D8> and
B@s one da@s !or (1D48:A0B;@s one da@s !or ( 1D8>01D48 (
8D24> ( 1D3>+en"e both !ill finish the !or in 3> das.
11; 4) men "an "omplete a !or in 1* das. ix das after the
started!oring,so more men Yoined them. +o! man das !ill the no! tae
to "ompletetheremaining !or6
ol$ C1=?1DI1(C2=?2DI2('4)=*D:*D1*;(9)=xD:1%:*D1*;;(' x(* das
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12;A is )>W as effi"ient as B. does half the !or done b
AXBtogether. /f alone does the !or n 4> das, then A,B and
together "an do the !orin$
ol$ A@s one da@s !or$B@s one das !or(1)>$1>> (3$2 das
and left the Yob. /n ho! man das A alone "an finishthe
remaining !or6
ol$ B@s 1> da@s !or(1>D1)(2D3 emaining
!or(:1%:2D3;;(1D3 #o!, 1D18 !or is done b A in 1 da.
Therefore 1D3 !or is done b A in 18=:1D3;(* das.
14;A "an finish a !or in 24 das, B n 5 das and in 12 das.
BXstart the!or but are for"ed to leave after 3 das. The remaining
!or done b Ain$
ol$ :B0;@s one da@s !or(1D501D12(9D3* Ior done b B X in 3
das(3=9D3*(9D12
emaining !or(1%:9D12;()D12#o! , 1D24 !or is done b A in 1 da.o,
)D12 !or is done b A in 24=)D12(1> das
1); and "an do a pie"e of !or n 2> das and 12 das
respe"tivel. startedthe !or alone and then after 4 das Yoined him
till the "ompletion of!or.+o! long did the !or last6
ol$ !or done b in 4 das (4D2> (1D)emaining !or( 1%1D)
(4D):0;@s one da@s !or (1D2>01D12 (8D*>(2D1)
#o!, 2D1) !or is done b and in one da. o, 4D) !or !ill be
done b and in 1)D2=4D)(* das +en"e Total time taen (:*04; das
( 1> das
1*;A does 4D) of !or in 2> das. +e then "alls in B and the
togetherfinishthe remaining !or in 3 das. +o! long B alone !ould
tae to do the!hole !or6
ol$ Ihole !or is done b A in 2>=)D4(2) das
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#o!, :1%:4D);; i.e 1D) !or is done b AX B in das.
Ihole !or !ill be done b AX B in 3=)(1) das
('B@s one da@s !or( 1D1)%1D2)(4D1)>(2D9)o, B alone !ould do
the !or in 9)D2( 39 K das.
19; A and B "an do a pie"e of !or in 4) das and 4> das
respe"tivel.Thebegan to do the !or together but A leaves after some
das and then B"ompletedthe remaining !or n 23 das. The number of
das after !hi"h A left the!or !as
ol$ :A0B;@s one da@s !or(1D4)01D4>(19D3*>Ior done b B in
23 das(23D4>emaining !or(1%:23D4>;(19D4>#o!, 19D3*> !or
!as done b :A0B; in 1 da.19D4> !or !as done b :A0B; in
:1=:3*>D19;=:19D4>;;( 5 daso, A left after 5 das.
18;A "an do a pie"e of !or in 1> das, B in 1) das. The !or
for )
das.The rest of !or finished b in 2 das. /f the get s 1)>>
for the!hole!or, the dail !ages of B and are
ol$ art of !or done b A( )D1>(1D2art of !or done b B(1D3art
of !or done b (:1%:1D201D3;;(1D*A@s share$ B@s share$ @s
share(1D2$1D3$1D*( 3$2$1A@s share(:3D*;=1)>>(9)>B@s
share(:2D*;=1)>>()>>@s share(:1D*;=1)>>(2)>A@s
dail !ages(9)>D)(1)>D%B@s dail
!ages()>>D)(1>>D%@s dail !ages(2)>D2(12)D%?ail !ages
of BX ( 1>>012)(22)D%
15;A alone "an "omplete a !or in 1* das and B alone "an "omplete
thesamein 12 das. tarting !ith A, the !or on alternate das. The
total !or!illbe "ompleted in ho! man das6
:a; 12 das :b; 13 das :"; 13 )D9 das :d;13 das
ol$ :A0B;@s 2 das !or ( 1D1* 0 1D12 (9D48!or done in * pairs of
das (:9D48; = * ( 9D8
remaining !or ( 1% 9D8 ( 1D8!or done b A on 13th da (
1D1*remaining !or ( 1D8 S 1D1* ( 1D1*on 14th da, it is Bs turn1D12
!or is done b B in 1 da.
1D1* !or is done b B in da.Total time taen( 13 das.o,
Ans!er is$ ?
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2>;A,B and "an do a pie"e of !or in 2>,3> and *> das
respe"tivel./n ho!man das "an A do the !or if he is assisted b B
and on ever thirdda6
ol$ A@s t!o da@s !or(2D2>(1D1>:A0B0;@s one da@s
!or(1D2>01D3>01D*>(*D*>(1D1>Ior done in 3
das(:1D1>01D1>;(1D)#o!, 1D) !or is done in 3 dasTherefore,
Ihole !or !ill be done in :3=);(1) das.
21;even men "an "omplete a !or in 12 das. The started the !or
andafter) das, t!o men left. /n ho! man das !ill the !or be
"ompleted b theremaining men6
:A; ) :B; * : ; 9 :?; 8 :E; none
ol$ 9=12 men "omplete the !or in 1 da.Therefore, 1 man@s 1 da@s
!or(1D849 men@s ) das !or ( )D12
('remaining !or ( 1%)D12 ( 9D12) men@s 1 da@s !or ( )D84)D84 !or
is don b them in 1 da9D12 !or is done b them in ::84D); = :9D12;; (
45D) das ( 5
4D) das.Ans$ E
22;.12 men "omplete a !or in 5 das. After the have !ored for *
das,* moremen Yoined them. +o! man das !ill the tae to "omplete the
remaining
!or6
:a; 2 das :b; 3 das :"; 4 das :d; )das
ol $ 1 man@s 1 da !or ( 1D1>812 men@s * das !or ( *D5 (
2D3remaining !or ( 1 S 2D3 ( 1D318 men@s 1 das !or ( 18D1>8 (
1D*1D* !or is done b them in 1 datherefore, 1D3 !or is done b them
in *D3 ( 2 das.
Ans $ A
23;.A man, a !oman and a bo "an "omplete a Yob in 3,4 and 12
dasrespe"tivel.+o! man bos must assist 1 man and 1 !oman to
"omplete the Yob in V ofa da6
:a;. 1 :b;. 4 :";. 15 :d;. 41
ol $ :1 man 0 1 !oman;@s 1 das !or ( 1D301D4(9D12Ior done b 1
man and 1 !omen n 1D4
da(::9D12;=:1D4;;(9D48emaining !or( 1% 9D48( 41D48
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Ior done b 1 bo in V da( ::1D12;=:1D4;; (1D48Therefore,
#umber of bos reuired( ::41D48;=48;( 41 daso,Ans!er$ ?
24;12 men "an "omplete a pie"e of !or in 4 das, !hile 1) !omen
"an"ompletethe same !or in 4 das. * men start !oring on the Yob and
after!oring for2 das, all of them stopped !oring. +o! man !omen
should be put on theYobto "omplete the remaining !or, if it is to
be "ompleted in 3 das.
:A; 1) :B; 18 :; 22 :?; data inadeuate
ol$ one man@s one da@s !or( 1D48one !oman@s one da@s
!or(1D*>* men@s 2 da@s !or(::*D48;=2;( Vemaining !or(3D4#o!,
1D*> !or s done in 1 da b 1 !oman.o, !or !ill be done in 3 das b
:*>=:3D4;=:1D3;;( 1)
!oman. o, Ans!er$ A
2);T!elve "hildren tae sixteen das to "omplete a !or !hi"h "an
be"ompletedb 8 adults in 12 das. ixteen adults left and four
"hildren Yoinedthem. +o!man das !ill the tae to "omplete the
remaining !or6
:A; 3 :B; 4 : ; * :?; 8
ol$ one "hild@s one da !or( 1D152one adult@s one da@s !or(
1D5*!or done in 3 das(::1D5*;=1*=3;( 1D2emaining !or( 1 S K(1D2:*
adults0 4 "hildren;@s 1 da@s !or( *D5*04D152( 1D121D12 !or is done
b them in 1 da.K !or is done b them 12=:1D2;( * daso, Ans!er(
2*;ixteen men "an "omplete a !or in t!elve das. T!ent four
"hildren"an"omplete the same !or in 18 das. 12 men and 8 "hildren
started !oringandafter eight das three more "hildren Yoined them.
+o! man das !illthe no!tae to "omplete the remaining !or6
:A; 2 das :B; 4 das : ; * das :?; 8 das
ol$ one man@s one da@s !or( 1D152one "hild@s one da@s !or(
1D432Ior done in 8 das(8=:12D1520 8D432;(8=:1D1*01D)4;
(3)D)4emaining !or( 1 %3)D)4( 15D)4:12 men011 "hildren;@s 1 da@s
!or( 12D152 0 11D432 (
15D21*#o!, 15D21* !or is done b them in 1 da.
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Therefore, 15D)4 !or !ill be done b them
in::21*D15;=:15D)4;;( 4 das o,Ans!er$ B
29;T!ent%four men "an "omplete a !or in 1* das. Thirt% t!o !omen
"an"omplete the same !or in t!ent%four das. ixteen men and
sixteen!omenstarted !oring and !ored for 12 das. +o! man more men
are to beadded to"omplete the remaining !or in 2 das6
:A; 1* men :B; 24 men : ; 3* men :?; 48 men
ol$ one man@s one da@s !or( 1D384one !oman@s one da@s
!or(1D9*8Ior done in 12 das( 12=: 1*D384 0 1*D9*8; (
12=:3D48;(3D4emaining !or(1 S (1D4:1* men01* !omen;@s t!o da@s
!or
(12=: 1*D38401*D9*8;(2D1*(1D8
emaining !or ( 1D4%1D8 (1D81D384 !or is done n 1 da b 1
man.Therefore, 1D8 !or !ill be done in 2 das in
384=:1D8;=:1D2;(24men
28;4 men and * !omen "an "omplete a !or in 8 das, !hile 3 men
and 9!omen"an "omplete it in 1> das. /n ho! man das !ill 1>
!omen "omplete it6
:A; 3) das :B; 4> das : ; 4) das :?; )> das
ol$ .olving these t!o euations, !e get$ x(11D4>> and (
1D4>>
Therefore, 1 !oman@s 1 da@s !or(1D4>>(' 1> !omen
!ill "omplete the !or in 4> das.
Ans!er$ B
25;Lne man,3 !omen and 4 bos "an do a pie"e of !or in 5*hrs, 2
men and8 bos"an do it in 8> hrs, 2 men X 3 !omen "an do it in
12>hr. )Cen X 12 bos"an doit in6
:A; 35 1D11 hrs :B; 42 9D11 hrs : ; 43 9D11 das :?; 44hrs
ol$ %%%%%%%%%%:2;adding :2; X :3; and subtra"ting :1;
3x04(1D5* %%%%%%%%%:4;-rom :2; and :4;, !e get
x(1D48>ubstituting, !e get $ (1D92> and ( 1D5*>
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:) men0 12 bo;@s 1 hour@s !or()D48>012D5*> (1D5*
01D8>(11D48> Therefore, ) men and 12 bos "an do the !or
in 48>D11 or43 9D11hours. o,Ans!er$
ipes an' Cisterns
Important #acts:
1./# (1D2> +en"e, both the pipes together !ill fill the
tan in 2> hours.
2;T!o pipes "an fill a tan in 1> hours X 12 hours
respe"tivel. Ihile3rd pipe empties the full tan n 2> hours. /f
all the three pipesoperate simultaneousl,in ho! mu"h time !ill the
tan be filled6
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ol$ #et part filled in 1 hour(1D1> 01D12 %1D2>
(8D*>(2D1) The tan be filled in 1)D2hours( 9 hrs 3>
min
3;A "istern "an be filled b a tap in 4 hours !hile it "an be
emptiedb another tap in 5 hours. /f both the taps are opened
simultaneousl,then after ho! mu"h time !ill the "istern get
filled6
ol$ #et part filled in 1 hour( 1D4 %1D5( )D3* Therefore
the "istern !ill be filled in 3*D) hours or 9.2 hours.
4;/f t!o pipes fun"tion simultaneousl, the reservoir !ill be
filled in12 das.Lne pipe fills the reservoir 1> hours faster
than the other.+o! man hours does it tae the se"ond pipe to fill
the reservoir.
ol$ hours to fill the reservoir.
);A "istern has t!o taps !hi"h fill it in 12 min and 1)
minrespe"tivel.There is also a !aste pipe in the "istern. Ihen all
the three areopened,the empt "istern is full in 2> min. +o! long
!ill the !aste pipe taetoempt the full "istern6
ol$ Ior done b a !aste pipe in 1 min (1D2>
%:1D1201D1);( %1D1> :%ve means empting;
*;A tap "an fill a tan in * hours. After half the tan is filled,
threemore similar taps are opened. Ihat is the total time taen to
fill thetan "ompletel6
ol$ Time taen b one tap to fill the half of the tan (3
hours art filled b the four taps in 1 hour(4D*(2D3
emaining part(1 %1D2(1D2 Therefore, 2D3$1D2$$1$x or
x(:1D2;=1=:3D2;(3D4 hours. i.e 4) min o, total time
taen( 3hrs 4)min.
9;A !ater tan is t!o%fifth full. ipe A "an fill a tan in 1>
min. AndB
"an empt it in * min. /f both pipes are open, ho! long !ill it
tae toempt or fill the tan "ompletel 6
ol$ learl, pipe B is faster than A and o, the tan !ill be
emptied. art to be emptied(2D). art emptied b :A0B; in
1 min( 1D* %1D1>(1D1) Therefore, 1D1)$2D)$$1$x or
x(::2D);=1=1);(* min. o, the tan be emptied in * min.
8;Bu"et has thri"e the "apa"it as Bu"et J. /t taes *> turns
for
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filling "apa"it of the pump6
ol$ ; m3Dmin. o,24>>Dx S 24>>D:x01>; (
8
on solving x()>.
1>;A lea in the bottom of a tan "an empt the full tan in 8
hr. Aninletpipe fills !ater at the rate of * lits a minute. Ihen
the tan is full,the inlet is opened and due to the lea, the tan is
empt in 12 hrs.+o! man liters does the "istern hold6
ol$ Ior done b the inlet in 1 hr( 1D8 %1D12(1D24 Ior done
b the inlet in i min( :1D24;=:1D*>;(1D144> Therefore,
Uolume of 1D144> part(* lit Uolume of !hole(:144>=*;
lit(8*4> lit.
11;T!o pipes A and B "an fill a "istern in 39 K min and 4)
minutesrespe"tivel. Both the pipes are opened. The "istern !ill be
filled in
Yust half an hour, if the pipe B is turned off after$
sol$ %x;min(1 Therefore,
x:2D9)01D4);0:3>%x;:2D9);(1 11xD22) 0
:*>%2x;D9)(1 11x0 18>%*x(22) x(5. o, B must
be turned off after 5 minutes.
Time an' Distance#ormulae:
/;peed ( ?istan"eDTime
//;Time ( ?istan"eDspeed
///; ?istan"e ( speed=time
/U; 1mDhr ( )D18 mDs
U;1 mDs ( 18D) MmDhr
U/;/f the ratio of the speed of A and B is a$b,then the ratio
ofthe time taen b them to "over the same distan"e is 1Da $ 1Dbor
b$a
U//; suppose a man "overs a distan"e at x mph and an
eualdistan"e at mph.then the average speed during the !holeYourne
is :2xDx0;mph
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Problems
1;A person "overs a "ertain distan"e at 9mph .+o! man metersdoes
he "over in 2 minutes.
olution$$speed(92mph(92=)D18 ( 2>mDsdistan"e "overed in 2min
(2>=2=*> ( 24>>m
2;/f a man runs at 3mDs. +o! man m does he run in 1hr
4>min
olution$$speed of the man ( 3=18D) mph ( )4D)mph?istan"e
"overed in )D3 hrs()4D)=)D3 ( 18m
3;Ialing at the rate of 4nph a man "overs "ertain distan"ein 2hr
4) min. unning at a speed of 1*.) mph the man !ill"over the same
distan"e in.
olution$$?istan"e(peed= time4=11D4(11m#e! speed
(1*.)mphtherefore Time(?D(11D1*.) ( 4>min Comple2
Problems
1;A train "overs a distan"e in )> min ,if it runs at a
speedof 48mph on an average.The speed at !hi"h the train must runto
redu"e the time of Yourne to 4>min !ill be.
olution$$
Time()>D*>
hr()D*hrpeed(48mphdistan"e(=T(48=)D*(4>mtime(4>D*>hr(2D3hr#e!
speed ( 4>= 3D2 mph( *>mph
2;Uias "an "over a distan"e in 1hr 24min b "overing 2D3 ofthe
distan"e at 4 mph and the rest at )mph.the totaldistan"e is6
olution$$
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3;!aling at of his usual speed ,a man is late b 2 K hr.the usual
time is.
olution$$&sual speed ( &sual time ( T?istan"e ( ?#e!
peed is #e! time is 4D3 T4D3 T S T ( )D2T(1)D2 ( 9 K
4;A man "overs a distan"e on s"ooter .had he moved 3mphfaster he
!ould have taen 4> min less. /f he had moved2mph slo!er he !ould
have taen 4>min more.the distan"e is.
olution$$.) mph?istan"e bet!een them is 8.) mTime( 8.)m D >.)
mph ( 19 hrs
9;2 trains starting at the same time from 2 stations
2>>mapart and going in opposite dire"tion "ross ea"h other
atadistan"e of 11>m from one of the stations.!hat is the ratio
oftheir speeds.
olution$$/n same time ,the "over 11>m X 5> m respe"tivelso
ratio of their speed (11>$5> ( 11$5
8;T!o trains start from A X B and travel to!ards ea"h other
atspeed of )>mph and *>mph resp. At the time of the meeting
these"ond train has traveled 12>m more than the first.the
distan"ebet!een them.
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olution$$mxD)> ( x012> D *>x( *>>so the distan"e
bet!een A X B is x 0 x 0 12> ( 132> m
5;A thief steals a "a r at 2.3>pm and drives it at
*>mph.thetheft is dis"overed at 3pm and the o!ner sets off in
another "arat 9)mph !hen !ill he overtae the thief
olution$$pmdistan"e "overed b the thief in x hrs ( distan"e
"overed bthe o!ner in x%1D2 hr*>x ( 9) : x% K;x( )D2 hrthief is
overtaen at 2.3> pm 0 2 K hr ( ) pm
1>;/n "overing distan"e,the speed of A X B are in the
ratio
of 3$4.A taes 3>min more than B to rea"h the destion.The
timetaen b A to rea"h the destinstion is.
olution$$atio of speed ( 3$4atio of time ( 4$3let A taes 4x
hrs,B taes 3x hrsthen 4x%3x ( 3>D*> hrx ( K hrTime taen b A
to rea"h the destination is 4x ( 4 = K ( 2 hr
11;A motorist "overs a distan"e of 35m in 4)min b moving at
aspeed of xmph for the first 1)min.then moving at double thespeed
for the next 2> min and then again moving at his originalspeed
for the rest of the Yourne .then x(6
olution$$Total distan"e ( 35 mTotal time ( 4) min? ( =Tx =
1)D*> 0 2x = 2>D*> 0 x = 1>D*> ( 35 mx ( 3* mph
12;A X B are t!o to!ns.Cr.-ara "overs the distan"e from A t>
Bon ""le at 19mph and returns to A b a tonga running at auniform
speed of 8mph.his average speed during the !holeYourne is.
olution$$Ihen same distan"e is "overed !ith different
speeds,then theaverage speed ( 2x D x0(1>.88mph
13;A "ar "overs 4 su""essive 3m stret"hes at speed
of1>mph,2>mph,3>mphX$*>mph resp. /ts average speed
is.
olution$$
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Average speed ( total distan"e D total timetotal distan"e ( 4 =
3 ( 12 mtotal time ( 3D1> 0 3D2> 0 3D3> 0 3D*>(
3*D*> hrspeed (12D3* = *> ( 2> mph
14;A person !als at )mph for *hr and at 4mph for 12hr.The
average speed is.
olution$$avg speed ( total distan"eDtotal time( )=* 0 4=12 D
18(4 1D3 mph
1);A bullo" "art has to "over a distan"e of 8>m in
1>hrs./f it "overs half of the Yournein 3D)th time.!ht should be
itsspeed to "over the remaining distan"e in the time left.
olution$$Time left ( 1> % 3D)=1>
( 4 hrspeed (4> m D4 hr(1> mph
1*;The ratio bet!een the speeds of the AX B is 2$3 antherefore A
taes 1> min more than the time taen b B to rea"hthe
destination./f A had !aled at double the speed ,he !ouldhave
"overed the distan"e in 6
olution$$atio of speed ( 2$3atio of time ( 3$2A taes 1> min
more
3x%2x ( 1> minA@s time(3> min%%%'A "overs the distan"e in
3> min ,if its speed is x%' +e !ill "over the same distan"e in
1) min,if its speeddoubles :i.e 2x;
19;A is t!i"e as fast as B and B is thri"e as fast as is.The
Yourne "overed b B in6
olution$$speed@s ratioa $ b ( 2$ 1b $ " ( 3$1Time@s ratio
b $ " ( 1$3b $ " ( 18$)4:if " "overs in )4 min i..e t!i"e to 18
min ;
18;A man performed 3D) of the total Yourne b ratio 19D2> bbus
and the remaining *)m on foot.!ht is his total Yourne.
olution$$
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x%:3D)x 0 19D2> x; (*.)x% 15xD2> ( *.)x(2> = *.)(13>
m
15;A train C leaves Ceerat at ) am and rea"hes ?elhi at 5am
.Another train # leaves ?elhi at 9am and rea"hes Ceerut at
1>3>amAt !hat time do the 2 trains "ross one another
olution$$
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relative speed is u%v mph
:4;/f t!o trains of length x m and m are moving in
oppositediredtions at u mph and vmph,then time taen b the train
to"ross ea"h other ( :x0;D:u0v; hr
:); uppose t!o trains or t!o bdies are moving in opposite
dire"tionat u mph and v mph then,their relative speed ( :u0v;
mph
:*;/f t!o train start at the same time from 2 points A X B
to!ardsea"h other and after "rossing the tae a X b hours in
rea"hing B X Arespe"tivel then A@s speed $ B@s speed ( :bj1D2 $
aj1D2 ;
Problems :1;-ind the time taen b a train 18>m
long,running at 92mph in"rossing an ele"tri" pole
olution$ peed of the train (92=)D18mDs (2> mDs
?istan"e move din passing the pole ( 18>m
euiredtime ( 18>D2> ( 5 se"onds
:2;A train 14> m long running at *>mph./n ho! mu"h time
!ill itpass a platform 2*>m long.
olution$ ?istan"e travelled (14> 0 2*> m (4>>
m, speed ( *> = )D18 ( )>DD3 m time(4>>=3 D
)> ( 24 e"onds
:3;A man is standing on a rail!a bridge !hi"h is 18> m.+e
findsthat a train "rosses the bridge in 2> se"onds but himself
in8 se". -ind the length of the train and its sppeed
olution$ i;?(18>0x T ( 2> se"onds (
18>0x D 2> %%%%%%%%%%%% 1 ii;?(x T(8
se"onds ?(T x(8 %%%%%%%%%%%%% 2 ubstitute 2 in
1 (18> 0 8 D 2> (1) mDs
m
:4;A train 1)>m long is running !ith a speed of *8 mph/n
!httime !ill it pass a man !ho is running at a speed of 8mph inthe
same dire"tion in !hi"h the train is going
olution$ elative peed ( *8%8(*>mph=)D18 ( )>D3
mDs time( 1)> = 3 D)> (5se"
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);A train 22>m long is running !ith a speed of )5 mph
D../n!hat time !ill it pass a man !ho is running at 9 mph in
thedire"tion opposite to that in !hi"h train is going.
olution$ elative peed ( )509(**mph=)D18 ( ))D3 mDs
time( 22>D)) = 3 (12se"
:*;T!o trains 139m and 1*3m in length are running to!ards
ea"hother on parallel lines,one at the rate of 42mph X another at48
mph./n !ht time !ill the be "lear of ea"h other from themoment the
meet.
olution$ elative speed (42048 ( 5> =)D18 ( 2)mDs
time taen b the train to pass ea"h other ( time taen to "over
:13901*3;m at 2) mDs ( 3>> D2) s (12 s
:9;A train running at )4 mph taes 2> se" to pass a
platform.#ext it taes 12 se" to pass a man !aling at *mph in the
samedire"tion in !hi"h the train is going.-ind length of the
trainand length of platform
olution$ elative speed !.r.t man ( )4%*(48mph the
length of the train is 48 = )D18 = 12 (1*>m time taen to
pass platform (2> se"
peed of the train ( )4 = )D18 (1)mDs 1*>0x (2>
=1) x(14>m length of the platform is 14>m
:8;A man sitting in a train !hi"h is travelling at )>mph
observesthat a goods train travelling in opposite ire"tion taes 5
se"to pass him ./f the goos train is 1)>m long fin its speed
olution$ elative speed (1)>D5 mDs (*> mph
speed of the train ( *>%)> (1>mph
:5;T!o trains are moving in the sam e dire"tion at *)mph
and4)mph. The faster train "rosses a man in slo!er train
in18se".thelength of the faster train is
olution$ elative speed (*)%4) mph ( )>D5 mDs
?istan""e "overed in18 s ()>D5 = 18 ( 1>>m the
length of the train is 1>>m
:1>;Atrain overtaes t!o persons !ho are !aling in the
samedire"tion in !hi"h the train is going at the rate of 2mph
an4mph and passes them "ompletel in 5 se" an 1> se"
respe"tivel.The length of train is
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olution$ 2mph ( )D5 mDs 4 mph (1>D5 mDs D5;
( 1> 5%) (x and 1>:5%1>;(5x
5%x() and 5>%5x(1>> on solving !e get
x()>,lenght of trains
:11; T!o stations A X B are 11> m apart on a straight
line.Lne train starts from A at 9am and travels to!ards B at
2>mph.Another train starts from B at 8am an travels to!ard A at
a speedof 2)mph.At !hat time !ill the meet
olution$ uppose the train meet x hr after 9am
?istan"e "overed b A in x hr(2>x m 2>x02):x%1; (
11>
4)x(13) x(3
o the meet at 1> am
:12;A traintravelling at 48mph "ompletel "rosses another
trainhaving half its length an travelling inopposite dire"tion at
42mphin12 se"./t also passes a rail!a platform in 4)se".the length
ofplatform is
olution$ mph J $ The too 5> se" to "ross ea"h
other:a; Either X J is suffi"ient:b;Both X J are not
suffi"ient:";onl J is suffi"ient:d;Both X J are neeedAns$ B
olution$ > x mt (13xD1> mt mph
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relative speed( 08> = )D18 mDs time taen b the
trains t "ross ea"h other is gven b 5> ( :x 0 13xD1>;D
:)04>> D 18; to find ,"learl xis also needed
so,both X J are not suffi"ient
:14;The speed of a train A,1>>m long is 4>W more than
then the speedof another train B,18>m long running in opposite
dire"tion.To fin outthe speed of B,!hi"h of the information given
in statements X J issuffi"ient $The t!o trains "rossed ea"h
other in * se"onds J $ The differen"e bet!een the spee of the
trains is 2*mph:a;Lnl is suffi"ient:b;Lnl J is suffi"ient:";Both X
J are needed:d;Both X J are not suffi"ientAns $ A
olution$ xD1>> mph (9xD) mph relative speed (
x 0 9xD) (2xD3 mDstime taen to "ross ea"h other (
:1>>018>;=3D2x s (42>Dx sno!,42>Dx ( * x(9>
mphthus,onl is suffi"ient
:1);The train running at "ertain speed "rosses astationar
enginein2> se"onds.to find out the sped of the train,!hi"h of
the follo!ing information is ne"essar:a;Lnl the length of the
train:b;onl the length of the engine:";Either the length of the
train or length of engine:d;Both the length of the train or length
of engineAns $ ?
olution$
in"e the sum of lengths of the tran and the engine is
needed, so both the length must be no!n
BOATS AND STREAMS
Important facts:
1;/n !ater, the dire"tion along the stream is "alleddo!n
stream.
2;?ire"tion against the stream is "alled upstream.
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3;The speed of boat in still !ater is & mDhr and thespeed of
stream is UmDhr then
speed do!n stream (& 0 U mDhrspeed up stream ( & [\] U
mDhr
#ormulas:
/f the speed do!n stream is A mDhr and the speed upstream is B
mDhr then
speed in still !ater ( K:A0B; mDhr
rate of stream (1D2:A%B; mDhrP$O!'(%"$ 1. /n one hour a boat
goes 11 m long thestream and ) m against the stream. The speedof
the boat in still !ater is6ol$peed in still !ater ( K : 110); mDhr(
8 mph
2.A man "an ro! 18 mph in still !ater. /t taes himthri"e as long
as ro! up as to ro!
do!n the river. find the rate of stream."ol:
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let the reuired disten"e x m. then ,xD50xD*()>D*> ( 2x03x(
)D*=18)x(1), x(3+en"e, the reuired disten"e is 3 m
4.A man "an ro! 3 uarters of a m aganist the streamis 111D4 min.
the speed of theman in still !ater is 6"ol:rate upstream (
9)>D*2) mDse" (1>D5 mDse"rate do!nstream (9)>D4)> mDse"
( )D3 mDse"rate in still !ater (1D2F1>D50)D3G ( 2)D18 mDse"(
2)D18=18D)() mph
).A boat "an travel !ith a speed of 13 mph in still
!ater. if the speed of stream is4 mph,find the time taen b the
boat to go *8 mdo!nstream6
"ol: peed do!n stream ( 1304( 19 mphtime taen to travel *8m
do!nstream (*8D19 hrs ( 4 hrs
*.A boat taes 5> min less to travel 3* miles
do!nstream then to travel the samedisten"e upstream. if the
speed of the boat in still!ater is 1> mph .the speed of the
stream is $"ol:0xGmphspeed upstream (F1>%xG mph3*DF1>0xG %
3*DF1>%xG ( 5>D*> (92x=*>(
5>F1>>%x2G:x0)>;:x%2; (>
x(2 mph
9.At his usual ro!ing rate, ahul 12 miles do!nstream in a
"ertain river in * hrsless than it taes him to travel the same
disten"eupstream. but if he "ould doublehis usual ro!ing rate for
his 24 miles roundthe do!n
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stream 12 miles !ould thentae onl one hour less than the up
stream 12 miles.!hat is the speed of the"urrent in miles per
hours6"ol:
04; ( 14 mphupstream (:1>%4; ( * mpho, average speed of the
boat sailor (F 2=14=*GD
F140*G mph(42D) mph (8.4 mphin"e, the average speed of the
""list is greater,he !ill return to A first.
5.A boat taes 15 hrs for travelling do!nstream from
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