tcs placement paper 2014

8/10/2019 TCS Placement Paper 2014

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1. ! 3" # $ % 2 an& $ # 2"≤3' (hat can )e sai& a)out the *alue o! "+

,. " -1

. " %-1. " -1

. " 1

,nswer:

ultipl" the secon& equation with -1 then it will )ecome - $ - 2"≥ - 3. ,&& the equations.

ou will get " % -1.

2. ! the price o! an item is &ecrease& )" 10 an& then increase& )" 10' the net e!!ect on the

price o! the item is

,. , &ecrease o! 99. o change

. , &ecrease o! 1

. ,n increase o! 1

,nswer:

! a certain num)er is increase& )" $ then &ecrease& )" $ or *ice *ersa' the net change is

alwa"s &ecrease. 6his change is gi*en )" a simple !ormula −( x 10)2=−(1010)2=−1%. egiti*e sign in&icates &ecrease.

3. ! m is an o&& integer an& n an e*en integer' which o! the !ollowing is &e!initel" o&&+,. 72m#n7m-n

. (m+n2)+(m−n2).m2+mn+n2. m #n

,nswer: an& 7riginal ,nswer gi*en as

ou ust remem)er the !ollowing o&& ± o&& e*en; e*en ± e*en e*en; e*en ± o&&

o&&

,lso o&& $ o&& o&&; e*en $ e*en e*en; e*en $ o&& e*en.

4. (hat is the sum o! all e*en integers )etween 99 an& 301+

,. 40000

. 20000

. 40400

. 20200

,nswer:

6he !irst e*en num)er a!ter 99 is 100 an& last e*en num)er )elow 301 is 300. (e ha*e to

!in& the sum o! e*en num)ers !rom 100 to 300. i.e.' 100 # 102 # 104 # ............... 300.

6a<e 2 ommon. 2 $ 7 50 # 51 # ...........1506here are total 101 terms in this series. =o !ormula !or the sum o! n terms when !irst term





an& last term is <nown is n2(a+l)=o 50 # 51 # ...........150 1012(50+150)=o 2 $ 1012(50+150) 20200

5. 6here are 20 )alls which are re&' )lue or green. ! > )alls are green an& the sum o! re& )alls an& green )alls is less than 13' at most how man" re& )alls are there+

,. 4

. 5

. ?

. >

,nswer:

@i*en A # # @ 1>; @ >; an& A # @ 13. =u)stituting @ > in the last equation' (e

get A ?. =o ma$imum *alue o! A ?

?. ! n is the sum o! two consecuti*e o&& integers an& less than 100' what is greatest

possi)ilit" o! n+

,. 98

. 94

. 9?

. 99

,nswer :

(e ta<e two o&& num)ers as 72n # 1 an& 72n - 1.

6heir sum shoul& )e less than 100. =o 72n # 1 # 72n - 1 100 ⇒ 4n 100.

6he largest 4 multiple which is less than 100 is 9?

>. x 2 1/100' an& $ 0 what is the highest range in which $ can lie+

,. -1/10 $ 0

. -1 $ 0

. -1/10 $ 1/10

. -1/10 $

,nswer: ,

Aemem)er:

7$ - a7$ - ) 0 then *alue o! $ lies in )etween a an& ).7$ - a7$ - ) % 0 then *alue o! $ &oes not lie in)etween a an& ). or 7 −∞' a an& 7)' −∞ i!

a )

x 2 1/100 ⇒( x 2−1/100)<0⇒7 x 2−(1/10)2)<0⇒7 x −1/10)( x +1/10)<0=o $ shoul& lie in)etween - 1/10 an& 1/10. ut it was gi*en that $ is -*e. =o $ lies in -1/10 to

0

8. 6here are 4 )o$es colore& re&' "ellow' green an& )lue. ! 2 )o$es are selecte&' how man"

com)inations are there !or at least one green )o$ or one re& )o$ to )e selecte&+

,. 1 . ?



. 9

. 5

,nswer: 5

6otal wa"s o! selecting two )o$es out o! 4 is 4C2 ?. ow' the num)er o! wa"s o!

selecting two )o$es where none o! the green or re& )o$ inclu&e& is onl" 1 wa". 7we select

"ellow an& )lue in onl" one wa". ! we su)stract this num)er !rom total wa"s we get 5

wa"s.

9. ,ll !aces o! a cu)e with an eight - meter e&ge are painte& re&. ! the cu)e is cut into

smaller cu)es with a two - meter e&ge' how man" o! the two meter cu)es ha*e paint on

e$actl" one !ace+

,. 24

. 3?

. ?0

. 48,nswer : ,

! there are n cu)es lie on an e&ge' then total num)er o! cu)es with one si&e painting is gi*en

)" 6×(n−2)2. Bere si&e o! the )igger cu)e is 8' an& small cu)e is 2. =o there are 4

cu)es lie on an e&ge. Bence answer 24

10. 6wo c"clists )egin training on an o*al racecourse at the same time. 6he pro!essional

c"clist completes each lap in 4 minutes; the no*ice ta<es ? minutes to complete each lap.

Bow man" minutes a!ter the start will )oth c"clists pass at e$actl" the same spot where the"

)egan to c"cle+

,. 10. 8

. 14

. 12

,nswer:

6he !aster c"cl"st comes to the starting point !or e*er" 4 min so his times are 4' 8' 12' .........

6he slower c"clist comes to the starting point !or e*er" ? min so his times are ?' 12'

18' ......... =o )oth comes at the en& o! the 12th min.

11. ' ' an& C are all &i!!erent in&i*i&uals; is the &aughter o! ; is the son o! ; is

the !ather o! C; ,mong the !ollowing statements' which one is true+

,. is the &aughter o! C

http://3.bp.blogspot.com/-D-VYm3SwaJ0/UFCdhmLvyFI/AAAAAAAAB1M/0Bxc1YX4Gpk/s1600/BR.png



. ! is the &aughter o! ' then an& are sisters

. ! is the gran&&aughter o! ' then an& are sisters

. C an& are )others.

,nswer:

Drom the &iagram it is clear that ! is the &aughter o! ' then an& are sisters.Aectangle in&icates ale' an& *al in&icates Demale.

12. n the a&oining &iagram' , an& ED@B are squres o!

si&e 1 unit such that the" intersect in a square o! &iagonal length 7E 1/2. 6he total area

co*ere& )" the squares is

,. annot )e !oun& !rom the in!ormation

. 1 1/2

. 1 >/8

. one o! these

,nswer:

Fet @ $ then using p"thogerous theorem CG2+GE2=CE2

⇒ x 2+ x 2=(1/2)2⇒2 x 2=1/4⇒ x 2=1/86otal area co*ere& )" two )igger squares , # ED@E - ,rea o! small square 2 - 1/8

15/8

http://2.bp.blogspot.com/-BQe346mBjew/UFC0uxFs6VI/AAAAAAAAB2M/gFIjvJIOh08/s1600/square+ans.png

http://4.bp.blogspot.com/-zVmJ6qn87Kw/UFCqNdxzsmI/AAAAAAAAB1g/NAMb_Ks8SRU/s1600/square.png



13. 6here are 10 stepping stones num)ere& 1 to 10 as shown at the si&e. , !l"

umps !rom the !irst stone as !ollows; E*er" minute it umps to the 4th stone !rom where it

starte& - that is !rom 1st it woul& go to 5th an& !rom 5th it woul& go to 9th an& !rom 9th it

woul& go to 3r& etc. (here woul& the !l" )e at the ?0th minute i! it starts at 1+

,. 1

. 5

. 4

. 9

,nswer : ,

,ssume these steps are in circular !ashion.

6hen the !l" umps are &enote& in the &iagram. t is clear that !l" came to the 1st position

a!ter 5th minute. =o again it will )e at 1st position a!ter 10th 15th .....?0th. min.

=o the !l" will )e at 1st stone a!ter ?0th min.

14. (hat is the remain&er when 617+1176 is &i*i&e& )" >+

,. 1

. ?

. 0

. 3

,nswer:

617 (7−1)17

17C0.717−17C1.716.11.....+17C16.71.116−17C17.117

! we &i*i&e this e$pansion e$cept the last term each term gi*es a remain&er 0. Fast term

gi*es a remain&er o! - 1.

http://4.bp.blogspot.com/-qM60VtZSif4/UFCuX5ZledI/AAAAAAAAB1w/ZwTI87iJcQc/s1600/Frog.png

http://3.bp.blogspot.com/-83jX1HwLKgY/UFCsilLxJPI/AAAAAAAAB1o/nos77S4P6QU/s1600/Steps.png



ow Drom Dermat little theorem' [ap−1p!"#=1=o [1767!"#=1

,&&ing these two remain&ers we get the !inal remain&er 0

15. n )ase >' a num)er is written onl" using the &igits 0' 1' 2' .....?. 6he num)er 135 in )ase

> is 1 $ 72 # 3 $ > # 5 >5 in )ase 10. (hat is the sum o! the )ase > num)ers 1234 an&

?543 in )ase >.

,. 11101

. 11110

. 10111

. 11011

,nswer:

n )ase > there is no >. =o to write > we use 10. !or 8 we use 11...... !or 13 we use 1?' !or 14

we use 20 an& so on.

=o !rom the column &' 4 # 3 > 10' we write 0 an& 1 carrie& o*er. now 1 # 3 # 4 8 11'

then we write 1 an& 1 carrie& o*er. again 1 # 2 # 5 8 11 an& so on

1?. 6he sequence $n& is &e!ine& )" 1 2 an& n+1=n+2n what is the *alue

o! 100

,. 9902

. 9900

. 10100

. 9904

,nswer: ,

(e <now that 1 2 so 2=1+1=1+2(1)=4'=2+1=2+2(2)=84='+1='+2(')=14=o the !irst !ew terms are 2' 4' 8' 14' 22' ......

6he &i!!erences o! the a)o*e terms are 2' 4' ?' 8' 10...

an& the &i!!erences o! &i!!erences are 2' 2' 2' 2. all are equal. so this series represents a

qua&ratic equation.

,ssume n an2+bn+c ow 1 a # ) # c 2

2 4a # 2) # c 4

http://3.bp.blogspot.com/-lA7YvCWgtA0/UFCyuNxpRWI/AAAAAAAAB2E/DU7h4NAn4Do/s1600/7+sum.png



' 9a # 3) # c 8

=ol*ing a)o*e equations we get a 1' ) - 1 an& 2

=o su)stituting in n n2+bn+c n2−n+2=u)stitute 100 in the a)o*e equation we get 9902.

1>.Din& the num)er o! rectangles !rom the a&oining !igure 7, square is also consi&ere& a

rectangle

,. 8?4. 32>?

. 1?38

. one

,nswer:

6o !orm a rectangle we nee& two horiGontal lines an& two *ertical lines. Bere there

are 13 *ertical lines an& > horiGontal lines. 6he num)er o! wa"s o! selecting 2 lines

!rom 13 *ertical lines is 1'C2 an& the num)er o! wa"s o! selecting 2 lines !rom >

horiGontals is 7C2. =o total rectangles 7C2 x 1'C2

18. ,' ' an& go !or a picnic. (hen , stan&s on a weighing machine' also

clim)s on' an& the weight shown was 132 <g. (hen stan&s' also clim)s on' an&

the machine shows 130 <g. =imilarl" the weight o! an& is !oun& as 102 <g an&

that o! an& is 11? <g. (hat is Hs weight

,. 58<g

. >8 <g

. 44 <g. one

,nswer :

@i*en , # 132; # 130; # 102' # 11?

Eliminate !rom 2n& an& 4th equation an& sol*ing this equation an& 3r& we get

*alue as 44.

19. Ao" is now 4 "ears ol&er than Eri< an& hal! o! that amount ol&er than ris. ! in 2

"ears' ro" will )e twice as ol& as Eri<' then in 2 "ears what woul& )e Ao"Hs agemultiplie& )" risHs age+

http://2.bp.blogspot.com/-QLxyvRU-a6I/UFC4u2HOIEI/AAAAAAAAB2g/5DmLo0y84UI/s1600/Rec.png



,. 28

. 48

. 50

. 52

,nswer: 48

20. I' ' I an& ( are integers. 6he e$pression I - - J is e*en an& the e$pression

- J - ( is o&&. ! I is e*en what must )e true+

,. ( must )e o&&

. - J must )e o&&

. ( must )e o&&

. J must )e o&&

,nswer: , or 7ut go !or

21. r an& rs =mith ha*e in*ite& 9 o! their !rien&s an& their spouses !or a part" at

the (ai<i<i each resort. 6he" stan& !or a group photograph. ! r =mith ne*er

stan&s ne$t to rs =mith 7as he sa"s the" are alwa"s together otherwise. Bow man"

wa"s the group can )e arrange& in a row !or the photograph+

,. 20K

. 19K # 18K

. 18 $ 19K

. 2 $ 19K

,nswer:

22. n a rectanglular coor&inate s"stem' what is the area o! a triangle whose *ertices

whose *ertices ha*e the coor&inates 74'0' 7?' 3 a&n 7? ' -3

,. ?

. >

. >.5

. ?.5

,nswer: ,

23. , &rawer hol&s 4 re& hats an& 4 )lue hats. (hat is the pro)a)ilit" o! getting

e$actl" three re& hats or e$actl" three )lue hats when ta<ing out 4 hats ran&oml" out

o! the &rawer an& imme&iatel" returning e*er" hat to the &rawer )e!ore ta<ing out the

ne$t+

,. 1/2

. 1/8

. 1/4

. 3/8

,nswer:

24. n how man" wa"s can we &istri)ute 10 i&entical loo<ing pencils to 4 stu&ents sothat each stu&ent gets at least one pencil+



,. 5040

. 210

. 84

. one o! these

,nswer:

25. 6he prime !actoriGation o! inteGer is , $ , $ $ ' where ,' an& are all

&istinct prine inteGers. Bow man" !actors &oes ha*e+

,. 12

. 24

. 4

. ?

,nswer: ,

2?. 6im an& Elan are 90 <m !rom each other.the" start to mo*e each other

simultanousl" tim at spee& 10 an& elan 5 <mph. ! e*er" hour the" &ou)le their spee&

what is the &istance that 6im will pass until he meet Elan

,. 45

. ?0

. 20

. 80

,nswer:

2>. , !ather purchases &ress !or his three &aughter. 6he &resses are o! same color )uto! &i!!erent siGe .the &ress is <ept in &ar< room .(hat is the pro)a)ilit" that all the

three will not choose their own &ress.

,. 2/3

. 1/3

. 1/?

. 1/9

,nswer:

28. is an integer an& %2' at most how man" integers among # 2' # 3' # 4'

# 5' # ?' an& # > are prime integers+

,. 1

. 3

. 2

. 4

,nswer:

29. , turtle is crossing a !iel&. (hat is the total &istance 7in meters passe& )" turtle+

onsi&er the !ollowing two statements

7I 6he a*erage spee& o! the turtle is 2 meters per minute7 Ba& the turtle wal<e& 1 meter per minute !aster than his a*erage spee& it woul&



ha*e !inishe& 40 minutes earlier

,. =tatement I alone is enough to get the answer

. oth statements I an& are nee&e& to get the answer

. =tatement alone is enough to get the answer

. ata ina&equate,nswer:

30. @i*en the !ollowing in!ormation' who is "oungest+

is "ounger than ,; , is talle& than

is ol&er than ; is "ounger than

is taller than ; , is ol&er than

,.

.

.

. ,

,nswer:

31. ! C7$ a4+'+*2++" has roots at $ 1' 2' 3' 4 an& C70

48' what is C75

,. 48

. 24

. 0

. 50

,nswer: ,

,- -() = (1++2+ x '+....... x 2012)2− x 2012

() = 1++2+ x '+....... x 2011

"n a3 3" "#an" "n -() " ()

9"3 : #:l3pl () 3 ;n 3" ;3 "

.() = +2+ x '+....... x 2012

a 1 ;n ;3 "

. () + 1 = 1++2+ x '+....... x 2012:33:3" 3 al:" n -()

3"n -() = ( x .g( x )+1)2− x 2012

-() = x 2.g( x )2+2.g( x )+1− x 2012

; -() l" () >3 3; 3"# a" "a*3l l" () an

" "3 1 ? x 2012

@:3 1 ? x 2012 = (1 ? )(1++2+ x '+....... x 2011)

- 3 "p";n l" () " "3 0 a "#an".

n:#" a "a*3l ' p#" -a*3;A 125 -a*3; ;- 3 n:#" a"



p"-"*3 B:a" an 27 -a*3; ;- 3 n:#" a" p"-"*3 *:". ;"all

; #an -a*3; ;" 3" n:#" a"

D" n; 3a3 3" 3;3al -a*3; ;- a n:#" = a p.bq.cr ....; 3" 3;3al -a*3; * a" p"-"*3 B:a" ;- a n:#"

= ([ p2+1).([q2+1).([r 2+1)...."" [ "a3"3 n3"F" l" 3an 3a3 ;- .

G"n ([ p2+1).([q2+1).([r 2+1).... = 125

; [ p2+1 = 5 [q2+1 = 5 [ p2+1 = 5

[ p2 = 4 ⇒ p = 8 ; HA #lal B = 8 ; HA = 8 ; H

G"n 3a3 27 -a*3; ;- 3 n:#" a" p"-"*3 *:"

; ([ p'+1).([q'+1).([r '+1).... = 27

;

[ p'+1 = '

⇒ =

[ p' = 2

⇒ p = 6A 7A 8

@ *;#nn " n; 3a3 p = B = = 8

; 3" "n n:#" ;:l " n 3" -;#a3 = a8.b8.c8 ....

:#" ;- -a*3; ;- 3 n:#" = (8+1).(8+1).(8+1) = 72H

,n a *la 3"" a" 60% ;- l ;- * 25% p;;. Da3 3" p;al3

3a3 a p;; l "l"*3" l"a"

:#" 3;3al 3:"n3 n 3" *la = 100

"n Gl = 60% (100) = 60

I;; l = 25% (60) = 15

; p;al3 3a3 a p;; l "l"*3" l"a" = I;; l / ;3al

3:"n3 = 15/100 = 15%

an @ a" :nnn a;:n a **:la 3a* ;- l"n3 120 #"3" 3

p"" 12 #/ an 6 #/ n 3" a#" "*3;n. D"n ll 3" #""3 -;

3" >3 3#"

#""3 @ "n *;" ;n" ;:n #;" 3an @.

J "la3" p"" = (12 ? 6) #/. ; " 3a" 120 / 6 "*;n 3; an ;n"

"3a ;:n.

; a-3" 20 "*;n #""3 @.

*;#pl"3" a ; n 20 a @ n 60 a C n 45 a. ll 3""

p";n ;n 3;"3" ;n a p;K"*3 ;3 a p;>3 ;- !.26000 a3 3"

p;>3 ;- @

D" n; 3a3 p;>3 #:3 " a" a 3" a3; ;- 3" "L*"n*". @:3

"L*"n*" a" n""l p;p;3;nal 3; 3" a. ; "L*"n*" ;- M @ M

C = 1/20 M 1/60 M 1/45 = H M ' M 4

; @ a" n 3" 3;3al p;>3 = ' / 1' N 26000 = !.6000

*;#pl"3" a p"*" ;- ; n '/4 ;- 3" 3#" n @ ;"A @ 3a" 4/5 ;- 3"



3#" n C ;". " ;3 a p;>3 ;- !. 40000 ; #:* @ "3

:#" C 3a" 20 Oa. ; @ 3a" 4/5 (20) = 16 a. 3a"

'/4(16) = 12

; 3" "L*"n*" a3; = 1/20 M 1/16 M 1/12 = 12 M 15 M 20

@J a" n 3" p;>3 ;- !.40000 = 15/47 (40000) = !.12765

n "#p3 3an " >ll" 3 an nl"3 pp" PQ n 42 #n:3". -3" 12

#n:3" an ;:3l"3 pp" P@Q ;p"n" * *an "#p3 3" 3an n '0

#n:3". -3" 6 #n:3" an;3" nl"3 pp" PCQ ;p"n" n3; 3" a#" 3anA

* *an >ll 3" 3an n '5 #n:3" an 3" 3an >ll". Rn 3" 3#"

3a"n 3; >ll 3" 3an

:#" 3;3al 3an *apa*3 = 210 93"

; *apa*3 ;- pp" = 210/42 = 5 93"

Capa*3 ;- @ = 210 / '0 = ? 7 93"

Capa*3 ;- C = 210 / '5 = 6 #n:#" 3an "3 >ll" n #n a-3" 3" 3 pp" ;3 ;p"n".

; x ×5+6×(−2)+4 x =210

⇒48+4 x =210⇒4 x =162⇒ x =40.5 ;3al 3#" 3a"n 3; >ll 3" 3an = 40.5 + 12 + 6 = 51.5

S;3"A a:3" an an n-an3 *;#n" a" 74A an #;3"J a"

46 #;" 3an a:3" an n-an3. ,- n-an3 a" 0.4 3#" ;- a:3"

a"A 3"n >n a:3" a".

:#" S + O + , = 74 .................(1)

l; "n S ? O ? , = 46 ⇒ S = O + , + 46

l; , = 0.4 O ⇒ , = 2/5 O

:33:3n S an , al:" n 3" >3 "B:a3;n " "3 O ? 25O ? 46 + O

+ 25O = 74

;ln O = 10

G;*" ;:3 24 *;T"" "an a3 p*" N p" . -3" a l" ;n"3 ;- 3;* ;3 p;l" ; " ;l 3" "3 -; U200 p" an #a" a

3;3al p;>3 ;- 3*" 3" *;3. Da3 #:3 " 3" p*" ;- N

;3al C;3 p*" = 24×N

1/' ;- 3" "an p;l"A "#ann "an a" 2/' (24) = 16

"lln p*" = 200 × 16 = '200

I;>3 = "lln p*" ? C;3 p*" = '200 ? 24×N

G"n I;>3 = 2 × C;3 p*"

'200 ? 24×N = 2 × (24×N)



;ln N = 44.44

@an: p"n '0% ;- n*;#" ;n p"3;l ;n *;;3" 20% ;- 3"

"#ann ;n ;:" "n3 an 3" alan*" ;n -;;. ,- " p"n !.'00 ;n

p"3;l 3"n a3 3" "p"n3:" ;n ;:" "n3

G"n '0% (,n*;#" ) = '00 ⇒ ,n*;#" = 1000

-3" an p"n3 !.'00 ;n p"3;lA " l"-3 3 !.700.

V p"nn ;n ;:" "n3 = 20% (700) = !.140

9"3 "p(#An) = # 3; 3" p;" n. ,- "p(10A #) = n "p(2A 2) "" 3; an

n a" n3"" 3"n n =

G"n 10m=n.22

⇒ 2m×5m=n.22⇒2m−2×5m=n

R; # = 2 " "3 l"a3 al:" ;- n = 25A an -; # W 2 " "3 n>n3"al:" a" p;l" -; n.

V; #an . ;- "a3 *;3n !. 5 p" #:3 " #" 3 45 ;-

*" *;3n !. 6.40 p" ; 3a3 20% an #a " ;3an" "lln

3" #3:" a3 !. 7.20 p"

,- 3" "lln p*" ;- 3" #3:" !.7.2 "n ;l a3 20% p;>3 3"n

CI ×120100 = 7.2 ⇒ CI = !.6

; appln "3" a"a" -;#:la

= K ×5+45×6.4K +45=6⇒ X = 18

" a;nal ;- a B:a" 3*" 3" " ;- "B:la3"al 3anl" 3"n 3"

a3; ;- "a ;- 3" anl" 3; 3" "a ;- B:a"

9"3 3" " ;- "B:la3"al 3anl" = 1 :n3.

D" n; 3a3 a"a ;- an "B:la3"al 3anl" = 'Y4a2

" = 1 :n3 a"a ;- 3" "B:la3"al 3anl" = 'Y4

; Oa;nal ;- 3" B:a" = 2 (" ;- 3" "B:la3"al 3anl") = 2

D" n; 3a3 a"a ;- 3" B:a" = 12D2 "" O = a;nal

; a"a ;- 3" B:a" = 12(22)=2

!a3; ;- 3" a"a ;- "B:la3"al 3anl" an B:a" = 'Y4 M 2 ⇒ 'YM8

!aK 3;" ' *" an 3"" ":l3 a" n;3" ;n 3"n a3 3"

p;al3 3a3 aK "3 10

la "#"#" "n ' *" a" ;ll" 3" n:#" ;- a ;- "33n n (

"" n 3" :# ;- -a*" ;n *")

= (n−1)C2 "" n = ' 3; 8



= 25 "" n = HA 12

= 27 "" n = 10A 11

= (20−n)C2 "" n = 1' 3; 18

" "B:" p;al3 = 276' = 27216

1 6he water !rom one outlet' !lowing at a constant rate' can !ill the swimming pool in 9

hours. 6he water !rom secon& outlet' !lowing at a constant rate can !ill up the same pool in

appro$imatel" in 5 hours. ! )oth the outlets are use& at the same time' appro$imatel" what is

the num)er o! hours require& to !ill the pool+

,ns: ,ssume tan< capacit" is 45 Fiters. @i*en that the !irst pipe !ills the tan< in 9 hours. =o

its capacit" is 45 / 9 5 Fiters/ Bour. =econ& pipe !ills the tan< in 5 hours. =o its capacit" is

45 / 5 9 Fiters/Bour. ! )oth pipes are opene& together' then com)ine& capacit" is 14

liters/hour. 6o !ill a tan< o! capacit" 45 liters' oth pipes ta<es 45 / 14 3.21 Bours.

2 ! >5 o! a class answere& the !irst question on a certain test correctl"' 55 percentanswere& the secon& question on the test correctl"' an& 20 percent answere& neither o! the

questions correctl"' what percentage answere& )oth correctl"+

t is a pro)lem )elongs to sets. (e use the !ollowing !ormula n7,Z n7, # n7 -

n7,

Bere n7,Z is the people who answere& atleast one o! the questions.

t was gi*en that 20 answere& neither question then the stu&ents who answere& atleast one

question is 100 - 20 80

ow su)stituting in the !ormula we get 80 >5 # 55 - n7,

⇒ n7, 50

3 , stu&entHs a*erage 7 arithmetic mean test score on 4 tests is >8. (hat must )e the

stu&ents score on a 5th test !or the stu&ents a*erage score on the 5th test to )e 80+

,ns: (e <now that ,*erage =:# ;- 3" ;"a3;n ; ;-;"a3;n=o =um o! 4 test scores >8×4312

=um o! 5 tests scores 80×5400

⇒ 5th test score400-31288

Alternative method: ! the stu&ent scores >8 in the !i!th test also' what coul& )e his a*erage+

o change. s it not+

ut to )ring the a*erage to 80' he must ha*e score& enough mar<s e$tra so that each o! the

!i*e su)ect scores increase upto 80. i.e.' he shoul& ha*e score& 2 $ 5 10 runs e$tra in the

!i!th su)ect. =o 5th su)ect score is >8 # 10 88

4 Aural househol&s ha*e more purchasing power than &o ur)an househol&s at the same

income le*el' since some o! the income ur)an an& su)ur)an househol&s use !or !oo& an&

shelter can )e use& )" the rural househol&s !or other nee&s. (hich o! the !ollowing

in!erences is )est supporte& )" the statement ma&e a)o*e+7, 6he a*erage rural househol& inclu&es more people than &oes the a*erage ur)an or



su)ur)an househol&.

7 Aural househol&s ha*e lower !oo& an& housing costs than &o either ur)an or su)ur)an

househol&s.

7 =u)ur)an househol&s generall" ha*e more purchasing power than &o either rural or ur)an

househol&s.

7 6he me&ian income o! ur)an an& su)ur)an househol&s is generall" higher than that o!rural househol&s.

7E ,ll three t"pes o! househol&s spen& more o! their income on housing than on all other

purchases com)ine&.

,ns: ! a*erage rural househol& inclu&es more people' then how come the" ha*e more

purchasing power+ n!act' the" ha*e less purchasing power as the" ha*e to !ee& more people.

ption , rule& out.

ption &oes not e$plain wh" rural househol&s ha*e more purchasing power than ur)an.

Aule& out.

! me&ian income o! ur)an an& su)ur)an househol&s is generall" higher than rural househol&s

the" are li<el" to ha*e more purchasing power' assuming other parameters constant. ut this

&oes not e$plain wh" rural househol&s ha*e more purchasing power. ptions rule& out.ption E &oes not pro*i&e an" e$planation wh" rural househol&s ha*e more purchasing

power. Aule& out.

ption is correct as' ! rural househol&s spen& less income on !oo& an& shelter &ue to less

prices the" &e!initel" ha*e more &isposa)le income to spen&.

5 Lose is a stu&ent o! horticulture in the Mni*ersit" o! Bose. n a horticultural e$periment in

his !inal "ear' 200 see&s were plante& in plot an& 300 were plante& in plot . ! 5> o! the

see&s in plot germinate& an& 42 o! the see&s in plot germinate&' what percent o! the

total num)er o! plante& see&s germinate&+

,ns: 6otal see&s germinate& in Clot 5> o! 200 114

6otal see&s germinate& in Clot 42 o! 300 12?

6otal germinate& see&s 114 # 12? 240

6he percentage o! germinate& see&s o! the total see&s 240500×100 48

? , close& c"lin&rical tan< contains 3?π cu)ic !eet o! water an& its !ille& to hal! its capacit".

(hen the tan< is place& upright on its circular )ase on le*el groun&' the height o! water in the

tan< is 4 !eet. (hen the tan< is place& on its si&e on le*el groun&' what is the height' in !eet'

o! the sur!ace o! the water a)o*e the groun&+

,ns: (e <now that the *olume o! c"lin&er πr 2h

@i*en tan< hight 4!t.⇒ πr 24 3?π ⇒ r 3

=o the ra&ius is 3 which means the &iameter is ?.



,s the c"lin&er is !ille& to initiall" e$actl" hal! o! the capacit"' (hen this c"lin&er is place&

on its si&e' (ater comes upto the height o! the ra&ius.

=o water comes upto 3 !t.

> 6he present ratio o! stu&ents to teachers at a certain school is 30 to 1. ! the stu&entenrollment were to increase )" 50 stu&ents an& the num)er o! teachers were to increase )" 5'

the ratio o! the teachers woul& then )e 25 to 1 (hat is the present num)er o! teachers+

,ssume the present stu&ents an& teachers are 30N' N

,!ter new recruitments o! stu&ents an& teachers the strength )ecomes 30N # 50' N # 5

respecti*el". ut gi*en that this ratio 25 : 1

⇒'0K +50K +5=251=ol*ing we get N 15

=o present teachers are 15.

8 ollege 6 has 1000 stu&ents. ! the 200 stu&ents maoring in one or more o! thesciences'130 are maoring in hemistr" an& 150 are maoring in iolog". ! at least 30 o! the

stu&ents are not maoring in either hemistr" or iolog"' then the num)er o! stu&ents

maoring in )oth hemistr" an& iolog" coul& )e an" num)er !rom

! we assume e$actl" 30 stu&ents are not maoring in an" su)ect then the stu&ents who ta<e

atleast one su)ect 200 - 30 1>0

(e <now that n7,Z n7, # n7 - n7,

⇒ 1>0 130 # 150 - n7,

=ol*ing we get n7, 110.

i.e.' =tu&ents who can ta<e )oth su)ects are 110

ut ! more than 30 stu&ents are not ta<ing an" su)ect' what can )e the ma$imum num)er o!

stu&ents who can ta<e )oth the su)ects+

,s there are 130 stu&ents are maoring in chemistr"' assume these stu&ents are ta<ing )iolog"

also. =o ma$imum stu&ents who can ta<e )oth the su)ects is 130

=o the num)er o! stu&ents who can ta<e )oth su)ects can )e an" num)er !rom 110 to 130.

9 Nell" an& hris are mo*ing into a new cit". oth o! them lo*e )oo<s an& thus pac<e&

se*eral )o$es with )oo<s. ! hris pac<e& ?0 o! the total num)er o! )o$es' what was the

ratio o! the num)er o! )o$es Nell" pac<e& to the num)er o! )o$es hris pac<e&+

=imple questions. ! chris pac<s ?0 o! the )o$es' <ell" pac<s remaining 40

=o Nell" : hris 40 : ?0 2 : 3

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10 ,mong a group o! 2500 people' 35 percent in*est in municipal )on&s' 18 percent in*est

in oil stoc<s' an& > percent in*est in )oth municipal )on&s an& oil stoc<s. ! 1 person is to )e

ran&oml" selecte& !rom 2500 people' what is the pro)a)ilit" that the person selecte& will )e

one who in*ests in municipal )on&s )ut not in oil stoc<s+

,ns: Bere 2500 is re&un&ant

Drom the &iagram we <now that onl" ones who in*este& in municipal )on&s are 28' the pro)a)ilit" is 28 / 100 >/25

11 achine , pro&uces )olts at a uni!orm rate o! 120 e*er" 40 secon&' an& achine

pro&uces )olts at a uni!orm rate o! 100 e*er" 20 secon&s. ! the two machines run

simultaneousl"' how man" secon&s will it ta<e !or them to pro&uce a total o! 200 )olts+

,ns: achine , pro&uces 120/40 3 )olts in 1 secon& an& machine pro&uces 100/20 5

)olts in one secon&.

Bence' )oth o! them will pro&uce 8 )olts per secon&.

Bence' the" wil ta<e 200/8 25 secon&s to pro&uce 200 )olts.

12 Bow man" prime num)ers )etween 1 an& 100 are !actors o! >150+

,ns: >' 150 2×52×11×1'=o there are 4 &istinct prime num)ers that are )elow 100

13 ,nal"sing the goo& returns that Balocircle nsurance C*t Ft& was gi*ing' Aati<a )ought a

1-"ear' As 10'000 certi!icate o! &eposit that pai& interest at an annual rate o! 8 compoun&e&

semi-annuall".(hat was the total amount o! interest pai& on this certi!icate at maturit"+

6his is a question on compoun& interest to )e calculate& semi annuall".

n the case o! semi annual compoun&ing' nterest rate )ecomes hal! an& um)er o! perio&s

)ecomes 2 per "ear.

=o , C(1+R100)n⇒ A=10A000(1+4100)2=10A000×2625 10'81?

nterest , - C 10' 81? - 10'000 81?

14 Luan is a gol& me&alist in athletics. n the month o! a"' i! Luan ta<es 11 secon&s to run "

"ar&s' how man" secon&s will it ta<e him to run $ "ar&s at the same rate+

,ns: ! uan ta<es 11 secon&s to run "ar&s' !or 1 "ar& he will ta<e 11 / " secon&s. 6o run $"ar&s his time will )e 11 / " × $ 11$/ "



15 , certain compan" retirement plan has a rule o! >0 pro*ision that allows an emplo"ee to

retire when the emplo"eeHs age plus "ears o! emplo"ment with the compan" total at least >0.

n what "ear coul& a !emale emplo"ee hire& in 198? on her 32n& )irth&a" !irst )e eligi)le to

retire un&er this pro*ision+

,ssume it has ta<en $ "ears to the !emale emplo"ee to reach the rule o! >0.=o her age shoul& )e 32 # $. ,lso she gains $ "ears o! e$perience.

⇒ 732 # $ # $ >0

⇒ $ 19.

Ber age at the time o! retirement 198? # 19 2005

1? ! the !ollowing' which is the closest appro$imation o! 750.2O0.49/199.8 +

ans: Dor appro$imation 750.2×0.49/199.8 can )e ta<en as

50×0.5/200 25/200 1/8 0.125

1> ,n&alusia has )een promoting the importance o! health maintenance. Drom Lanuar"

1'1991 to Lanuar" 1'1993' the num)er o! people enrolle& in health maintenance organiGations

increase& )" 15 percent. 6he enrollment on Lanuar" 1'1993 was 45 million. Bow man"

million people7to the nearest million was enrolle& in health maintenance organiGations on

Lanuar" 1'1991+

,ns: ! a num)er N is to )e increase& )" $ it shoul& )e multiplie& )" (100+ x )100=o (hen the enrollment in Lanuar" 1' 1991 is multiplie& )" (100+ x )100 we got 45

million.K ×(100+15)100=45N 45×100115 39.13

18 (hat is the lowest possi)le integer that is &i*isi)le )" each o! the integers 1 through >'

inclusi*e+

,ns: ! a num)er has to )e &i*isi)le )" each num)er !rom 1 to >' that num)er shoul& )e

F.. o!71'2'3'4'5'?'> 420

19 ! the area o! a square region ha*ing si&es o! length ? cms is equal to the area o! a

rectangular region ha*ing wi&th 2.5 cms' then the length o! the rectangle' in cms' is,ns: @i*en ,rea o! the square ,rea o! rectangle

⇒a2=l.b=u)stituting the a)o*e *alues in the !ormula

⇒62=l.2.5⇒ l 14.4 cm

20 , tan< contains 10'000 gallons o! a solution that is 5 percent so&ium chlori&e )" *olume.

! 2500 gallons o! water e*aporate !rom the tan<' the remaining solution will )e

appro$imatel" what percentage o! so&ium chlori&e+

,ns: =o&ium chlori&e in the original solution 5 o! 10'000 500(ater in the original solution 10'000 - 500 9'500



! 2'500 Fiters o! the water is e*aporate& then the remaining water 9'500 - 2'500 >'000

=o&ium chlori&e concentration 500500+7000×100 ?.?>

7concentration shoul& )e calculate& alwa"s on the total *olume

21 ,!ter loa&ing a &oc<' each wor<er on the night crew loa&e& 3/4 as man" )o$es as each

wor<er on the &a" o! the crew. ! the night crew has 4/5 as man" wor<ers as the &a" crew'

what !raction o! all the )o$es loa&e& )" two crews &i& the &a" crew loa&+

,ssume the num)er o! )o$es loa&e& in &a"shi!t is equal to 4' then the num)er o! )o$e&

loa&e& in night shi!t 3

,ssume the wor<e& on &a"shi!t 5' then wor<ers on night shi!t 4

=o )o$es loa&e& in &a" shi!t 4 $ 5 20' an& )o$es loa&e& in night shi!t 3 $ 4 12

so !raction o! )o$es loa&e& in &a" shi!t 2020+12=58

22 , )a<er" opene& "ester&a" with its &ail" suppl" o! 40 &oGen rolls. Bal! o! the

rolls were sol& )" noon an& 80 o! the remaining rolls were sol& )etween noon an&

closing time. Bow man" &oGen rolls ha& not )een sol& when the )a<er" close&

"ester&a"+,ns: ! hal! o! the rolls were sol& )" noon' the remaining are 50 740 20.

@i*en 80 o! the remaining were sol& a!ter the noon to closing time

⇒ 80 720 1?

Mnsol& 20 - 1? 4

23 ! 4C' where C is a prime num)er greater than 2' how man" &i!!erent positi*e

e*en &i*isors &oes n ha*e inclu&ing n+

,ns: 22×I1

(e <now that total !actors o! a num)er which is in the !ormat o! aP×Q×*R...

7C # 1. 7P # 1. 7A # 1 .... 72 # 1.71 # 1 ?

,lso o&& !actors o! an" num)er can )e calculate& easil" )" not ta<ing 2 an& its

powers.

=o o&& !actors o! 22×I1 the !actors o! I1 71 # 1 2

E*en !actors o! the num)er ? - 2 4

24 , &ealer originall" )ought 100 i&entical )atteries at a total cost o! q rupees. !

each )atter" was sol& at 50 percent a)o*e the original cost per )atter"' then' in termso! q' !or how man" rupees was each )atter" sol&+



,ns: Cer )atter" cost q / 100

! each )atter" is sol& !or 50 gain' then selling price

C;3I*"×(100+Gain100)

⇒ q100×(100+50100)='q200

25 6he price o! lunch !or 15 people was 20> poun&s' inclu&ing a 15 percent gratuit"

o! ser*ice. (hat was the a*erage price per person' EIFM@ the gratuit"+

,ns: Fet the net price e$clu&ing the gratuit" o! ser*ice $ poun&s

6hen' total price inclu&ing 15 gratuit" o! ser*ice x ×(100+15100)

1.15 $ poun&s

=o' 1.15 $ 20> poun&s

⇒ $ 20> / 1.15 180 poun&s

et price o! lunch !or each person 180 / 15 12 poun&s

1 ! the !ollowing' which is the closest appro$imation o! 750.2O0.49/199.8 +

,ns: Dor appro$imation 750.2×0.49/199.8 can )e ta<en as

50×0.5/200 25/200 1/8 0.125

2 Bow man" prime num)ers )etween 1 an& 100 are !actors o! >150+

,ns: >' 150 2×52×11×1'=o there are 4 &istinct prime num)ers that are )elow 100

3 ,mong a group o! 2500 people' 35 percent in*est in municipal )on&s' 18 percent in*est in

oil stoc<s' an& > percent in*est in )oth municipal )on&s an& oil stoc<s. ! 1 person is to )e

ran&oml" selecte& !rom 2500 people' what is the pro)a)ilit" that the person selecte& will )e

one who in*ests in municipal )on&s )ut not in oil stoc<s

,ns: Bere 2500 &oes not require.

Drom the &iagram we <now that onl" ones who in*este& in municipal )on&s are 28'

the pro)a)ilit" is 28 / 100 >/25

4 ountr" lu) has an in&oor swimming clu). 6hirt" percent o! the mem)ers o! a

swim clu) ha*e passe& the li!esa*ing test. ,mong the mem)ers who ha*e not passe&the test' 12 ha*e ta<en the preparator" course an& 30 ha*e not ta<en the course. Bow

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man" mem)ers are there in the swim clu)+

,ns: 30 # 12 42 &i& not pass the test. 6his is equal to >0 o! the total mem)ers. =o

total mem)ers 100/ >0 $ 42 ?0

5 , nec<lace is ma&e )" stringing in&i*i&ual )ea&s together in the repeating pattern re& )ea&' green )ea&' white )ea&' )lue )ea& an& "ellow )ea&. ! the nec<lace

)egins with a re& )ea& an& en&s with a white )ea&' then coul& )e:

,ns: 6he pattern is A @ ( A @ ( A .......

=o' (hite )ea& comes at these positions 3r&' 8th' 13th' 18th...

! we ta<e this as a arithmetic progression' then this series can )e e$presse& as 3 # 7n -

1 5. 7 Drom the !ormula !or general term o! ,C a # 7n-1&.

6his can )e e$presse& as 5n - 2

(e chec< the answer options so onl" ?8 satis!" the con&ition.

? , &og ta<en !our leaps !or e*er" !i*e leaps o! hare )ut three leaps o! the &og is

equal to !our leaps o! the hare. ompare spee&+

,ns: n terms o! num)er o! leaps' the ratio o! the og an& hare spee&s are 4 : 5

ut @i*en that 3 leaps o! &og 4 leaps o! hare'. i.e.' Feap lengths 4 : 3 7! og is

co*ering in 3 leaps what hare as co*ere& in 4 leaps then Feap lengths are in*ersel"

proportional

=o og spee& 4 $ 4 1?

Bare spee& 5 $ 3 15

=o spee&s ratio 1? : 15

> 6here are two )o$es'one containing 39 re& )alls Q the other containing 2? green

)alls."ou are allowe& to mo*e the )alls )/w the )o$es so that when "ou choose a )o$

ran&om Q a )all at ran&om !rom the chosen )o$'the pro)a)ilit" o! getting a re& )all is

ma$imiGe&.this ma$imum pro)a)ilit" is

,ns: Rer" interesting question.

,s we are allowe& to mo*e the )alls' we <eep onl" one re& )all in !irst )o$ an& mo*e

all the remaining )alls to the secon& )o$

=o !ist )o$ contains 1 re&)all' secon& )o$ contains 38 re& # 2? green ?4 )alls

Cro)a)ilit" o! choosing an" )o$ is 1/ 2.

=o pro)a)ilit" o! ta<ing one re& )all 12×(1)+12('864)\0.8

8 n how man" wa"s can 3 postcar&s can )e poste& in 5 post)o$es+

,ns: Dirst car& can go into an" o! the !i*e )o$es' =econ& can go into an" o! the !i*e

)o$es' 6hir& can go into an" o! the !i*e )o$es 5×5×5=125

9 ,pple costs F rupees per <ilogram !or !irst 30<gs an& P rupees per <ilogram !or

each a&&itional <ilogram. ! the price o! 33 <ilograms is 11.?>an& !or 3?<gs o! ,pples

is 12.48 then the cost o! !irst 10 <gs o! ,pples is,ns: " !raming equations we get



30F#3P11.?>

30F#?P12.48

Eliminate P )" multipl"ing the !irst equation )" 2 an& su)tracting secon& equation

!rom the !irst

6hen we get F 0.3?2ost o! 10 <gs o! apples 0.3?2 $ 10 3.?2

10 letters in the wor& ,M=EA are permute& in all possi)le wa"s an& arrange& in

alpha)etical or&er then !in& the wor& at position 49 in the permute& alpha)etical

or&er+

a ,A=EM

) ,AE=M

c ,A=ME

& ,AEM=

,ns: 6he )est wa" to sol*e this pro)lems is Lust as< how man" wor&s starts with ,. !

we !i$ ,' then the remaining letters can )e arrange& in 5K wa"s 120. =o the as<e&

wor& must start with ,.

,rrange all the gi*en letters in alpha)etical or&er. ,EA=M

Fet us !in& all the wor&s start with ,. ,OOOO 4K 24 wa"s

ow we !in& all the wor&s start wit ,E. ,EOOOO 4K 24 wa"s

=o ne$t wor& start with ,A an& remaining letters are E=M

=o option

11 , is twice e!!icient than . , an& can )oth wor< together to complete a wor< in> &a"s. 6hen !in& in how man" &a"s , alone can complete the wor<+

,ns: Fet us assume , can &o 2 units o! wor< each &a"' then can &o onl" 1 unit a

&a". ! )oth can complete the wor< in > &a"s' total wor< &one )" these two togeter

72 # 1 $ > 21 units

! these 21 units to )e &one )" , alone' then he will ta<e 21 / 2 10.5 &a"s.

12 n a 8 $ 8 chess )oar& what is the total num)er o! squares.

,ns: 6he total num)er o! squares in a n $ n chess )oar& is equal to Sthe sum o! !irst n

natural num)er squaresS

i.e.'n(n+1)(2n+1)6=o =u)stituting 8 in the a)o*e !ormula we get 204

13 I' ' ( an& J are inteGers an& the e$pressing I - - J is e*en an& - ( - J is

o&&. ! I is e*en then which o! the !ollowing is true+

7a must )e o&&

7) -J must )e o&&

7c ( must )e o&&

7& J must )e o&&

,ns. ! I is e*en an& I - - J is e*en then an& J )oth shoul& )e o&& or )oth



shoul& )e e*en.

! - ( - J is o&&' an& an& J are also o&& ( shoul& )e o&&

! - ( - J is e*en' an& an& J are e*en then ( shoul& )e o&&.

=o option is correct. i.e.' ( must )e

14 6he remain&er when 1K#2K#3K...#50K &i*i&e& )" 5K will )e

6he remain&er when the terms greater than 5K are &i*i&e& )" 5K )ecomes 0 so we nee&

to consi&er the terms upto 4K.

=o remain&er will )e whate*er is o)taine& )" &i*i&ing 1K#2K#3K#4K with 5K.

=o remain&er is o)taine& )" &i*i&ing 71#2#?#24 33 with 5K 7 120

=o remain&er is 33.

15 ! there are =i$ perio&s in each wor<ing &a" o! a school' n how man" wa"s can

one arrange 5 su)ects such that each su)ect is allowe& at least one perio&+

,ns. 6o arrange ? perio&s with 5 su)ects' then one su)ect can )e arrange& in two

slots.

5 =u)ects can )e arrange& in ? perio&s in 6I5 wa"s an& now we ha*e 1 perio&

which we can !ill with an" o! the 5 su)ects in 5 wa"s. so 6I5×5=3?00

,lternate metho&:

,ssume the su)ects are I1' I2' ,' ' ' '. Bere I is the su)ect which repeats. =o

arranging ? o)ects in ? places will )e equal to ?K >20 7here no nee& to &i*i&e this

num)er with 2K as e*en though the su)ect is same' )ut not i&entical

ut this repeate& su)ect can )e an" o! the !i*e. =o total arrangements are >20 $ 5 3?00

1? ,n article manu!acture& )" a compan" consists o! two parts I an& . n the

process o! manu!acturing o! part I' 9 out 100 parts man" )e &e!ecti*e. =imilarl" ' 5

out o! 100 are li<el" to )e &e!ecti*e in the manu!acturer o! . alculate the pro)a)ilit"

that the assem)le& pro&uct will not )e &e!ecti*e+

a 0.?485

) 0.?5?5

c 0.8?45

& none o! these

,ns: Cro)a)ilit" that the part I is non&e!ecti*e is 1 - 9/100.91

Cro)a)lit" that the part is non&e!ecti*e is 1 - 5/100.95

so' Cro)a)lit" o! non&e!ecti*e pro&uct0.91×0.950.8?45

1. Adam sat with his friends in the Chinnaswamy stadium at Madurai towatch the 100 metres running race organized by the Asian athleticsAssociation. Five rounds were run. After every round half the teams wereeliminated. Finally, one team wins the game. How many teams

articiated in the race!Ans" #otal five rounds were run. $o in the final round % teams must have



articiated. &n the enultimate round ' teams, and (rd round ), %ndround 1* and in the first round (% teams must have articiated as ineach round half of the teams got eliminated.

%. From the to of a + metres high building A, the angle of elevation of

the to of a tower C- is (0 and the angle of deression of the foot of thetower is *0. /hat is the height of the tower!Ans"

Ans" /e have to find the value of C-. /e use $ine rule to find the

answer easily. $ine rule is aSinA=bSinB=cSinC

&n triangle -, HSin60= xSin'0

$o H'Y2= x 12⇒ x =H'Y

&n triangle C-, CDSin'0=H'YSin60

CD12=H'Y'Y2⇒CD='$o height of the tower + 2 ( 1%

(. '+ members attended the arty. &n that %% are males, %3 are females.#he sha4e hands are done between males, females, male and female.#otal 1% eole given sha4e hands. How many such 4inds of such sha4ehands are ossible!Ans" &f only 1% eole sha4ed their hands, then total hand sha4es

are 12C2 **

'. Ferrari $.5.A is an &talian sorts car manufacturer based in Maranello,&taly. Founded by nzo Ferrari in 1+%) as $cuderia Ferrari, the comanysonsored drivers and manufactured race cars before moving intoroduction of street6legal vehicles in 1+'3 as Ferrari $.5.A. #hroughout itshistory, the comany has been noted for its continued articiation in

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racing, esecially in Formula 7ne where it has emloyed great success.8ohit once bought a Ferrari. &t could go ' times as fast as Mohan9s oldMercedes. &f the seed of Mohan9s Mercedes is (: 4m;hr and the distancetraveled by the Ferrari is '+0 4m, find the total time ta4en for 8ohit todrive that distance.

Ans" As Ferrari<s seed is four times that of the mercedes, &ts seed is (:= ' 1'0$o time ta4en by the ferrari '+0 ; 1'0 (.: Hours

:. A sheet of aer has statements numbered from 1 to '0. For all valuesof n from 1 to '0, statement n says" >=actly n of the statements on thissheet are false.9 /hich statements are true and which are false!a? #he even numbered statements are true and the odd numberedstatements are false.b? #he odd numbered statements are true and the even numbered

statements are false.c? All the statements are false.d? #he (+th statement is true and the rest are falseAns" Assume there is only one statement is there. #he statement shouldread @=actly 1 statement on this sheet is false@ . &f the truth value of thestatement is true, then given statement should be false. #his iscontradiction. &f the statement is false, #hen the given statement is true.but there is not other true statement.Assume there are two statements. y the above logic, %nd statementshould not be true. ut 1st statement is true as it truthfully says the

truthfulness. y this logic we 4now that &f there are @n@ statements, n61?th statement is the only true statement And all other are false

*. &f there are (0 cans out of them one is oisoned if a erson tastes verylittle he will die within 1' hours so if there are mice to test and %' hoursto test, what is the minimum no. of mice9s reBuired to find oisoned can!Ans"

&f only ( erson are used, by giving wine dros suggested by the

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diagram, we can find the oisoned cas4s uto ).for e=amle, &f the %nd and (rd ersons die, then 3th cas4 is oisoned. Asa rule of thumb, &f we have n mice, we can easily find the oison cas4s

uto 2n. As the number of cas4s are less than (% we can use only :

mice.

3. How many + digit numbers are ossible by using the digits 1, %, (, ',: which are divisible by ' if the reetition is allowed!Ans" &f A number has to be divisible by ', the last two digits must bedivisible by '. $o ossibilities are, 1%, %', (%, '', :%. And the of theremaining 3 laces, each lace got filled by any of the five digits. $o

these 3 laces got filled by : = : = .....3 times? 57 ways. $o total ways

are : = 57 58

). A hare and a tortoise have a race along a circle of 100 yards diameter.

#he tortoise goes in one direction and the hare in the other. #he harestarts after the tortoise has covered 1;: of its distance and that tooleisurely. #he hare and tortoise meet when the hare has covered only 1;)of the distance. y what factor should the hare increase its seed so as totie the race!

Assume the circumference of the circle is %00 meters. Hare and

tortoise started at the same oint but moves in the oosite

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#otal 1% ersons the reamining ersons must have sha4ed hand

with atleast one erson. $o answer is 1%.

11. #here are two bo=es, one containing 10 red balls and the other

containing 10 green balls. Kou are allowed to move the ballsbetween the bo=es so that when you choose a bo= at random and a

ball at random from the chosen bo=, the robability of getting a red

ball is ma=imized. #his ma=imum robability is

&f rearrangement is not allowed, then actual robability of ic4ing

u a red ball 12(10)+12(0)=12As we are allowed to move the ball, we 4ee only 1 red in the first

bo=, and shirt the remaining + to the second.

$o

12(1)+H1H(0)=141H1%. #he difference between two no is + and the roduct of the two is

1'. /hat is the sBuare of their sum!

/e 4now that (a+)2=(a−)2+4a$ubstituting a 6 b +, and ab

1', (a+b)2=(H)2+4(14)=1'7

1(. #here are two water tan4s A and , A is much smaller than .

/hile water fills at the rate of one liter every hour in A, it gets filled

u li4e 10, %0, '0, )0, 1*0 in tan4 . At the end of first hour, has

10 liters, second hour it has %0, third hour it has '0 and so on?. &f

tan4 is 1;(% filled after %1 hours, what is the total duration

reBuired to fill it comletely!

Ans" #he data related to the first tan4 A is not necessary. As you

can see, the caacity that gets filled in the tan4 after each hour is

doubled. $o &f the tan4 is 1;(%nd art is full after %1 hours, it is

1;1*th art full after %% hours, 1;)th art full after %( hours, 1;'th

art full after %' hours, 1;% full after %: hours, comletely full after

%* hours.

1'. ( friends A, , C went for wee4 end arty to Mc-onald9s

restaurant and there they measure there weights in some order &n 3

rounds. A, , C, A, C, AC, AC. Final round measure is 1::4g

then find the average weight of all the 3 rounds!

Average weight La 2 b 2 c 2 a2b? 2 b2c? 2 c2a?2a2b2c? ;

3 ' a2b2c? ;3 ' = 1::;3 )).: 4gs



#here are only % cases. ither left one is guilty or one of the

remaining + to his right is guilty.

$o &f the left most is guilty, All the statements including the guilty

one are lies. A and C are correct.

7r &f Any one e=cet left most one is guilty, #hen one of thestatements given by the erson should be true. &n this case all the

susects are lying does not hold. $o &f is correct, A is not correct.

i.e., only A or is correct. 7tion C is correct.

1+. A hollow cube of size : cm is ta4en, with a thic4ness of 1 cm. &t

is made of smaller cubes of size 1 cm. &f 1face of the outer surface

of the cube are ainted, totally how many faces of the smaller cubes

remain unainted!

#he Hallow cube volume n'−(n−2)2, Here n is the number of

small cubes lie on the big cube edge.

ow n : so Hallow cube volume

5'−(5−2)2=125−27=H8

$o +) small cubes reBuired to ma4e a hallow cube of size : cm.

ow total surfaces * = +) :))

ow if the bigger cube is ainted ' sides, total ' = %: small faces

got aint. $o remaining small faces which does not have aint after

cutting is :)) 6 100 '))

%0. My flight ta4es of at %am from a lace at 1) 10 and landed

10 Hrs later at a lace with coordinates (*30/. /hat is the local

time when my lane landed!

a? 1% noon

b? *" '0 AM

c? :" %0 5M

d? *":0 AM

8emember, while moving from east to west countries lag in time.8emember when #est cric4et starts in ngland! (. (0 in afternoon.

8ight! ie., /e are in after noon means they are in morning.

&f the coordinates change from 10 to 30/, the lane has moved a

total of )0 degrees. /e 4now that with each degree time increases

by ' minutes while going from east to west. How! %' = *0 min ;

(*0 degrees, $o 1 degree ' min?

$o total time change ' = )0 (%0 min : hrs 2 %0 minutes.

After 10 hours local time is % am 2 10 6 :.%0 hrs? *.'0 AM.



1. Ray writes a two digit number. He sees that the number exceeds 4 times the sumof its digits by 3. If the number is increased by 18, the result is the same as thenumber formed by reversing the digits. ind the number.a! 3"

b! 4#c! 4$d! "%&olution' (et the two digit number be xy.4)x * y! *3 + 1x * y .......)1!1x * y * 18 + 1 y * x ....)#!&olving 1st e-uation we get #x y + 1 .....)3!&olving #nd e-uation we get y x + # .....)4!&olving 3 and 4, we get x + 3 and y + "

#. a, b, c are non negitive integers such that #8a*3b*31c + 3/". a * b * c + 0

a! reater than 14b! less than or e-ual to 11c! 13d! 1#In a calender,2umber of months having #8 days + 12umber of months having 3 days + 42umber of months having 31 days + %#8 x 1 * 3 x 4 * 31 x % + 3/"Here, a + 1, b + 4, c + %.

a*b*c + 1#

3. eorge can do a iece of wor in 8 hours. 5aul can do the same wor in 1 hours,Hari can do the same wor in 1# hours. eorge, aul and hari start the same worat $ am, while george stos at 11 am, the remaining two comlete the wor. 6hattime will the wor comlete0a! 11.3 amb! 1# noonc! 1#.3 md! 1 m

(et the total wor + 1# units. 7s eorge comletes this entire wor in 8 hours, his caacity is 1" units hour &imilarly, the caacity of aul is 1# units hour the caacity of Hari is 1 units hour 7ll 3 started at $ am and wored uto 11 am. &o total wor done uto 11 am + # x)1" * 1# * 1! + %4Remaining wor + 1# %4 + 4/2ow this wor is to be done by aul and hari. 4/ )1# * 1! + # hours )arox!

&o wor gets comleted at 1 m

4. If x9y denotes x raised to the ower y, ind last two digits of )114193843! *)1$/194181!



a! #b! 8#c! 4#d! ##

Remember 1 raised to any ower will give 1 as unit digit.:o find the digit in the 1th lace, we have to multily, 1th digit in the base x unitdigit in the ower.

&o the (ast two digits of the given exression + #1 * /1 + 8#

". ; can dig a well in 1/ days. 5 can dig a well in #4 days. ;, 5, H dig in 8 days. H

alone can dig the well in How many days0a! 3#b! 48c! $/d! #4 7ssume the total wor + 48 units.<aacity fo ; + 48 1/ + 3 units day<aacity of 5 + 48 #4 + # units day<aacity of ;, 5, H + 48 8 + / units dayrom the above caacity of H + / # 3 + 1&o H taes 48 1 days + 48 days to dig the well

/. If a lemon and ale together costs Rs.1#, tomato and a lemon cost Rs.4 and anale costs Rs.8 more than a lemon. 6hat is the cost of lemon0( * 7 + 1# ...)1!: * ( + 4 .....)#!( * 8 + 7:aing 1 and 3, we get 7 + 1 and ( + #

%. 3 mangoes and 4 ales costs Rs.8". " ales and / eaches costs 1##. /mangoes and # eaches costs Rs.144. 6hat is the combined rice of 1 ale, 1

each, and 1 mango.a! 3%b! 3$c! 3"d! 3/2ote' It is 114 not 144.3m * 4a + 8" ..)1!"a * / + 1## ..)#!/m * # + 114 ..)3! )1! x # += /m * 8a + 1% )3! += /m * # + 114

&olving we get 8a # + "/ ..)4!)#! += "a * / + 1##

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3 x )4! + #4a / + 1/8&olving we get a + 1, + 1#, m + 1"&o a * * m + 3%

8. 7n organisation has 3 committees, only # ersons are members of all 3 committee

but every air of committee has 3 members in common. what is the least ossiblenumber of members on any one committee0a! 4b! "c! /d! 1

:otal 4 members minimum re-uired to serve only on one committee.

$. :here are " sweets ;ammun, a>u, 5eda, (adu, ;ilebi which can be consumed in" consecutive days. ?onday to riday. 7 erson eats one sweet a day, based on thefollowing constraints.)i! (adu not eaten on monday)ii! If ;amun is eaten on ?onday, (adu should be eaten on friday.)iii! 5eda is eaten the day following the day of eating ;ilebi

)iv! If (adu eaten on tuesday, a>u should be eaten on monday

based on above, eda can be eaten on any day exceta! tuesdayb! mondayc! wednesdayd! friday

rom the )iii! clue, eda must be eaten after >ilebi. so 5eda should not be eaten onmonday.

1. If @6A&B is #" #3 #1 1$ 1%, :hen ?CIa! 13 11 8 % /b! 1 #3"%c! $ 8 % / "d! % 8 " 3?CI + 13 11 $ % /2ote' this is a dummy -uestion. Dont answer these -uestions

11. 7ddition of /41 * 8"# * $%3 + #4"/ is incorrect. 6hat is the largest digit that canbe changed to mae the addition correct0

a! "b! /

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c! 4d! %

/41 8"#

$63#4//

largest among tens lace is %, so % should be relaced by / to get #4"/

1#. Aalue of a scooter dericiates in such a way that its value at the end of each year is 34th of its value at the beginning of the same year. If the initial value of scooter is4,, what is the value of the scooter at the end of 3 years.a! #31#"b! 1$

c! 1343"d! 1/8%"

value of the scooter at the end of the year + 40000×('4)' + 1/8%"

13. 7t the end of 1$$4, R was half as old as his grandmother. :he sum of the yearsin which they were born is 3844. How old R was at the end of 1$$$a! 48b! ""c! 4$d! "3

In 1$$4, 7ssume the ages of ? and R + #, then their birth years are 1$$4 #, 1$$4 .Eut given that sum of these years is 3844.&o 1$$4 # * 1$$4 + 3844C + 48In 1$$$, the age of R is 48 * " + "3

14. 6hen numbers are written in base b, we have 1# x #" + 333, the value of b is0a! 8b! /c! 2one

d! %(et the base + b

&o, )b*#!)#b*"! + (b+2)(2b+5)='b2+'b+'2b2+Hb+10='b2+'b+'b2−6b−7=0&olving we get b + % or 1&o b + %

1". How many olynomials of degree =+1 satisfy f ( x 2)=[f ( x )2=f (f ( x )

a! more than #b! #



c! d! 1

(et f)x! + x 2f ( x 2)=[ x 22= x 4

(f ( x ))2=[ x 22= x 4f (f ( x ))=f ( x 2)=[ x 22= x 4

Fnly 1

1/. igure shows an e-uilateral triangle of side of length " which is divided intoseveral unit triangles. 7 valid ath is a ath from the triangle in the to row to themiddle triangle in the bottom row such that the ad>acent triangles in our ath share acommon edge and the ath never travels u )from a lower row to a higher row! orrevisits a triangle. 7n examle is given below. How many such valid aths are

there0a! 1#b! 1/c! #3d! #4

&ol'2umber of valid aths + )n1! G + )"1!G + #4

1%. In the -uestion, 79E means, 7 raised to ower E. If xy9# J , then which oneof the following statements must be true0)i! x J )ii! J )iii! xy J a! )i! and )iii!b! )iii! onlyc! 2oned! )i! only 7s y9# is always ositive, xy9# J is ossible only when x J . Ftion d iscorrect.

18. :he mared rice of a coat was 4K less than the suggested retail rice. Leshaurchased the coat for half the mared rice at the fiftieth anniversary sale. 6hatercentage less than the suggested retail rice did Lesha ay0a! /

b! #c! %

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d! 3(et the retail rice is Rs.1. then maret rice is )14! K of 1 + /. Leshaurchased the coat for half of this rice. ie., 3 only. which is % less than the retailrice. &o Ftion < is correct.

. , cow an& horse are )ought !or As.2'00'000. 6he cow is sol& at a pro!it o! 20 an& the

horse is sol& a t a loss o! 10. 6he o*erall gain is As.4000' the ost price o! cow+

a 130000

) 80000

c >0000

& 120000

,ns: *erall pro!it 4000200000×100=2%" appl"ing alligation rule' we get

=o cost price o! the cow 2/5 $ 200000 80'000

2. , circle has 29 points arrange& in a cloc< wise manner !rom o to 28. , )ug mo*es

cloc<wise manner !rom 0 to 28. , )ug mo*es cloc<wise on the circle accor&ing to !ollowing

rule. ! it is at a point i on the circle' it mo*es cloc<wise in 1 sec )" 71 # r places' where r is

the remain&er 7possi)l" 0 when i is &i*i&e& )" 11. ! it starts in 23r& position' at what

position will it )e a!ter 2012 sec.

,ns: ,!ter 1st secon&' it mo*es 1 # 723/11r 1 # 1 2' =o 25th position

,!ter 2n& secon&' it mo*es 1 # 25/11 1 # 3 4' =o 29th position 0

,!ter 3r& secon&' it mo*es 1 # 0/11 1 # 0 1' =o 1st position

,!ter 4th secon&' it mo*es 1 # 1 3r& position

a!ter 5th' 1 # 3/11 4 =o >th

,!ter ?th' 1 # >/11 8 so 15th

,!ter >th' 1 # 15/11 5 so 20th

,!ter 8th' 1 # 20/11 10th' =o 30th 1st

=o it is on 1st a!ter e*er" 3 # 5n secon&s. =o it is on 1st position a!ter 2008 secon&s 73 # 5 $

401 =o on 20th a!ter 2012 position.

3. n a cit" 100 *otes are registere&' in which ?0 *ote !or congress an& 40 *ote !or LC.

6here is a person ,' who gets >5 o! congress *otes an& 8 o! LC *otes. Bow man" *otes

got )" ,+,ssume total *otes are 100. =o , got

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>5 o! ?0 45

8 o! 40 3.2

, total o! 48.2

4. ean o! 3 num)ers is 10 more than the least o! the num)ers an& 15 less than greatest o!

the 3. ! the me&ian o! 3 num)ers is 5' Din& the sum o! the 3 num)ers+,ns: e&ian is when the gi*en num)ers are arrange& in ascen&ing or&er' the mi&&le one. Fet

the num)ers are $' 5' " where $ is the least an& " is greatest.

@i*en that x +5+ y '= x +10an& x +5+ y '= y −15=ol*ing we get $ 0 an& " 25.

=o sum o! the num)ers 0 # 5 # 25 30

5. , an& start !rom house at 10am. 6he" tra*el !ro their house on the @ roa& at 20<mph

an& 40 <mph. there is a Lunction 6 on their path. , turns le!t at 6 unction at 12:00 noon'

reaches 6 earlier' an& turns right. oth o! them continue to tra*el till 2pm. (hat is the&istance )etween , an& at 2 pm.

istnace )etween Bouse an& 6 unction 20 $ 2 40.

ie.' reache& 6 at 11 am.

continue& to right a!ter 11 am an& tra*elle& upto 2. =o &istance co*ere& )" him 3 $ 40

120

, reache& 6 at 12 noon an& tra*elle& upto 2 =o &istance& tra*elle& )" him 2 $ 20 40

=o total &istance )etween them 120 # 40 1?0 <m

?. n a particular "ear' the month o! anuar" ha& e$actl" 4 thurs&a"s' an& 4 sun&a"s. n

which &a" o! the wee< &i& anuar" 1st occur in the "ear.

a mon&a" ) tues&a"

c we&nes&a"

& thurs&a"

,ns: ! a month has 31 &a"s' an& it starts with sun&a"' 6hen =un&a"s' on&a"s' tues&a"s are

5 !or that month. ! this month starts with mon&a"' then mon&a"s' tues&a"s' an& we&nes&a"s

are 5 an& remaining &a"s are 4 each. so this month start with on&a".

>. ,' E' D' an& @ ran a race.

, sai& S &i& not !inish 1st /4th

E sai& S &i& not !inish 4thS

D sai& S !inishe& 1stS

@ sai& S !inishe& 4thS

! there were no ties an& e$actl" 3 chil&ren tol& the truth' when who !inishes 4th+

a ,

) E

c D

& @

,ns: ption

8. , chil& was loo<ing !or his !ather. Be went 90 m in the east )e!ore turning to his right. he

went 20 m )e!ore turning to his right a!ain to lo< !or his !ather at his uncles place 30 m !romthis point. Bis !ather was not there. Drom there he went 100m north )e!ore meeting hiss



!ather in a street. Bow !ar &i& the son meet his !ather !rom the starting point.

a 90

) 30

c 80

& 100

Drom the &iagram' , 90 - 30 ?0 an& 100 - 20 80

AD= AB2+BD2−−−−−−−−−−Y=602+802−−−−−

−−−Y=100

9. n an o!!ice' at *arious times &uring the &a" the )oss gi*es the secretar" a letter to

t"pe' each time putting the letter on top o! the pile in the secretar"Hs in)o$. =ecretar"

ta<es the top letter an& t"pes it. oss &eli*ers in the or&er 1' 2' 3' 4' 5 which cannot

)e the or&er in which secretar" t"pes+

a 2' 4' 3' 5' 1

) 4' 5' 2' 3' 1

c 3' 2' 4' 1' 5

& 1' 2' 3' 4' 5

,ns: ption

10. ,t 12.00 hours' L starts to wal< !rom his house at ? <mph. ,t 13.30' C !ollows him

!rom LHs house on his )ic"cle at 8 <mph. (hen will L )e 3 <m )ehin& C+

" the time C starts L is 1.5 hr $ ? 9 <m awa" !rom his house.

L is 3 <m )ehin& when C is 3 <m ahea& o! him. ie.' C has to co*er 12 <m. =o he ta<es

12 / 78 - ? ? hrs a!ter 13.30. =o the require& time is 19.30Brs

11. L is !aster than C. L an& C each wal< 24 <m. =um o! the spee&s o! L an& C is >

<mph. =um o! time ta<en )" them is 14 hours. 6hen L spee& is equal to

a > <mph

) 3 <mph

c 5 <mph

& 4 <mph

@i*en L % CL # C >' onl" options are 7?' 1' 75' 2' 74' 3

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Drom the gi*en options' ! L 4 the C 3. 6imes ta<en )" them

244+24'=14

12. n a @? summit hel& at lon&on. , !rench' a german' an italian' a )ritish' a spanish'

a polish &iplomat represent their respecti*e countries.7i Colish sits imme&iatel" ne$t to )ritish

7ii @erman sits imme&iatel" ne$t to italian' ritish or )oth

7iii Drench &oes not sit imme&iatel" ne$t to italian

7i* ! spanish sits imme&iatel" ne$t to polish' spanish &oes not sit imme&iatel" ne$t

to talian

(hich o! the !ollowing &oes not *iolate the state& con&itions+

a DC=@

) D@C=

c D@=C& D=C@

e D@=C

,ns: ption

13. Aa &ri*es slowl" along the perimeter o! a rectangular par< at 24 <mph an&

completes one !ull roun& in 4 min. ! the ratio o! length to )re&th o! the par< is 3 : 2'

what are the &imansions+

a 450 m $ 300 m

) 150 m $ 100 m

c 480 m $ 320 m

& 100 m $ 100 m

24 <mph 24×100060=400 m / min

n 4 minutes he co*ere& 4 $ 400 1?00 m

6his is equal to the perimeter 2 7 l # ) 1?00

ut l : ) 3:2

Fet l 3<' ) 2<

=u)stituting' we get 2 7 3< # 2< 1?00 % < 180

=o &imensions are 480 $ 320

14. is 30 o! P' P is 20 o! C an& is 50 o! C. (hat is /

ans: 6a<e C 100' then 50' P 20' ?. =o / 3/25

15. ,t what time )etween ? an& > are the han&s o! the cloc< coinci&e+

,ns. 6otal '600

Dor hour 3?0/12 '00/hr

Dor inute !ull rotation '600/hr

Fet the line is HtH ' !or ? ?O301800



then

30 t # 1803?0 t

330t 180

t 180/330

t ?/11 hr ?/11O?03?0/1132611,ns. is ?:32

1?. =eries 1' 4' 2' 8' ?' 24' 22' 88 +

=ol : 6he gi*en series is in the !ormat: $ 4' -2' $4' -2' $4' -2' $4....

1$4 4

4-22

8-2?

?$424

24-222

22$488

88-28?

,ns: 8?

1>. 4 (omen Q ? men ha*e to )e seate& in a row gi*en that no two women can sit

together. Bow man" &i!!erent arrangements are there.

=ol : Fet us !irst sit all the ? men in ? positions in ?K wa"s. ow there are > gaps

)etween them in which 4 women can sit in 7P4 wa"s.

=o total wa"s are ?K $ 7

P4

18. x y + y x =46 Din& $ Q " *alues +

=ol: 145+451=46 Bence $ 1' " 45

19. n 10 "ears' , will )e twice as ol& as was 10 "ears ago. ! , is now 9 "ears

ol&er than the present age o! is

=oln: , #1027-10 ........71

, # 9 ......... 72 !rom equations. 1 Q 2

we get 39 , will )e 39#948 "ears ol&.

20. , stu&ent can select one o! ? &i!!erent math )oo<' one o! 3 &i!!erent chemistr"

)oo< Q one o! 4 &i!!erent science )oo<.n how man" &i!!erent wa"s stu&ents can

select )oo< o! math' chemistr" Q science.

=ol: 6C1×'C1×4C1 ?$3$4>2 wa"s

21. =um o! two num)er is 50 Q sum o! three reciprocal is 1/12 so !in& these twonum)ers



; 3 ;n ;3 !.1A20A000

,n a B:aa3* "B:a3;nA (;" *;"L*"n3 a" n;3 n"*"al "al) 3"

*;n3an3 3"# n;3 0. " *:" ;- 3" :# ;- 3" B:a" ;- 3 ;;3

"B:al 3; 3" B:a" ;- 3" :# ;- 3" *:" ;- 3 ;;3. D* ;- 3"

-;ll;n 3:"a)@;3 ;;3 a" "al

) "3" ;- 3" ;;3 "al

*) 3 l"a3 ;n" ;;3 n;n?"al

)3 l"a3 ;n" ;;3 "al

nM :#" 3" "n B:aa3* "B:a3;n ax 2+bx +c=0 ;"

;;3 a" pA B.

; "n 3a3 (α2+ β2)'=(α'+ β')2

@ "pann "

"3A α6+'.α4. β2+'.α2. β4+ β6=α6+ β6+2.α'. β'

'.α2. β2(α2+ β2)=2.α'. β'

'.(α2+ β2)=2.α. β

'.(α2+ β2)+6.α. β−6.α. β=2.α. β

'.(α+ β)2=8.α. β ...(1)

D" n; 3a3 :# ;- 3" ;;3 = α+ β=−ba

p;:*3 ;- 3" ;;3 = α. β=ca

:33:3n n 3" "B:a3;n (1) " "3 '.

(−ba)2=8.ca⇒'.b2=8.a.c

" na3:" ;- 3" ;;3 *an " "3"#n" >nn 3" #an3:" ;- 3"

"3"#nan3 = b2−4ac

@:3 " n; 3a3 a* = 'b28

; b2−4ac = b2−4.'b28=−b88<0

; 3" ;;3 a" #ana.

'. #an ;l 12 *an" n 10U a l; ;- % 3"n aan ;l 12 *an"

a3 12U a p;>3 ;- % >n 3" al:" ;- .

nM V"" 12 *an" ##a3"al.

9; % = CP−SPCP×100; V"" I = 10 an l;% = %

CP−10CP×100=b⇒CP−10CP=b100

,n 3" "*;n *a" " ;3 a p;>3 ;- %



; I;>3 % = SP−CPCP×100

; V"" I = 12 an p;>3% = %

12−CPCP×100=b⇒12−CPCP=b100

;ln 1 an 2 " "3 = 1/11 ; H.0H%

4. >n 3" 3;3al n:#" ;- *;#na3;n ;- 5 l"33" aAAaAA 3an ;#"

; all a3 a 3#"

nM 1 l"33" *an " *;"n n 2 a. a ;

2 l"33" *an " *;"n n ' a. aaA aA

' l"33" *an " *;"n n ' a. A aaA a

4 l"33" *an " *;"n n 2 a. aaA a5 l"33" *an " *;"n n 1 a.

; 3;3al a a" 11

5. a3 3" :# ;- all 3" 4 3 n:#" 3a3 *an " -;#" :n all

;- 3" 3 2A'A5 an 7

nM :" -;#:la (n?1)] (111..n 3#") (:# ;- 3" 3)

"" n n:#" ;- T""n3 l"33"

; an" ' ] 1111 17

6. '0^72^87 " 11 " "#an"

nM R"#a3 l33l" 3";"# aA a p−1 p "#an" 1.

".A '010 ; 810"n " 11 "#an" 1.

" :n3 3 ;- 7287 8 (:n **l*3 ;- :n3 3) Cl* ""

; 7287 = 10X + 8

'0(10K +8)11=('010)K .'0811=1k .'0811

8811=22411=(25)4.2411=1611=5

7. 12'45678H1011121'14151617181H20......424'44 a3 "#an"

"n " 45

nM 9"3 = 12'45678H1011121'14151617181H20......424'44

!"#an" "n " 5 4. ; = 5X + 4 .....(1)

!"#an" "n " H :# ;- 3" 3 ;- " H.

D" n; 3a3 1+2+'+...44 = HH0 D* " 3 :# a H. ;

"#an" "n " H 0.

; = H9 .....(2)

EB:a3;n (1) an (2) " H9 = 5X + 4

R; X = 1 3 "B:a3;n "3 a3>". ; l"a3 p;l" n:#" a3>"3" *;n3;n H

http://www.campusgate.co.in/2011/10/finding-unit-and-last-two-digits-of.html

http://www.campusgate.co.in/2011/10/finding-unit-and-last-two-digits-of.html



; " "n"al -;#a3 ;- = (9CS ;- (HA 5)) + 9"a3 n:#" a3>"

3" *;n3;n.

; = .45 + H

D"n " 45A " "3 H a "#an".

tcs placement paper 2014

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