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1. Ray writes a two digit number. He sees that the numberexceeds 4 times the sum of its digits by 3. If the number isincreased by 18, the result is the same as the number formed byreversing 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 + 0a! reater than 14b! less than or e-ual to 11c! 13

d! 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 thesame wor in 1 hours, Hari can do the same wor in 1# hours. eorge, aul and hari start the same wor at $ am, while georgestos at 11 am, the remaining two comlete the wor. 6hat timewill the wor comlete0



a! 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 is1" 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 doneuto 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 inthe base x unit digit 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, Hdig in 8 days. H alone can dig the well in How many days0



a! 3#b! 48c! $/d! #4

7ssume the total wor + 48 units.<aacity of ; + 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 an ale costs Rs.8 more than a lemon. 6hat isthe 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 thecombined rice of 1 ale, 1 each, and 1 mango.a! 3%b! 3$c! 3"d! 3/3m * 4a + 8" ..)1!"a * / + 1## ..)#!/m * # + 144 ..)3! )1! x # += /m * 8a + 1%4 x )3! += /m * # + 144&olving we get 8a # + "/ ..)4!

)#! += "a * / + 1##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 3members in common. what is the least ossible number ofmembers on any one committee0a! 4b! "c! /d! 1

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

$. :here are " sweets ;ammun, a>u, 5eda, (adu, ;ilebi whichcan be consumed in " consecutive days. ?onday to riday. 7erson eats one sweet a day, based on the following 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 shouldnot be eaten on monday.

1. If @6A&B is #" #3 #1 1$ 1%, :hen ?CI

a! 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 can be changed to mae the addition correct0a! "b! /c! 4d! %

/41 8"#

$63

#4//

(argest among tens lace is %, so % should be relaced by / toget #4"/

1#. Aalue of a scooter dereciates in such a way that its value atthe end of each year is 34th of its value at the beginning of thesame year. If the initial value of scooter is 4,, what is thevalue of the scooter at the end of 3 years0a! #31#"b! 1$c! 1343"



d! 1/8%"

Aalue of the scooter at the end of the year +40000×(34)3+

1/8%"

13. 7t the end of 1$$4, R was half as old as his grandmother. :he sum of the years in which they were born is 3844. How oldR was at the end of 1$$$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! 2oned! %(et the base + b

&o, )b*#!)#b*"! +(b+2)(2b+5)=3b2+3b+3

2b2+9b+10=3b2+3b+3b2−6b−7=0

&olving we get b + % or 1&o b + %



1". How many olynomials of degree =+1

satisfyf(x2)=[f(x)]2=f(f(x)

a! more than #

b! #c! d! 1

(et f)x! +x2

f(x2)=[x2]2=x4

(f(x))2=[x2]2=x4

f(f(x))=f(x2)=[x2]2=x4

Fnly 1

1/. igure shows an e-uilateral triangle of side of length " whichis divided into several unit triangles. 7 valid ath is a ath fromthe triangle in the to row to the middle triangle in the bottom rowsuch that the ad>acent triangles in our ath share a common edgeand the ath never travels u )from a lower row to a higher row!

or revisits a triangle. 7n examle is given below. How many suchvalid 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 one of 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 is correct.

18. :he mared rice of a coat was 4K less than the suggestedretail rice. Lesha urchased the coat for half the mared rice atthe fiftieth anniversary sale. 6hat ercentage less than thesuggested retail rice did Lesha ay0a! /b! #c! %

d! 3(et the retail rice is Rs.1. then maret rice is )14! K of1 + /. Lesha urchased the coat for half of this rice. ie., 3only. which is % less than the retail rice. &o Ftion < is correct.

1. A cow and horse are bought for Rs.2,00,000. The cowis sold at a profit of 20% and the horse is sold a t a lossof 10%. The overall gain is Rs.4000, the Cost price ofcow

a! 1"0000b! #0000c! $0000d! 120000

Ans &verall profit ' 4000200000×100=2%

() appl)ing alligation rule, we get



*o cost price of the cow ' 2+ - 200000 ' #0,000

2. A circle has 2 points arranged in a cloc/ wise annerfro o to 2#. A bug oves cloc/wise anner fro 0 to2#. A bug oves cloc/wise on the circle according tofollowing rule. f it is at a point i on the circle, it ovescloc/wise in 1 sec b) 1 3 r! places, where r is thereainder possibl) 0! when i is divided b) 11. f itstarts in 2"rd position, at what position will it be after

2012 sec.Ans After 1st second, it oves 1 3 2"+11!r ' 1 3 1 '2, *o 2th positionAfter 2nd second, it oves 1 3 2+11 ' 1 3 " ' 4, *o2th position ' 0After "rd second, it oves 1 3 0+11 ' 1 3 0 ' 1, *o 1stpositionAfter 4th second, it oves 1 3 1 ' "rd position

after th, 1 3 "+11 ' 4 *o $thAfter th, 1 3 $+11 ' # so 1thAfter $th, 1 3 1+11 ' so 20thAfter #th, 1 3 20+11 ' 10th, *o "0th ' 1st*o it is on 1st after ever) " 3 n seconds. *o it is on 1stposition after 200# seconds " 3 - 401! *o on 20th



after 2012 position.

". n a cit) 100% votes are registered, in which 0%vote for congress and 40% vote for (56. There is aperson A, who gets $% of congress votes and #% of (56votes. 7ow an) votes got b) AAssue total votes are 100. *o A got$% of 0 ' 420% of 40 ' #A total of "%

4. 8ean of " nubers is 10 ore than the least of thenubers and 1 less than greatest of the ". f theedian of " nubers is , 9ind the su of the "nubersAns 8edian is when the given nubers are arranged inascending order, the iddle one. :et the nubers are -,, ) where - is the least and ) is greatest.

;iven that x+5+y3=x+10

and x+5+y3=y−15

*olving we get - ' 0 and ) ' 2.*o su of the nubers ' 0 3 3 2 ' "0

. A and ( start fro house at 10a. The) travel frotheir house on the 8; road at 20/ph and 40 /ph. there is a 5unction T on their path. A turns left at T

<unction at 1200 noon, ( reaches T earlier, and turnsright. (oth of the continue to travel till 2p. =hat isthe distance between A and ( at 2 p.>istnace between 7ouse and T <unction ' 20 - 2 ' 40.ie., ( reached T at 11 a.( continued to right after 11 a and travelled upto 2. *o



distance covered b) hi ' " - 40 ' 120A reached T at 12 noon and travelled upto 2 *o distancedtravelled b) hi ' 2 - 20 ' 40*o total distance between the ' 120 3 40 ' 10 /

. n a particular )ear, the onth of <anuar) had e-actl)4 thursda)s, and 4 sunda)s. &n which da) of the wee/did <anuar) 1st occur in the )ear.a! onda)b! tuesda)c! wednesda)d! thursda)Ans f a onth has "1 da)s, and it starts with sunda),Then *unda)s, 8onda)s, tuesda)s are for that onth.f this onth starts with onda), then onda)s,tuesda)s, and wednesda)s are and reaining da)s are4 each. so this onth start with 8onda).

$. A, ?, 9, and ; ran a race.

A said @ did not finish 1st +4th? said @ did not finish 4th@9 said @ finished 1st@; said @ finished 4th@f there were no ties and e-actl) " children told the truth,when who finishes 4tha! Ab! ?

c! 9d! ;Ans &ption >

#. A child was loo/ing for his father. 7e went 0 in theeast before turning to his right. he went 20 before





b! 4, , 2, ", 1

c! ", 2, 4, 1,

d! 1, 2, ", 4,

Ans &ption (

10. At 12.00 hours, 5 starts to wal/ fro his house at

/ph. At 1"."0, 6 follows hi fro 5Bs house on his

bic)cle at # /ph. =hen will 5 be " / behind 6

() the tie 6 starts 5 is 1. hr - ' / awa) fro his

house.

5 is " / behind when 6 is " / ahead of hi. ie., 6 hasto cover 12 /. *o he ta/es 12 + # ! ' hrs after

1"."0. *o the reuired tie is 1."07rs

11. 5 is faster than 6. 5 and 6 each wal/ 24 /. *u of

the speeds of 5 and 6 is $ /ph. *u of tie ta/en b)

the is 14 hours. Then 5 speed is eual to

a! $ /ph

b! " /ph

c! /ph

d! 4 /ph

;iven 5 D 6

5 3 6 ' $, onl) options are , 1!, , 2!, 4, "!

9ro the given options, f 5 ' 4 the 6 ' ". Ties ta/en

b) the ' 244+243=14

12. n a ; suit held at london. A french, a geran,

an italian, a british, a spanish, a polish diploat



represent their respective countries.

i! 6olish sits iediatel) ne-t to british

ii! ;eran sits iediatel) ne-t to italian, (ritish or

bothiii! 9rench does not sit iediatel) ne-t to italian

iv! f spanish sits iediatel) ne-t to polish, spanish

does not sit iediatel) ne-t to talian

=hich of the following does not violate the stated

conditions

a! 96(*;

b! 9;6(*c! 9;*6(

d! 9*6(;

e! 9(;*6

Ans &ption (

1". Ra< drives slowl) along the perieter of a rectangular

par/ at 24 /ph and copletes one full round in 4 in.

f the ratio of length to bredth of the par/ is " 2, what

are the diansions

a! 40 - "00

b! 10 - 100

c! 4#0 - "20

d! 100 - 100

24 /ph ' 24×100060=400 + in

n 4 inutes he covered 4 - 400 ' 100

This is eual to the perieter 2 l 3 b! ' 100

(ut l b ' "2



:et l ' "/, b ' 2/

*ubstituting, we get 2 "/ 3 2/ ! ' 100 'D / ' 1#0

*o diensions are 4#0 - "20

14. 8 is "0% of E, E is 20% of 6 and F is 0% of 6.

=hat is 8 + F

ans Ta/e 6 ' 100, then F ' 0, E ' 20, 8 ' . *o 8+F

' "+2

1. At what tie between and $ are the hands of the

cloc/ coincide

Ans. Total ' 3600

9or hour ' "0+12 ' 300 +hr

9or 8inute ' full rotation ' 3600 +hr

:et the line is BtB , for ' G"0'1800

then"0 t 3 1#0'"0 t

""0t ' 1#0

t ' 1#0+""0

t ' +11 hr +11G0'"0+11'"2611

Ans. is "2

1. *eries 1, 4, 2, #, , 24, 22, ##

*ol The given series is in the forat - 4, 2, -4, 2,

-4, 2, -4....

1-4 ' 4



42'2

#2'

-4'24

242'22 22-4'##

##2'#

Ans #

1$. 4 =oen H en have to be seated in a row given

that no two woen can sit together. 7ow an) different

arrangeents are there.*ol :et us first sit all the en in positions in I

wa)s. Fow there are $ gaps between the in which 4

woen can sit in 7P4wa)s.

*o total wa)s are I - 7P4

1#. xy+yx=469ind - H ) values

*ol 145+451=46

7ence - ' 1, ) ' 4

1. n 10 )ears, A will be twice as old as ( was 10 )ears

ago. f A is now )ears older than ( the present age of

( is*oln A 310'2(10! ........1!

A '( 3 ......... 2!

fro euations. 1 H 2

we get ( ' " A will be "3'4# )ears old.



20. A student can select one of different ath boo/,

one of " different cheistr) boo/ H one of 4 different

science boo/.n how an) different wa)s students canselect boo/ of ath, cheistr) H science.

*ol 6C1×3C1×4C1' -"-4'$2 wa)s

21. *u of two nuber is 0 H su of three reciprocal is

1+12 so find these two nubers

*ol -3) ' 0 .....1! -'0) ....2!

1x+1y=112⇒y+xxy=112⇒12(y+x)=xy..."!

put 2! in 4!

⇒ 12)30)!'0)!)

⇒ 12)30012)'0)y2

⇒ y20)300'0

⇒ y2"0)20)300'0

⇒ ))"0!20)"0!'0

⇒ )20! )"0!'0

)'20 or )'"0

if )'20 then - ' "0

or )'"0 then - ' 20

two nubers are "0 H 20

1.>inalal divides his propert) aong his four sons after

donating Rs.20,000 and 10% of his reaining

propert). The aounts received b) the last three

sons are in arithetic progression and the aount



received b) the fourth son is eual to the total

aount donated. The first son receives as his share

R*.20,000 ore than the share of the second son.

The last son received R*.1 la/h less than the eldestson. 10. 9ind the share of the third son.

a! Rs.#0,000

b! Rs.1,00,000

c! Rs.1,20,000

d! Rs.1,0,000

Ans Assue the aounts received b) the 2nd, "rd,

and 4th sons are a3d, a, ad as the) are in A6!Fow ?ldest son received Rs.20,000 ore than the

2nd son. *o 7e gets a3d320,000

:ast son received 1 la/h less than the eldest son. *o

a3d320,000! ad! ' 1,00,000 ⇒ 2d '

#0,000 ⇒ d ' 40,000

*o Aounts received b) the 4 sons are a 3 0,000,

a340,000, a, a 40,000

t was given that the )oungest sonBs share is eual

to 20,000 3 127is propert)!

Assue 7is propert) ' J rupees.

Then 20,000 3 12J! ' a 40,000 ...........1!

and the Reaining propert) ' *u of the properties

received b) all the four sonBs together.

Reaining propert) ' 910J20,000!

⇒910J20,000! ' a 3 0,000 ! 3 a340,000!

3 a 3 a 40,000! ..2!



*olving =e get J ' 40,000 and a ' 1,20,000

*o third son got Rs.1,20,000

n a uadratic euation, whose coefficients are notnecessaril) real! the constant ter is not 0. The

cube of the su of the suares of its roots is eual

to the suare of the su of the cubes of its roots.

=hich of the following is true

a!(oth roots are real

b! Feither of the roots is real

c! At least one root is nonreald!At least one root is real

Ans Assue the given uadratic euation

is ax2+bx+c=0whose roots are p, .

Fow given that (α2+β2)3=(α3+β3)2

() e-panding weget, α6+3.α4.β2+3.α2.β4+β6=α6+β6+2.α3.β3

3.α2.β2(α2+β2)=2.α3.β3

3.(α2+β2)=2.α.β

3.(α2+β2)+6.α.β −6.α.β =2.α.β

3.(α +β)2=8.α.β...1!

=e /now that su of the roots ' α +β =−ba

product of the roots ' α.β =ca

*ubstituting in the euation 1! we get 3.



(−ba)2=8.ca⇒3.b2=8.a.c

The nature of the roots can be deterined b) finding

the agnitude of the deterinant ' b2−4ac

(ut we /now that ac ' 3b28

*o b2−4ac' b2−4.3b28= −b88<0

*o the roots are iaginar).

". A an sold 12 candies in 10K had loss of b% then

again sold 12 candies at 12K had profit of b% find

the value of b.

Ans 7ere 12 candies is iaterial.

:oss % ' CP−SPCP×100

*o 7ere *6 ' 10 and loss% ' b%

CP−10CP×100=b⇒CP−10CP=b100

n the second case he got a profit of b%

*o 6rofit % ' SP−CPCP×100

*o 7ere *6 ' 12 and profit% ' b%

12−CPCP×100=b⇒12−CPCP=b100



*olving 1 and 2 we get b ' 1+11 or .0%

4. find the total nuber of cobinations of lettersa,b,a,b,b ta/ing soe or all at a tie

Ans 1 letter can be chosen in 2 wa)s. a or b

2 letters can be chosen in " wa). aa, ab, bb

" letters can be chosen in " wa)s. bbb, aab, bba

4 letters can be chosen in 2 wa)s. aabb, bbba

letters can be chosen in 1 wa).

*o total wa)s are 11

. what is the su of all the 4 digit nubers that can

be fored using all of the digits 2,", and $

Ans use forula n1!I - 111..n ties! - *u of

the digits!

here n is nuber of different letters

*o answer is " I - 1111 - 1$

. "0L$2L#$ divided b) 11 gives reainder

Ans 9erat little theore sa)s, ap−1 preainder is

1.

ie., 3010 or 810when divided b) 11 reainder is 1.

The unit digit of 7287is # using c)clicit) of unit

digits! Clic/ here

*o 7287' 10J 3 #

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




30(10K+8)11=(3010)K.30811=1k.30811

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

$.

12"4$#1011121"14111$1#120......424"44

what is reainder when divided b) 4

Ans :et F '

12"4$#1011121"14111$1#120......424"44

Reainder when F is divided b) is 4. *o F ' J 3

4 .....1!

Reainder when F is divided b) is *u of the

digits of F divided b) . =e /now that 1323"3...44

' 0 =hich gives digit su as . *o reainder

when F is divided b) is 0.

*o F ' : .....2!

?uation 1! and 2! we : ' J 3 4

9or J ' 1 this euation gets satisfied. *o least

possible nuber satisfies the condition is

*o The general forat of F ' w:C8 of , !! 3

:east nuber satisfies the condition.

*o F ' w.4 3

=hen F is divided b) 4, we get as reainder.1.The wages of 24 en and 1 woen aounts to

Rs.1100 per da). 7alf the nuber of en and "$

woen earn the sae aount per da). =hat is

the dail) wage of a an



:et the wage of a an is and woan be w.

2431w'1100

123"$w ' 1100

*olving we get ' "0

2. The su of three digits a nuber is 1$. The

su of suare of the digits is 10. f we substract

4 fro the nuber, the nuber is reversed.

9ind the nuber.

:et the nuber be abc.

Then a 3 b 3 c' 1$ .....1!

a2+b2+c2=109.....2!

100a310b3c 4 ' 100c310b3a ......"!

9ro ", we get a c '

*o the possibilities for a, c, b! are ,1,10!,

$,2,#!, #,",!, ,4,4!

9ro the above, #,",! satisfies the condition.

". A calculator has a /e) for suaring and another

/e) for inverting. *o if - is the displa)ed nuber,

then pressing the suare /e) will replace - b) -L2

and pressing the invert /e) will replace - b) 1+-.

f initiall) the nuber displa)ed is and one

alternativel) presses the invert and suare /e) 1ties each, then the final nuber displa)ed

assuing no roundoff or overflow errors! will be

?vern nuber of inverse /e) has no effect on the

nuber.



() pressing the suare /e), the value got

increased li/e 2, 4, #, .... =hich are in the forat

of 2n. *o after the 1 pressings the power

becoes 216

*o the final nuber will be 6216=665536

4. 7ow an) two digit nubers are there which

when substracted fro the nuber fored b)

reversing itBs digits as well as when added to the

nuber fored b) reversing its digits, result in a

perfect suare.

:et the nuber -) ' 10- 3 )

;iven that, 10-3) 10) -! ' -)! is a perfect

suare

*o -) can be 1, 4, . 1!

*o given that 10-3) 310) 3-! ' 11-3)! is a

perfect suare.

*o -3) be 11. 6ossible options are ,2!, #,"!,

$,4!,,! 2!

9ro the above two conditions onl) ,! satisfies

the condition

&nl) 1 nuber satisfies.

. 9ind the th word of *7MNAFJ in dictionar)

*ol Arranging the letters in alphabetical order we

get A 7 J F * M N

Fow Total words start with A are I



Total words start with A7 are I ' 120

Fow

Total words start with A7J are 4I ' 24

Total words start with A7F are 4I ' 24Total words start with A7*J are "I '

Fow A7*FJMN will be the last word reuired.

. Car A leaves cit) C at p and is driven at a

speed of 40/ph. 2 hours later another car (

leaves cit) C and is driven in the sae direction as

car A. n how uch tie will car ( be /s aheadof car A if the speed of car is 0/ph

Relative speed ' 0 40 ' 20 /ph

nitial gap as car ( leaves after 2 hours ' 40 - 2 '

#0 /s

Car ( should be / ahead of the A at a reuired

tie so it ust be # / awa)

Tie ' # + 20 ' 4.4 hrs or 2$ ins

$. 9ind the average of the ters in the series 1

23"43....31200

*ol12! 3"4! 3 ! 3........1200! ' 100

Average ' 100 + 200 ' 0.

#. n is a natural nuber and nL" has 1 factors.

Then how an) factors can nL4 have

Total factors of a nuber F'ap.bq.cr...is p31!

31!r31!...



As n3has 1 factors n3can be one of the two

forats given below

n3'a15

n3' a3.b3

f n3'a15then n ' a5and nuber of factors

of n4' 21

n3' a3.b3then n ' ab and nuber of factors n4'

2

. Two cars start fro the sae point at the sae

tie towards the sae destination which is 420

/ awa). The first and second car travel at

respective speeds of 0 /ph and 0 /ph. After

travelling for soetie the speeds of the two cars

get interchanged. 9inall) the second car reachesthe destination one hour earlier than the first.

9ind the tie after which the speeds get

interchanged

:et the total tie ta/en b) the cars be a and b

:et the tie after which the speed is interchanged

be t

9or car A, 0t30at! ' 420, 0a "0t '420 .......1!

9or car (, 0t 3 0bt! ' 420, 0b 3 "0t '

420 ....2!

Msing both 1! and 2!, we get 0a 3 0b ' #40



(ut as a b '1, 0a 3 0a1! ' #40.

*olving a ' .

*ubstituting in euation 1, we get t ' 4

1.A and ( run a 1 / race. f A gives ( a start of0, A wins b) 14 seconds and if A gives ( a

start of 22 seconds, ( wins b) 20 eters. 9ind

the tie ta/en b) A to run 1 /.

To solve these t)pe of uestions, alwa)s /eep in

)our ind that, the ratio of the speeds of two

contestents never change.

A gives ( a start of 0 eans, A runs 1000 and ( runs onl) 0. () the tie A reaches the

target, ( has to ta/e 22 seconds to reach the

target.

ab=1000950−14b=9801000−22b

0,000 1100b ' 40 #b

*olving we get b ' 2+"

Fow Assue ABs speed ' -

1000950−14(25/3)=x25/3

- ' 10

*o - ta/es 1000+10 ' 100 seconds.

2. A owes ( Rs.0. 7e agrees to pa) ( over a

nuber of consecutive da)s on a 8onda),pa)ing single note or Rs.10 or Rs.20 on each

da). n how an) different wa)s can A repa) (.

7e can pa) b) all 10 rupee notes ' 1 wa)



" Ten rupee 3 1 twent) rupee ' 4!3!×1!' 4

wa)s

1 Ten rupee 3 2 twent) rupee notes ' 3!2!×1!'

" wa)s

Total wa)s ' 1 3 4 3 " ' #

". =, O, P, Q are integers. The e-pression O P

Q is even and the e-pression P Q = is odd.

f O is even what ust be true

a! = ust be oddb! P Q ust be odd

c! Q ust be even

d! Q ust be odd

*ol O is even so P, Q both are even or both are

odd.

Fow P Q in both cases even. *o P Q ! = '

odd happens onl) when w is oddAns = is odd

4. Ra< writes a nuber. 7e sees that the

nuber of two digits e-ceeds four ties the su

of its digits b) ". f the nuber is increased b)

1#, the results is the sae as the nuber

fored b) reversing the digits. 9ind the ne-tiediate prie greater than the nuber.

:et the nuber be -) ' 10- 3 )

10- 3 ) ' 4-3)! 3 " ⇒ 2- ) 1 1!

Also 10- 3 ) 31# ' 10) 3- , )-! ' 1#, )- '



2 2!

*olving we get - ' ", ) '

The nuber is ". *o ne-t iediate prie is

"$

. Jate wanted to bu) 2/gs of apples. The

vendor /ept the 2/g weight on the right side and

weighed 4 apples for that. *he doubted on the

correctness of the balance and placed 2 /g

weight on the left side and she could weight 14

apples for 2 /gs. f the balance was correct howan) apples she would have got

As she got less apples when the weight put on

the right side, the left pan has ore weight sa)

w /gs.

Fow w 3 4a ' 2

and w3 2 ' 14a

*olving we get a ' 2+ Jgs.

*o she gets, 2+2+! ' apples

. 9ind the reainder when "2L""L"4 is

divided b) 11

=e /now that when the divisor is a prie

nuber, 9erat little theore sa)s, ap−1

when

divided b) p, reainder is 1

*o 3210gives reainder 1.

Fow we have to write "2L""L"4 in this forat.

*o we have to find the reainder ""L"4 when



divided b) 10. The reainder is nothig but unit

digit of the nuber. Clic/ here to learn this

concept

""L"4 gives unit digit of .*o ""L"4 ' 10 J 3

323334=32(10K+9)=(3210)K.329Fow this e-pression when divided b) 11 leaves a

reainder of 329which inturn is eual

to (−1)9= −1=10

$. 9ind the option to replace the uestion ar/

in the series below

1 $ 2 4$2

*ol - 1 '

- " ' 1

1 - ' $

$ - $ ' 2

2 - ' 4$2

*o '

#. There are several bags of sae weight. A

bag is /gs plus three fourth of the weight of an

other bag. =hat is the weight of a bag

:et the bags weight is -

Then 3 34- ' -,

*olving we get - ' 24





. 9ind the reainder when L0 is divided b)

21

Ans 650=(63)16.62=21616.62*o this e-pression gives a reainder of "

10. 9ind last two digits of the following

e-pression

201G202G20"G204G24G24$G24#G24!L2

To find the last two digits of a product ta/e the

last two digits in each nuber and ultipl).

01G02G0"......4#G4 use onscreen calculator!

this gives 24. *o 242' $ *o last two digits are

$

1. Ahed, (abu, Chitra, >avid and ?esha each

choose a large different nuber. Ahed sa)s, 8)nuber is not the largest and not the sallestS.(abu sa)s, 8) nuber is not the largest and notthe sallestS. Chitra sa)s, 8) nuber is thelargestS. >avid sa)s, 8) nuber is the sallestS.?esha sa)s, 8) nuber is not the sallestS. ?-actl) one of the five children is l)ing. Theothers are telling the truth. =ho has the largestnuber2.a! ?esha".b! >avid4.c! Chitra.d! (abu



.$.Ans A

:argestD

A ( C > ?

A 9 T+9 T+9 T+9 T+9( T+9 9 T+9 T+9 T+9C 9 9 T 9 9> T+9 T+9 T+9 9 T+9? T+9 T+9 T+9 T+9 T

#..9ro the above table, f we assue that A has

the largest then A and C both are l)ing. *iilarl)if we find the truthfullness of the reainingpeople, it is clear that ? has the largest and Clied. &nl) one 9 in the last colun!

10.11. 2. n the euation A 3 ( 3 C 3 > 3 ? ' 9;

where 9; is the two digit nuber whose value is109 3 ; and letters A, ( , C , > , ?, 9 and ;

each represent different digits. f 9; is as largeas possible. =hat is the value of ;12. a! 41". b! 214. c! 11. d! "1. Ans (1$. 9; is as large as possible and all the $

nubers should be different.1#. () trial and ?rror ethod,1. 3 # 3 $ 3 3 ' " is getting

repeated twice.20. 3 # 3 $ 3 3 4 ' "44 is getting

repeated



21. 3 # 3 $ 3 3 4 ' """ repeats22. 3 # 3 3 3 4 ' "22". Fone of the nubers repeat in the above

case and "2 is the a-iu nuber 9; canhave. The value of ; is 2.

24.2. ". A farer has a rose garden. ?ver) da) he

either pluc/s $ or or 24 or 2" roses. The roseplants are intelligent and when the farerpluc/s these nubers of roses, the ne-t da) "$or " or or 1# new roses bloo in the gardenrespectivel). &n 8onda), he counts 1# roses inthe garden. 7e pluc/s the roses as per his planon consecutive da)s and the new roses bloo asper intelligence of the plants entioned above.After soe da)s which of the following can bethe nuber of roses in the garden

2. a! 42$. b! $

2#. c! "02. d! "$ "0. Ans A"1. f he pluc/s 2", then onl) 1# grows the ne-t

da). This eans total roses get decreases b) .*o after n da)s assue the nuber of roses gotdecreased 1# where n ' "$, then 4 roses left.

"2.

"". 4. =hat is the value of 4444444G#######G444444423444444"#!+44444444L2

"4. a! #######"". b! #######4". c! ########



"$. d! 4444444""#. Ans A". :et - ' 44444444

40. (x+1)×(2x−3)×(x−2)+(x−6)x241. (x2−x−2)×(2x−3)+(x−6)x2

42. 2x3−2x2−4x−3x2+3x+6+x−6x2

4". 2x3−5x2x2=2x−5

44. *ubstituting the value of - in 2- , weget #######"

4.

4. 4. 9or which of the following nS is thenuber 2L$4 32L20#32L2n is a perfectsuare

4$. a! 20124#. b! 21004. c! 20110. d! 20201. Ans >

2. 2L$4 32L20#32L2n ' K2

". 2L$4 32L20#32L2n

' (237)2+22058+(2n)24. =e tr) to write this e-pression

as (a+b)2=a2+2ab+b2

. Fow a ' 237, 2ab ' 22058and b ' 2n. *ubstituting the value of a in 2ab, we get b

' 2020$.#. . Ra< writes a nuber. 7e sees that the

nuber of two digits e-ceeds four ties the su



of its digit b) ". f the nuber is increased b)1#, the result is the sae as the nuber foredb) reversing the digit. 9ind the nuber

. a! "0. b! $1. c! 422. d! 4". Ans A4. ;oing b) the options, " ' #4! 3 "... . =eight of 8, > and is $4. *u of >

and is 4 greater than 8. is 0% less than>. =hat is >Bs weight.

$. Ans 10#. 8 3 > 3 ' $4 ⇒ 8 ' $4 >

. 8 ' > 3 3 4

$0. ' 410>

$1. $4 > ' > 3 3 4

$2. $4 > 410> ' > 3 410> 3 4

$". ⇒ > ' 10

$4.$. $. 9ather is ties faster than son. 9ather

copletes a wor/ in 40 da)s before son. f bothof the wor/ together, when will the wor/ getcoplete

$. a. # da)s

$$. b. # 1+" da)s$#. c. 10 da)s$. d. 20 da)s#0. Ans (#1. As efficienc) is inversel) proportional to

da)s, f 9ather sonBs efficienc) is 1, then



>a)s ta/en b) the should be 1 . Assue,the da)s ta/en b) the are /, /.

#2. ;iven that father ta/es 40 da)s less. *o / / ' 40 ⇒ / ' 10

#". 9ather ta/es 10 da)s to coplete the wor/.Total wor/ is 10 - ' 0 units.

#4. f both of the wor/ together, the)coplete 3 1 units a da). +da). To coplete0 units, the) ta/e 0+ ' # 1+" da)s.

#.#. #. A bea/er contains 1#0 liters of alcohol.

&n 1st da), 0 l of alcohol is ta/en out andreplaced b) water. 2nd da), 0 l of i-ture issta/en out and replaced b) water and the processcontinues da) after da). =hat will be theuantit) of alcohol in bea/er after " da)s

#$. Ans "."##. Mse the forula,

#. FinalAlcohol=InitialAlcohol

(1−Replacementquant

ityFinalVolume)n0. FinalAlcohol=180(1−60180)3=180×(23)3=53.31.2. . f ffn!! 3 fn! ' 2n3", f0! ' 1 then

f2012! ' ". Ans 201"4. f f0!! 3 f0! ' 20! 3 " ⇒ f1! ' "1 ' 2,

f1! ' 2



. ff1!! 3 f1! ' 21! 3 " ⇒ f2! ' 2 ' ",

f2! ' ". ff2!! 3 f2! ' 22! 3 " ⇒ f"! ' $" ' 4,

f"! ' 4$. ..............#. f2012! ' 201".100. 10. =hat will be in the ne-t series101. 1, $, #, 4, , $, "4", ...102. Ans "4410". 1 ' 1

104. $ ' 1 - $10. # ' 1 - $ 3 110. 4 ' $ - $ 3 110$. 0 ' $ - $ 3 110#. ' # - $10. $ ' # - $ 3 1110. "4" ' 4 - $111. Fe-t ter should be 4 - $ 3 1 ' "44

112.11". 11. n a " - " grid, coprising tiles can bepainted in red or blue. =hen tile is rotated b)1#0 degrees, there is no difference which can bespotted. 7ow an) such possibilities are there

114. a. 111. b. "211. c. 4

11$. d. 211#. Ans (

This grid even rotated 1#0 degreesthe relative positions of the tiles do not change.



*o we paint tile nuber 1Bs with red or blueonl) one color should be used! , 2Bs with red orblue.....tile red or blue. Then total possibilities

are 25' "21. n a staircase, there ar 10 steps. A child is atteptingto clib the staircase. ?ach tie she can either a/e 1step or 2 steps. n how an) different wa)s can sheclib the staricasea! 10b! 21c! "

d! Fone of theseAns dMse fibonnacci series, with starting two ters as 1, 2. *one-t ters are ", , #, 1", 21, "4, , #

2. A bo) bu)s 1# sharpners, (rown+white! for Rs.100. 9or ever) white sharpener, he pa)s one rupee ore thanthe brown sharpener. =hat is the cost of white sharpener

and how uch did he bu)a! , 1"b! , 10c! , 10d! Fone of theseAns CAssue that he bought b, brown sharpeners and w, whitesharpeners and the cost of brown sharpener is - and

white sharpener is - 3 1*o w-31! 3 b- ' 100w 3 b ' 1#b ' 1# w*ubstituting in euation 1, we get w-31! 3 1# w!- '



100 so w 3 1# - ' 100Ta/e option 1 f white sharpners are 1", - ' 100 1"! +1# ' 4.#""&ption 2, f white sharpeners are 10, - ' 100 10!+1#' *o white sharpeners cost is .&ption " *atisfies this condition.

". :etters of alphabets no fro 1 to 2 are consecutivel)with 1 assigned to A and 2 to Q. () 2$th letter weean A, 2#th (. n general 23n, andn negative inteUers is sae as the letters nubered n.:et 6 ' , strange countr) ilitar) general sends thissecret essage according ot the following codificationschee. n codif)ing a sentence, the 1st tie a letteroccurs it is replaced b) the pth letter fro it. 2nd tieif occurred it is replaced b) 6L2 letter fro it. "rd tieit occurred it is replaced b) pL" letter fro it. =hat isthe code word for A((ATA:a! ;7FFQ&&R

b! ;7J5Q&7Rc! ;77;Q&;Rd! ;7:JQ&RAns >A should be coded as 13 ' ; it occurred for first tie!( should be coded as 23 ' 7 it occurred for first tie!( *hould be coded as 2 3 " ' "# 2 ' 12 ' : itoccurred for second tie!

&ption > is correct

4. &f a set of "0 nubers, average of 1st 10 nubers iseual to average of last 20 nubers. The su of last 20nubers isa! 2 - su of last 10 nubers



b! 2 - su of 1st 10 nubersc! su of 1st 10 nubersd! Cannot be deterinedAns (:et average of first 10 nubers is a. Then su ' 10aAverage of last 10 nbers also a. Then their su ' 20a9ro the options ( correct

. n how an) wa)s a tea of 11 ust be selected atea en and 11 woen such that the tea ustcoprise of not ore than " en.a! 1b! 22c! 24d! 124"Ans (8a-iu " en can be pla)ed which eans there canbe 0, 1, 2, " en in the tea.

(5C0×11C11)+(5C1×11C10)+(5C2×11C9)+(5C3×11C8)=2256

. The wages of 24 en and 1 woen aount to 1100per da). 7alf the nuber of en and "$ woen hassae one). The dail) wages paid to each an isa! "$b! 400c! "0d! "2

Ans C24 3 1w ' 110012 3 "$ w ' 1100*olving we get 12 ' 21w*ubstituting in the first euation we get, 42w 3 1 w '



1100 ⇒ w ' 200

8 ' "0

$. A nuber when successivel) divided b) , ", 2 givesreainder 0, 2, 1 respectivel) in that order. =hat will bethe reainder when the sae nuber is dividedsuccessivel) b) 2, ", in that ordera! 4, ", 2b! 1, 0,4c! 2, 1, "d! 4, 1, 2

Ans (

use this siple techniue.V 1 - "! 3 2W ' V - ! 3 0W ' 2

6rocedure

:et the nuber ' FFow F ' JJ ' ": 3 2: ' 28 3 1J ' "28 3 1! 3 2 ' 8 3

F ' 8 3 ! ' "0 8 3 29or 8 ' 0 we get the least nuber as 2. Fowwhen 2 is divided b) 2, we get 12 as uotientand 1 as reainder. =hen 12 is divided b) " weget 4 as uotient, and 0 as reainder. =hen 4 isdivided b) we gt 4 as reainder.



#. a,b,c,d,e are distinct nubers. if $a!$b!$c!$d!$e!'22 then a3b3c3d' 7int22 is divisible b) 11.

22 ' 11×11×19×1×1=11× −11×19× −1×1=

Two of the ters in the given e-pression shouldeual to 1. As all the digits are distinct, two ofthe ters should be negative.&ne possible solution ' $ 4!$ !$ #!$ $4!$ $!Then a 3 b 3 c 3 d 3 e ' 4 3 3 # 3 $4 3

$ ' "(ut as the su of onl) 4 ters was as/ed, wehave to subtract one ter.*o given answer can be one of 22, "0, 2$0,2#2, 2#0

. f A L( eans A raised to the power of (, inwhich of the following choices ust 6 be greater

than Ea! 0.L6'0.LEb! 0.L6'0.2LEc! 0.L6D0.L&ption A is wrong as 6 ' E

&ption ( is wrong as PQ=Log0.92Log0.9=0.79139

&ption C is also wrong as aP>aQ then 6DE if a D

1

10. 2 gears one with 12 teeth and other onewith 14 teeth are engaged with each other. &neteeth in saller and one tooth in bigger arear/ed and initiall) those 2 ar/ed teeth are in



contact with each other. After how an)rotations of the saller gear with the ar/edteeth in the other gear will again coe intocontact for the first tiea!$b! 12c! >ata insufficientd! #4Correct &ption AAssue the distance between the teeth is 1 c.Then the circuference of first gear is 12 cand the second is 14 c.Fow :C8 12, 14! ' #4. *o to cover #4 c, the

first gear has to rotate 8412' $ rounds the

second gear rotates #4 + 14 ' rounds as it isbigger!

1. One day Eesha started 30 min late from home andreached her oce 50 min late while driving 25% slowerthan her usual seed. !ow much time in min does eesha

usually ta"e to reach her oce from home#$ns &e "now that 'eed is inversely roortional to time

&hile she drives 25% slower means she drove at 34(')

&e "now that * + ' , -

&hen seed ecame 34(') then -ime ta"en should

e 43(-)

i.e/ 'he has ta"en 43(-) - e,tra to cover the distance.

E,tra -ime + T3+ 20 min (as 20 min late due to slow

driving)$ctual time - + 0 inutes



2. n 2003 there are 24 days in eruary and 35 days ina year in 2006 there are 27 days in eruary and 3days in the year. f the date march 11 2003 is -uesday/then which one of the following would the date march 11

2006 would e#Ans: If 11-3-2003 is Tuesday, Then 11-3 - 2004 is Thursday

The number of odd days between the two dates are[3667]Rem= 2.

3) !ow many ositive integers less than 500 can eformed using the numers 1/2/3/and 5 for digits/ each

digit eing used only once.

Ans: Sin!e diit numbers = 4

"oub!e diit numbers = 4 # 3 = 12

Three diit numbers = 3 # 3 # 2 # 1 = 1$

Tota! = 34

6) $ circular swimming ool is surrounded y a concretewall 6 feet wide.if the area of the wall is 11825 of the areaof the ool/ then the radius of the ool in feet is#

%et the radius of the &oo! be r. Then area of the wa!! and &oo! =π(r+4)2

Area of the &oo! =π(r)2

Area of the wa!! =π(r+4)2−π(r)2



'i(enπ(r+4)2−π(r)2=1125(πr2)

r2+8r+16−r2=1125r2

11r2−200r−400=0

So!(in r = 20

5) $ survey of n eole in the town of adaville foundthat 50% of them refer rand $. $nother survey of 100eole in the town of chottaville found that 0% referrand $.n total 55% of all the eole surveyed togetherrefer 9rand $.&hat is the total numer of eolesurveyed#

So!: )0* +n 0* +100 = ))* +n 100

So!(in we et n = 200

) n the simle sutraction rolem elow some singledigits are relaced y letters .ined the value of:$;5*;<*# $5<5194: :6*

So!: 1) - / = $ So " = $

10 + -1 - $ = 4 So = 3

10 +)-1 - = / So = /



+A-1 - 1 = So A = $

/A )" " = ) 40 144 = 240

:) -wo full tan"s one shaed li"e the cylinder and theother li"e a cone contain li=uid fuel the cylindrical tan"held 500 lts more then the conolical tan" $fter 200 lts offuel is umed out from each tan" the cylindrical tan"now contains twice the amount of fuel in the canonicaltan" !ow many lts of fuel did the cylindrical tan" have

when it was full#

Ans: %et the y!indria! tan a&aity # )00 then the onia! tan

a&aity = #

After 200 !ts &um&ed out, then remainin fue! with the tans = # 300, #

- 200

'i(en that first term is doubt the seond.

x+300x−200=21So!(in we et # = /00

y!indria! tan a&aity = 1200 !ts

4. $ sho sells chocolates t is used to sell chocolates for>s.2 each ut there were no sales at that rice.&hen itreduced the rice all the chocolates sold out enaling the

sho"eeer to reali?e >s 16.70 from the chocolatesalone f the new rice was not less than half the originalrice =uoted !ow many chocolates were sold#

So!: 140 = 2 5 ) 5 1/ 5 /



6ow now hoo!ate &rie shou!d be reater than 1 and !ess than 2. So 2

# ) # 1/ = 1/0

So Tota! hoo!ates so!d = / and 6ew hoo!ate &rie = 7s.1./

7) Eesha ought two varities of rice costing 50>s er "gand 0 >s er "g and mi,ed them in some ratio.-hen shesold that mi,ture at :0 >s er "g ma"ing a ro@t of 20 %&hat was the ratio of the mic,ture#

So!: Se!!in &rie of the mi#ture = /0 and &rofit = 20*

ost &rie of the mi#ture =70×100120=70×56

y a&&!yin a!!iation ru!e:

So ratio =60−1753:1753−50= 1 : )

10. 'tar =uestion f f(1)+6 and f(,;y)+f(,);f(y);:,y;6/then f(2);f(5)+#



So!: %et # =1 and y = 1

f+1 1 = f+1 f+1 / # 1 # 1 4 ⇒ f+2 = 1

:et - '2 and ) ' 2f2 3 2! ' 1 3 1 3 $ - 2 - 2 3 4 ⇒ f4! ' $0

:et - ' 1 and ) ' 4f 1 3 4! ' 4 3 $0 3 2# 3 4 ' 10f2! 3 f! ' 12

1.f ffn!!3fn!'2n3" and f0!'1, what is the valueof f2012!a! 2011b! 2012c! 201"d! 40Ans: Option C6ut n ' 0Then ff0!!3f0! ' 20! 3 " ⇒ f1! 3 1 ' " ⇒ f1! '

26ut n ' 1ff1!! 3 f1! ' 21! 3 " ⇒ f2! 3 2 ' ⇒f2! ' "

6ut n ' 2ff2!! 3 f2! ' 22! 3 " ⇒ f"! 3 " ' $ ⇒ f"! ' 4

......f2012! ' 201"

2. f 3"32'11022, 3234'1#"2, then$323'Ans: 143547

f the given nuber is a 3 b 3 c then a.b X a.c Xa.b 3 a.c b⇒ 3"32 ' ." X .2X ." 3 .2 " ' 11022

⇒ 3234 ' .2 X .4 X .2 3 .4 2 ' 1#"2

$323' $.2 X $. X $.2 3 $. 2 ' 14"4$



". The savings of eplo)ee euals incoe inuse-penditure.f the incoe of A,(,C are in the ratio12" and their e-pense ratio "21 then what is theorder of the eplo)ees in increasing order of theirsiUe of their savings

Ans: A < B < CAs the the ratio of their incoes are in ascendingorder, and their e-penses are in descending order,their savings also in their incoes order.*o savings order ' A Y ( Y C

4. ?ntr) fee is Re.1.there are " rides each is of Re.1.total bo)s entering are "000.total incoe is Rs.$200.#00 students do all " rides. 1400 go for atleast 2rides.none go the sae ride twice. then no ofstudents who do not go an) ride isAns: 1000 Total entries are "000 *o fee collected through entr)fee ' "000 - 1 ' Rs."000

ncoe generated through rides ' $200 "000 '4200Fow #00 went for " rides so total fee paid b) these#00 ' #00 - " ' 24001400 #00! went for 2 rides so fee paid b) these00 ' 00 - 2 ' 1200Assue J went for e-actl) 1 rideThen J - 1 3 1200 3 2400 ' 4200 ⇒ J ' 00

*o nuber of bo)s who did not go for an) ride '"000 00 3 00 3 #00 ! ' 1000

. The average ar/ obtained b) 22 candidates in ane-aination is 4. The average of the first ten is while the last eleven is 40 .The ar/s obtained b)



the 11th candidate is Ans: 0t is clear that 22 - 4 ' 10 - 3 J 3 11 - 40 KZRightarrow J ' 0

. =hat is the largest positive integer n for which"Ln divides 44L44Ans: n = 0

The digit su of 4444is when reainder

obtained 4444divided b)

4444' (45−1)44?ach ter is a ultiple of but the last ter which

is (−1)44' 1

*o the digit su of 4444is 1.

Fow the divisibilit) rule for ", , 2$... is the su ofthe digits should be divisible b) ", , 2$ respectivel).

n each case the digit su is either ultiple of " or.*o for an) value of n D 1, the given e-pression is not

divisible b) 3n

$. 11I!322I!3""I!....20122012I! ' Ans: 2013!-111I!'1 ⇒ 2I1

11I!322I!'134' ⇒ "I1

11I!322I!3""I!'13431#'2" ⇒ 4I1

........................

.......................11I!322I!3""I!3........320122012I!'201"I1



2. 1. A two digit nuber is 1# less than the square of

the sum of its iits. 7ow an) such nubers arethere1!12!2"!"4!4Ans &ption 1Ta/e F ' 10a3b.

;iven that, 10a3b31# ' (a+b)2

for a ' 1 to , the :.7.*. will be, 2#3b, "#3b,

4#3b,.....,10#3b.As :7* is perfect suare for the values of b ' 1 to ,onl) 2#3b, 4#3b, #3b, $#3b can be eual to ",4, 4, #1 for b ' #, 1, , " respectivel). (ut onl)$#3b ' #1 for b ' " *o onl) one such nuber ispossible. .e, "

2. A two digit nuber is 1# less than the sum of the

squares of its iits" 7ow an) such nubers arethere1!12!2"!"4!4Ans &ption 2&nl) 4$ and $ satisf) the condition

". 9or real nuber -, int-! denotes integer part of-.int-! is the largest integer less than or eual to-.int1,2!'1,int2,4!'". 9ind the value ofint1+2!3int1+23 100!3int1+232+100!3....



3int1+23+100!*ol int 1+2! ' 0int 1+2 3 100 ! ' 100into 1+2 3 2+100! ' 0......int 1+2 3 0+100 ! ' 1int 1+2 3 1 +100! ' 1.......int 1+2 3 +100! ' 1*o 100 3 1 3 1 3 .....0 ties ' 10

4. ;iven a suare of length 2. ts corners are cutsuch that to represent a regular octagon. 9ind thelength of side of octagon*ol

".4. :et - is the side of the octagon and - 3 2) is the side

of the suare.

). n the given octagon, y2+y2=x2⇒2y2=x2⇒y=x2√

(ut x2√+x+x2√=2

⇒2√x+x=2

⇒x=22√ +1=22√ +1×2√ −12√ −1=2(2√ −1)

. 9ind the nuber of wa)s a batsan can score a



double centur) onl) in ters of 4Bs H BsAssue the batsan scored - 4Bs and ) Bs.

4- 3 ) ' 200 ⇒2x+3y=100⇒x=100−3y2=50−32y

As - is an integer, ) should be a ultiple of 2.f ) ' 0, - ' 0) ' 2, - ' 4$) ' 4, - ' 44...) ' "2, - ' 2

*o total wa)s are "20!+2 3 1 ' 1$ if 0 Bs arepossible! otherwise 1

. 000 voted in an election between twocandidates.14% of the votes were invalid.The winnerwon b) a argin appro-iatel) closer to 1%.9indthe nuber of votes secured b) the personnvalid Notes ' 14 % 000! ' $00

Nalid Notes ' 000 $00 ' 4"00Assue the looser got - votes. Then the winnerust have got - 3 1% -!(ut - 3 - 3 1% -! ' 4"00*olving - ' 2000*o :ooser got 2000 and winner got 2"00

$. There are 100 wine glasses. offered ) servant

to " paise for ever) bro/en glass to be deliveredsafel) and forfeit paisa for ever) glass bro/en atthe end of da). 7e recieved Rs.2.40 .how an) glassdid he brea/.a. 20 b. $" c. d. #



f a glass has been bro/en, he has to loose " paisa 3 paise ' 12 paiseAssue J glasses got bro/en

100 - " 12 - J ' 240 ⇒K=5

#. A is 20 percent ore efficient than (. f the twoperson can coplete a piece of wor/ in 0 da)s.inhow an) da)s. A wor/ing alone can coplete thewor/a. #0 b. 0 c. 100 d. 110As A is 20% ore efficient than (, f (Bs per da)

wor/ is 100 units then ABs 120.(oth persons together copletes 100 3 120! units' 220 units a da).The) too/ 0 da)s to coplete the wor/. *o totalwor/ ' 0 - 220f A alone set to coplete the wor/, he ta/es

' 60×220120=110da)s

. A propert) was originall) on a )ears lease andtwo thirds of the tie passed is eual to the fourfifth of the tie to coe.how an) )ears are thereto go.a. 4 b. 0 c. 0 d. Assue - )ears have passed and ) )ears to go

;iven 23x=45y⇒x=32×45y=65y

(ut - 3 ) ' *o 65y+y=99

*olving we get ) ' 4 )ears

10. n how an) different wa)s can the letters of the



word @:?A>F;@ be arranged in such a wa) that thevowels alwa)s coe together.a. "0b. $20c. 4#0d. 040;iven letters are A, ?, , >, :, F, ;&f which A? are vowels. :et us cobine the into asingle letter -. Fow total letters are -, >, :, F, ;These letter are arranged in I wa)s. (ut " vowelscan arrange theselves in "I wa)s. *o total wa)s I- "I ' $20

11. There is a plane contains "2 points.all the "2points have eual distance fro point -. which of thefollowing is true .a. all "2 points lie in circleb. the distance fro - to all "2 points is less than thedistance between each other

c. both a and bd. none of these*ol &ption "O ust be the center of the circle and "2 points are on the circuference. *o &ption A is correct

Fuber of diagnols of a regular pol)gon ' n(n−3)2

*o for a pol)gon of "2 sides, Fuber of diagnols '44. Fow the iniu distance between an) two

points ' 2πr32=1156r

Fow total lengh of all the distances fro "2 points

' 2πr3 *u of the lengths of all the 44 diagnols.

*u of the lengths of - to all the "2 points ' "2radius ' "2r



(ut the 44 diagnols have 1 diaeters connecting 2oposite points connecting via center. *o *u of thelengths of distances fro point to point is clearl)greater than su of the length fro - to all "2ponts. &ption ( is correctCorrect &ption "

12. =hen as/ed what the tie is,a person answeredthat the aount of tie left is 1+ of the tiealread) copleted.what is the tie.1. # p2. # a". 12 p4. 12 a*ol A da) has 24 hrs. Assue - hours have passed.Reaining tie is 24 -!

24−x=15x⇒x=20

Tie is # 68

1". 6erieter of the bac/wheel ' feet,frontwheel'$ feet on a certain distance ,the front wheelgets 10 revolution ore than the bac/ wheel.what isthe distance:et the bac/wheel ade - revolutions then frontwheel a/es - 3 10- - ' - 3 10! - $- ' "

*o distance traveled ' " - ' "1

14. There are 2 groups naed brown and red. The)can nBt arr) in the sae group. f the husband orwife dies then the person will revert to their own



group. f a person is arried then the husband willhave to change his group to his wifeBs group.Children will own the otherBs group. f an is redthen his otherBs brother belong to which group if heis arrieda. redb. brownc. red and brownd. none&ption bf a an is Red, his other ust be red, his othersbrother also red but after arriage, he getsconverted to (rown.

1. A rectangular par/ 0 long and 40 wide hasconcrete crossroads running in the iddle of thepar/ and rest of the par/ has been used as a lawn.ifthe area of the lawn is 210 s.,then what is thewidth of the road.

a. 2.1 b. "c. .#2 d. none

Option : B.



$.:et us shift the path to the left hand side and top.This does not change the area of the lawn.

#.Fow lawn area ' 0 -! 40 -! for - ' ", we get lawn area ' 210. 1. A an is

/nown to spea/ truth " out of 4 ties. 7e throws die and

reports that it is a . The probabilit) that it is actuall) a

is

*ol f actuall) appeared, he can report it with the

probabilit) of "+4. f has not appeared, still he can

report it wrongl) with the probabilit) of 1+4

*o the probabilit) that it is actuall) a ' 6robabilit) toappear - 7is truthfulness to report 3 6robabilit) to

appear an) other nuber - 7is lieing probabilit) !

' 16×34+56×14=13

The probabilit) that it is actuall) ' Probability that he

reports 6Total probability to appear 6=34×1634×16+14×56=38

2. n how an) wa)s can we distribute 10 pencils to 4

children so each child gets atleast one pencil

Fuber of wa)s of distributing r identical ob<ects to n

distinct ob<ects so that each get atleast one

' (n−1)C(r−1)' (10−1)C(4−1)=9C3

". A drawer holds 4 red hats and 4 blue hats. what is

probabilit) of getting e-actl) " red hats or " blue hats

when ta/ing out 4 hats randol) out of drawer and

iediatel) returning ever) hat to drawer before ta/ing



out ne-t

As the ob<ects are replaced, the probabilit) of drawing

red or blue is eual.

6robabilit) to draw " red hats consecutivel)' 12×12×12=18

*iilarl) probabilit) to draw " blue hats consecutivel)

' 12×12×12=18

Total probabilit) ' 12×12×12=18312×12×12=18' 14

4. A father purchased dress for his " daughters. The

dresses are of sae color but diff siUe and the) are /ept

in dar/ roo. what is probabilit) that all the " will not

choose their own dress

This is a case of dearrangeents ' Dn=n!(12!−13!+14!

−....)*o nuber of wa)s that none of the chooses their own

dress ' D3=3!(12!−13!)=2*o probabilit) ' 23!=13

. 0% of ale in a town and $0% of feale in a town

are eligible to vote. out of which $0% of ale and 0%

of feale who are eligible to vote voted for candidate A.

what is the value of votes in % did A get



:et the ratio of en and woen are 100 /

8ale eligible votes ' 0 and feale eligible votes ' $0%

/!

Fuber of ales who voted for A ' $0% 0! ' 42Fuber of feales who voted for A ' 0%$0% J! '

42% /!

6ercentage of votes got b) A

' 42+42100(K)60+70100(K)×100=4200+42K6000+70K×100

*o this value cannot be deterined as the value of J is

not /nown

. ;eorge and 8ar/ can paint $20 bo-es in 20 da)s. 8ar/

and 7arr) in 24 da)s and 7arr) and ;eorge in 1 da)s.

;eorge wor/s for 4 da)s, 8ar/ for # da)s and 7arr) for #

da)s. The total nuber of bo-es painted b) the is

Capacit) of ; 3 8 ' $20 + 20 ' "

8 3 7 ' $20 + 24 ' "07 3 ; ' $20 + 1 ' 4#

Cobined capacit) ' 2 ; 3 7 3 8! ' 114

; 3 7 3 8 ' 114 + 2 ' $

Fow capacit) of ; ' ;3738! 7 3 8! ' $ "0 ' 2$

8 ' ;3738! 7 3 ;! ' $ 4# '

7 ' ;3738! ; 3 8! ' $ " ' 21

;iven that ; wor/ed for 4 da)s, and ar/ for # and harr)

for # da)s

*o total wor/ b) the ' 4 - 2$ 3 # - 3 # - 21 ' "4#



$. Two euilateral triangle of side 12c are placed one on

top another, such a pionted star is fored if the si-

vertices lie on a circle what is the area of the circle not

enclosed b) the *tara!1

b!$

c!#

d!#"

*ol ;iven that two euilateral triangles of length 12 has

inscribed in a circle.

Altitude of the triangle ' 3√2a' 3√2(12)' 63√

=e /now that centroid divides the altitude in the ratio 2

1 and 23Altitude! ' Circu radius

Circu radius ' 23(63√)=43√

Area of the circle ' πr2=3.14×(43√)2

Fow the two triangles in the circle fors 12 salleuilateral triangles with side 4. *o their total area

' 12×3√4a2' 12×3√442

Area which is not covered b) the euilateral triangles

' 3.14×(43√)2 12×3√442' $. A68



#. There are 4 different letters and 4 addressedenvelopes.n how an) wa)s can the letters be put inthe envelopes so that atleast one letter goes to thecorrect address a!1 b!1 c!1# d!12Total wa)s of putting r letters to r covers ' rI ' 4I ' 24Fuber of wa)s that none of the goes into the right

envolope ' D4=4!(12!−13!+14!) '

*o atleast one envolope goes into the right one ' 24

' 1

.There are 20en and 10 woen in a coittee, ifall will wor/ the) will coplete 12 units per da), if allen wor/ the) will coplete 1 units per da), how an)units will woen coplete per da) thin/ there is a ista/e in this uestion. f all en andwoen together coplete 12 units, how onl) en can do

1 Mnits of wor/ a da)9orgetting about the realit), =oen can do " units ada).

10. 7ow an) odd and even nubers are there between42 and 400 9ind the su of odd nubers and the suof even nubersI*ol &dd nubers are fro 4" to ". Fuber of odd

nubers ' l−ad+1=399−432+1=179

Their su ' n2(l+a)' "

?ven nubers are fro 44 to "#. Fuber of even



nubers ' l−ad+1=398−442+1=178

Their su ' 1782(398+44)=39338

11. The faous church in the cit) of Juba/onna has abig cloc/ tower and is said to be over "00 )ears old.?ver) 8onda) 10.00 A 8 the cloc/ is set b) Anton), doingservice in the church. The Cloc/ loses ins ever) hour.=hat will be the actual tie when the fault) cloc/ shows" 6.8 on 9rida)a. 4 A8b.".1 68c. 4.4 A8d. " A8Total tie passed in the fault) cloc/ ' 8onda) 10 a to9rida) " p ' 24 - 4 3 hours ' and hours ' 101hrs4 in in the fault) cloc/ ' 0 inutes of the correctcloc/101 hrs in the faulct) cloc/ '

10154×60' 112.2 7rs.

7rs 3 1.2 7rs9rida) 10 a 3 1 hrs ' *aturda) 2a0.2 - 0 in ' 12 in*o *aturda) 2.12 in A8

12. *uresh Raina and ;auta ;abhir after a

scintillating 6: atch decide to travel b) c)cle to theirrespective villages. (oth of the start their <ourne)travelling in opposite directions. ?ach of their speeds is iles per hour. =hen the) are at a distance of 0 iles,a housefl) starts fl)ing fro *uresh RainaBs c)cle towards;auta ;abhir at a relative speed of 1$ iles per hour



with respect to RainaBs speed. =hat will be the tieta/en b) housefl) to reach ;abhira. 10 hrsb. 1 hrsc. 20 hrsd. 2 hrs*ol

9l) speed is 1$ /ph w.r.t to suresh as fl) is oving inopposite direction to suresh, its actual speed is 1$ ' 11.Fow relative speed of fl) and gabhir ' 11 ' /ph

*o fl) ta/es ' 5011−6' 10 7rs

1. The value of diaond varies directl) as the suare of

its weight. f a diaond falls and brea/s into two pieceswith weights in the ratio 2". what is the loss percentagein the value*ol :et weight be -S the cost of diaond in the original state is proportional

to x2

when it is fallen it brea/s into two pieces 2) and the ")- ' )

&riginal value of diaond ' (5y)2' 25y2

Nalue of diaond after brea/age ' (2y)2 +(3y)2 =13y2



so the percentage loss will be ' 25y2 −13y225y2×100=48%

2. 9ive college students et at a part) and e-changed

gossips. Ma said, &nl) one of us is l)ingS. >avid said, ?-actl) two of us are l)ingS. Thara said, ?-actl) " of usare l)ingS. Euerishi said, ?-actl) 4 of us are l)ingS.Chitra said All of us are l)ingS. =hich one was telling thetrutha!>avidb!Euerishic!Chitra

d!Thara*ol As all are contradictor) stateents, it is clear that&F:P one of the is telling the truth. *o reaining 4 ofthe are l)ing. Euerishi entioned that e-actl) 4 arel)ing. *o, he is telling the truth.?-planation :et us 1st assue that Ma is telling thetruth. Then according to her onl) one is l)ing. (ut if onl)one is l)ing then all the others[ stateents are

contradicting the possibilit). n the sae wa) all theother stateents should be chec/ed. f we assue theEuerishi is telling the truth, according to hi e-actl) 4ebers are l)ing. *o all the others are telling lies andhe is the one who is telling the truth. This case fitsperfectl).

". Cara, a blue whale participated in a weight loss

progra at the biggest office. At the end of ever) onth,the decrease in weight fro original weight waseasured and noted as 1, 2, , 21, #, 44, 2$. =hileCara ade a steadfast effort, the weighing achineshowed an erroneous weight once. =hat was that.



a! 2$b! 2c! 44d! #*&: This is a nuber series proble nothing to do withthe data given.1- 131'22 - 232' - "3"'2121 - 434'## and not ### - 3 ' 4444G3 ' 2$ 4. The letters in the word A>&6T* are peruted in allpossible wa)s and arranged in alphabetical order thenfind the word at position 42 in the peruted alphabeticalordera! A&T>*6b! A&T6>*

c! A&T>6*d! A&*T6>*&:n alphabetical order A > & 6 * TA \ \ \ \ \ the places filled in I wa)s ' 120, (ut weneed a ran/ less than 120. *o the word starts with A.A > \ \ \ \ ept) places can be filled in 4I'24A & \ \ \ \ the places filled with 4I wa)s ' 24. f we

add 24 3 24 this total crosses 42. *o =e should notconsider all the words starting with A&.A & > \ \ \ "I' A & 6 \ \ \ "I'Till this " words are obtained, we need the 42nd word.A&* \ \ \ "I'



?-actl) we are getting the su 42. *o last " letters inthe descending order are T6>.*o given word is A&*T6>

4. A an who goes to wor/ long before sunrise ever)orning gets dressed in the dar/. n his soc/ drawer hehas blac/ and # blue soc/s. =hat is the probabilit) thathis first pic/ was a blac/ soc/, but his second pic/ was ablue soc/*&: This is a case of without replaceent. =e have toultipl) two probabilities. 1. 6robabilit) of pic/ing up ablac/ soc/, and probabilit) of pic/ing a blue soc/, giventhat first soc/ is blac/.

6C114C1×8C113C1=2491

. There are red balls,# blue balls and $ green balls in abag. f are drawn with replaceent, what is theprobabilit) at least three are red*ol At least " reds eans we get either " red or 4 red

or red. And this is a case of replaceent.case 1 " red balls +21 - +21 - +21 - 1+21 -1+21case 2 4 red balls +21 - +21 - +21 - +21 - 1+21case " red balls +21 - +21 - +21 - +21 - +21

Total probabilit) ' ' +21 - +21 - +21 - 1+21 -1+21!3+21 - +21 - +21 - +21 - 1 !+21!3 +21 -

+21 - +21 - +21 - +21! ' "12+1#0$

. Total nuber of 4 digit nuber do not having the digit" or .



*olconsider 4 digits \ \ \ \

1st blan/ can be filled in 7C1wa)s 0,", are neglected

as the first digit should not be 0!2st blan/ can be filled in 8C1wa)s 0 considered along

with 1,2,4,,$,#,!

"st blan/ can be filled in 8C1wa)s

4st blan/ can be filled in 8C1 wa)s

Therefore total 4 digit nuber without " and is $ - # -# - #'"#4

$. 9ind the issing in the series $0, 4, 4, 41,\\\\.*ol 40

$04 ' 1 ' 42

44 ' ' 32

441 ' 4 ' 22

4140 ' 1 ' 12

.A school has 120, 12 and 144 students enrolled forits science, arts and coerce courses. All studentshave to be seated in roos for an e-a such thateach roo has students of onl) the sae course andalso all roos have eual nuber of students. =hatis the least nuber of roos needed

*ol =e have to find the a-iu nuber whichdivides all the given nubers so that nuber ofroots get iniiUed. 7C9 of 120,12 H 144 is 24. ?ach roo have 24 students of the sae course.

Then roos needed 12024+19224+14424 ' 3# 3



' 1 . A farer has a rose garden. ?ver) da) he pic/seither $,,24 or 2" roses. =hen he pluc/s thesenuber of flowers the ne-t da) "$,", or 1# newflowers bloo. &n 8onda) he counts 1# roses. f hecontinues on his plan each da), after soe da)swhat can be the nuber of roses left behind 7int Consider nuber of roses reaining ever) da)!a!$b!4c!"0d!"$*&:let us consider the case of 2". when he pic/s up 2"roses the ne-t da) there will be 1# new, so in thiscase., flowers will be less ever) da). *o when hecounts 1#, the ne-t da) 1#4,1$,1$4,1,................

finall) the no. of roses left behind will be 4.

10. =hat is the "2nd word of @=ATF;@ in adictionar)*ol Arranging the words of waiting in Alphabetical&rder A,;,,,F,T,=

*tart with A\ \ \ \ \ \ This can be arranged in I+2I

wa)s'$20+2'"0 wa)sso canBt be arranged starting with A alone as it isas/ing for "2nd word so it is out of range

A;\ \ \ \ \then the reaining letters can bearranged in I+2I wa)s so,120+2'0 wa)s. &ut of



range as it has to be within "2 words.A;\ \ \ \ Fow the reaining letters can bearranged in 4I wa)s '24A;F \ \ \ \ can be arranged in 4I+2I wa)s or 12wa)sso,24312 '"th word so out of range. *o we shouldnot consider all the words start with A;Fnow A;F\ \ \can be arranged in "I wa)s ' wa)sso 243'"0 within rangeFow onl) two word left so, arrange in alphabeticalorder.A;FT= "1st wordA;FT= "2nd word

10. 1. A anufacturer of chocolates a/es different flavors of chocolates. The chocolates aresold in bo-es of 10. 7ow an) differentS bo-es ofchocolates can be ade*olf n siilar articles are to be distributed to r

persons, x1+x2+x3......xr=neach person is eligible tota/e an) nuber of articles then the total wa)s

are n+r−1Cr−1

n this case x1+x2+x3......x6=10

in such a case the forula for non negative integral

solutions is n+r−1Cr−1

7ere n ' and r'10. *o total wa)s

are 10+6−1C6−1' "00"

2. n a single throw with two dice, find theprobabilit) that their su is a ultiple either of " or4.



a. 1+"b. 1+2c. +d. 1$+"*ol Their su can be ",4,,#,,129or two dice, an) nuber fro 2 to $ can be get inn1! wa)s and an) nuber fro # to 12 can be getin 1" n! wa)s.Then possible wa)s are 2 3 " 3 3 3 4 3 1 ' 20possible cases.*o probabilit) is 20+"!'+!

". ( alone can do piece of wor/ in 10 da)s. A alonecan do it in 1 da)s. f the total wages for the wor/is Rs 000, how uch should ( be paid if the) wor/together for the entire duration of the wor/a. 2000b. 4000c. 000

d. "000*olTie ta/en b) A and ( is in the ratio of ' "2Ratio of the =or/ ' 2 " since, tie and wor/ areinversel) proportional!Total one) is divided in the ratio of 2 " and ( getsRs."000

4. &n a 2 uestion test, points were deducted foreach wrong answer and # points were added for rightanswers. f all the uestions were answered howan) were correct if the score was Uero.a. 10b. 11



c. 1"d. 12*ol:et - ues were correct. Therefore, 2 -! werewrong

8x−5(26−x)=0

*olving we get -'10

. Arun a/es a popular brand of ice crea in arectangular shaped bar c long, c wide and 2cthic/. To cut costs, the copan) had decided to

reduce the volue of the bar b) 1%. The thic/nesswill reain sae, but the length and width will bedecreased b) soe percentage. The new width willbe,a. .b. 4.c. $.d. .

*olNolue 'l×b×h' 6×5×2 ' 0 cm3

Fow volue is reduced b) 1%.

Therefore, new volue ' (100−19)100×60=48.6

Fow, thic/ness reains sae and let length andbreadth be reduced to -%

so, new volue (x100×6)(x100×5)2=48.6*olving we get - '0thus length and width is reduced b) 10%Few width ' 10% of !'4.



. f all the nubers between 11 and 100 are writtenon a piece of paper. 7ow an) ties will the nuber4 be used*ol =e have to consider the nuber of 4Bs in twodigit nubers. \ \f we fi- 4 in the 10th place, unit place be filled with10 wa)s. f we fi- 4 in units place, 10th place befilled with wa)s 0 is not allowed!*o total 1 wa)s.A#ternati$e#%:There are total 4Bs in 14, 24, "4...,4H total 10 4Bs in 40,41,42....4thus, 310'1. $. f twent) four en and si-teen woen wor/ on ada), the total wages to be paid is 11,00. f twelveen and thirt) seven woen wor/ on a da), thetotal wages to be paid reains the sae. =hat isthe wages paid to a an for a da)[s wor/

*ol :et an dail) wages and woan dail) wages be8 and = respectivel)24831='11001283"$='1100solving the above euations gives 8'"0 and='200 #. The cost price of a cow and a horse is Rs " la/hs.

The cow is sold at 20% profit and the horse is sold at10% loss. &verall gain is Rs 4200. =hat is the costprice of the cow*ol6rofit ' 42006rofit '*6 C6



4200'*6 "00000 therefore *6'"04200-3) ' "000001.2- 3 0.) ' "04200*olving for - ' 114000 ' C6 of cow.

. 1, 2, 2, ", ", ", 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, ", ", ",", ", ", 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2,", ", ", ", ", ", ", ", ", ", 4......n the above seuence what is the nuber of theposition 2### of the seuence.a! 1b! 4c! "d! 2*ol 9irst if we count 122"""4444. the) are 10n the ne-t ter the) are 20Fe-t the) are "0 and so on

*o Msing n(n+1)2×10≤2888

11. 9or n ' 2" we get :7* as 2$0. Reainingters 12#.

12.

1". Fow in the 24th ter, we have 24 1Bs, and ne-t4# ters are 2Bs. *o ne-t $2 ters are "Bs.

14. The 2### ter will be "S. 10. 7ow an) 4digit nubers contain no.2

*ol Total nuber of four digit nubers '000 i.e1000 to !=e tr) to find the nuber of nubers not havingdigit 2 in the.Fow consider the units place it can be selected in wa)s i.e 0,1,",4,,,$,#,!



Tens place it can be selected in wa)s i.e0,1,",4,,,$,#,!7undreds place it can be selected in wa)s i.e0,1,",4,,,$,#,!Thousands place can be selected in # wa)s i.e1,",4,,,$,#,! here B0B cannot be ta/enTotal nuber of nubers not having digit 2 in it ' - - - # '#"2Total nuber of nubers having digit 2 in it ' 000#"2 '"1#

1. 1. 2ab is a four digit nuber divisible b) 2. fa nuber fored fro the two digits ab is a ultipleof 1", then ab isa. 2b. 4c.10d.2*ol 9or a nuber to be divisible b) 2, last twodigits of that nuber should be divisible b) 2. *o b

ust be either 2 or $it is given that ab ust be divisible b) 1" and in theoptions onl) 2 is divisible b) 1". 2. The average teperature of Tuesda) =ednesda)and Thursda) was "$ C. The average teperature of=ednesda) and Thursda) and 9rida) was "# C. if theteperature on 9rida) was " C.

9ind the teperature on Tuesda).a. "$.""b. "#.""c. "d. Fone of the above*ol



tues 3 wed 3 thurs!+"'"$tues 3 wed 3 thurs'111...1!wed 3 thurs 3 fri!+"'"#wed 3 thurs 3 fri! '114...2!;iven frida) is ".then, 2! 1! 9ri Tues ' "*o " Tues ' "Tuesda) '"

". There are bo-es in a cargo. The weight of the1st bo- is 200 J;, the weight of the 2nd bo- is 20%higher than the third bo-, whose weight is 2%higher than the 1st bo- weight. The 4th bo- whichweighs "0 J; is "0% lighter than the th bo-. 9indthe difference in average weight of the 4 heaviestbo-es and the four lightest bo-es.*ol weight of 1st bo-'200weight of "rd bo-'12+100!G200'20weight of 2nd bo-'120+100!G20'"00

weight of 4th bo- '"0weight of th bo-'10+$!G"0'00average of 4 highest weightedbo-es'003"03"00320!+4'"0average of 4 lightestbo-es'"03"003203200!+4'2$therefore difference'"02$'$

4. The length, breadth and height of a roo are inthe ratio "21. f the breadth and height are halved,while the length is doubled. Then the total area ofthe 4 walls of the roo will be decreased b)a. "0%b. 1#.$%



c. 1%d. 1".%*ol ;iven lbh'"21let h'10, b ' 20, and l ' "0

area ' 2(l+b)h

area' 2G"-32-!G- ' 2(30+20)10=1000

Fow after those ad<ustents in the easureents,l'0, b'10, h'

area' 2(l+b)h' 2(60+10)5=700

6ercentage decrease' 1000−7001000×1000=30%

. A circle circuscribes three unit circles that toucheach other. =hat is the area of the larger circleFote that p is the ratio of the circuference to thediaeter of a circle ".1412!.*ol

1.1$. () <oining centres of " unit circles we will get an

euilateral triangle of length 2 unit. =e have to findthe length of the orange line.And center of the euilateral triangle will be thecenter of the big circle.*o radius of the big circle will be ' 1 3 Circu



radius of the euilateral triagle!Circu radius of euilateral triangle

' 23×3√2×2=23√

Area of big circle will be ' πr2=3.14×(1+23√)

. Ra<esh calculated his average over the last 24tests and found it to be $. 7e finds out that thear/s for three tests have been inverted b) ista/e.The correct ar/s for these tests are #$, $ and #.=hat is the appro-iate percentage differencebetween his actual average and his incorrectaverage*ol Fo Changencorrect value is $#, $, #correct values are #$, $, #difference between correct and incorrect value is' 3 1#'0 $. 5o/e is faster than 6aul, 5o/e and 6aul each wal/24 J8. The su of their speed is $ J per hour. Andthe su of ties ta/en b) the is 14 hours. Then,5o/e speed isa. " J8+7rb. 4 J8+7rc. J8+7rd.$ J8+7r

*olSpeed=Timedistance

let the speed of <o/e - then speed of paul will be $-

24x+247−x=14

Tr) to plugin the values fro the options. f 5o/e



speed is 4 the paul is ".

#. The crew of a rowing tea of # ebers is to bechosen fro 12 en 81, 82, ., 812! and #woen =1, =2,., =#!, such that there are tworows, each row occup)ing one the two sides of theboat and that each side ust have 4 ebersincluding at least one woen. 9urther it is also/nown =1 and 8$ ust be selected for one of itssides while 82, 8" and 810 ust be selected forother side. =hat is the nuber of wa)s in whichrowing tea can be arranged.*o:=e need two person for one side and 1 woen forthe another side. =e select that woen in $ wa)s.Fow that second side people can sit in $-4I wa)s.Fow for the first side we need two people fro the

reaining 14. *o this can be done in 14C2wa)s and

this side people can sit in 4C2×4!wa)s.

Again the first group a) ta/e an) of the two sides.

*o total wa)s are 2×7×4!×14C2×4!

. n a certain cit), 0% of the registered voters arecongress supporters and the rest are (56 supporters.n an assebl) election, if $% of the registeredcongress supporters and 20% of the registered (56

supporters are e-pected to vote for candidate A,what percent of the registered voters are e-pected tovote for candidate A*ol let the people in the cit) be 100Congress supporters ' 0% of 100 ' 040% are (56'40% of 100 ' 40



out of 0,$% voted for congress'$%0!'4out of 40%,20% voted for congress'20%40!'#Total'4 3 # ' "Total percent' "%

10. Anusha, (anu and ?sha run a running race of100 eters. Anusha is the fastest followed b) (anuand then ?sha. Anusha, (anu and ?sha aintainconstant speeds during the entire race. =henAnusha reached the goal post, (anu was 10behind. =hen (anu reached the goal post ?sha was10 behind. 7ow far was behind Anusha when thelatter reached the goal post.optiona! $0b! #1c! 0d! #0*ol

() that tie Anusha covered 100, (hanu covered0. *o ratio of their speeds ' 10 () that tie (hanu reached 100, ?sha covered0. *o ratio of their speeds ' 10 Ratio of the speed of all the three ' 100 0 #1() that tie Anusha covered 100, ?sha Coversonl) #1.

11. *even different ob<ects ust be divided aongthree persons. n how an) wa)s this can be done if at least one of the gets e-actl) one ob<ect.*ol >ivision of 3n3p ob<ects into three groups is

given b) (m+n+ p)!m!×n!× p!



(ut $ ' 1 3 " 3 " or 1 3 2 3 4 or 1 3 1 3

*o The nuber of wa)s are (7)!1!×3!×3!×12!+(7)!1!×2!

×4!+(7)!1!×1!×5!×12!' $0 3 10 3 21 ' 1

12. ;eorge while driving along the highwa) saw roadar/ers which are at eual distances fro eachother. 7e crosses the ar/ers ever) 20 seconds. fhe increases his speed b) - eters per second, hecrosses the ar/ers at ever) 1 seconds. (ut if heincreases his speed b) ) eters per second, hecrosses the ar/er at ever) 10th second. f )- ' 40eters per second, then what is the distancebetween two ar/ers.*ol :et speed be 'U +s then >istance' 20U U3-!1'20U] U3)!10'20UAlso given that ) - ' 40solving we get 20U'1200 1". 7ow an) different digit nubers can befored fro the nuber 22""### b) rearrangingits digits so that the odd digits occup) even position*ol &dd places are 4 and these are occupied b)"". *o this can be done in 4I+ 2I 2I! ' There are even nubers which have to be placedat odd places. *o I+2I"I! ' 10 wa)sso total nuber of wa)s of arranging all these

nubers are 10 G ' 0 wa)s

14. n a vessel, there are 10 litres of alcohol. Anoperation is defined as ta/ing out five litres of whatis present in the vessel and adding 10 litres of purewater to it. =hat is the ratio of alcohol to water after



two operationsa! 1 b! 2 "c! 1 d! " 2

*ol Final concentration = Initial

concentration(1−replacement quantityFinal volume)Final concentration = 1×(1−1015)=13Final concentration = 13×(1−1020)=16*o ratio of alcohol water ' 1

1. Ada sat with his friends in the Chinnaswa) stadiuat 8adurai to watch the 100 etres running raceorganiUed b) the Asian athletics Association. 9ive roundswere run. After ever) round half the teas wereeliinated. 9inall), one tea wins the gae. 7ow an)teas participated in the raceAns Total five rounds were run. *o in the final round 2teas ust have participated. n the penultiate round4 teas, and "rd round #, 2nd round 1 and in the firstround "2 teas ust have participated as in each roundhalf of the teas got eliinated.

2. 9ro the top of a etres high building A(, the angleof elevation of the top of a tower C> is "0^ and the angleof depression of the foot of the tower is 0^. =hat is theheight of the towerAns



Ans =e have to find the value of C>. =e use *ine rule

to find the answer easil). *ine rule is aSinA=bSinB=cSinC

n triangle (>?, 9Sin60=xSin30

*o 93√2=x12⇒x=93√

n triangle (C>, CDSin30=93√Sin60

CD12=93√3√2⇒CD=3

*o height of the tower ' 3 " ' 12

". 4 ebers attended the part). n that 22 are ales,2$ are feales. The sha/e hands are done betweenales, feales, ale and feale. Total 12 people givensha/e hands. 7ow an) such /inds of such sha/e hands

are possibleAns f onl) 12 people sha/ed their hands, then total

hand sha/es are 12C2'



4. 9errari *.6.A is an talian sports car anufacturerbased in 8aranello, tal). 9ounded b) ?nUo 9errari in12# as *cuderia 9errari, the copan) sponsored driversand anufactured race cars before oving intoproduction of streetlegal vehicles in 14$ as 9errari*.6.A. Throughout its histor), the copan) has beennoted for its continued participation in racing, especiall)in 9orula &ne where it has eplo)ed great success.Rohit once bought a 9errari. t could go 4 ties as fast as8ohan[s old 8ercedes. f the speed of 8ohan[s 8ercedesis " /+hr and the distance traveled b) the 9errari is40 /, find the total tie ta/en for Rohit to drive thatdistance.Ans As 9errariBs speed is four ties that of theercedes, ts speed is " - 4 ' 140*o tie ta/en b) the ferrari ' 40 + 140 ' ". 7ours

. A sheet of paper has stateents nubered fro 1 to40. 9or all values of n fro 1 to 40, stateent n sa)s

_?-actl) n of the stateents on this sheet are false.[=hich stateents are true and which are falsea! The even nubered stateents are true and the oddnubered stateents are false.b! The odd nubered stateents are true and the evennubered stateents are false.c! All the stateents are false.d! The "th stateent is true and the rest are false

Ans Assue there is onl) one stateent is there. Thestateent should read @?-actl) 1 stateent on this sheetis false@ . f the truth value of the stateent is true, thengiven stateent should be false. This is contradiction. fthe stateent is false, Then the given stateent is true.but there is not other true stateent.



Assue there are two stateents. () the above logic,2nd stateent should not be true. (ut 1st stateent istrue as it truthfull) sa)s the truthfulness. () this logicwe /now that f there are @n@ stateents, n1!thstateent is the onl) true stateent And all other arefalse

. f there are "0 cans out of the one is poisoned if aperson tastes ver) little he will die within 14 hours so ifthere are ice to test and 24 hours to test, what is theiniu no. of ice[s reuired to find poisoned canAns

f onl) " person are used, b) giving wine dropssuggested b) the diagra, we can find the poisonedcas/s upto #.for e-aple, f the 2nd and "rd persons die, then $th

cas/ is poisoned. As a rule of thub, f we have n ice,we can easil) find the poison cas/s upto 2n. As the

nuber of cas/s are less than "2 we can use onl) ice.



$. 7ow an) digit nubers are possible b) using thedigits 1, 2, ", 4, which are divisible b) 4 if therepetition is allowedAns f A nuber has to be divisible b) 4, the last twodigits ust be divisible b) 4. *o possibilities are, 12, 24,"2, 44, 2. And the of the reaining $ places, eachplace got filled b) an) of the five digits. *o these $

places got filled b) - - .....$ ties! ' 57wa)s. *o

total wa)s are - 57' 58

#. A hare and a tortoise have a race along a circle of 100)ards diaeter. The tortoise goes in one direction and thehare in the other. The hare starts after the tortoise hascovered 1+ of its distance and that too leisurel). Thehare and tortoise eet when the hare has covered onl)1+# of the distance. () what factor should the hareincrease its speed so as to tie the race



Assue the circuference of the circle is 200eters. 7are and tortoise started at the sae pointbut oves in the opposite direction. t is given thatb) that tie tortoise covered 40 1+th of thedistance!, 7are started and both et after hare hascovered 2. This iplies, in the tie hare hascovered 2, hare has covered 200 40 2 ' 1"eters.*o 7are tortoise speeds ' 2 1" ' 2$

Fow 7are and tortoise has to reach the starting pointeans, 7are has to cover 1$ eters and Tortoisehas to cover onl) 2 eters in the sae tie.

As tie 'DistanceSpeed=2527=1755×K

e., 7are has to increase its speed b) a factor J.*olving we get J ' "$.#

. 9or the 99A world cup, 6aul the octopus has been

predicting the winner of each atch with aaUingsuccess. t is ruored that in a atch between 2teas A and (, 6aul pic/s A with the saeprobabilit) as A[s chances of winning. :et[s assuesuch ruors to be true and that in a atch between



;hana and (olivia] ;hana the stronger tea has aprobabilit) of 2+" of winning the gae. =hat is theprobabilit) that 6aul will correctl) pic/ the winner ofthe ;hana(olivia gaea! 1+b! 4+c! +d! 2+"The probabilit) that 6aul correctl) pic/s the winner 'ABs Chances of winning!-6auls pic/ing the winnercorectl)! 3 ABs chances of loosing! - 6aul pic/s

wrongl)! ' 23×23+13×13=59

10. " people à1, a2 a" eet and sha/e handsin a circular fashion. n other words, there are totall)" handsha/es involving the pairs, à1, a2, à2,a", , à", a", à", a1. Then siUe of thesallest set of people such that the rest have sha/en

hands with at least one person in the set isa! 12b! 11c! 1"d! 1#Ans à1, a2, à2, a",à", a4, à4, a,à, a, à, a$ , à", a", à", a19ro the above arrangeent, f we separate a", a,

a, .....a". Total 12 persons the reaining personsust have sha/ed hand with atleast one person. *oanswer is 12.

11. There are two bo-es, one containing 10 red balls



and the other containing 10 green balls. Pou areallowed to ove the balls between the bo-es so thatwhen )ou choose a bo- at rando and a ball atrando fro the chosen bo-, the probabilit) ofgetting a red ball is a-iiUed. This a-iuprobabilit) isf rearrangeent is not allowed, then actual

probabilit) of pic/ing up a red ball ' 12(10)+12(0)=12

As we are allowed to ove the ball, we /eep onl) 1red in the first bo-, and shirt the reaining to thesecond.

*o ' 12(1)+919(0)=1419

12. The difference between two no is and theproduct of the two is 14. =hat is the suare of theirsu

=e /now that (a+b)2=(a−b)2 + 4ab

*ubstituting a b ' , and ab '

14, (a+b)2=(9)2+4(14)=137

1". There are two water tan/s A and (, A is uchsaller than (. =hile water fills at the rate of oneliter ever) hour in A, it gets filled up li/e 10, 20, 40,#0, 10 in tan/ (. At the end of first hour, ( has 10liters, second hour it has 20, third hour it has 40 andso on!. f tan/ ( is 1+"2 filled after 21 hours, what isthe total duration reuired to fill it copletel)Ans The data related to the first tan/ A is notnecessar). As )ou can see, the capacit) that getsfilled in the tan/ ( after each hour is doubled. *o fthe tan/ is 1+"2nd part is full after 21 hours, it is



1+1th part full after 22 hours, 1+#th part full after2" hours, 1+4th part full after 24 hours, 1+2 full after2 hours, copletel) full after 2 hours.

14. " friends A, (, C went for wee/ end part) to8c>onald[s restaurant and there the) easure thereweights in soe order n $ rounds. A, (, C, A(, (C,AC, A(C. 9inal round easure is 1/g then find theaverage weight of all the $ roundsAverage weight ' Va 3 b 3 c 3 a3b! 3 b3c! 3c3a!3a3b3c!W + $ ' 4 a3b3c! +$ ' 4 - 1+$ '##. /gs

1. A grand father has " grand children. Agedifference of two children aong the is ". ?ldestchild age is " ties the )oungest child[s age and theeldest child age is two )ear ore than the su ofage of other two children. =hat is the age of theeldest child

Ans As the eldest sonBs age is " ties that of the)oungest, eldest sonBs age should be a ultiple of ".9ro the given options ta/e 1 as the eldest sonBsage. Then )oungest sonBs age becoes . (ut ?ldestsons age is 2 ore than the su of the reainingtwo sons. *o *u of the reaining two sons is 1".*o the age of the iddle son is 1" ' #. =hichsatisfies another condition in the uestion that the

difference between the two sons age is ". *o answeris 1.

1. n a i-ture of a, b and c, if a and b are i-edin " ratio and b and c are i-ed in # ratio and ifthe final i-ture is " liters, find the aount of b



Ans As b is coon in both ratios, we shouldeuate b in both ratios b) ultipl)ing suitablenubers.ab ' " ' 24 40bc ' # ' 40 2Fow a b c ' 24 40 2.

Aount of b in the i-ture ' 4089×35' 1.$"

1$. After the t)pist writes 12 letters and addresses12 envelopes, she inserts the letters randol) intothe envelopes 1 letter per envelope!. =hat is the

probabilit) that e-actl) 1 letter is inserted in aniproper envelopeAns Tric/) one but siple. 7ow do )ou put e-actl)1 letter in the wrong envelope we need iniutwo. *o answer is 0.

1#. 10 suspects are rounded b) the police anduestioned about a ban/ robber). &nl) one of the

is guilt). The suspects are ade to stand in a lineand each person declares that the person ne-t tohi on his right is guilt). The rightost person is notuestioned. =hich of the following possibilities aretrueA. All suspects are l)ing.(. leftost suspect is innocent.C. leftost suspect is guilt)

a! A onl)b! A or Cc! A or (d! ( onl)There are onl) 2 cases. ?ither left one is guilt) or



one of the reaining to his right is guilt).*o f the left ost is guilt), All the stateentsincluding the guilt) one are lies. A and C are correct.&r f An) one e-cept left ost one is guilt), Thenone of the stateents given b) the person should betrue. n this case all the suspects are l)ing does nothold. *o f ( is correct, A is not correct. i.e., onl) Aor ( is correct. &ption C is correct.

1. A hollow cube of siUe c is ta/en, with athic/ness of 1 c. t is ade of saller cubes of siUe1 c. f 1face of the outer surface of the cube arepainted, totall) how an) faces of the saller cubesreain unpainted

The 7allow cube volue ' n3−(n−2)2, 7ere n is the

nuber of sall cubes lie on the big cube edge.Fow n ' so 7allow cube volue

' 53−(5−2)2=125−27=98

*o # sall cubes reuired to a/e a hallow cube of siUe c. Fow total surfaces ' - # ' ##Fow if the bigger cube is painted 4 sides, total 4 -2 sall faces got paint. *o reaining sall faceswhich does not have paint after cutting is ## 100' 4##

20. 8) flight ta/es of at 2a fro a place at 1#F

10? and landed 10 7rs later at a place withcoordinates "F$0=. =hat is the local tie when )plane landeda! 12 noonb! 40 A8c! 20 68



d! 0 A8Reeber, while oving fro east to west countrieslag in tie. Reeber when Test cric/et starts in?ngland ". "0 in afternoon. Right ie., =e are inafter noon eans the) are in orning.f the coordinates change fro 10 ? to $0=, theplane has oved a total of #0 degrees. =e /nowthat with each degree tie increases b) 4 inuteswhile going fro east to west. 7ow 24 - 0 in +"0 degrees, *o 1 degree ' 4 in!*o total tie change ' 4 - #0 ' "20 in ' hrs 320 inutes.After 10 hours local tie is 2 a 3 10 .20 hrs! '.40 A8.

1! &f the following, which is the closest appro-iation of0.2G0.4!+1.#

Ans 9or appro-iation 0.2×0.4!+1.# can be ta/en

as

0×0.+200 ' 2+200 ' 1+# ' 0.12

2! 7ow an) prie nubers between 1 and 100 arefactors of $10

Ans $, 10 ' 2×52×11×13

*o there are 4 distinct prie nubers that are below 100

"! Aong a group of 200 people, " percent invest in

unicipal bonds, 1# percent invest in oil stoc/s, and $percent invest in both unicipal bonds and oil stoc/s. f 1person is to be randol) selected fro 200 people,what is the probabilit) that the person selected will beone who invests in unicipal bonds but not in oil stoc/sAns 7ere 200 does not reuire.



9ro the diagra we /now that onl) ones whoinvested in unicipal bonds are 2#%, the probabilit)is 2# + 100 ' $+2

4! Countr) Club has an indoor swiing club. Thirt)percent of the ebers of a swi club have passedthe lifesaving test. Aong the ebers who havenot passed the test, 12 have ta/en the preparator)course and "0 have not ta/en the course. 7ow an)ebers are there in the swi clubAns "0 3 12 ' 42 did not pass the test. This iseual to $0 % of the total ebers. *o totalebers ' 100+ $0 - 42 ' 0

! A nec/lace is ade b) stringing F individual beadstogether in the repeating pattern red bead, greenbead, white bead, blue bead and )ellow bead. f thenec/lace begins with a red bead and ends with awhite bead, then F could beAns The pattern is R ; = ( P R ; = ( P R .......*o, =hite bead coes at these positions "rd, #th,1"th, 1#th...f we ta/e this as a arithetic progression, then thisseries can be e-pressed as " 3 n 1! . 9ro theforula for general ter of A6 ' a 3 n1!d!.



This can be e-pressed as n 2=e chec/ the answer options so onl) # satisf) thecondition.

! A dog ta/en four leaps for ever) five leaps of harebut three leaps of the dog is eual to four leaps ofthe hare. Copare speedAns n ters of nuber of leaps, the ratio of the>og and hare speeds are 4 (ut ;iven that " leaps of dog ' 4 leaps of hare,. i.e., :eap lengths ' 4 " f >og is covering in "leaps what hare as covered in 4 leaps then :eaplengths are inversel) proportional!*o >og speed ' 4 - 4 ' 17are speed ' - " ' 1*o speeds ratio ' 1 1

$! There are two bo-es,one containing " red ballsH the other containing 2 green balls.)ou are

allowed to ove the balls b+w the bo-es so thatwhen )ou choose a bo- rando H a ball at randofro the chosen bo-,the probabilit) of getting a redball is a-iiUed.this a-iu probabilit) isAns Ner) interesting uestion.As we are allowed to ove the balls, we /eep onl)one red ball in first bo- and ove all the reainingballs to the second bo-

*o fist bo- contains 1 redball, second bo- contains"# red 3 2 green ' 4 balls6robabilit) of choosing an) bo- is 1+ 2.*o probabilit) of ta/ing one red ball



' 12×(1)+12(3864)A0.8

#! n how an) wa)s can " postcards can be postedin postbo-esAns 9irst card can go into an) of the five bo-es,*econd can go into an) of the five bo-es, Third can

go into an) of the five bo-es ' 5×5×5=125

! Apple costs : rupees per /ilogra for first "0/gsand E rupees per /ilogra for each additional/ilogra. f the price of "" /ilogras is 11.$and for"/gs of Apples is 12.4# then the cost of first 10 /gsof Apples isAns () fraing euations we get"0:3"E'11.$"0:3E'12.4#?liinate E b) ultipl)ing the first euation b) 2and subtracting second euation fro the firstThen we get : ' 0."2Cost of 10 /gs of apples ' 0."2 - 10 ' ".2

10! letters in the word A(M*?R are peruted in allpossible wa)s and arranged in alphabetical orderthen find the word at position 4 in the perutedalphabetical order

a! AR(*?Mb! AR(?*Mc! AR(*M?d! AR(?M*Ans The best wa) to solve this probles is 5ust as/how an) words starts with A. f we fi- A, then the



reaining letters can be arranged in I wa)s ' 120.*o the as/ed word ust start with A.Arrange all the given letters in alphabetical order.A(?R*M:et us find all the words start with A(. A(GGGG ' 4I' 24 wa)sFow we find all the words start wit A?. A?GGGG' 4I' 24 wa)s*o ne-t word start with AR and reaining letters are(?*M*o option (

11! A is twice efficient than (. A and ( can both wor/together to coplete a wor/ in $ da)s. Then find inhow an) da)s A alone can coplete the wor/Ans :et us assue A can do 2 units of wor/ eachda), then ( can do onl) 1 unit a da). f both cancoplete the wor/ in $ da)s, total wor/ done b)these two togeter ' 2 3 1 ! - $ ' 21 units

f these 21 units to be done b) A alone, then he willta/e 21 + 2 ' 10. da)s.

12! n a # - # chess board what is the total nuberof suares.Ans The total nuber of suares in a n - n chessboard is eual to @the su of first n natural nubersuares@

i.e., n(n+1)(2n+1)6*o *ubstituting # in the above forula we get 204

1"! O, P, = and Q are inteUers and the e-pressing O P Q is even and P = Q is odd. f O is even then



which of the following is truea! P ust be oddb! PQ ust be oddc! = ust be oddd! Q ust be oddAns. f O is even and O P Q is even then P and Qboth should be odd or both should be even.f P = Q is odd, and P and Q are also odd =should be oddf P = Q is even, and P and Q are even then =should be odd.*o option C is correct. i.e., = ust be &>>

14! The reainder when 1I32I3"I...30I divided b)I will beThe reainder when the ters greater than I aredivided b) I becoes 0 so we need to consider theters upto 4I.*o reainder will be whatever is obtained b)

dividing 1I32I3"I34I with I.*o reainder is obtained b) dividing 1323324!'"" with I 120!*o reainder is "".

1! f there are *i- periods in each wor/ing da) of aschool, n how an) wa)s can one arrange sub<ects such that each sub<ect is allowed at least

one periodAns. To arrange periods with sub<ects, then onesub<ect can be arranged in two slots.

*ub<ects can be arranged in periods in 6P5wa)s

and now we have 1 period which we can fill with an)



of the sub<ects in wa)s. so 6P5×5="00

Alternate ethodAssue the sub<ects are O1, O2, A, ( , C, >,. 7ere O

is the sub<ect which repeats. *o arranging ob<ectsin places will be eual to I ' $20 here no need todivide this nuber with 2I as even though thesub<ect is sae, but not identical!(ut this repeated subect can be an) of the five. *ototal arrangeents are $20 - ' "00

1! An article anufactured b) a copan) consists

of two parts O and P. n the process of anufacturingof part O, out 100 parts an) be defective.*iilarl) , out of 100 are li/el) to be defective inthe anufacturer of P. Calculate the probabilit) thatthe assebled product will not be defectivea! 0.4#b! 0.c! 0.#4

d! none of theseAns 6robabilit) that the part O is nondefective is ' 1 +100'.16robablit) that the part P is nondefective is ' 1 +100'.so, 6robablit) of nondefective

product'0.1×0.'0.#4

1! The water fro one outlet, flowing at a constant rate,can fill the swiing pool in hours. The water frosecond outlet, flowing at a constant rate can fill up thesae pool in appro-iatel) in hours. f both the outletsare used at the sae tie, appro-iatel) what is thenuber of hours reuired to fill the pool



Ans Assue tan/ capacit) is 4 :iters. ;iven that thefirst pipe fills the tan/ in hours. *o its capacit) is 4 + ' :iters+ 7our. *econd pipe fills the tan/ in hours. *o its capacit) is 4 + ' :iters+7our. f both pipesare opened together, then cobined capacit) is 14liters+hour. To fill a tan/ of capacit) 4 liters, (oth pipesta/es 4 + 14 ' ".21 7ours.

2! f $ % of a class answered the first uestion on acertain test correctl), percent answered the seconduestion on the test correctl), and 20 percent answeredneither of the uestions correctl), what percentageanswered both correctl)t is a proble belongs to sets. =e use the following

forula nAB(! ' nA! 3 n(! nA∩(!

7ere nAB(! is the people who answered atleast one of

the uestions.t was given that 20% answered neither uestion thenthe students who answered atleast one uestion is 100%

20% ' #0%Fow substituting in the forula we get #0% ' $% 3

% nA∩(!

⇒ nA∩(! ' 0%

"! A studentBs average arithetic ean! test score on 4tests is $#. =hat ust be the students score on a th

test for the students average score on the th test to be#0

Ans =e /now that Average =Sum of the observations No of

observations



*o *u of 4 test scores ' $#×4'"12

*u of tests scores ' #0×'400

⇒ th test score'400"12'##

A#ternati$e metho: f the student scores $# in the fifthtest also, what could be his average Fo change. s itnot(ut to bring the average to #0, he ust have scoredenough ar/s e-tra so that each of the five sub<ectscores increase upto #0. i.e., he should have scored 2 - ' 10 runs e-tra in the fifth sub<ect. *o th sub<ect scoreis $# 3 10 ' ##

4! Rural households have ore purchasing power than dourban households at the sae incoe level, since soeof the incoe urban and suburban households use forfood and shelter can be used b) the rural households forother needs. =hich of the following inferences is best

supported b) the stateent ade aboveA! The average rural household includes ore peoplethan does the average urban or suburban household.(! Rural households have lower food and housing coststhan do either urban or suburban households.C! *uburban households generall) have orepurchasing power than do either rural or urbanhouseholds.

>! The edian incoe of urban and suburbanhouseholds is generall) higher than that of ruralhouseholds.?! All three t)pes of households spend ore of theirincoe on housing than on all other purchases cobined.Ans f average rural household includes ore people,



then how coe the) have ore purchasing powernfact, the) have less purchasing power as the) have tofeed ore people. &ption A ruled out.&ption C does not e-plain wh) rural households haveore purchasing power than urban. Ruled out.f edian incoe of urban and suburban households isgenerall) higher than rural households the) are li/el) tohave ore purchasing power, assuing other paraetersconstant. (ut this does not e-plain wh) rural householdshave ore purchasing power. &ptions > ruled out.&ption ? does not provide an) e-planation wh) ruralhouseholds have ore purchasing power. Ruled out.&ption ( is correct as, f rural households spend lessincoe on food and shelter due to less pricesthe) definitel) have ore disposable incoe to spend.

! 5ose is a student of horticulture in the Mniversit) of7ose. n a horticultural e-perient in his final )ear, 200seeds were planted in plot and "00 were planted in plot

. f $% of the seeds in plot gerinated and 42% ofthe seeds in plot gerinated, what percent of the totalnuber of planted seeds gerinatedAns Total seeds gerinated in 6lot ' $% of 200 '114Total seeds gerinated in 6lot ' 42% of "00 ' 12Total gerinated seeds ' 114 3 12 ' 240The percentage of gerinated seeds of the total seeds

' 240500×100' 4#%

! A closed c)lindrical tan/ contains "πcubic feet of

water and its filled to half its capacit). =hen the tan/ isplaced upright on its circular base on level ground, the



height of water in the tan/ is 4 feet. =hen the tan/ isplaced on its side on level ground, what is the height, infeet, of the surface of the water above the ground

Ans =e /now that the volue of c)linder ' πr2h;iven tan/ hight ' 4ft.

⇒ πr24' "π

⇒ r ' "

*o the radius is " which eans the diaeter is .

As the c)linder is filled to initiall) e-actl) half of thecapacit), =hen this c)linder is placed on its side, =atercoes upto the height of the radius.*o water coes upto " ft.

$! The present ratio of students to teachers at a certainschool is "0 to 1. f the student enrollent were toincrease b) 0 students and the nuber of teachers wereto increase b) , the ratio of the teachers would then be2 to 1 =hat is the present nuber of teachersAssue the present students and teachers are "0J, JAfter new recruitents of students and teachers thestrength becoes "0J 3 0, J 3 respectivel). (utgiven that this ratio ' 2 1

⇒30K+50K+5=251

*olving we get J ' 1*o present teachers are 1.



#! College T has 1000 students. &f the 200 studentsa<oring in one or ore of the sciences,1"0 are a<oringin Cheistr) and 10 are a<oring in (iolog). f at least"0 of the students are not a<oring in either Cheistr)or (iolog), then the nuber of students a<oring in bothCheistr) and (iolog) could be an) nuber frof we assue e-actl) "0 students are not a<oring inan) sub<ect then the students who ta/e atleast onesub<ect ' 200 "0 ' 1$0

=e /now that nAB(! ' nA! 3 n(! nA∩(!

⇒ 1$0 ' 1"0 3 10 nA∩(!

*olving we get nA∩(! ' 110.

i.e., *tudents who can ta/e both sub<ects are 110

(ut f ore than "0 students are not ta/ing an) sub<ect,what can be the a-iu nuber of students who canta/e both the sub<ects

As there are 1"0 students are a<oring in cheistr),assue these students are ta/ing biolog) also. *oa-iu students who can ta/e both the sub<ects is 1"0

*o the nuber of students who can ta/e both sub<ectscan be an) nuber fro 110 to 1"0.

! Jell) and Chris are oving into a new cit). (oth of



the love boo/s and thus pac/ed several bo-es withboo/s. f Chris pac/ed 0% of the total nuber of bo-es,what was the ratio of the nuber of bo-es Jell) pac/edto the nuber of bo-es Chris pac/ed*iple uestions. f chris pac/s 0% of the bo-es, /ell)pac/s reaining 40%*o Jell) Chris ' 40% 0% ' 2 "

10! Aong a group of 200 people, " percent invest inunicipal bonds, 1# percent invest in oil stoc/s, and $percent invest in both unicipal bonds and oil stoc/s. f 1person is to be randol) selected fro 200 people,what is the probabilit) that the person selected will beone who invests in unicipal bonds but not in oil stoc/sAns 7ere 200 is redundant

9ro the diagra we /now that onl) ones who investedin unicipal bonds are 2#%, the probabilit) is 2# + 100 '$+2

11! 8achine A produces bolts at a unifor rate of 120ever) 40 second, and 8achine ( produces bolts at aunifor rate of 100 ever) 20 seconds. f the twoachines run siultaneousl), how an) seconds will it



ta/e for the to produce a total of 200 boltsAns 8achine A produces 120+40 ' " bolts in 1 secondand achine ( produces 100+20 ' bolts in one second.7ence, both of the will produce # bolts per second.7ence, the) wil ta/e 200+# ' 2 seconds to produce 200bolts.

12! 7ow an) prie nubers between 1 and 100 arefactors of $10

Ans $, 10 ' 2×52×11×13

*o there are 4 distinct prie nubers that are below 100

1"! Anal)sing the good returns that 7alocircle nsurance6vt :td was giving, Rati/a bought a 1)ear, Rs 10,000certificate of deposit that paid interest at an annual rateof #% copounded seiannuall).=hat was the totalaount of interest paid on this certificate at aturit)

This is a uestion on copound interest to be calculatedsei annuall).n the case of sei annual copounding, nterest ratebecoes half and Fuber of periods becoes 2 per )ear.

*o A ' 6(1+R100)n⇒A=10,000

(1+4100

)2=10,000×2625

' 10,#1nterest ' A 6 ' 10, #1 10,000 ' #1

14! 5uan is a gold edalist in athletics. n the onth of8a), if 5uan ta/es 11 seconds to run ) )ards, how an)



seconds will it ta/e hi to run - )ards at the sae rateAns f <uan ta/es 11 seconds to run P )ards, for 1 )ardhe will ta/e 11 + ) seconds. To run - )ards his tie will

be 11 + ) × - ' 11-+ )

1! A certain copan) retireent plan has a rule of $0provision that allows an eplo)ee to retire when theeplo)eeBs age plus )ears of eplo)ent with thecopan) total at least $0. n what )ear could a fealeeplo)ee hired in 1# on her "2nd birthda) first beeligible to retire under this provision

Assue it has ta/en - )ears to the feale eplo)ee toreach the rule of $0.*o her age should be "2 3 -. Also she gains - )ears ofe-perience.⇒ "2 3 -! 3 - ' $0

⇒ - ' 1.

7er age at the tie of retireent ' 1# 3 1 ' 200

1! &f the following, which is the closest appro-iationof 0.2G0.4!+1.#

ans 9or appro-iation 0.2×0.4!+1.# can be ta/en

as

0×0.+200 ' 2+200 ' 1+# ' 0.12

1$! Andalusia has been prooting the iportance ofhealth aintenance. 9ro 5anuar) 1,11 to 5anuar)1,1", the nuber of people enrolled in healthaintenance organiUations increased b) 1 percent. The



enrollent on 5anuar) 1,1" was 4 illion. 7ow an)illion peopleto the nearest illion! was enrolled inhealth aintenance organiUations on 5anuar) 1,11Ans f a nuber J is to be increased b) - % it should be

ultiplied b) (100+x)100

*o =hen the enrollent in 5anuar) 1, 11 is ultiplied

b) (100+x)100we got 4 illion.

K×(100+15)100=45

J ' 45×100115' ".1"

1#! =hat is the lowest possible integer that is divisible b)each of the integers 1 through $, inclusiveAns f a nuber has to be divisible b) each nuber fro1 to $, that nuber should be :.C.8 of1,2,",4,,,$! '420

1! f the area of a suare region having sides of length cs is eual to the area of a rectangular region havingwidth 2. cs, then the length of the rectangle, in cs,isAns ;iven Area of the suare ' Area of rectangle

⇒a2=l.b

*ubstituting the above values in the forula

⇒62=l.2.5

⇒ l ' 14.4 c

20! A tan/ contains 10,000 gallons of a solution that is percent sodiu chloride b) volue. f 200 gallons ofwater evaporate fro the tan/, the reaining solutionwill be appro-iatel) what percentage of sodiu



chlorideAns *odiu chloride in the original solution ' % of10,000 ' 00=ater in the original solution ' 10,000 00 ' ,00f 2,00 :iters of the water is evaporated then thereaining water ' ,00 2,00 ' $,000

*odiu chloride concentration ' 500500+7000×100' .$

%concentration should be calculated alwa)s on thetotal volue!

21! After loading a doc/, each wor/er on the night crewloaded "+4 as an) bo-es as each wor/er on the da) ofthe crew. f the night crew has 4+ as an) wor/ers asthe da) crew, what fraction of all the bo-es loaded b) twocrews did the da) crew load Assue the nuber of bo-es loaded in da)shift is eualto 4, then the nuber of bo-ed loaded in night shift ' "Assue the wor/ed on da)shift ' , then wor/ers on

night shift ' 4

*o bo-es loaded in da) shift ' 4 - ' 20, and bo-esloaded in night shift ' " - 4 ' 12

so fraction of bo-es loaded in da) shift ' 2020+12=58

22! A ba/er) opened )esterda) with its dail) suppl)of 40 doUen rolls. 7alf of the rolls were sold b) noon



and #0 % of the reaining rolls were sold betweennoon and closing tie. 7ow an) doUen rolls hadnot been sold when the ba/er) closed )esterda)Ans f half of the rolls were sold b) noon, thereaining are 0 % 40! ' 20.;iven #0% of the reaining were sold after the noonto closing tie⇒ #0% 20! ' 1

Mnsold ' 20 1 ' 4

2"! f F'46, where 6 is a prie nuber greater than

2, how an) different positive even divisors does nhave including n

Ans F ' 22×P1

=e /now that total factors of a nuber which is in

the forat of aP×bQ×cR...' 6 3 1!. E 3 1!. R 3

1! .... ' 2 3 1!.1 3 1! ' Also odd factors of an) nuber can be calculated

easil) b) not ta/ing 2 and its powers.*o odd factors of 22×P1' the factors of P1' 1 3 1!

' 2?ven factors of the nuber ' 2 ' 4

24! A dealer originall) bought 100 identical batteriesat a total cost of rupees. f each batter) was soldat 0 percent above the original cost per batter),

then, in ters of , for how an) rupees was eachbatter) soldAns 6er batter) cost ' + 100f each batter) is sold for 0% gain, then selling



price ' CostPrice×(100+Gain100)⇒ q100×

(100+50100

)=3q200

2! The price of lunch for 1 people was 20$ pounds,including a 1 percent gratuit) of service. =hat wasthe average price per person, ?OC:M>F; thegratuit)Ans :et the net price e-cluding the gratuit) ofservice ' - poundsThen, total price including 1% gratuit) of service

' x×(100+15100) ' 1.1 - pounds

*o, 1.1 - ' 20$ pounds ⇒ - ' 20$ + 1.1 ' 1#0 pounds

Fet price of lunch for each person ' 1#0 + 1 ' 12pounds

f f-! ' (1+x+x2+x3+.......x2012)2−x2012

g-! ' 1+x+x2+x3+.......x2011

Then what is the reainder when f-! is divided b)g-!

:et us ultipl) g-! with - on the both sides

-.g-! ' x+x2+x3+.......x2012

add 1 on both sides

-. g-! 3 1 ' 1+x+x2+x3+.......x2012

*ubstitute this value in f-!

then f-! ' (x.g(x)+1)2−x2012



f-! ' x2.g(x)2+2.g(x)+1−x2012

Fow f-! is divisible b) g-! first two ters are

e-actl) divisible b) g-! and we get 1 x2012

(ut 1 x2012' 1 -!1+x+x2+x3+.......x2011!

if this e-pression is divisible b) g-! we get 0 asreainder.

A nuber has e-actl) " prie factors, 12 factors of this nuber are perfect suares and 2$ factors ofthis nuber are perfect cubes. overall how an)

factors does the nuber have=e /now that the total factors of a nuber F

' ap.bq.cr....

Fow the total factors which are perfect suares of a

nuber F ' ([ p2]+1).([q2]+1).([r2]+1)....where V-W is greatest inteUer less than that of -.

;iven ([ p2]+1).([q2]+1).([r2]+1)....' 12

*o [ p2]+1' ] [q2]+1' ] [ p2]+1'

[ p2]' 4 ⇒ p ' # or , siilarl) ' # or , r ' # or

;iven that 2$ factors of this nuber are perfectcubes

so ([ p3]+1).([q3]+1).([r3]+1)....' 2$



*o [ p3]+1' " ⇒ ' [ p3]' 2

⇒ p ' , $, #

() cobining we /now that p ' ' r ' #*o the given nuber should be in the forat

' a8.b8.c8....

Fuber of factors of this nuber ' #31!.#31!.#31! ' $2

n a class there are 0% of girls of which 2% poor. =hat is the probabilit) that a poor girl is selected isleaderAssue total students in the class ' 100Then ;irls ' 0% 100! ' 06oor girls ' 2% 0! ' 1*o probabilit) that a poor girls is selected leader '6oor girls + Total students ' 1+100 ' 1%

A and ( are running around a circular trac/ of length120 eters with speeds 12 +s and +s in thesae direction. =hen will the) eet for the firsttieA eets ( when A covers one round ore than (.ABs relative speed ' 12 ! +s. *o he ta/es 120 + seconds to gain one e-tra round.

*o after 20 seconds A eets (.

A copletes a wor/ in 20 da)s ( in 0 da)s C in 4da)s. All three persons wor/ing together on apro<ect got a profit of Rs.2000 what is the profit of(



=e /now that profits ust be shared as the ratio oftheir efficiencies. (ut efficiencies are inversel)proportional to the da)s. *o efficiencies of A ( C' 1+20 1+0 1+4 ' " 4*o ( share in the total profit ' " + 1" O 2000 'Rs.000

A copletes a piece of wor/ in "+4 of the tie in (does, ( ta/es 4+ of the tie in C does. The) got aprofit of Rs. 40000 how uch ( getsAssue C ta/es 20 >a)s. Fow ( ta/es 4+ 20! '1 da)s. A ta/es "+41! ' 12Fow their efficiencies ratio ' 1+20 1+1 1+12 '12 1 20(Bs share in the profit of Rs.40000 ' 1+4$ 40000!' Rs.12$

An ept) tan/ be filled with an inlet pipe _A[ in 42

inutes. After 12 inutes an outlet pipe _([ isopened which can ept) the tan/ in "0 inutes.After inutes another inlet pipe _C[ opened into thesae tan/, which can fill the tan/ in " inutes andthe tan/ is filled. 9ind the tie ta/en to fill the tan/Assue total tan/ capacit) ' 210 :itersFow capacit) of pipe A ' 210+42 ' :itersCapacit) of ( ' 210 + "0 ' $ :iters

Capacit) of C ' 210 + " ' inAssue tan/ gets filled in - in after the third pipegot opened.

*o x×5+6×(−2)+4x=210

⇒48+4x=210⇒4x=162⇒x=40.5



Total tie ta/en to fill the tan/ ' 40. 3 12 3 '1.

8other, daughter and an infant cobined age is $4,and otherBs age is 4 ore than daughter andinfant. f infant age is 0.4 ties of daughter age,then find daughters age.Assue 8 3 > 3 ' $4] .................1!Also given 8 > ' 4 ⇒ 8 ' > 3 3 4

Also ' 0.4 > ⇒ ' 2+ >

*ubstituting 8 and values in the first euation we

get > 25> 4 3 > 3 25> ' $4*olving > ' 10

A ;rocer bought 24 /g coffee beans at price O per/g. After a while one third of stoc/ got spoiled so hesold the rest for K200 per /g and ade a total profitof twice the cost. =hat ust be the price of O

Total Cost price ' 24×O

As 1+"rd of the beans spoiled, reaining beans are2+" 24! ' 1 /gs

*elling price ' 200 × 1 ' "200

6rofit ' *elling price Cost price ' "200 24×O

;iven 6rofit ' 2 × Cost price

"200 24×O ' 2 × 24×O!

*olving O ' 44.44

(hanu spends "0% of his incoe on petrol on



scooter 20% of the reaining on house rent and thebalance on food. f he spends Rs."00 on petrol thenwhat is the e-penditure on house rent ;iven "0% ncoe ! ' "00 ⇒ ncoe ' 1000

After having spent Rs."00 on petrol, he left withRs.$00.7is spending on house rent ' 20% $00! ' Rs.140

:et e-p,n! ' to the power n. f e-p10, ! ' ne-p2, 2! where to and n are integers then n '

;iven 10m=n.22

⇒ 2m×5m=n.22⇒2m−2×5m=n

9or ' 2 we get least value of n ' 2, and for D2 we get infinite values are possible for n.

7ow an) /gs. of wheat costing Rs. per /g ustbe i-ed with 4 /g of rice costing Rs. .40 per /g

so that 20% gain a) be obtained b) selling thei-ture at Rs. $.20 per /g f the selling price of the i-ture is Rs.$.2 when soldat 20% profit then

C6 ×120100' $.2 ⇒ C6 ' Rs.

Fow b) appl)ing weighted average forula

' K×5+45×6.4K+45=6

⇒ J ' 1# /gs

The diagonal of a suare is twice the side ofeuilateral triangle then the ratio of Area of theTriangle to the Area of *uare is:et the side of euilateral triangle ' 1 unit.



=e /now that area of an euilateral triangle ' 3√4a2

As side ' 1 unit area of the euilateral triangle

' 3√4

Fow >iagonal of the suare ' 2 side of theeuilateral triangle! ' 2

=e /now that area of the suare ' 12D2where > '

diagonal

*o area of the suare ' 12(22)=2

Ratio of the areas of euilateral triangle and suare

' 3√4 2 ⇒ 3√:8

Ra< tossed " dices and there results are noted downthen what is the probabilit) that ra< gets 10Alwa)s reeber when " dice are rolled the nuberof wa)s of getting n where n is the su of faces ondice!

' (n−1)C2where n ' " to #

' 2 where n ' , 12' 2$ where n ' 10, 11

' (20−n)C2where n ' 1" to 1#

The reuired probabilit) ' 2763' 27216

1. f ") 3 - D 2 and - 3 2)≤", =hat can be said about

the value of )

A. ) ' 1(. ) D1C. ) Y1>. ) ' 1Answer (



8ultipl) the second euation with 1 then it will becoe

- 2)≥ ". Add the euations. Pou will get ) D 1.

2. f the price of an ite is decreased b) 10% and thenincreased b) 10%, the net effect on the price of the iteisA. A decrease of %(. Fo changeC. A decrease of 1%>. An increase of 1%Answer C

f a certain nuber is increased b) -% then decreasedb) -% or vice versa, the net change is alwa)s decrease.

This change is given b) a siple forula −

(x10)2= −(1010)2= −1%. Fegitive sign indicates

decrease.

". f is an odd integer and n an even integer, which ofthe following is definitel) oddA. 23n!n!

(. (m+n2)+(m−n2)

C. m2+mn+n2

>. 3n

Answer C and > &riginal Answer given as >!

Pou <ust reeber the following odd ± odd ' even]

even ± even ' even] even ± odd ' odd

Also odd - odd ' odd] even - even ' even] even - odd '



even.

4. =hat is the su of all even integers between and"01A. 40000(. 20000C. 40400>. 20200Answer >

The first even nuber after is 100 and last evennuber below "01 is "00. =e have to find the su ofeven nubers fro 100 to "00. i.e., 100 3 102 3 104 3............... "00.Ta/e 2 Coon. 2 - 0 3 1 3 ...........10!There are total 101 ters in this series. *o forula forthe su of n ters when first ter and last ter is

/nown is n2(a+l)

*o 0 3 1 3 ...........10 ' 1012(50+150)

*o 2 - 1012(50+150)' 20200

. There are 20 balls which are red, blue or green. f $balls are green and the su of red balls and green ballsis less than 1", at ost how an) red balls are thereA. 4(.

C. >. $Answer (

;iven R 3 ( 3 ; ' 1$] ; ' $] and R 3 ; Y 1". *ubstituting ; ' $ in the last euation, =e get R Y .



*o a-iu value of R '

. f n is the su of two consecutive odd integers andless than 100, what is greatest possibilit) of nA. #(. 4C. >. Answer C

=e ta/e two odd nubers as 2n 3 1! and 2n 1!.Their su should be less than 100. *o 2n 3 1! 3 2n 1! Y 100 ⇒ 4n Y 100.

The largest 4 ultiple which is less than 100 is

$. x2Y 1+100, and - Y 0 what is the highest range in

which - can lieA. 1+10 Y - Y 0(. 1 Y - Y 0

C. 1+10 Y - Y 1+10>. 1+10 Y -Answer A

Reeber- a!- b! Y 0 then value of - lies in between a andb.- a!- b! D 0 then value of - does not lie inbetween a

and b. or −∞, a! and b, −∞! if a Y b

x2Y 1+100 ⇒

(x2−1/100)<0⇒+x2−(1/10)2)<0⇒+x−1/10)(x+1/10)<0

*o - should lie inbetween 1+10 and 1+10. (ut it was



given that - is ve. *o - lies in 1+10 to 0

#. There are 4 bo-es colored red, )ellow, green andblue. f 2 bo-es are selected, how an) cobinationsare there for at least one green bo- or one red bo- to beselectedA. 1( . C. >. Answer

Total wa)s of selecting two bo-es out of 4 is 4C2' .

Fow, the nuber of wa)s of selecting two bo-es wherenone of the green or red bo- included is onl) 1 wa). weselect )ellow and blue in onl) one wa)!. f we substractthis nuber fro total wa)s we get wa)s.

. All faces of a cube with an eight eter edge are

painted red. f the cube is cut into saller cubes with atwo eter edge, how an) of the two eter cubeshave paint on e-actl) one faceA. 24(. "C. 0>. 4#Answer A

f there are n cubes lie on an edge, then total nuber of

cubes with one side painting is given b) 6×(n−2)2. 7ere

side of the bigger cube is #, and sall cube is 2. *othere are 4 cubes lie on an edge. 7ence answer ' 24



10. Two c)clists begin training on an oval racecourse atthe sae tie. The professional c)clist copletes eachlap in 4 inutes] the novice ta/es inutes to copleteeach lap. 7ow an) inutes after the start will bothc)clists pass at e-actl) the sae spot where the) beganto c)cleA. 10(. #C. 14>. 12Answer >

The faster c)cl)st coes to the starting point for ever) 4in so his ties are 4, #, 12, ......... The slower c)clistcoes to the starting point for ever) in so his tiesare , 12, 1#, ......... *o both coes at the end of the12th in.

11. 8, F, & and 6 are all different individuals] 8 is thedaughter of F] F is the son of &] & is the father of 6]Aong the following stateents, which one is true

A. 8 is the daughter of 6(. f ( is the daughter of F, then 8 and ( are sistersC. f C is the granddaughter of &, then C and 8 aresisters



>. 6 and F are bothers.Answer (

9ro the diagra it is clear that f ( is the daughter ofF, then 8 and ( are sisters. Rectangle indicates 8ale,and &val indicates 9eale.

12. n the ad<oining diagra, A(C>and ?9;7 are sures of side 1 unit such that the)intersect in a suare of diagonal length C?! ' 1+2. Thetotal area covered b) the suares is

A. Cannot be found fro the inforation

(. 1 1+2C. 1 $+#>. Fone of theseAnswer C

:et C; ' - then using p)thogerous

theore CG2+GE2=CE2

⇒ x2+x2=(1/2)2⇒2x2=1/4⇒x2=1/8



Total area covered b) two bigger suares ' A(C> 3?9;? Area of sall suare ' 2 1+# ' 1+#

1". There are 10 stepping stones nubered 1 to10 as shown at the side. A fl) <ups fro the first stoneas follows] ?ver) inute it <ups to the 4th stone frowhere it started that is fro 1st it would go to th andfro th it would go to th and fro th it would go to"rd etc. =here would the fl) be at the 0th inute if itstarts at 1

A. 1(. C. 4>.

Answer A

Assue these steps are in circular fashion.Then the fl) <ups are denoted in the diagra. t isclear that fl) cae to the 1st position after th inute.



*o again it will be at 1st position after 10th1th .....0th. in.

*o the fl) will be at 1st stone after 0th in.

14. =hat is the reainder when 617+1176 is divided b)

$A. 1(.

C. 0>. "Answer C

617' (7−1)17'

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

f we divide this e-pansion e-cept the last ter eachter gives a reainder 0. :ast ter gives a reainderof 1.

Fow 9ro 9erat little theore, [ap−1p]Rem=1

*o [1767]Rem=1Adding these two reainders we get the final reainder' 0



1. n base $, a nuber is written onl) using the digits 0,

1, 2, ...... The nuber 1" in base $ is 1 - 723 " - $ 3

' $ in base 10. =hat is the su of the base $

nubers 12"4 and 4" in base $.A. 11101(. 11110C. 10111>. 11011Answer (

n base $ there is no $. *o to write $ we use 10. for #we use 11...... for 1" we use 1, for 14 we use 20 and soon.*o fro the colun d, 4 3 " ' $ ' 10, we write 0 and 1

carried over. now 1 3 " 3 4 ' # ' 11, then we write 1and 1 carried over. again 1 3 2 3 ' # ' 11 and so on

1. The seuence {An}is defined b) A1' 2

and An+1=An+2nwhat is the value of A100

A. 02(. 00C. 10100>. 04Answer A



=e /now that A1' 2 so A2=A1+1=A1+2(1)=4

A3=A2+1=A2+2(2)=8

A4=A3+1=A3+2(3)=14

*o the first few ters are 2, 4, #, 14, 22, ......The differences of the above ters are 2, 4, , #, 10...and the differences of differences are 2, 2, 2, 2. all areeual. so this series represents a uadratic euation.

Assue An' an2+bn+c

Fow A1' a 3 b 3 c ' 2

A2' 4a 3 2b 3 c ' 4A3' a 3 "b 3 c ' #

*olving above euations we get a ' 1, b ' 1 and C ' 2

*o substituting in An' n2+bn+c' n2−n+2

*ubstitute 100 in the above euation we get 02.

1$.9ind the nuber of rectangles fro the ad<oining

figure A suare is also considered a rectangle!

A. #4

(. "2$C. 1"#>. Fone

Answer C



To for a rectangle we need two horiUontal lines andtwo vertical lines. 7ere there are 1" vertical linesand $ horiUontal lines. The nuber of wa)s of

selecting 2 lines fro 1" vertical lines is 13C2and thenuber of wa)s of selecting 2 lines fro $

horiUontals is 7C2. *o total rectangles ' 7C2x13C2

1#. A, (, C and > go for a picnic. =hen A stands ona weighing achine, ( also clibs on, and theweight shown was 1"2 /g. =hen ( stands, C alsoclibs on, and the achine shows 1"0 /g. *iilarl)the weight of C and > is found as 102 /g and that of( and > is 11 /g. =hat is >Bs weightA. #/g(. $# /gC. 44 /g>. FoneAnswer C

;iven A 3 ( ' 1"2] ( 3 C ' 1"0] C 3 > ' 102, ( 3> ' 11?liinate ( fro 2nd and 4th euation and solvingthis euation and "rd we get > value as 44.

1. Ro) is now 4 )ears older than ?ri/ and half ofthat aount older than ris. f in 2 )ears, ro) will be

twice as old as ?ri/, then in 2 )ears what would beRo)Bs age ultiplied b) risBs ageA. 2#(. 4#C. 0>. 2



Answer 4#

20. O, P, O and = are integers. The e-pression O P Q is even and the e-pression P Q = is odd. f Ois even what ust be trueA. = ust be odd(. P Q ust be oddC. = ust be odd>. Q ust be oddAnswer A or C (ut go for C!

21. 8r and 8rs *ith have invited of their friendsand their spouses for a part) at the =ai/i/i (eachresort. The) stand for a group photograph. f 8r*ith never stands ne-t to 8rs *ith as he sa)sthe) are alwa)s together otherwise!. 7ow an)wa)s the group can be arranged in a row for thephotographA. 20I

(. 1I 3 1#IC. 1# - 1I>. 2 - 1IAnswer C

22. n a rectanglular coordinate s)ste, what is thearea of a triangle whose vertices whose vertices havethe coordinates 4,0!, , "! adn , "!

A. (. $C. $.>. .Answer A



2". A drawer holds 4 red hats and 4 blue hats. =hatis the probabilit) of getting e-actl) three red hats ore-actl) three blue hats when ta/ing out 4 hatsrandol) out of the drawer and iediatel)returning ever) hat to the drawer before ta/ing outthe ne-tA. 1+2(. 1+#C. 1+4>. "+#Answer (

24. n how an) wa)s can we distribute 10 identicalloo/ing pencils to 4 students so that each studentgets at least one pencilA. 040(. 210C. #4>. Fone of these

Answer C

2. The prie factoriUation of inteUer F is A - A - ( -C, where A, ( and C are all distinct prine inteUers.7ow an) factors does F haveA. 12(. 24C. 4

>. Answer A

2. Ti and ?lan are 0 / fro each other.the)start to ove each other siultanousl) ti at speed10 and elan /ph. f ever) hour the) double their



speed what is the distance that Ti will pass until heeet ?lanA. 4(. 0C. 20>. #0Answer (

2$. A father purchases dress for his three daughter.The dresses are of sae color but of different siUe.the dress is /ept in dar/ roo .=hat is theprobabilit) that all the three will not choose theirown dress.A. 2+"(. 1+"C. 1+>. 1+Answer (

2#. F is an integer and FD2, at ost how an)integers aong F 3 2, F 3 ", F 3 4, F 3 , F 3 ,and F 3 $ are prie integersA. 1(. "C. 2>. 4Answer C

2. A turtle is crossing a field. =hat is the totaldistance in eters! passed b) turtle Consider thefollowing two stateentsO! The average speed of the turtle is 2 eters perinute



P! 7ad the turtle wal/ed 1 eter per inute fasterthan his average speed it would have finished 40inutes earlierA. *tateent O alone is enough to get the answer(. (oth stateents O and P are needed to get theanswerC. *tateent P alone is enough to get the answer>. >ata inadeuateAnswer (

"0. ;iven the following inforation, who is)oungestC is )ounger than A] A is talled than (C is older than (] C is )ounger than >( is taller than C] A is older than >A. >(. (C. C>. A

Answer (

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