easy ways in mathematics

8/18/2019 Easy Ways in Mathematics

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EASY WAYS IN MATHEMATICS

GATE ACADEMY

THALAYOLAPARAMBU


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AVERAGE


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Average is one of the important topics in

Quantitative Aptitude section.

To or three independent !uestionsusua""# come from this topic in each test.

$ut it%s importance does not end here.

&or so"ving ten to fifteen !uestions indata interpretation re!uires thorough

'no"edge in Average( )ercentage( and

*atio + )roportion.

AVERAGE


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A,E*A-EAVERAGE or more accurately an arithmetic

mean, in

crude terms, is the sum of ‘n’ dierent datadivided by ‘n’• &or Eg. If a atsman scores /0( 10 and /2 runs in

the first( second and third innings respective"#(then the average runs in three innings 3/04104/2 3 552 3 /6

• / / 33 runs.• Therefore( the to most"# used

formu"ae in average are78i. Average = Total of dataNo. of data

and ii. Total = Avera e x No. of data


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I""ustrative Question85

• 59 The average of 5:: numers as ca"cu"ated as;:.


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A,E*A-EA,E*A-Ei. To find the average of some figures; divide the sum tota" # the numer

of figures.

• Eg. The age of a"" memers of a fami"# is given e"o. &ind out theaverage age of a fami"# memer. ;:( 00(0(:(50

• Therefore( average age is ;:400404:450 3 520 3 /0• 0 0 33ii. If average age and numer of memers are given( the tota" of the age

can e arrived # mu"tip"#ing the average age ith numer ofmemers.• Eg.( The average age of a 5 memer Cric'et Team is >. What

is the tota" age of these 5 memersB• > ? 5 3 //;• 3333

iii. The average age of /: students in a c"ass is 5/. When the age of theTeacher is rec'oned( the average age increases # one. What is theage of the TeacherB

• The tota" age of the /: students 3 /: ? 5/ 3 /6:• The age of /5 memers inc"uding Teacher 3 /5 ? 51 3 1/1• Therefore( the age of Teacher 3 1/1 @ /6: 3 11

Short cut7 /: 4 51 3 11


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Short cut7 /: 4 51 3 11

• E)


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E?.No. iv. The average age of : students in a c"ass is 5:. When the age

of the Teacher is rec'oned( the average age increases # to. What is

the age of the TeacherB

• The tota" age of the : students 3 : ? 5: 3 ::• The age of 5 memers inc"uding Teacher 3 5 ?

5 3 0

• Therefore( the age of Teacher 3 0 @ :: 3 0

Short cut• The contriution of age in the average of each

student is .

• That means ? : 3 1: #ears have eencontriuted from the age of Teacher ma'ingaverage age of students from 5: to 5. So 5more age re!uire for the teacher hen she is

a"so inc"uded .


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When average va"ue is decreasing

E?.No. ,. The average eight /: passengers in a oat is 10gs. When one more passenger is entered the oat theaverage of the passengers( decreases # one. What is

the t. of the ne passengerBThe tota" t. of the /: passengers 3 /: ? 10 3 5/0:• The ne Wt. of /5 passengers inc"uding the ne

passenger 3 11 ? /5 3 5/;1Therefore( the t. of the ne passenger 3 5/;1 @ 5/0: 3 51

gs

Short cut• As the average t. is reduced # one( the ne passenger

has ta'en over one g each from a"" passengers. So( hegot /: gs and the average reduced # one to 11.

• The ne passenger shou"d have the average t. of 11gs.

• /: gs a"read# contriuted # /: passengers.• The additiona" t. re!uired is thus 51 to ma'e 11

• Therefore( t. of the ne passenger is 11 8 /:3 51


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When average va"ue is increasing

The average eight /: passengers in a oat is 10 gs.

When one more passenger entered the oat the averageof the passengers( increases # to. What is the t. ofthe ne passengerB

The tota" eight of the /: passengers 3 /: ? 10 3 5/0:

The t. of /5 memers inc"uding nepassenger 3 /5 ? 12 3 5102

• Therefore( the t. of ne passenger 3 5102 @ 5/0: 3 5:2• 33

Dr ;: 4 12

35:2 gs

Dr ;: 4 12

35:2 gsShort cut 7 Wt. contriuted to

/: passengers 3;:4 ne average 12 3 5:2


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vii. The average eight of a c"ass for /0 students is 12.0 'g. If

the eight of the teacher e inc"uded( the average rises #

0:: gm. What is the eight of the teacherB

Ans7 =a9;:.0 =9 1> =c9 ;; =d9 ;0.0 =e9 ;2

• Inc"uding the teacher( there are /; persons• i.e.( =/0 students 4 5 teacher9• And then( the average eight 12.0 'g 4 0:: gm.

3 1> 'g.• Tota" eight for /; persons 3 1> /; 352> 'g.• Tota" eight for /0 persons 3 12.0 /035;;.0 'g•

Weight of the teacher 3 52>F5;;.0 3 ;0.0 'g.

Short cut 7 Wt. contriuted to /0 students 3 52.0 'g

4 ne average 1> 3 ;0.0 g Ans7 =d9 ;0.0


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When average va"ue is decreasing

E?.No. ,iii. The average age of a group of 0 picnic group is >#ears. When one more person Goined the group( theaverage age of the group( decreases # one. What is

the age of the ne personBThe tota" age of the 0 persons 3 0 ? > 3 2::• The ne age. of ; persons inc"uding the ne one 3 2 ?

; 3 2:Therefore( the age of the ne person 3 2: @ 2:: 3 #ears

Short cut• As the average age is reduced # one( the ne person

has ta'en over one #ear each from a"". So( he got 0#ears and the average reduced # one to 2.

• The ne passenger shou"d have the average age of 2#ears.• 0 #ears a"read# contriuted # 0 passengers.• The additiona" age re!uired is thus to ma'e 2• Therefore( age of the ne person is 2 @ 0 3

#ears


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When average va"ue is increasing due to sustitution

E?.No. i?. The average eight a team of persons is 10 gs.When one person of 0: g has gone and a ne person Goined

the team( the average t. of the team memers increases # g. What is the t. of the ne memerB The t. of ne memer 3 t. of person rep"aced 4 no. of o"d team

memers ? difference in the average =diff. ma# e 4 or 89JKetai"ed or'ing7 The tota" t. of the memers 3 ? 10 3 66:• The ne Wt. of after sustituting the ne memer 3 ? 10

3 5::5 'gsThe additiona" t. contriuted # the ne memer 3 5::5 @ 66:3 55

gsTherefore( actua" t. of the ne memer 3 0: 4 55 3 ;5'gsShort cut7 i. 0: 4 ? 3 ;5

• Dr ii. As the average t. is increasing ithout increasing thenumer of memers( e can arrive the anser as under78

• In order to increase g in the average of memers( ? 'g ie.( 55 'g must e added to the t. of the memer "eft.

Therefore( the t. of the ne passengeris 0: 4 553 ;5


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When average va"ue is decreasing due to sustitution

E?.No. ?. The average age of a team of 1: students is 51#ears.When one student of 5; #ears has gone and a nestudent Goined the team( the average age of the team

memers decreases # L #ears. What is the age of thene memerB

The age of ne memer 3 Age of person rep"aced 4 no. ofo"d team memers ? difference in the average =diff. ma#e 4 or 89J

Ketai"ed or'ing7 The tota" age of the 1: memers 3 1: ? 5; 3;1:

• The ne age of 1: students after sustituting the nememer 3 1: ? 50 3 ;/:

The age decreased due to sustituting the ne memer 3 ;1:

@ ;/:3 5: gsTherefore( actua" age of the ne memer 3 5; 8 5: 3 ;#ears

Short cutThe age of the ne memer 3 5; 4 1: ? @ L

3 5; @ 5: 3 ; #ears


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When average va"ue is decreasing due to sustitution

E?.No.?i. The average mar's scored in Maths # a team of 0students in a ench is >>. When one student ho got5:: mar's has gone and a ne student Goined the team(

the average mar's of the team memers decreases #. What is the mar' in Maths scored # the nestudentB

The mar's of ne memer 3 Mar's of person rep"aced 4 no.of o"d team memers ? difference in the average =diff.

ma# e 4 or 89JKetai"ed or'ing7 The tota" mar's of the 0 studets 3 >> ? 0 3

11:• The ne mar's of 0 students after sustituting the ne

memer 3 >; ? 0 3 1/:

The mar's decreased due to sustituting the ne memer 311: @ 1/:3 5: mar's

Therefore( actua" mar's of the ne memer 3 5:: 8 5: 3 6:

Short cutThe mar's of the ne memer 3 5:: 4 0 ? @

3 5:: @ 5: 3 6: mar's


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&inding the ne Average due to sustitution

=9 5/.6

• Q.No.?ii. The average age of : students in a c"ass is 51

#ears. A 50 #ear o"d student "eaves the c"ass ut a 5/ #earo"d student gets admission in his p"ace. What i"" e the

average age in #ears of the students in the c"ass noB

• =a9 51 =9 5/.6 =c9 1 =d9 /: =e9 5>

• So"ution7• Tota" age : students in #ears7 : ? 51 3 >:

#ears

• The difference in age eteen the nestudent and the o"d student 35085/3

• The ne average i"" e >:8 3 2> : 3

5/.6 #ears Ans7 =9 5/.6


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• Q.No.?iii. The average eight of 6 Mangoesincreases # : grams if one of them

eighing 5: grams is rep"aced # another.

The eight of the ne Mango isO.

• =a9 51: gm =9 56 gm =c9 1: gm =d9 /:: gm=e9 5>: gm

• Ans. =d9 /:: gm

• E?p"anation7• 5: 4 6 ? : 3 /:: gm

&inding the eight of sustituted item

= i9 Th ' t i d 5: t d t i i ti i /0


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=?i9 The average mar's otained # 5: students in an e?amination is /0.

If average mar's of the passed students is /6 and that of the fai"ed

students is 50( hat i"" e the numer of passed studentsB

• So"ution7•


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Qn.?ii. A goods train in five successive minutes from the start runs ;> m 52 m

:> m /5 m 0/0 m and for the ne?t five minutes maintains an average speed of //

'm per hour. &ind the ho"e distance covered and the average speed of the train

in 'm per hour.

• 5'm 3 5::: m• 5 hr 3 ;: minutesJ• Average speed in the "ast 0 minutes 3 //'mhour 3 //:::

m ;: minutes 3 00: mminute

• Kistance covered in the "ast 0 minutes 3 00: 0 minutes320:m

• Kistance covered in a"" the 5: minutes 3=20:4;>4524:>4 /5 40/09m

• 3 1::: m =or 1 'm.9• Average speed in 5: minutes 3 1::: m 5: minutes• 3 1:: mminute• 3 1::;: mhour 3 1::: mhour

• 3 1'mhour

Q iii Th " h d f "" th ' f i tit ti i


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Qn.?iii7 The average sa"ar# per head of a"" the or'ers of an institution is

*s ;: . The average sa"ar# head of 5 officers is *s 1::. The average

sa"ar# per head of the rest is *s. 0;. &ind the tota" numer of or'ers in

the institution.

=a95:/ =9 5:// =c9 5:/1 =d9 5:/0 =e9 5:/;

• ::

• Sa"ar# of the rest 3 =?F59 ? 0;• Hence( tota" sa"ar# of the or'ers =?F59 0; 4 1>:: OO=9• E!uating =59 and =9• ;:? 3 1>:: 4 0;?F;2

• 3 ;:?80;? 3 1>::8;2• 3 1? 3 15>• ? 3 15> 1 35:/• So( the tota" numer of or'ers in the institution 3 5:/

Q i D G M i t i : ' h f


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Qn.?iv7 Dn a Gourne# across Mumai a ta?i averages : 'm p h for

2:R of the distance( 0 'm p h for 5: R of the distance and > 'm p

h for the remainder &ind the average speed of the ho"e Gourne#.

• 'm p h 3 :> 3 .0 hours

• Tota" time ta'en 3 /.0 4:.14.0 3 ;. 1hoursor =; 1 5: 3

;1 5: 3/ 09

• Average speed 35:: / 0 3 5:: 0

/ 3 50.;0 'mph.

Th t t f 6th t th 5;th f th


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?v. The average temperature from 6th to the 5;th of a month(

oth da#s inc"usive( as /1.; and 5:th to the 52th it as

/0.>. The temperature on the 6th as /:.0. What as it on

52thB

• Average temperature from the 6th to 5;th =i.e.( in > da#s of amonth9 is /1.;

• Tota" temperature from the 6th to 5;th =i.e.( in > da#s of amonth9 3 /1.; > 3 3 2;.>

• $ut the temp. on 6th is given to e /:.0• Tota" temperature( from 5:th to 5;th of a month 3 2;.>F

/:.0 31;./

• No the average temperature from the 5:th to 52th of amonth is /0.>

• .P. The tota" temperature from the 5:th to 52th of a month 3/0.> > 3 >;.1

• The temperature on 52th i"" e 3 >;.1F1;./ 3( 1:.5

i & fi f hi i ti G t t i " di


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?vi. &or five of his e?amination suGects not inc"uding

Eng"ish and Histor#( a student%s average mar' is ;0. &or

Eng"ish and Histor#( the mar's are e!ua" and hen these are

inc"uded( his average mar' is reduced to ;5. Ca"cu"ate his

mar' for Eng"ish.

• Average mar's for the 2 suGects 3 ;5• .P. Tota" mar's for the 2 suGects 3 ;52312

• Average mar's for 0 suGects =other than Eng"ishand Histor#9 3 ;0• .P. Tota" mar's for the 0 suGects 3 ;003/0• Tota" mar's in Eng"ish and Histor# 3 12F/0 3

5:

• Mar's for Eng"ish and Histor# are given to ee!ua"

• His mar's in Eng"ish 3 5: 3 05.

?vii A man spends on an average *s 5;6 12 for the first 2 months


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?vii. A man spends on an average *s. 5;6.12 for the first 2 months

and *s. 5>5.:0 for the ne?t 0 months. &ind his month"# sa"ar# if he

saves *s. /:>.1; during the ho"e #ear.

• =a9 566 =9 :: =c9 :5 =d9 : =e9 :/• Average e?penditure for the first 2 months 3 *s. 5;6.12

• .P. Tota" e?penditure in that period 3 *s. 5;6.12 23 *s. 55>;.6

• Average e?penditure for the ne?t 0 months 3 *s. 5>5.:0• Tota" e?penditure in the period 3 *s.5>5.:0 .0 3 *s.6:0.0

• His savings in the ho"e #ear 3 *s./:>.1;• .P. Tota" income for the ho"e #ear 3 55>;.6 4

6:0.0 4 /:>.1; 3 *s.1::.• Average income for each month 3 *s.1:: 5 *s. ::

• i.e.( His month"# income 3 *s.::.

?viii An esta"ishment is permitted an average month"# contingenc#


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?viii. An esta"ishment is permitted an average month"# contingenc#

e?penditure of *s. 0:: per month during the financia" #ear. When a tria"

chec' as made at the end of the first nine months of the #ear it as

found that the average month"# contingenc# e?penditure or'ed out to

e *s. 055. What average month"# e?penditure for the ne?t three months

shou"d e aimed at in order to attain the permissi"e average of *s. 0::per month for the ho"e #earB =a9 1;0 =9 1;; =c9 51:5=d9 1;2 =e9 055

• Average month"# contingenc# e?penditure a""otted for the ho"e#ear 3 *s. 0::

• .P.Tota" annua" contingenc# e?penditure 3 *s. 0:: ? 5 3 *s.;:::• Average month"# contingenc# e?penditure during the first ninemonths 3 *s. 055

• .P. Tota" e?penditure in the first nine months 3 *s. 055 ? 63 *s. 1066

• As per a""otment( amount "eft for the ne?t three months3 *s.=;:::F10669 3 *s.51:5

• Average month"# contingenc# e?penditure for the "ast / months3 *s(51:5 / 3 *s. 1;2

?i? The average of 55 numers is ;/ that of the first ;


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?i?. The average of 55 numers is ;/( that of the first ;

numers is ;: and that of the "ast ; is ;0. Which among the

fo""oing is the si?th numerB =a9 ;0 =9 ;/ =c9 ;: =d9 ;2 =e9 02

• Average for the first ; numers 3 ;:• .P. Tota" for the first ; numers 3 ;: ; 3 /;:• Average for the "ast ; numers 3 ;0• Tota" for the "ast ; numers 3 ;0 ; 3 /6:

• Tota" for the aove to =si?th numer inc"udedtice9 3 /;: 4 /6: 3 20:

• Average for a"" the 55 numers 3 ;/

• Tota" for a"" the 55 numers 3 ;/ 55 3 ;6/• The si?th numer 3 20:F;6/ 3 02• Ans7 =e9 02

Qn The average temperature for Monda# Tuesda# Wednesda# and


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Qn. .The average temperature for Monda#( Tuesda#( Wednesda# and

Thursda# is ;: and the average for Tuesda#( Wednesda#( Thursda# and

&rida# is ;/. If the ratio of temperature for Monda# and &rida# e 570(

hat is the temperature on Monda# and &rida#B

Ans7 =a9 ;0(20 =9 ;/(20 =c9 ;/(2: =d9 ;2(20 =e9 02( /;

• The tota" temperature% for Monda#( Tuesda#( Wednesda#and Thursda# 3 ;: 1 3 1:

• The tota" temperature for Tuesda#( Wednesda#( Thursda#and &rida# 3 ;/ 1 3 0

•


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Qn7??i. The average of /5 numers is /5. When one numer

is added( their comined average increases # :.0. What is

the va"ue of the ne numerB Ans7 =a9 ;2 =9 02 =c9 12 =d9 22 =e9 /2

• Inc"uding the ne numer( there are / numers• )resent"#( the average 3 /5 4 :.0 3 /5.0• Tota" of a"" the / numers 3 /5.0 / 3 5::>• Tota" of the first /5 numers 3 /5 /5 3 6;5

• The ne numer 3 5::>F6;5312.• .P. Ans7 =c9 12• Short cut• The contriution of the ne numer to the average of each numer is

:.0.• That means :.0 ? /5 3 50.0 have een contriuted from the ne

numer ma'ing average of the numers from /5 to /5.0. So /5.0 more

re!uire for the numer hen it is a"so inc"uded .

• Therefore( the va"ue of of the ne numer is /5.0450.0 3 12

Qn7??ii A atsman has a certain average of runs for 5;


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Qn7??ii. A atsman has a certain average of runs for 5;

innings. In the 52th innings( he ma'es a score of >0 runs(

there# increasing his average # /. What is the average after

the 52th inningsB Ans7 =a9 /1 =9 02 =c9 12 =d9 22 =e9 /2

•


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??iii. The average height of 0 students of a c"ass is 5m 1:cm. &ive ne

entrants increase the average height to 5m 10 cm. Ketermine the average

height of the 0 ne entrants. =5 metre 3 5:: cm9Ans7 =a9 /m1:cm =9 5m2: cm=c9 1m2:cm =d9 2m2:cm =e9 /m2:cm

• Inc"uding 0 ne entrants the average height 5 m 10 cm 3510 cm

• Tota" height of these /: students 3 510 /: 3 1/0: cm

• Average height of 0 students 3 51:cm =5m 1: cm9

• Tota" height of 0 students 3 51: 0 3 /0:: cm

• Tota" height of 0 ne entrants 3 1/0:F/0::• 3 >0:cm

• Average height of 0 ne entrants3 >0: 0352:cm 3 5m 2: cm

Ans7=9 5m2: cm

1 The average age of a c"ass as 50 #ears hen 0 o#s hose average


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• Suppose there ere ? students in the eginning.

• Their tota" age 3 50?• Tota" age of 0 ne o#s 3 =5 #ears9 0• With the inc"usion of 0 o#s Average age 3 =50 #ears 8 ;

months9 3 "1#ears

• Then the tota" age 3 51 =?809• &rom the pro"em• 50? 4 =59 =09 3 =51 9=?409• 50? 4 =59 =09 3 =519 ? 4 =519=09• 50?F=519? 3 0 =51F59• ? 3 0 3 5:• .P. 3 5: 4 5: 3 :

• There ere : students in the eginning.

1. The average age of a c"ass as 50 #ears( hen 0 o#s hose average

age as 5 #ears( ; months ere admitted( the c"ass average as

reduced # ; months. Ho man# o#s ere thereB

0 The average age of > men is increased # #ears hen


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• men e • *ep"aced # to omen( average age of > persons 3 ?4• After rep"acing( the tota" age of > persons 3 > = ?49• The tota" age of > men 3 >?

• Tota" age of ; men( after to men ( hose ages are : and1 #ears "eft 3 >? @ =:419• 3 >? F11• Tota" age of omen 3 >=? 4 9 @ =>? 8119

• 3 >? 4 5; @ >? 411• 3 ;:• .P. Average age of omen 3 ;: • 3 /: #ears.

0. The average age of > men is increased # #ears( hen

to of them( hose ages are : and 1 #ears are rep"aced

# to omen. What is the average age of the omenB

; Th i ht f " f /0 t d t i 12 0 ' If


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;. The average eight of a c"ass for /0 students is 12.0 'g. If

the eight of the teacher e inc"uded( the average rises #

0:: gm. What is the eight of the teacherB

• Inc"uding the teacher( there are /; persons =/0 students 4 5teacher9• And then( the average eight 12.0 'g 4 0:: gm. 3 1>'g.• Tota" eight for /; persons 3 1> /; 3 52> 'g.• Tota" eight for /0 persons 3 12.0 /0 3 5;;.0 'g7

• .P. Weight of the teacher 3 52>F5;;.0• 3 ;0.0 'g.Short cut• The contriution of eight in the average of each

student is 0::gm.• That means /0 ? .0 3 52.0 g have een contriuted

from the eight of the Teacher ma'ing average eight of

students from 12.0 to 1> g. So 1> more gs re!uire for

the teacher hen she is a"so inc"uded . • Therefore( age of the teacher is 1>452.0 3 ;0.0

2. E"even persons contriuted a certain sum. Nine of them gave *s. each and


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2. E"even persons contriuted a certain sum. Nine of them gave *s. each and

the other to gave *s. and *s. .0: more respective"# than the average

suscription of a"" the e"even suscriers. What is the sum the to persons

suscriedB Ans7 =a9 *s.6.0: =9 0.0: =c9 1.0: =d9 .0: =e9 0.::

• So"ution7• Average of 5/ numers is ;> the average of 5st 2


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>. Average of 5/ numers is ;>( the average of 5st 2

numers is ;/ and the average of "ast 2 numers is 2:. &ind

the 2th numer. Ans7 =a9 /1 =9 02 =c9 12 =d9 22 =e9 /2

• So"ution7• Average of 5/ numers 3 ;>• Tota" of 5/ numers 3 5/ ;> 3 >>1• Average of 5st 2 numers 3 ;/• Tota" of 5st 2 numers 3 2 ;/ 3 115• Average of "ast 2 numers 3 2:• Tota" of "ast 2 numers 3 2 2: 3 16:

• Tota" of 5st ; numers 3 >>18 16: 3 /61• 2th numer 3 Tota" of 5st 2 numers 8 Tota" of 5st ;

numers

• 3 115 F/61 3 12

6. An officers pension on retirement of services is e!ua" to ha"f the


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6. An officers pension on retirement of services is e!ua" to ha"f the

average month"# sa"ar# during the "ast /; months of his service. His

sa"ar# from 5.5.:5: is *s. />::: per month ith increments of *s. 1:: as

from 5.5:.:5:( 5.5:.:55 and 5.5:.:5. If he retires on 5.5.:5/( ho

much pension does he draB =a9 /1::: =9 56::: =c9 560: =d9 522::=e9 NDTA

• His income from 5858:5: to /:.6.:5: 3 *s. />::: 6 3*s. /1:::

• His income from 5.5:.:5: to /:.6.:55 3 *s. =/>::: 4 1::9 53 />1:: ? 5 3 *s. 1;:>::

• His income from 5.5:.:55 to /:.6.:5 3 *s.=/>1::41::953 />>:: 5 3 *s. 1;0;::

• His income from 5.5:.:5 to /5.5.:5 3 *s.=/>>::41::9/• 3 *s.552;::

• His income for the "ast /; months Average month"# sa"ar# 3/1:::4 1;:>:: 4 1;0;:: 4 552;:: 3 5/>;:::

• Average month"# sa"ar# 3 *s. 5/>;::: /; 3 *s. />0::

• )ension amount 3 =*s. />0::9 3 *s. 560:

/:. The average age of a c"ass of / students is 50 #ears. A ne student


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/:. The average age of a c"ass of / students is 50 #ears. A ne student

Goined the c"ass and the average age increased # ; months. &ind the age

of the ne student.

• Tota" numer of students inc"uding the nestudent 3 / 4 5 3 1

• Average increase in the age of 1 students 3 ; months 3 #ear

• Tota" increase in the age of 1 students 3 1 3 "#ears• .P. The age of the ne student 3 50 4 5• 3 2 #ears.• Dr •

The tota" age of the / students 3 / ? 50 3 /10• The age of 1 memers inc"uding ne student 3 1 ? 50 3 /2

• Therefore( the age of ne student 3 /2 @ /10 3 2 #rs

/5. The tota" of the present ages of A( $( C and K is 6; #ears. What is $%s


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p g ( ( #

present ageB I. The average age of A( $ and K is : #ears. II. The average

age of C and K is 0 #ears.

• -ive anser V• =a9 if the data in Statement I a"one are sufficient to anser the !uestion( hi"e

the data in Statement II a"one are not sufficient to anser the !uestionV -iveanser =9 if the data in Statement II a"one are sufficient to anser the

!uestion( hi"e the data in Statement I a"one are not sufficient to anser the

!uestionV -ive anser =c9 if the data either in Statement I or in Statement II

a"one are sufficient to anser the !uestionV -ive anser =d9 if the data even in

oth Statements( I and II together are not sufficient to anser the !uestionV

-ive anser =e9 if the data in oth Statements I and II together are necessar# to

anser the !uestion.

I. Tota" age of A( $( C and K is 6;

II. Statement I gives A4 $ 4 K 3 : / 3 ;:#ears

III. Statement II gives C 4 K 3 0 3 0:#ears•. &rom I(II and III statements $%s age a"one cannot e

determined.•. .P. The the anser is d%. 3 =d9 if the data even in oth

Statements( I and II together are not sufficient to anser

the !uestion

)ractice Questions


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59 The average of 5:: numers as ca"cu"ated as ;:.


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Anser 'e#

• 5. =a9 3 Kiff 30: ie.( contriuted to 5:: :.0'g each ma'ing the average 06.0• . =c9 3 0 45: 3/0 ? 32: 40 320 / 3 0 •

/. =d9 3 : ? / 3;:• 1. =9 3 0 friends gave ?035:Va"ance to ma'e ne average of 685:356

Method of A""egation7

/: 50

1

6 ;

ie.( 67; 3 /730: ? 3:=passed 7 fai"ed9 /4

0. =a9


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TIME +KISTANCE

&D*MU


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&D*MU 'mhr 3 5> ? 0 3 0 msec 5>E?.No.. E?press 5:msec in 'mhr So"ution7 5: msec 3 5: ? 5> 3 /; 'mhr 0


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I""ustrative E?.No. /

• Qn. trave"ed 5; Ms in > hours # car. Athat speed as he drivingB

• Here trave"ed 5; Ms ta'ing > hours.• The distance trave""ed in one hour is the

speed in MHr.

• The formu"a to find out the speed is• Speed = Distance = 1! = " #$s Time 8

I""ustrative E? No 1


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I""ustrative E?.No. 1• Qn. ,anchinad E?press Train started from

Thiruvananthapuram is running at a speed of >:

M per hour. If the distance eteen

Thiruvananthapuram Centra" and )iravom *oad is

1: Ms( ho much time it i"" ta'e to reach

)iravom *oad stationB• Here the distance given is 1: Ms.• The speed is givenV >: MHr• The formu"a to find out the time is• Time = Distance Speed

• .P. Time = %&

*+ = 3 hours

I""ustrative E? No 0


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I""ustrative E?.No. 0• Qn. Ahi"ash is c#c"ing at a speed of > M per

hour. Ho much distance i"" he cover in 0 hours

and /: minutesB

• Here the time is 0 hrs and /: minutes.• ie.( 0 hours.

• The speed is givenV > MHr• The formu"a to find out the distance is• !istance = Speed " #ime

• .P. 8 x 5.5 = 44 KMs• Shortcut: distance covered in 1 hr = 8• So, in 5 hours, 5 x 8 = 40• In 1 hr 8KM covered; so in ½ hour 4 KM.

• .P. 1:41 3 11

A d h th di t i t t


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Solution:• By the theorem, x"& x 55 = - 'm)hr • "& ' 55

(o) the average s*eed is !1.! km/hr

THEOREM: If a certain distance is covered at ‘X km/hr’and the same distance is covered at ‘Ykm/hr’, then the

averae s!eed in the "ho#e $o%rne& is ' XY km/hr X ( Y

E?. No.;. A man covers a certain distance # car driving at 2:Mhr and he returns ac' to the starting point riding on a

scooter at 00 'mhr . &ind his average speed for the ho"e

Gourne#.

Average speed hen the distance remain constant

and the first and second speeds varies

Average Speed


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Average SpeedE?.No. 2. A man trave"s in a car from A to $ at a speed of ;:

Ms per hour and returns to A at a speed of 1: Ms perhour. In his entire trave"( hat as his average speed in

M per hourB The average speed per hour 3 ' XY km/hr

X ( Y here ?% is thespeed in M per hour of trave" from A to $ and #% is thespeed in M per hour of trave" from $ to A J

Sustituting 7 ? ;: ? 1: 3 1>:: 3 1> ;: 4 1: 5::Ca"cu"ating average speed hen e!ua" distance trave"ed

thrice

The average speed per hour hen e!ua" distance trave"edthrice at a speed of a%( % and c% Mhr i"" e or'edout using the formu"a /ac a 4 c 4 ca7 here a% isthe speed in M per hour of trave" from A to $ and % isthe speed in M per hour of trave" from $ to A and c% is

the speed in M per hour of trave" from A to CJ

Average Speed hen e!ua" distance trave"ed thrice at a


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g p !

speed of a%( % and c%

E?. No.>. A man trave"s in a car from A to $ at a speed of >:Ms per hour and returns to A at a speed of ;: Ms per

hour. Again he trave"s to $ at a speed of /: M per hour.In his entire trave"( hat as his average speed in Mper hourB

The average speed per hour hen e!ua" distance trave"edthrice at a speed of a%( % and c% Mhr i"" e or'ed

out using the formu"a /ac a 4 c 4 ca7 here a% isthe speed in M per hour of trave" from A to $ and % isthe speed in M per hour of trave" from $ to A and c% isthe speed in M per hour of trave" from A to CJ

Sustituting 7

/ac 3 / ? >: ?;: ? /: 3 1/::: 31/:::a 4c 4 ca >: ? ;: 4 ;: ? /: 4 /: ? >: 1>:: 45>::41:: 6:::

3 1/6 3 1>

To find the distance


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To find the distance

E?. No.6. A man trave"s a certain distance on a scooter from his houseand office. Having an average speed of /: Ms per hour( he is "ate# 5: minutes. Hoever( ith a speed of 1: Ms per hour( hereaches his office 0 minutes ear"ier. What is the distance eteenhis house and officeB

Solution:


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DISTANCE, SPEED & TIME


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KISTANCE( S)EEK + TIME

T*AINS


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T*AINSQns of Trains% is actua""# the same as ZTime + Kistance[

The on"# difference is that the "ength of the moving oGect

=Train9 is a"so considered in this.• Some important things to e noted are781. When the two trains are moving in opposite directions

their speed should be added to find the relative speed.

2. When the two trains are moving in the same directiontheir the relative speed is the difference of their speeds.

3. When the train passes a Plat form it should travel thelength equal to the sum lengths of the train & Plat formboth.

4. ength of a telegraph post! signal post! a stationar" manor a lamp post or telephone post etc.! shall be ta#en as $.

0. To convert mhr in to metersecond mu"tip"# ith 5:::;: ?;: 3 0 5>

;. To convert metersecond in to mhr mu"tip"# ith ;: ?;:5::: 35>

E? No i Ho man# seconds i"" a train of 5::meters "ength running at


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E?.No. i. Ho man# seconds i"" a train of 5::meters "ength running at

the rate of /; M an hour ta'e to pass a te"egraph postB

• So"ution7

• In passing the post the train must trave" its on "ength.=The "ength of the Te"egraph post is :9

• No( /; MHr 3 /; ? 0 3 5: msec• 5>

• .+. The time re!uired for the train to pass the post 3 5::3 5: seconds• 5:

E?.ii. Ho "ong does a train 55: m "ong running at the rate of /; 'mhr

ta'e to cross the ridge 5/ m in "engthB

• So"ution7 In crossing the ridge the train must trave" its on "engthp"us the "ength of the ridge.

No( /; MHr 3 /; ? 0 3 5: msec

5>

.P. The time re!uired for the train to cross the ridge3 55:45/ 3 1.

5: seconds

E? No iii A Train of "ength :: meters ta'es ; seconds to pass a signa"


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E?.No. iii. A Train of "ength :: meters ta'es ; seconds to pass a signa"

post. If so( ho much time it i"" ta'e to pass a p"atform of 1:: metersB

• The "ength of the train 3 :: m

• =The "ength of the signa" post is :9• The "ength of the train 4 )"at form 3 :: 4 1:: 3

;:: m

• The time ta'en to pass ::m 3 ; seconds

• So time ta'en to cover ;::m 3 ; ? ;::• ::

• 3 5> seconds• Dr ;:: is / times that of :: • To pass :: m( train ta'es ; seconds• Hence( to cover ;:: m( / ? ; 3 5> seconds

To convert mhr in to metersecond mu"tip"# ith 5::: ;: ?;: 3 05>


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To convert mhr in to metersecond mu"tip"# ith 5::: ;: ?;: 05>

To convert metersecond in to mhr mu"tip"# ith ;: ?;: 5::: 35>0

E?.No. iv. A Train of "ength 0: meters hi"e running at a

speed of /; 'm per hour ta'es ho much time to pass a"amp post eside the trac' B

The "ength of the train 3 0: m

• =The "ength of the 35:msec• Train runs 5: ms in a second

• Time ta'en to cover 0:m 3 0:• 5:• 3 0 seconds• So the train ta'es 0 seconds to pass a "amp post

eside the trac'.

E? No v A Train of "ength 50: meters is running at a speed of


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E?.No. v. A Train of "ength 50: meters is running at a speed of

01 'mhr. Ho much time it i"" ta'e to tide over a ridge of

"ength 10: metersB

• The "ength of the train 3 50: m• The "ength of the train 4 "ength of ridge 3 50: 4

10: 3 ;:: m

• The time ta'en in msec. to cover the ridge at aspeed of 01 'ms hour 3 01 ? 05> 3 01 ? 0• 5>• 3 50 seconds

• So time ta'en to cover ;::m 3 ;:: 50• 3 1: seconds• The train i"" ta'e 1: seconds to tide over a ridge of

"ength 10: meters.

E?.No. vi. A Train of "ength 5/: meters is running at a speed


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of 1: 'mhr. Another train of 6:m "ength is running at a

speed of / 'mhr in a para""e" trac' in the opposite direction.

Ho much time i"" e ta'en # the trains to cross mutua""#B

• The trains are running in opposite directions andhence( the re"ative speed of the trains together 31: 4 / 3 2 'mhr.

• The time ta'en in msec. at a speed of 2 'ms hour 3 2 ? 05> 3 2 ? 0 3 1 ? 0

5>

3 : m second

Kistance to e trave"ed to cross each other3 5/: 4 6: 3 : m

• Time ta'en to cover :m 3 : : 3 55 seconds

• So( the trains i"" ta'e 55 seconds to cross each other


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E?.No. viii. A Trains overta'es a person standing in the


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p"atform in 1 seconds and a p"atform of "ength 5:: meters in

51 seconds. &ind out the "ength and speed of the trainB

Here( in order to pass the p"at form( the train shou"drun the distance of the p"atform 4 "ength of the train.

In order to pass the person standing in the p"at form(the train need to run the "ength of the train on"# forhich it ta'es 1 seconds.

Hence( the time ta'en to run 5::ms "ength( p"at forma"one( 3 51 8 1 3 5: seconds

That means( in 5: seconds the train runs 5:: m

So( the speed of the train in Mhr i"" e 5: ? 5>

0

3 /; 'mhr.

The train runs 5: m in one sec. Hence( in 1 seconds iti"" run 1: m. The "ength of the train is( thus( 1:m.

E?.No. i?. A Train of "ength 0: meters is running at a speed


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of 1: 'mhr. A person is running at a speed of 1 'mhr in the

same direction. Ho much time i"" e ta'en # the train to

cross himB

The "ength of the train 3 0: m

• =The "ength of the person running is ta'en as :9When the train and the man are running in the same

direction( the re"ative speed together i"" e 31: 8 1 3 /; 'mhr on"#.

• The distance the train to run for overta'ing therunning man 3 0: 4 : 3 0:

• The time ta'en in msec. at a speed of /; 'ms hour 3 /; ?05> 3 /; ? 0 3 ? 0 3 5: 5> ie.(5: msecond

So time ta'en to cover 0: m 3 0: 5: 3 0seconds

E?.No. ?. If the man in the ear"ier !uestion is running in the


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opposite direction of the train( ho much time i"" e ta'en

# the train to over ta'e the manB

• The train and the man are running in opposite directionsand hence( the re"ative speed of the train and mantogether 3 1: 4 1 3 11 'mhr.

• The time ta'en in msec. at a speed of 11 'ms hour 3 11? 05> 3 11 ? 0 m second

• 5>• Kistance to e trave"ed to cross each other 3 0: 4 : 30: m

• Time ta'en to cover 0:m 3 0: 11 ? 0 msec

• 5>• 3 0: ? 5>• 11 ? 0 3 : 0 m second• 55

So the trains i"" ta'e : and 055 seconds to cross the man.

E?.No. ?i. A passenger sitting in a train hich is running at a speed of ;: 'mhr

oserves that another train running in the opposite direction is passing him in ;


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oserves that another train running in the opposite direction is passing him in ;

seconds. The train running to the opposite direction has "ength of 5:m. If so(

hat is speed of the trainB

• The train running in opposite direction has 5:m "engthand it crosses the train in hich the man is sitting in ;seconds.

• Therefore( the re"ative speed 3 5:• ; 3 : msecond

• ie.( hen e convert the speed in to 'mhr 3 : ? 5>• 0• 3 2 'mhr hich is the sum tota" of the speeds of oth

the trains.•

Speed of the first train is given ie.( 3 ;: 'mhr • Hence( the Speed of the second train is 3 2@ ;:'mhr

• 3 5 'mhr

E?.No. ?ii. A Train running at a speed of 1: 'mhr ta'es one


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minute to overta'e another train hich is running in the

same direction in another trac' at a speed of 'mhr. The

"ength of the first train is 50:m. What is the "ength of the

second trainB

• The "ength of the first train 3 50: m• The trains are running in the same direction( the

proportionate speed together i"" e 3 1: 8 35> 'mhr on"#.

• The time ta'en in msec. at a speed of 5> 'ms hour 3 5> ?0 3 0 msecond

5> ie.( in one second gains 0m• The distance the train runs in one minute 3 0 ?;:3 /:: m

• So the "ength of the second train 3 /::850:

• 350: m


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TIME &

WORK

TIME ANK WD*


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TIME ANK WD*• Wor' and Time is an important area in a""

competitive e?aminations inc"uding $an'.• There are traditiona" methods of anseringsuch !uestions.

• $ut overa"" time management ecomesdifficu"t.• It is possi"e to arrive at the correct

ansers of the pro"ems in or' and time%ver# !uic'"# through short cut methods.

•


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If M 1 persons can do W 1 works in D 1 days and M 2 persons

can do W 2 work in D 2 days, then we have a eneral

formula in the relationship of M 1 D 1W 2 ! M 2 D 2W 1 • The aove re"ationship can e ta'en as a ver#

asic and a""8in8one formu"a. • We a"so derive7

i9. %ore men less da"s and converse"# more da"s less menii9 %ore men more wor# and converse"# more wor# more men

iii9. %ore da"s more wor# and converse"# more wor# moreda"s

If e inc"ude the or'ing hours =sa# T5 and T9 for the togroups( then the re"ationship is M 1 D 1 1 W 2 ! M 2 D 2 2 W 1

Again( if efficienc# =sa# E5 and E9 of the persons in the to

groups is different( then the re"ationship is M

1

D 1

1

' 1

W 2

! M 2

D 2

2

' 2

W 1

E?.No.5. A% can do a piece of or' in 0 da#s. Ho man# da#s i"" he ta'eto comp"ete / or's of the same t#peB


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to comp"ete / or's of the same t#peB

Solution:

• We reca"" the formu"a ’more .or' more

days’ • It simp"# means that e i"" get the anser #

mu"tip"ication.

• Thus( our anser 3 0 ? / 3 50 da#s.This is the ver# simp"e a# of so"ving the !uestion(

ut #ou shou"d 'no ho the ver# asic and a""8in8one formu"a can e used in this !uestion.

• *eca"" the asic formu"a 7 M 1 D 1 W 2 ! M 2 D 2 W 1

As A% is the on"# person to do the or' in oth thecases( M 1 ! M 2 ! 1 "So useless to carry it#

(ubstituting) D 1 ! $ days, W 1 ! 1, D2!% &nd W 2 !'

Putting the values in the formula we have ) * x 3 + D2 ? 5

E?.No.. 5; men can do a piece of or' in 5: da#s. Ho

man men are needed to comp"ete the or's in 1: da sB


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man# men are needed to comp"ete the or's in 1: da#sB

Solution:

• We reca"" the formu"a ’more days less men’ • It simp"# means that to do a or' in 5: da#s( 5;

men are needed.

Thus( to do the or' in one da#( 5; ? 5: men are needed.

• So( to do the or' in 1: da#s( 5; ? 5: 3 1 menare needed. 1:We shou"d 'no ho the asic formu"a or's here7 • *eca"" the asic formu"a 7 M 1 D 1 W 2 ! M 2 D 2 W 1

M 1 ! 1(, D 1 ! 1), W 1 ! 1, and M 2 ! *, D2! +), W 2 !1

(ubstituting) M 1 D 1 W 2 ! M 2 D 2 W 1,e getV5; ? 5: 3 M2 x 40

Or M2 3 5; ? 5: 3 1 men are needed.

1:

THEOREM: If ‘)’ can do a !iece of "ork in ‘X’ da&s and‘*’ can do it in ‘Y’ da&s then ) and * "orkin toether


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* can do it in Y da&s, then ) and * "orkin toether"i## do the same "ork in XY da&s.X ( Y

• E?. No./.Qn. A can do a piece of or' in 0 da#s and $

can do it in ; da#s. Ho "ong i"" the# ta'e if

oth or' togetherB• Solution:• By the theorem, & B can do the work in 5 x ! days• 5 '

!• = ,& = 8 da-s 11 11r A0 can do 1/5 ork in a da- and 20 can do 1/! ork in a da-.

(o) A and 2 together can do 1/5 ' 1/! ork in a da-. Therefore A and2 can do the ork in 1 da-s = ,& = 5

E?. No.1. A can do a piece of or' in 5 da#s and $ can do itin : da#s Ho "ong i"" the# ta'e if oth or' togetherB


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Solution:

• By the theorem, & B can do the work in 1 x & days• 1 '

&

= %& = 15 = ".5 da-s

, r A0 can do 1/1 ork in a da- and 20 can do 1/&ork in a

da-.

(o) A and 2 together can do 1/1 ' 1/& ork in a da-.Therefore A and 2 can do the ork in 1 da-s =%& = ". 5 da-s

THEOREM: If ‘)’ can do a !iece of "ork in ‘X’ da&s and‘*’ can do it in ‘Y’ da&s, then ) and * "orkin toether

"i## do the same "ork in XY da&s.X ( Y

in : da#s. Ho "ong i"" the# ta'e if oth or' togetherB

E?. No.0. ) can do a piece of or' in ; da#s and Q can do itin 5> da#s Ho man# da#s i"" the# ta'e to comp"ete the


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Solution:• By the theorem, & B can do the work in ! x 18 days• !

'18

= ! x 18 = 3 = %.5 da-s % r A0 can do 1/! ork in a da- and 20 can do 1/18ork in a

da-.

(o) A and 2 together can do 1/!' 1/18 ork in a da-.

Therefore A and 2 can do the ork in 1 da-s =

THEOREM: If ‘)’ can do a !iece of "ork in ‘X’ da&s and

‘*’ can do it in ‘Y’ da&s, then ) and * "orkin toether"i## do the same "ork in XY da&s.

X ( Y

in 5> da#s. Ho man# da#s i"" the# ta'e to comp"ete the

or' if oth or' togetherB

E?. No.;. *aGu can do a piece of or' in :da#s and SaGeevcan do it in /: da#s Ho man# da#s i"" the# ta'e to


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Solution:• By the theorem, & B can do the work in & x ,&

days• &

',& = & x ,& = !&& = 1 da-s 4ence) Ans.,6

5& 5&r 7a90 can do 1/& ork in a da- and (aeev0 can do

1/,&ork in a da-.

(o) A and 2 together can do 1/&' 1/,& ork in a da-.

THEOREM: If ‘)’ can do a !iece of "ork in ‘X’ da&s and‘*’ can do it in ‘Y’ da&s, then ) and * "orkin toether

"i## do the same "ork in XY da&s.X ( Y

can do it in /: da#s. Ho man# da#s i"" the# ta'e to

comp"ete the or' if oth or' togetherB

=59 ; da#s =9 6 da#s=/9 5 da#s =19 51 da#s =09 None of these

THEOREM: If ‘)’, * and + can do a !iece of "ork in ‘X’,‘Y’ and‘ ’ da&s res!ective#& then a## of them "orkin


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Y and da&s res!ective#&, then a## of them "orkintoether "i## do the same "ork in XY da&s. XY ( Y ( X

• E?. No.2.Qn. A can do a piece of or' in 0 da#s and $

can do it in ; da#s. If C ho can do the or'

in 5 da#s Goins them( ho "ong i"" the#ta'e to comp"ete the or' togetherB• Solution:• By the theorem, & B- can do the work in

• 5 x ! x 1 days• 5x! ' !x1 ' 5x1• = ,!& = da-s 1! 3

E?. No.>. e"an ta'es 6 da#s to p"ough a )add# fie"d( ,e"u ta'es 5:da#s and Muthu can do it in 50 da#s. Ho man# da#s i"" the# ta'e to


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Solution:

• By the theorem, & B- can do thework in

• 3x 1& x 15 days 3x1& ' 1&x15 ' 15x3• = 18 = , , da-s

5 5

THEOREM: If ‘)’, * and + can do a !iece of "ork in ‘X’,

‘Y’ and‘ ’ da&s res!ective#&, then a## of them "orkintoether "i## do the same "ork in XY da&s. XY ( Y ( X

da#s and Muthu can do it in 50 da#s. Ho man# da#s i"" the# ta'e to

comp"ete the or' if a"" of them or' togetherB

E?. No.6. Keepa can do a piece of or' in 5: da#s and Indu can do it in 50da#s and Nandu i"" do it in /: da#s. Ho man# da#s i"" the# ta'e to


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Solution:• By the theorem, & B- can do the work in• 1& x 15 x ,& days• 1&x15 ' 15x,& ' ,&x1&

• = %5&& = 5 da-s 3&&



# # # # #


E?. No.5:. -eetha can do a piece of or' in 5: da#s and Smitha can do itin 5 da#s and Suma i"" do it in 50 da#s. Ho man# da#s i"" the# ta'e to


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Solution:• By the theorem, & B- can do the work in• 1& x 1 x 15 days• 1&x1 ' 1x15 ' 15x1&

• = 1&x1x15 = % da-s %5&



# # # # #


Practice Questions


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5. Mohan can do a piece of or' in 5: da#s and*amesh can do it in 5 da#s. Ho man# da#s i""

the# ta'e to comp"ete the or' if oth or'togetherB. In the aove !uestion if Suresh ho can do the

or' in 50 da#s Goins them( ho "ong i"" the#ta'e to comp"ete the or' togetherB

/. Mohan can do a piece of or' in : da#s andSohan can do it in /: da#s. Ho "ong ou"d the#ta'e to comp"ete the it( if oth or' togetherB

1. *aGu( *in'u and *am can do a or' in ;( 5 and

1 da#s respective"#. In hat time i"" the# do itor'ing togetherB 0. A can do a or' in 2 da#s. If A does tice as

much or' as $ in a given time( find ho "ong Aand $ ou"d ta'e to do the or' together.

THEOREM: If ‘)’ and * can do a !iece of "ork in ‘X’ da&sand ) a#one can do it in ‘&’ da&s, then * a#one can do the


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& & ,same "ork in XY da&sY-X

• E?. No.5.Qn. ama"a and ,ima"a can do a piece of

or' in ; da#s. ama"a a"one can do it in 6

da#s. In ho man# da#s can ,ima"a a"onedo the or'B• Solution:• By the theorem, .imala alone can do the work in

• ! x 3 days 3 : ! = 5% = 18 da-s

,

E?. No.. A and $ can do a piece of or' in5 da#s( $ and C in 50 da#s(C and A in : da#s. Ho "ong ou"d each ta'e separate"# to do the same


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Solution:

• By the theorem, 2"& B-# can do the work in• 1 x 15 x & = 5 da-s• 1x15 ' 15x& ' &x1• = ,!&& = 5 da-s "&

Therefore) "& B-# can do the work in 2 / $ ! 1)days0 " less men more days’ 9

No( A a"one can do the or' in 1& x 15 = ,& days

/As A= "& B-# " B-#3 15 :1&

$ a"one can do the or' in 1& x & = & days/As 0= "& B-# " &-#3 & :1&

C a"one can do the or' in 1& x 1 = !& days

/As c= "& B-# " &B#3 5 :1&

C and A in : da#s. Ho "ong ou"d each ta'e separate"# to do the same

or'B

E?. No.. Suresh and -anesh can do a piece of or' in 5: da#s(-anesh and Sreenadh in 5 da#s( Sreenadh and \agadeesh in 50 da#s.


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Solution:• By the theorem, 2"& B-# can do the work in• x1& x 1 x 15

1&x1 ' 1x15 ' 15x1&

• = x1& x 1 x 15%5&

• = ,!&& = 8 da-s %5&

Therefore) "& B-# can do the work in 4 days0" more men less days’ 9

Hence( the anser is 7 =19 > da#s

# ( g #

Then( numer of da#s re!uired to do the same or' if the# or'ed

togetherB

=59 ; da#s =9 6 da#s=/9 5 da#s =19 > da#s =09 None of these

E?. No./. A and $ ta'es ; da#s to construct a a"" . A a"one can do it in6 da#s. Ho "ong ou"d $ ta'e separate"# to do the same or'B


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Solution:• THEOREM: If ‘)’ and * can do a !iece of "ork in ‘X’ da&s and ) a#one can do it in

‘&’ da&s, then * a#one can do the same "ork in XY da&s

Y- X

! x 3 = 5%• 3:! , = 18 da-s• = ,!&& = 5 da-s "&

E?. No.1. Mohammed and Mathe together does a or' in ; da#s.Mohammed a"one can do it in 5: da#s. Ho "ong ou"d Mathe ta'e

separate"# to do the same or'B XY da&s

Y- X

! x 1& = !&

• 1& : ! % = 15 da-sE?. No.0. A( $ and C can do a piece of or' in 5 da#s( A and $ together

ta'es 5; da#s. If so( Ho "ong ou"d C a"one ta'e to do the sameor'B

XY da&s 1 x 1! = 13

Y- X 1! : 1 % = %8 da-s

# g p #

t

E?. No.;. *aman can do a piece of or' in ; da#s. rishnan needs 5da#s to do the same or'. The or' done Goint"# # these to in a da#


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Solution:• THEOREM: If ‘)’ and * can do a !iece of "ork

in ‘X’ da&s and ) a#one can do it in ‘&’ da&s,then * a#one can do the same "ork in XY da&s

Y- X To do the "ork Raman 0rishnan toethertakes

! x 1 = "! ' 1 18 =% da-s ie.) ;ovind alone can do

it in % da-s.

Then) if #rishnan and ;ovind together does

that ork the- ill take

# G # # #

i"" e done # -ovind a"one in a da#. Then( ho "ong ou"d e ta'en

to do the same or' Goint"# # rishnan and -ovindB

t

E?. No.2. If / men or 1 oman can reap a fie"d in 1/ da#s( ho man#da#s i"" 2 men and 0 oman ta'e to reap the same fie"dB


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Solution: 1irst Method• Three men reap 5 of the fie"d in a da# 1/

.+. 1 man rea*s 1

%, x, of the


84/130

Solution: 5 men and 5; o#s can do the or' in 0 da#s O=59

5/ men and 1 o#s can do the or' in 1 da#s O=9No( it is eas# to find that if the no. of or'ers is mu"tip"ied # an#

numer( the time must e divided # the same numer. =derived from7

1ore .or'ers less time %9• Hence( mu"tip"#ing the no. of or'ers in =59 and =9 # 0 and 1

respective"#( e getV• 0 =5 men 4 5; o#s 9 can do or' in 5 = 1 da- 5 • 1 =5/ men 4 1 o#s 9 can do or' in % = 1 da- %

r) 51m'1!?6 = %1,m ' %?6r) !&m'8&? = 5m ' 3!?

r) !&m:5 m= 3!?:8&?

r) 8m = 1!?

.+. 1 man = ?o-s

Th9s)@..

# # g # p

the same fie"dB

t

E?. No.>OO


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• 1 men ' 1! ?o-s = % ?o-s ' 1! ?o-sand

• " men ' 1& ?o-s = 1% ?o-s ' 1& ?o-s• The 9estion no ?ecomes B• Cf %& ?o-s can do a ork in 5 da-s ho

long ill % ?o-s take to do it0• No) the ?asic form9la M5K5 W 3 MK W5 ( ehave

• %& x 5 = % x D 2

• r K 3 1: ? 01

= 8 1, da-s

2 men and 5: o s i"" ta'e 8 1

E?. No.6. A can do a piece of or' in ; da#s. $ needs > da#s to do thesame or'. C ta'es as "ong as A and $ ou"d ta'e or'ing together.


86/130

Solution:"& B# can do the work in ! x 8 = %da-s

! ' 8

".+. E takes % da-s to com*lete the ork.

"

.+. 2 and E takes % x 8

"FFF = % x 8 = da-s

% x 8 % ' 5!

5

same or'. C ta'es as "ong as A and $ ou"d ta'e or'ing together.

Then( ho "ong ou"d e ta'en to do the same or' Goint"# # $ and CB

t

E?.5:. A is tice as good a or'man as $. Together( the# finish the or' in 51da#s. In ho man# da#s can it e done # each separate"#B


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• So"ution7

• et - finish the wor# in 2x da"s. (ince is twice active

as -! finish the wor# in x da"s. /0- finish the wor# in2 ? 3 51 da#s

/? Dr ? 3 5

.P. A ta'es 5 da#s to comp"ete the or' and $ finishes theor' in 5 ? 3 1 da#s

Quic'er Approach7 Tice 4 Dne time 3 Thrice active person does the

or' in 51 da#sThen one time active person $( i"" do it in 51 ? / 3

1 da#s.

.P. Tice active person A i"" do it in 1 3 5 da#s


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E?. No.5. If 0 men can ma'e a road in /0 da#s( then 2 men

can do the same or' in B


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Solution:Short cut

If M1 men do a work in D1 days and

M2 men can do a work in D2 daysthen, M1D1 ! M2D25 x ,5 = " x D da-s

.+. D = 5 x ,5 = 15 da-s "

Hence( Ans7 =19 50 da#s

can do the same or' inOOB=59 5:: da#s =9 60 da#s=/9 5: da#s =19 50 da#s =09 None of these

E?. No.5/. TaGmaha" as ui"t # :::: men or'ing for :

#ears. Then( ho man# men ou"d have een re!uired to


90/130

Solution:

Short cut

If M1 men do a work in 51 years and M2 men can do a work in 52 yearsthen, M151 ! M252

&&&& x & = $ x 1&

.+. $ = &&&& x & = %&&&& men

1&

Hence( Ans7 =19 1::::

#ears. Then( ho man# men ou"d have een re!uired to

ui"t it in 5: #earsB=59 5:::: =9 0::: =/9 /:::: =19 1::::=09 None of these

E?. No.51. 5: men can finish a or' in 5 da#s # or'ing >hours dai"#. Then the numer of da#s re!uired to finish the


91/130

Solution:Short cut 6If M1 men do a work in D1 days 7y workin 81

hours daily and M2 men can do the samework in D2 days 7y workin 82 hours dailythen, M1D181! M2D2829

= 1& x 8 x 1 = 1! x D x " da-s

.+. D = 1& x8 x 1 = 15 da-s 1! x "

Hence( Ans7 =59 500 da#s

or' # 5; men or'ing 2 hours dai"# OOB=59 50 da#s =9 5; da#s=/9 : da#s =19 1 da#s =09 0

E?. No.50. 5: carpenters can ma'e 51 ta"es in 5 da#s #or'ing > hours dai"#. Then the numer of ta"es that can e


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Solution:Short cut 6If M1 men do W1 work in D1 days 7y workin

81 hours daily and M2 men can do W2 workin D2 days 7y workin 82 hours daily then, M1D181! M2D2829

W1 W2

= 1& x 1 x 8 = 8 x 15 x " 1% G

.+. G = 8 x15 x "x1% = " ta?les

1& x 1 x 8Hence Ans7 " ta?les

made # > carpenters in 50da#s # or'ing 2 hours dai"#OB=59 > =9 2 =/9 5: =19 5; =09 0

E?. No.5;. men or 1 omen can do a or' In > da#s. Then1 men and > omen together can finish the or' in..B=59 > =9 2 =/9 5: =19 5; =09 0


93/130

Solution:

Short cut 6If men or ; women can finish a work in /

days, then M men and < women can finish the

work in, ; / 9 < ;M

= x % x 8 = x % x 8

x8'%x% 1!'1!

.+. 1 men and > omen together can finish the or'in = x% x 8 = " da-s

,Hence( Ans7 =9 " da-s

=59 > =9 2 =/9 5: =19 5; =09 0

E?. No.52. -opu can do a or' in > da#s( Suman can do it in5> da#s. The# together or'ed and finished a or'. Their

ti h "d di id d i th ti B


94/130

Solution:Short cut 6 can do a work in / days, ; in y days and = in

> days0 If they worked toether theirremuneration should 7e divided in the ratio , 1: 1 : 1 9 / y >

= 1 B 1 B 1 HE$ of 8)1)18 " 8 1 18

.+. " x 1 B "x 1 B "x 1

8 1 18 = 3B!B%

Hence( Ans7 =19 67;71

remuneration shou"d e divided in the ratio OB=59 577/ =9 /7172 =/9 5:75575 =19 67;71 =09 None of these

)ractice Questions 5 A and $ can do a piece of or' in /: da s and 1: da s


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5. A and $ can do a piece of or' in /: da#s and 1: da#srespective"#. The# egan the or' together ut A "eftafter some da#s and $ finished the remaining or' in 5

da#s. After ho man# da#s did A "eaveB. A certain numer of men comp"ete a piece of or' in ;:

da#s. If there ere > men more( the or' cou"d efinished in 5: da#s "ess. Ho man# men ere thereorigina""#B

/. > chi"dren and 5 men comp"ete a piece of or' in 6da#s. If each chi"d ta'es tice the time ta'en # a man tofinish the or'( in ho man# da#s i"" 5 men finish thesame or'B

1. men and / oman can finish a piece of or' in 5: da#s(

hi"e 1 men can do it in 5: da#s. In ho time man# da#si"" / men and / omen or'ing together finish itB 0. / men and 1 o#s do a or' in > da#s hi"e 1 men and 1

o#s finish it in ; da#s. men and 1 o#s i"" finish it inO.da#sB

Anser e#5 $# direct formu"a7


96/130

5. $# direct formu"a7• *e!uired no. of da#s = ,& x %& {%& : 1 I = 1 da-s ,& ' %& { %& I. Het there ?e x men originall-. Then 1 man ill do the ork in !&x da-s.

Cn the second case 1 man ill do the ork in x ' 865&da-s

No) !&x = 5& x ' 86. .P. 3 1:: = %& men.5:

/. If each chi"d ta'es tice the time ta'en # a man( > chi"dren 3 1 men..P. > chi"dren 4 1 men 3 5; men do the or' in 6 da#s.

.P. 5 men finish the or' in 6?5; = 1 da-s

51. 1 men do in 5: da#s. .P. men do in : da#s .P. / omen do in 5:?: = & da-s

: @ 5:And / men do it in 1: da#s .P. / men 4 / omen do it in :?1: / 3 :?1: = 8 da-s

/ :4 1: / 5::

0. / men 4 1 o#s do in > da#s O.=59 1 men 4 1 o#s do in ; da#s O.=9

Sutracting =59 from =9 e have 7 5 man does in >?; 3 1 da#s O.=/9 > @ ;.P. / men do the or' in 1 3 > da#s O..=19 &rom =59 and =19 e conc"ude that o#s do

no or'. / Hence( men 4 1 o#s 3 men on"# and do in 1 5da#s


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PIPES & CISTERNS

)I)ES ANK CISTE*NS

i d i 7l l h h f


98/130

• ipes and cisterns pro7lems are almost the same as those of?ime and Work pro7lems0

• The on"# difference ith )ipes and cisterns pro"ems is that there

are out"ets as e"" as in"ets.• In"et7 A pipe connected ith a tan' =or a cistern or reservoir9 is

ca""ed an inlet) if it fi""s it. • Dut"et7 A pipe connected ith a tan' =or a cistern or reservoir9 is

ca""ed an o9tlet) if it empties it.

i9. If a pipe can fi"" a tan' in x hours( then the part fi""ed in 5 hour is 5 /

ii9 If a pipe can empt# a tan' in " hours( then the part emptied in 5 hour is5

y

iii9. If a pipe can fi"" a tan' in x hours and another pipe can empt# a tan' in" hours( then the part fi""ed in 5 hour( hen oth the pipes areopened 3 5 8 5J

x " J

.P. Time ta'en to fi"" the tan' hen oth the pipes are opened 3 ?#

y-x

)I)ES ANK CISTE*NSO&ormu"aei 9 If i fi"" t ' i h d th i fi"" th


99/130

iv9. If a pipe can fi"" a tan' in x hours and another pipe can fi"" the sametan' in " hours( then the net part fi""ed in 5 hour( hen oth thepipes are opened 3 5 4 5 J

x " J.P. Time ta'en to fi"" the tan' hen oth the pipes are opened 3 ?#

y+x

v9. If a pipe can fi"" a tan' in x hours and another pipe can fi"" the sametan' in " hours( ut a third one empties the fu"" tan' in ] hours( and

a"" of them are opened together( the net part fi""ed in 5 hour 3 5 4 5 8 5 J

x " J

.P. Time ta'en to fi"" the tan' hen a"" the pipes are opened 3 ?#]

yz+xz-xy

vi9. A pipe can fi"" a tan' in x hours. Kue to a "ea' in the ottom it is fi""edin " hours. If the tan' is fu""( the time ta'en # the "ea' to empt# thetan' 3 ?#

" 8 x

E?. No.5. To pipes A and $ can fi"" a tan' in /; hours and 10 hoursrespective"#. If oth the pipes are opened simu"taneous"# ho much time

ou"d e ta'en to fi"" the tan'B


100/130

Solution:

• 1orm%#a : XY hours That is; Y( X

,! x %5 = 1!&• ,! '%5 81 = & ho9rsE?. No.. A pipe can fi"" a tan' in 50 hours. Kue to a "ea' in the ottom it is

fi""ed in 2$ hours. If the tan' is fu""( ho much time the "ea' ou"dta'e to empt# itB

Solution: 1orm%#a : XY hrs Y- X

15 x & = ,&&

• & :15 5 = !& hrsE?. No./. A tap can fi"" the cistern in > hours and another can empt# it in

5; hours. If oth the taps are opened simu"taneous"# ho much timei"" e ta'en to fi"" the cisternB

1orm%#a : XY da&s 8 x 1! = 18 Y- X 1! : 8 8 = 1! hrs

ou"d e ta'en to fi"" the tan'B

t

ILLUSTRATIVE EXAMPLE-4


101/130

iv9. A pipe can fi"" a tan' in 2$ hours hi"e pipe $ a"one can fi"" the sametan' in 3$ hours( ut a third one empties the fu"" tan' in 1: hours. If a""the pipes are opened together( ho much time i"" e needed to ma'ethe tan' fu""B

Solution: :: 8 ;:: 51:: 2 2 hrs



102/130

v9. To pipes A and $ can fi"" a cistern in 5 hour and 20minutes respective"#. There is a"so an out"et C. If a"" the

pipes are opened together( the tan' ecomes fu"" in 0:minutes. Ho much time i"" e ta'en # C to empt# thefu"" tan'B

Solution:Work done 7y - in 1 minute 3 5 4 5 8 5 J 3 5 4 5 8 5 J

7$ 6* *$ J 2$ 3$ 4$5+ / 3 5 /:: 5:: .P. The C can empt# the tan' in 5:: minutes

Dr direct method # using formu"a7 1orm%#a : ?#] 3 ;: ? 20 ? 0: 30:: 3 5:: min

yz+xz-xy 20?0: 4;:?0:8 ;:?20 0:.P. Time ta'en # C to empt# the tan' 3 5:: min.



103/130

vi9. In hat time ou"d a cistern e fi""ed # three pipeshose diameters are 5 cm( 55 /cm( cm( running together(

hen the "argest a"one i"" fi"" it in ;5 minutes( the amountof ater f"oing in # each pipe eing proportiona" to thes!uare of it%s diameterB

Solution: In 1 minute the pipe of 2 cm diameter fills 5 of the cistern

;5 In 1 minute the pipe of 1 cm diameter fills 5 ? 5 of the cistern ;5 1 O.=9

In 1 minute the pipe of 55 /cm diameter fills 5 ? 1 of the cistern

;5 O=9 .P. In 5 minute 5 4 5 4 1 J 3 5 ;5 ;5 ? 1 ;5 ? 6J /; of the cistern is fi""ed.

.P. The ho"e cistern is fi""ed in /; minutes=9 Amount of ater f"oing is proportiona" to the s!uare of the diameter

of the pipe. Since cm diameter fi""s

5

;5 of the cistern( 5cm fi""s

I


104/130

vii9. There is a "ea' in the ottom of the cistern. When thecistern is thorough"# repaired( it ou"d fi""ed in / hours.

It no ta'es an hour "onger. If the cistern is fu""( ho"ong the ou"d the "ea' ta'e to empt# the cisternB Solution: =e@uired time ! /.0 ? 1 1 @ /.0 3 > hours

viii. A tan' has a "ea' hich ou"d empt# it in > hours. A tapis turned on hich admits ; "iters a minute into the tan'. Itis no emptied in 5 hours. Ho man# "iters does the tan'ho"dB

Solution:

?he filler tap can fill the tank in ! 5 ? > 5 @ > 3 1 hours

.P. The capacit# of the tan' 3 1 ? ;: ? ; 3 >;1: "itres

)ractice Questions 5 )ipes A and $ can fi"" a tan' in 5: hours and 50


105/130

5. )ipes A and $ can fi"" a tan' in 5: hours and 50hours respective"#. $oth together can fi"" it inOO.hrsB

. A tap can fi"" the cistern in > hours and another canempt# it in 5; hours. If oth the taps are openedsimu"taneous"#( the time in hours to fi"" the tan' is..B

/. A pipe can fi"" a tan' in ? hours and another canempt# it in # hours. The# can together fi"" it in =#^?9O..B

1. A cistern can e fi""ed # pipes A and $ in 1 hoursand ; hours respective"#. When fu""( the tan' can eemptied # a third pipe C in > hours. If a"" the taps e

opened together( the cistern i"" e fu"" in OO..B 0. To taps can separate"# fi"" a cistern in 5: minutesand 50 minutes respective"# and hen the astepipe is open ( the# can together fi"" it in 5> minutes.The aste pipe can empt# the fu"" cistern in

OO.minutesB

Anser e#5 $# direct formu"a7


106/130

5. $# direct formu"a7• A4$ together i"" fi"" the tan' in =1& x 15 = ! ho9rs 1& ' 15

. > ? 5; = 1! hrs.5;8 >

/. ?# = ho9rs#8?

1. A4$4C i"" fi"" the tan' in1 ? ; ? > = % x ! x 8 = % = , ,/" ho9rs

; ? > 4 1 ? > @ 1 ? ; 5! "

0. To pipes =fi""ers9 fi"" the tan' in 1& x 15 = !min. O.=591& ' 15

To fi""ers 4 a "ea' i"" in 5> minutes =given9 O.=9.P. The "ea' i"" empt# the tan' in from =59 and 5> ? ; 3 6 min.

5> @ ;


107/130

PROBLEM BASED ON AGES

PROBLEM BASED ON AGES


108/130

• TD SD


109/130

• The age of the father / #ears ago as 2 times the age of hisson. At present the father%s age is 0 times that of his son.

What are the age of the father and his sonB Solution: Aet the present ae of son ! / years?hen the present ae of father is ! $/ years

' years ao , * "/ '# !$/'r */ C 21 ! $/ C 'r 2/ !14.P. 3 6 #rs ie.( Son%s age 3 6 #ears&ather%s age 3 6 ? 0 3 10 #ears

Quic'er Method

Theorem7 t 1 "ears earlier ! the father8s age was x times that of his son. t

present the 9ather8s age is " times that of his son! then present age of son will

be t 1 /x : 1

x : " and father8s age can also be wor#ed out.

(on8s age + 3 /6 : 1 + 1; + "ears .P. &ather%s age 3 6 ? 0 3 10 #ears

6 : * #

I


110/130

• At present the age of the father 0 times that of the age ofhis son. Three #ears hence( the father%s age ou"d e 1

times that of his son. What are the present age of thefather and his sonB Solution: Aet the present ae of son ! / years?hen the present ae of father is ! $/ years

' years hence , + "/ '# !$/'r +/ 12 ! $/ 'r / ! years.P. 3 6 #rs ie.( Son%s age 3 6 #ears. &ather%s age 3 6 ? 0 3 10 #ears

Quic'er Method

Theorem7 he present age of the father is " times the age of his son. t 2 "ears hence! the 9ather8s age become times the age of his son! then

present age of son will be / : 1 t 2 " : and father8s age can also be wor#ed

out.

(on8s age + /4 : 1 x 3 + + "ears .P. &ather%s age 3 6 ? 0 3 10 #ears

I


111/130

• Three #ears ear"ier the father as 2 times as o"d as his son.Three #ears hence( the father%s age ou"d e 1 times that

of his son. What are the present age of the father and hissonB Solution:Let the present age of son = x yearsThen the present age of father is = y years3 years earlier , 7 (x -3 =y-3 or 7x-y = !"##(!3 years hen$e , 4 (x +3 =y+3Or 4x + !2 = y+ 3 or 4x % y = -& #'(2ol)ing (! an* (2 e get x =& years an* y = 4 years

Quic'er Method

Theorem7 t 1 "ears earlier the age of the father was x times the age ofhis son. t 2 "ears hence! the 9ather8s age become times the age of his

son! then present age of son will be t 2 / : 1 0 t 1 /x : 1

x : and father8s age can also

be wor#ed out.

(on8s age + 3/4 : 1 03/6


112/130

in the tau"ar form as fo""os

t5#earsear"ier

)resent t #earshence

&ather%s agethat of the age

of son

? times # times ] times

(on8s age + / : 1 t 2

" :

(on8s age + t 1/x: 1

x : "

(on8s age + t 2 / : 1 0 t 1/x: 1

x :

I


113/130

• The age of a man is 1 times that of his son. &ive #earsago( the man as 1 times age as o"d as his son as at that

time. What is the present age of the manB Solution:y the ta.le, e see that /or1la ! ill .e 1se*'

• (on8s age + */ : 1 + ; + ; "rs

: 4 5• .P. &ather%s age 3 1 ? > 3 / #rs• Note 7 The re"ation eteen ear"ier and present

ages are given. So e "oo' for the formu"a derivedfrom the to corresponding co"umns of the ta"e.

• That gives the formu"a =59

(on8s age + t 1/x: 1

x : "

I


114/130

• After 0 #ears the age of a father i"" e thrice the age ofhis son hereas five #ears ago( he as 2 times age as o"d

as his son as at that time. What are their present agesB Solution:y the ta.le, /or1la 3 ill .e 1se*'

• (on8s age + */6 : 10*/3


115/130

• 5: #ears ago( Sita%s mother as 1 times o"der than herdaughter. After 5: #ears( the mother i"" e tice o"der

than her daughter. What is the present age of SitaB Solution:y the ta.le, /or1la 3 ill .e 1se*'

• =aughter8s age + 1$/4 : 101$/2


116/130

• The age of the father 0 #ears ago as 6 times the age of hisson. At present the father%s age is 1 times that of his son.What is the age of the father and his sonB

Solution: Aet the present ae of son ! / years?hen the present ae of father is ! +/ years

$ years ao , fatherEs ae ! "/ $# !+/$ r / C +$ ! +/ C$

r $/ !+).P. 3> #rs ie.( Son%s age 3 > #ears&ather%s age 3 > ? 13 / #ears

Quic'er Method

Theorem7 t 1 "ears earlier ! the father8s age was x times that of his son. t present the 9ather8s age is " times that of his son! then present age of

son will be t 1 /x : 1

x : " and father8s age can also be wor#ed out.

(on8s age + * / : 1 + 4$ + ; "ears .P. &ather%s age 3 > ? 1 3 / #ears

: 4 0

I


117/130

• The age of the grand father 1 #ears ago as 55 times theage of his grand son. At present the grand father%s age is 2times that of his grandson. What is the age of thegrandfather and his grandsonB

Solution: Aet the present ae of randson ! / years?hen the present ae of randfather is ! */ years

$ years ao , fatherEs ae ! 11"/ +# !*/+r 11/ C ++ ! */ C+r +/ !+).P. 35: #rs ie.( grand Son%s age 3 5: #ears-rand &ather%s age 3 5: ? 23 2: #ears

Quic'er MethodTheorem7 t 1 "ears earlier ! the grandfather8s age was x times that of

grand son. t present the grand 9ather8s age is " times that of his son!

then present age of grandson will be t 1 /x : 1

x : " and grandfather8s age can also be

wor#ed out.8 %

I


118/130

The sum of ages of a mother and her daughter is 0: #ears. A"so 0 #rsago( mother%s age as 2 times the age of her daughter. What are thepresent age of the mother and daughterB

Solution: Aet the present ae of dauhter ! / years?hen the present ae of mother is ! $)/ years

$ years ao , ! *"/ $# !$) /$ ! */'$ !$)/$ r 4/ ! $) C$'$ !4)

r 4/ !4) ie0, / !4)

F 4 3 5: #ears.P. 35: #rs ie.( daughter%s age 3 5: #earsSum of ages of mother and daughter 3 0:. .P. Age of mother 3 0:85:31:#rs

Quic'er Method

&ormu"a7 otal ages 0 ?o. of "rs ago /imes : 1

imes 0 1

and mother8s age can also be wor#ed out.

=aughter8s age + *$ 0* /6 : 1 + ;$ + 1$ "ears

601 >

Sum of ages of mother and daughter 3 0:.

.P. Mother%s age 3 0:85: 3 1: #rs

I


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The sum of ages of a father and his daughter is 0; #ears. After 1 #rsthe father%s age i"" e / times the age of his daughter. What is thepresent age of daughterB

Solution:Let the present age of *a1ghter = x yearsThen the present age of father is = -x years

fter 4 years = 3(x +4 = % x+4 = 4x =+4-!2 = 4"Or 4x = 4$

Dr ? 3 1>

1 3 5 #ears.P. 35 #rs ie.( daughter%s age 3 5 #earsSum of ages of father and daughter 3 0;. .P. Age of father 3 0;85311#rs

Quic'er Method

&ormu"a7 otal ages < ?o. of "rs after /imes : 1 imes 0 1

and father8s age can also be wor#ed out.

=aughter8s age + *7 < 4 /3 : 1 + 4; + 12 "ears

301 1

Sum of ages of father and daughter 3 0; #rs.

P % 3 8 3

RATIO & PROPORTION


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*ATID• The numer of times one !uantit# contains in another !uantit#


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of the same 'ind is ca""ed the ratio of the to !uantities.

• C"ear"#( the ratio of the to !uantities is e!uiva"ent to the

fraction that one !uantit# is of the other.• Dserve carefu""# that the to !uantities must e of the same

'ind.

• There can e a *atio eteen *s.: and /:( ut there can e no

*atio eteen *s.: and /: Mangoes.• The *atio to / is ritten as 7/ or /

• and / are ca""ed the terms of the ratio.• is the first term and / is the second term

• The first term of a ratio is ca""ed the antecedent and the secondthe conse!uent.• In the ratio 7/( is the antecedent and / is the conse!uent.• The "itera" meaning of antecedent is Z that hich goes efore[• The ord conse!uent "itera""# means Zthat hich goes after.[

*ATID• The *atio eteen to concrete !uantities of the same 'ind is


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an astract numer.

• Thus the ratio eteen *s.0 and *s.2 is 072

• Since a fraction is not a"tered # mu"tip"#ing or dividing oth itsnumerator and denominator # the same numer( a ratio hich

is a"so a fraction is not a"tered # mu"tip"#ing or dividing oth its

terms # the same numer.

• Thus /70 is same as ;75: and 507: is the same as /71.• CDM)DUNK *ATID• The *atios are compounded # mu"tip"#ing together the

antecedent for a ne and the conse!uent for a ne conse!uent.

• E?amp"e7 &ind the *atio compounded of the fo""oing 1 ratios. 17/( 675/( ;70 and 750

• So"ution7 The re!uire ratio 3 1?6?;? 3 5;• / ?5/ ?0 ? 50 0

*ATIDO• When t17/ is compounded ith itse"f the resu"ting *atio is


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1 7 /. It is ca""ed Kup"icate *atio of 17/(

• Simi"ar"#( 1/7 // is the Trip"icate *atio of 17/. • 1 7 / is ca""ed Sudup"icate *atio of 1 7 /. • a5/ 7 5/ is Sutrip"icate *atio of a and . • IN,E*SE *ATID

• If 7/ e the given ratio( then 7 5 / or /7 is ca""ed its inverse orreciproca" ratio.

• If the antecedent 3 the conse!uent( the ratio is ca""ed the *atioof e!ua"it#( such as 1 7/ .

• Thus /70 is same as ;75: and 507: is the same as ;:7>:

• If the antecedent ^ the conse!uent( the ratio is ca""ed the *atioof greater ine!ua"it#( such as ;7/ .

• If the antecedent _ the conse!uent( the ratio is ca""ed the *atioof "ess ine!ua"it#( such as /71 .

I""ustrative e?amp"e85

Kivide 510> in to to parts such that one ma# e to the other as


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72.

So"ution7

5st )art 3 ? 510> 3 ? 510> 3 /1

42 6

• Simi"ar"#( nd part 7 2 ? 510> 3 2? 510> 3 55/1

42 6

I""ustrative e?amp"e8&ind three numers in the ratio of 57 7 / so that the sum of their s!uare is

e!ua" to 0:1.

So"ution7

Henc( the re!uired numers are ;(5 and 5>

I""ustrative e?amp"e8/

A($(C and K are four !uantities of the same 'ind such that A7$3/71(


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$7C3>76( C7K 3 5075;. =i9 &ind the ratio A7K( =ii9. &ind A7$7C

and =iii9 find A7$7C7K.

So"ution7

=i9 A 3 / ( $ 3 6 ( C 3 50

$ 1 C > K 5;

(•) .P. A 3 A ? $ ? C 3 / ? > ? 50 3 0 K $ C K 1 6 5; >

.P. A7 K 3 0 7 >

=ii9 A7$ 3 /71 3 ; 7 >

$7C3>76

.P. A7$7C 3 /71 7 >=iii9 We put don the first ratio in its origina" form and change

the terms of the other ratios so as to ma'e each

I""ustrative e?amp"e8/O=iii9

=iii9 We put don the first ratio in its origina" form and change


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the terms of the other ratios so as to ma'e each

antecedent e!ua" to the preceding conse!uent.

A7$ 3 /71 OOOOOO=59

$7C 3 >76 OOOOOO=9

3 5 7 6 =divided # >9

>

3 1 7 6 ? 1 =mu"tip"ied # 19 3 1 7 6

> OOOOO=9i

C7K 3 50 7 5; OOOO=/9

3 5 7 5; =divided # 509 3 6 7 50 ? 6 `mu"tip"ied # 6X 3 6 7 1 ....=/9i

50 50 ` X 0

.P. A7$7C7K 3 /717 6 7 1 3 /: 7 1: 7 10 7 1> 0

I""ustrative e?amp"e81A($(C( K and E are five !uantities of the same 'ind such that

A $ 5 $ C / 1 C K ; 6 d K E 5 5; &i d A $ C K E


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A7$357( $7C3/71( C7K 3 ;76 and K7E 3 575;. &ind A7$7C7K7E.

So"ution7

A7$ 3 57 3 /7;

$7C 3 /71 3 ;7>

C7K 3 ;76 3 > 7 5

K7E 3 5 7 5;

.P. A7$7C7K7E 3 /7;7 > 7 5 7 5;In this e?amp"e( e moved from e"o( ecause it made the ca"cu"ations

easier.?heorem: If the ratio 7etween the first and the second@uantities is a:7 and the ratio 7etween the second and

third @uantities is c:d, then the ratio amon first, secondand third @uantities is iven 7y ac:7c:7d0 a . $ * a$ .$ .*

The aove ratio can e

diagrammatica""#

represented as7

Qn.2. • A ag contains an e!ua" numer of one


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• A ag contains an e!ua" numer of one

rupee coin( 0: paise coins and 0 paisecoins. The tota" amount in the ag is *s./0.

Ho man# coins of each t#pe are there in

the agB

• =a9 : =9 0 =c9 /: =d9 50 =e9 5:• Ans. =a9 : • E?p"anation7 Tota" va"ue of coins 3 *s./0• If coins 5: each the tota" va"ue i"" e

3*s.52.0

• So( : each the tota" va"ue i"" e 3*s./0

Q.No.>. A ag contains coins of *s.5( 0: paise and.0 paise in the ratio

/707;. If there are ;: *s.5 coins in the ag( hat i"" e tota" amount in

the agB


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the agB

• So"ution7• Rs.".%% Ps.5% &s

#5

• Ratio ' 5 6

• So no. o( coins 6% "%%"#%

• Va)ue Rs. 6% 5%

'%• T*ere(ore+ Tota) ,a)ue is Rs. 6%5%'% "4%.%%

•


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easy ways in mathematics

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