apptitude zooming

7/26/2019 Apptitude Zooming

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1. A manufacturer of chocolates makes 6 dierent avors of chocolates. Thechocolates are sold in boxes of 10. How many dierent! boxes of chocolatescan be made"#ol$

%f n similar articles are to be distributed to r &ersons' x1+x2+x3......xr=neach

&erson is eli(ible to take any number of articles then the total ways are

n+r)1Cr)1

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

in such a case the formula for non ne(ative inte(ral solutions is n+r)1Cr)1

Here n *6 and r*10. #o total ways are 10+6)1C6)1* +00+

,. %n a sin(le throw with two dice' -nd the &robability that their sum is amulti&le either of + or .a. 1/+

b. 1/,c. /d. 12/+6#ol$ Their sum can be +''6'3''1,4or two dice' any number from , to 2 can be (et in 5n17 ways and any numberfrom 3 to 1, can be (et in 51+ n7 ways.

Then &ossible ways are , 8 + 8 8 8 8 1 * ,0 &ossible cases.#o &robability is 5,0/+67*5/7

+. 9 alone can do &iece of work in 10 days. A alone can do it in 1 days. %f thetotal wa(es for the work is :s 000' how much should 9 be &aid if they work

to(ether for the entire duration of the work"a. ,000b. 000c. 000d. +000#ol$

Time taken by A and 9 is in the ratio of * +$,:atio of the ;ork * , $ + 5since' time and work are inversely &ro&ortional7

Total money is divided in the ratio of , $ + and 9 (ets :s.+000

. <n a ,6 =uestion test' &oints were deducted for each wron( answer and 3&oints were added for ri(ht answers. %f all the =uestions were answered howmany were correct if the score was >ero.a. 10b. 11c. 1+d. 1,#ol$?et x =ues were correct. Therefore' 5,6 x7 were wron(

8x)5(26)x)=0

#olvin( we (et x*10

. Arun makes a &o&ular brand of ice cream in a rectan(ular sha&ed bar 6cmlon(' cm wide and ,cm thick. To cut costs' the com&any had decided to



reduce the volume of the bar by 1@. The thickness will remain same' but thelen(th and width will be decreased by some &ercenta(e. The new width will be'a. .b. .c. 2.d. 6.

#ol$olume *l×b×h* 6×5×2 * 60 cm3

Bow volume is reduced by 1@.

Therefore' new volume * (100)19)100×60=48.6

Bow' thickness remains same and let len(th and breadth be reduced to x@

so' new volume$ (x100×6)(x100×5)2=48.6#olvin( we (et x *0thus len(th and width is reduced by 10@

Bew width * 510@ of 7*.

6. %f all the numbers between 11 and 100 are written on a &iece of &a&er. Howmany times will the number be used"#ol$ ;e have to consider the number of Cs in two di(it numbers. D D %f we -x in the 10th &lace' unit &lace be -lled with 10 ways. %f we -x inunits &lace' 10th &lace be -lled with ways 50 is not allowed7#o total 1 ways.Alternatively:

There are total Cs in 1' ,' +...'E total 10 Cs in 0'1',....

thus' 810*1.

2. %f twenty four men and sixteen women work on a day' the total wa(es to be&aid is 11'600. %f twelve men and thirty seven women work on a day' the totalwa(es to be &aid remains the same. ;hat is the wa(es &aid to a man for adayFs work"#ol$ ?et man daily wa(es and woman daily wa(es be G and ; res&ectively,G816;*116001,G8+2;*11600solvin( the above e=uations (ives G*+0 and ;*,00

3. The cost &rice of a cow and a horse is :s + lakhs. The cow is sold at ,0@&ro-t and the horse is sold at 10@ loss. <verall (ain is :s ,00. ;hat is thecost &rice of the cow"#ol$ro-t * ,00ro-t *# I,00*# +00000 therefore #*+0,00x8y * +000001.,x 8 0.y * +0,00#olvin( for x * 11000 * I of cow.

. 1' ,' ,' +' +' +' ' ' ' ' 1' 1' ,' ,' ,' ,' +' +' +' +' +' +' ' ' ' ' ' ' ' '1' 1' 1' ,' ,' ,' ,' ,' ,' +' +' +' +' +' +' +' +' +' +' ......



%n the above se=uence what is the number of the &osition ,333 of these=uence.a7 1b7 c7 +d7 ,

#ol$ 4irst if we count 1,,+++. they are 10%n the next term they are ,0Bext they are +0 and so on

#o Jsin( n(n+1)2×10K2888

4or n * ,+ we (et ?H# as ,260. :emainin( terms 1,3.

Bow in the ,th term' we have , 1Cs' and next 3 terms are ,Cs. #o next 2,terms are +Cs.

The ,333 term will be +!.

10. How many di(it numbers contain no.,"#ol$ Total number of four di(it numbers *000 5i.e 1000 to 7;e try to -nd the number of numbers not havin( di(it , in them.Bow consider the units &lace it can be selected in ways 5i.e 0'1'+'''6'2'3'7

Tens &lace it can be selected in ways 5i.e 0'1'+'''6'2'3'7Hundreds &lace it can be selected in ways 5i.e 0'1'+'''6'2'3'7

Thousands &lace can be selected in 3 ways 5i.e 1'+'''6'2'3'7 here C0C cannotbe taken

Total number of numbers not havin( di(it , in it * x x x 3 *3+, Total number of numbers havin( di(it , in it * 0003+, *+163

***********************************************1. The value of diamond varies directly as the square of its weight. If a diamond falls and breaksinto two pieces with weights in the ratio 2:3. what is the loss percentage in the value!ol: "et weight be #$%

the cost of diamond in the original state is proportional tox2

when it is fallen it breaks into two pieces 2y and the 3y$ & 'y

(riginal value of diamond &(5y)2&25y2

)alue of diamond after breakage &(2y)2+(3y)2=13y2

so the percentage loss will be & 25y2*13y225y2×100=48%

2. +ive college students met at a party and e$changed gossips. ,ma said- #(nly one of us is lying%.avid said- #/$actly two of us are lying%. Thara said- #/$actly 3 of us are lying%. 0uerishi said-#/$actly of us are lying%. hitra said #ll of us are lying%. 4hich one was telling the trutha5avidb50uerishic5hitrad5Thara!ol: s all are contradictory statements- it is clear that (6"7 one of them is telling the truth. !o

remaining of them are lying. 0uerishi mentioned that e$actly are lying. !o- he is telling thetruth./$planation: "et us 1st assume that ,ma is telling the truth. Then according to her only one is



lying. 8ut if only one is lying then all the others9 statements are contradicting the possibility. In thesame way all the other statements should be checked. If we assume the 0uerishi is telling the truth-according to him e$actly members are lying. !o all the others are telling lies and he is the onewho is telling the truth. This case fits perfectly.

3. ara- a blue whale participated in a weight loss program at the biggest office. t the end of every

month- the decrease in weight from original weight was measured and noted as 1- 2- - 21- ;- '-2<. 4hile ara made a steadfast effort- the weighing machine showed an erroneous weight once.4hat was that.a5 2<b5 2c5 'd5 ;!(": This is a number series problem nothing to do with the data given.1$ 1=1&22 $ 2=2& $ 3=3&21

21 $ =&;; and not ;;; $ '=' & ''>= & 2<

. The letters in the word (?T! are permuted in all possible ways and arranged in alphabeticalorder then find the word at position 2 in the permuted alphabetical ordera5 (T!?b5 (T?!c5 (T?!d5 (!T?!(":In alphabetical order : ( ? ! T @ @ @ @ @ : the places filled in 'A ways & 12B- 8ut we need a rank less than 12B. !o the word startswith . @ @ @ @ : empty places can be filled in A&2 ( @ @ @ @ : the places filled with A ways & 2. If we add 2 = 2 this total crosses 2. !o 4eshould not consider all the words starting with (. ( @ @ @ : 3A& ( ? @ @ @ : 3A&Till this 3 words are obtained- we need the 2nd word.(! @ @ @ : 3A&

/$actly we are getting the sum 2. !o last 3 letters in the descending order are T?.!o given word is (!T?

. man who goes to work long before sunrise every morning gets dressed in the dark. In his sockdrawer he has black and ; blue socks. 4hat is the probability that his first pick was a black sock-but his second pick was a blue sock!(": This is a case of without replacement. 4e have to multiply two probabilities. 1. ?robabilityof picking up a black sock- and probability of picking a blue sock- given that first sock is black.

6C114C1×8C113C1=2491

'. There are red balls-; blue balls and < green balls in a bag. If ' are drawn with replacement-what is the probability at least three are red!ol: t least 3 reds means we get either : 3 red or red or ' red. nd this is a case of replacement.case 1 : 3 red balls : C21 $ C21 $ C21 $ 1'C21 $ 1'C21



case 2 : red balls : C21 $ C21 $ C21 $ C21 $ 1'C21case 3 : ' red balls : C21 $ C21 $ C21 $ C21 $ C21

Total probability & & DC21 $ C21 $ C21 $ 1'C21 $ 1'C215=DC21 $ C21 $ C21 $ C21 $ D1' 5C215=DC21 $ C21 $ C21 $ C21 $ C215 & 312C1;B<

. Total number of digit number do not having the digit 3 or .!ol:consider digits @ @ @ @

1st blank can be filled in7C1ways DB-3- are neglected as the first digit should not be B5

2st blank can be filled in8C1ways DB considered along with 1-2--'-<-;-E5

3st blank can be filled in8C1ways

st blank can be filled in8C1 ways

Therefore total digit number without 3 or is < $ ; $ ; $ ;&3';

<. +ind the missing in the series: <B- '- '- 1-@@@@.!ol: B

<BF' & 1 &42

'F' & E &32

'F1 & &22

1FB & 1 &12

;. school has 12B- 1E2 and 1 students enrolled for its science- arts and commerce courses. llstudents have to be seated in rooms for an e$am such that each room has students of only the samecourse and also all rooms have equal number of students. 4hat is the least number of roomsneeded!ol: 4e have to find the ma$imum number which divides all the given numbers so that number ofroots get minimiGed. H+ of 12B-1E2 1 is 2. /ach room have 2 students of the samecourse.

Then rooms needed12024+19224+14424 & ' =; = & 1E

E. farmer has a rose garden. /very day he picks either <--2 or 23 roses. 4hen he plucks thesenumber of flowers the ne$t day 3<-3-E or 1; new flowers bloom. (n Jonday he counts 1;E roses.

If he continues on his plan each day- after some days what can be the number of roses left behindDHint : onsider number of roses remaining every day5a5<b5c53Bd53<!ol:let us consider the case of 23. when he picks up 23 roses the ne$t day there will be 1; new- so inthis case.- ' flowers will be less every day. !o when he counts 1;E- the ne$t day 1;-1<E-1<-1E-................finally the no. of roses left behind will be .

1B. 4hat is the 32nd word of K4ITI6LK in a dictionary!ol: rranging the words of waiting in lphabetical (rder : -L-I-I-6-T-4



!tart with @ @ @ @ @ @ This can be arranged in AC2A ways&<2BC2&3B waysso canMt be arranged starting with alone as it is asking for 32nd word so it is out of rangeL@ @ @ @ @then the remaining letters can be arranged in 'AC2A ways so-12BC2&B ways. (ut ofrange as it has to be within 32 words.LI@ @ @ @ 6ow the remaining letters can be arranged in A ways &2L6 @ @ @ @ can be arranged in AC2A ways or 12 ways

so-2=12 &3th word so out of range. !o we should not consider all the words start with L6now L6I@ @ @can be arranged in 3A ways & waysso 2=&3B within range6ow only two word left so- arrange in alphabetical order.L6TII4 F 31st wordL6TI4I F 32nd word***********************************************1. man is known to speak truth 3 out of times. He throws die and reports that it is a . Theprobability that it is actually a is!ol: If actually appeared- he can report it with the probability of 3C. If has not appeared- still hecan report it wrongly with the probability of 1C

!o the probability that it is actually a & D?robability to appear $ His truthfulness to report =?robability to appear any other number $ His lieing probability 5 &16×34+56×14=13

The probability that it is actually &Probability that he reports 6Total probability to appear 6

&34×1634×16+14×56=38

2. In how many ways can we distribute 1B pencils to children so each child gets atleast onepencil6umber of ways of distributing r identical obNects to n distinct obNects so that each get atleast one &

(n*1)C(r*1)&(10*1)C(4*1)=9C3

3. drawer holds red hats and blue hats. what is probability of getting e$actly 3 red hats or 3blue hats when taking out hats randomly out of drawer and immediately returning every hat todrawer before taking out ne$ts the obNects are replaced- the probability of drawing red or blue is equal.

?robability to draw e$actly 3 red hats and 1 blue hat &12×12×12×12=116

!imilarly probability to draw e$actly 3 blue hats and 1 red hat &12×12×12×12=116

Total probability &116+116=18

. father purchased dress for his 3 daughters. The dresses are of same color but diff siGe and theyare kept in dark room. what is probability that all the 3 will not choose their own dress

This is a case of deFarrangements &Dn=n!(12!*13!+14!*....)!o number of ways that none of them chooses their own dress &D3=3!(12!*13!)=2!o probability &23!=13

'. BO of male in a town and <BO of female in a town are eligible to vote. out of which <BO of

male and BO of female who are eligible to vote voted for candidate . what is the value of votes inO did get"et the ratio of men and women are 1BB : k



Jale eligible votes & B and female eligible votes & <BO Dk56umber of males who voted for & <BO DB5 & 26umber of females who voted for & BOD<BO DP5 & 2O Dk5

?ercentage of votes got by &42+42100(K)60+70100(K)×100=4200+42K6000+70K×100

!o this value cannot be determined as the value of P is not known

. Leorge and Jark can paint <2B bo$es in 2B days. Jark and Harry in 2 days and Harry andLeorge in 1' days. Leorge works for days- Jark for ; days and Harry for ; days. The totalnumber of bo$es painted by them isapacity of L = J & <2B C 2B & 3J = H & <2B C 2 & 3BH = L & <2B C 1' & ;ombined capacity & 2 DL = H = J5 & 11L = H = J & 11 C 2 & '<6ow capacity of L & DL=H=J5 F DH = J5 & '< F 3B & 2<J & DL=H=J5 F DH = L5 & '< F ; & E

H & DL=H=J5 F DL = J5 & '< F 3 & 21Liven that L worked for days- and mark for ; and harry for ; days!o total work by them & $ 2< = ; $ E = ; $ 21 & 3;

<. Two equilateral triangle of side 12cm are placed one on top another- such a pionted star isformed if the si$ vertices lie on a circle what is the area of the circle not enclosed by the !tara51b5'<c5;d5;3!ol: Liven that two equilateral triangles of length 12 has inscribed in a circle.

ltitude of the triangle &3Q2a&3Q2(12)&63Q

4e know that centroid divides the altitude in the ratio 2 : 1 and23Dltitude5 & ircum radius

ircum radius &23(63Q)=43Q

rea of the circle &πr2=3.14×(43Q)2

6ow the two triangles in the circle forms 12 small equilateral triangles with side . !o their totalarea &12×3Q4a2& 12×3Q442

rea which is not covered by the equilateral triangles & 3.14×(43Q)2F12×3Q442& <.' ≃68

;. There are different letters and addressed envelopes.In how many ways can the letters be putin the envelopes so that atleast one letter goes to the correct address a51' b51 c51; d512Total ways of putting r letters to r covers & rA & A & 2

6umber of ways that none of them goes into the right envolope &D4=4!

(12!*13!+14!

)& E

!o atleast one envolope goes into the right one & 2 F E & 1'



E.There are 2'Bmen and 1'B women in a committee- if all will work they will complete 12 units perday- if all men work they will complete 1' units per day- how many units will women complete perdayI think there is a mistake in this question. If all men and women together complete 12 units- howonly men can do 1' ,nits of work a day+orgetting about the reality- 4omen can do F3 units a day.

1B. How many odd and even numbers are there between 2 and BB +ind the sum of oddnumbers and the sum of even numbersA

!ol: (dd numbers are from 3 to 3EE. 6umber of odd numbers &l*ad+1=399*432+1=179

Their sum &n2(l+a)& 3E''E

/ven numbers are from to 3E;. 6umber of even numbers &l*ad+1=398*442+1=178

Their sum &1782(398+44)=39338

11. The famous church in the city of Pumbakonnam has a big clock tower and is said to be over 3BB

years old. /very Jonday 1B.BB J the clock is set by ntony- doing service in the church. Thelock loses mins every hour. 4hat will be the actual time when the faulty clock shows 3 ?.J on+ridaya. Jb.3.1 ?Jc. .' Jd. 3 JTotal time passed in the faulty clock & Jonday 1B am to +riday 3 pm & 2 $ = ' hours & E and 'hours & 1B1 hrs' min in the faulty clock & B minutes of the correct clock1B1 hrs in the faulcty clock &

10154×60& 112.2 Hrs.

E Hrs = 1.2 Hrs+riday 1B am = 1 hrs & !aturday 2amB.2 $ B min & 12 min!o !aturday 2.12 min J

12. !uresh Raina and Lautam Lambhir after a scintillating I?" match decide to travel by cycle totheir respective villages. 8oth of them start their Nourney travelling in opposite directions. /ach oftheir speeds is miles per hour. 4hen they are at a distance of 'B miles- a housefly starts flyingfrom !uresh RainaMs cycle towards Lautam Lambhir at a relative speed of 1< miles per hour with

respect to RainaMs speed. 4hat will be the time taken by housefly to reach Lambhira. 1B hrsb. 1' hrsc. 2B hrsd. 2' hrs!ol:

+ly speed is 1< kmph w.r.t to suresh as fly is moving in opposite direction to suresh- its actual speedis 1< F & 11.6ow relative speed of fly and gambhir & 11 F & ' kmph



!o fly takes &5011*6& 1B Hrs

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&1. two digit number is 1; less than the square of the sum of its digits. How many such numbersare thereD15 1D25 2D35 3D5 ns: (ption 2Take 6 & 1Ba=b.

Liven that- D1Ba=b5=1; & P2 & Da=b52

Liven number & P2 F 1; & D1Ba=b5

That means- when we add 1; to the given number it should be a perfect square. !o P2 takes thefollowing values. 1- - E- 1- 2'- 3- E- - ;1- 1BB- 121- ....1 to 1 are ruled out as if we subtract 1; from them- the resulting number is a single digit number.6ow 2' F 1; & <

3 F 1; & 1;E F 1; & 31 F 1; & ;1 F 1; & 31BB F 1; & ;2121 F 1; & 1B36ow 3- ;2 satisfies.

2. two digit number is 1; less than the sum of the squares of its digits. How many such numbersare thereD15 1D25 2D35 3D5 ns: (ption 2(nly < and < satisfy the condition

3. +or real number $- intD$5 denotes integer part of $.intD$5 is the largest integer less than or equal to$.intD1-25&1-intDF2-5&F3. +ind the value of intD1C25=intD1C2= 1BB5=intD1C2=2C1BB5=....=intD1C2=EEC1BB5!ol: int D1C25 & B

int D1C2 = 1BB 5 & 1BBinto D1C2 = 2C1BB5 & B......int D 1C2 = 'BC1BB 5 & 1int D1C2 = '1 C1BB5 & 1.......int D1C2 = EEC1BB5 & 1!o 1BB = 1 = 1 = .....'B times & 1'B

. Liven a square of length 2m. Its corners are cut such that to represent a regular octagon. +indthe length of side of octagon

!ol:



"et $ is the side of the octagon and $ = 2y is the side of the square.

In the given octagon-y2+y2=x2⇒2y2=x2⇒y=x2Q

8utx2Q+x+x2Q=2

⇒2Qx+x=2

⇒x=22Q+1=22Q+1×2Q*12Q*1=2(2Q*1)

'. +ind the number of ways a batsman can score a double century only in terms of Ms Msssume the batsman scored $ Ms and y Ms.

$ = y & 2BB ⇒2x+3y=100⇒x=100*3y2=50*32y

s $ is an integer- y should be a multiple of 2.Ify & B- $ & 'By & 2- $ & <y & - $ & ...y & 32- $ & 2

!o total ways are D32FB5C2 = 1 & 1< D if B Ms are possible5 otherwise 1

. 'BBB voted in an election between two candidates.1O of the votes were invalid.The winner wonby a margin appro$imately closer to 1'O.+ind the number of votes secured by the person

Invalid )otes & 1 O D'BBB5 & <BB)alid )otes & 'BBB F <BB & 3BB & R Dsay5ssume the looser got M"K votes and winner got K4K votes.4 F " & 1'O DR54 = " & R!olving we get 4 & '<.'O and " & 2.'O!o 4inner got '<.'OD3BB5 & 2<2

<. There are 1BB wine glasses. I offered my servant to 3 paise for every broken glass to be deliveredsafely and forfeit E paisa for every glass broken at the end of day. He recieved Rs.2.B .how manyglass did he break.

a. 2B b. <3 c. ' d. ;If a glass has been broken- he has to loose 3 paisa = E paise & 12 paisessume P glasses got broken

1BB $ 3 F 12 $ P & 2B ⇒K=5

;. is 2B percent more efficient than 8. If the two person can complete a piece of work in Bdays.in how many days. working alone can complete the worka. ;B b. EB c. 1BB d. 11Bs is 2BO more efficient than 8- If 8Ms per day work is 1BB units then Ms 12B.8oth persons together completes D1BB = 12B5 units & 22B units a day.

They took B days to complete the work. !o total work & B $ 22BIf alone set to complete the work- he takes &60×220120=110days



E. property was originally on a EE years lease and two thirds of the time passed is equal to thefour fifth of the time to come.how many years are there to go.a. ' b. 'B c. B d. ''ssume $ years have passed and y years to go

Liven23x=45y⇒x=32×45y=65y

8ut $ = y & EE

!o65y+y=99

!olving we get y & ' years

1B. In how many different ways can the letters of the word K"/I6LK be arranged in such a waythat the vowels always come together.a. 3Bb. <2Bc. ;Bd. 'BBLiven letters are - /- I- - "- 6- L

(f which /I are vowels. "et us combine them into a single letter $. 6ow total letters are $- - "-6- LThese letter are arranged in 'A ways. 8ut 3 vowels can arrange themselves in 3A ways. !o totalways 'A $ 3A & <2B

11. There is a plane contains 32 points.all the 32 points have equal distance from point $. which ofthe following is true .a. all 32 points lie in circleb. the distance from $ to all 32 points is less than the distance between each otherc. both a and bd. none of these!ol: (ption 3S must be the center of the circle and 32 points are on the circumference. !o (ption is correct

6umber of diagnols of a regular polygon &n(n*3)2

!o for a polygon of 32 sides- 6umber of diagnols & . 6ow the minimum distance between any

two points &2πr32=1156r

6ow total lengh of all the distances from 32 points &2πr= !um of the lengths of all the

diagnols.!um of the lengths of $ to all the 32 points & 32 radius & 32r8ut the diagnols have 1 diameters connecting 2 oposite points connecting via center. !o !um

of the lengths of distances from point to point is clearly greater than sum of the length from $ to all32 ponts. (ption 8 is correctorrect (ption 3

12. 4hen asked what the time is-a person answered that the amount of time left is 1C' of the timealready completed.what is the time.1. ; pm2. ; am3. 12 pm. 12 am!ol: day has 2 hrs. ssume $ hours have passed. Remaining time is D2 F $5

24*x=15x⇒x=20

Time is ; ?J



13. ?erimeter of the backwheel &E feet-front wheel&< feet on a certain distance -the front wheel gets1B revolution more than the back wheel.what is the distance"et the backwheel made $ revolutions then front wheel makes $ = 1B$ $ E & D$ = 1B5 $ <$ & 3'

!o distance traveled & 3' $ E & 31'

1. There are 2 groups named brown and red. They can nMt marry in the same group. If the husbandor wife dies then the person will revert to their own group. If a person is married then the husbandwill have to change his group to his wifeMs group. hildren will own the motherMs group. If man isred then his motherMs brother belong to which group if he is marrieda. redb. brownc. red and brownd. none(ption: b

If a man is Red- his mother must be red- his mothers brother also red but after marriage- he getsconverted to 8rown.

1'. rectangular park B m long and B m wide has concrete crossroads running in the middle ofthe park and rest of the park has been used as a lawn.if the area of the lawn is 21BE sq.m-then whatis the width of the road.a. 2.E1 mb. 3mc. '.;2 md. noneOption : B

"et us shift the path to the left hand side and top. This does not change the area of the lawn.6ow lawn area & DB F $5 DB F $5for $ & 3- we get lawn area & 21BE.&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&1. If fDfDn55=fDn5&2n=3 and fDB5&1- what is the value of fD2B125a5 2B11

b5 2B12c5 2B13d5 BE'Ans: Option C?ut n & BThen fDfDB55=fDB5 & 2DB5 = 3 fD15⇒ = 1 & 3 fD15 & 2⇒

?ut n & 1fDfD155 = fD15 & 2D15 = 3 fD25⇒ = 2 & ' fD25 & 3⇒

?ut n & 2fDfD255 = fD25 & 2D25 = 3 fD35⇒ = 3 & < fD35 & ⇒

......

fD2B125 & 2B13

2. If '=3=2&1'1B22- E=2=&1;3'2- then <=2='&



Ans: 143547 If the given number is a = b = c then a.b a.c a.b = a.c F b

'=3=2 & '.3 '.2 '.3⇒ = '.2 F 3 & 1'1B22 E=2= & E.2 E. E.2⇒ = E. F 2 & 1;3'2

<=2='& <.2 <.' <.2 = <.' F 2 & 13'<

3. The savings of employee equals income minus e$penditure.If the income of -8- are in the ratio1:2:3 and their e$pense ratio 3:2:1 then what is the order of the employees in increasing order oftheir siGe of their savingsAns: A < B < Cs the the ratio of their incomes are in ascending order- and their e$penses are in descending order-their savings also in their incomes order.!o savings order & U 8 U

. /ntry fee is Re.1.there are 3 rides each is of Re.1. total boys entering are 3BBB.total income isRs.<2BB. ;BB students do all 3 rides. 1BB go for atleast 2 rides.none go the same ride twice. then noof students who do not go any ride is

Ans: 1000 Total entries are 3BBB !o fee collected through entry fee & 3BBB $ 1 & Rs.3BBBIncome generated through rides & <2BB F 3BBB & 2BB6ow ;BB went for 3 rides so total fee paid by these ;BB & ;BB $ 3 & 2BBD1BB F ;BB5 went for 2 rides so fee paid by these BB & BB $ 2 & 12BBssume P went for e$actly 1 rideThen P $ 1 = 12BB = 2BB & 2BB P & BB⇒

!o number of boys who did not go for any ride & 3BBB F DBB = BB = ;BB 5 & 1BBB

'. The average mark obtained by 22 candidates in an e$amination is '. The average of the first tenis '' while the last eleven is B .The marks obtained by the 11th candidate is Ans: 0It is clear that 22 $ ' & 1B $ '' = P = 11 $ B ⇒ P & B

. 4hat is the largest positive integer n for which 3Vn divides VAns: n = 0

The digit sum of4444is when remainder obtained4444divided by E

4444&(45*1)44

/ach term is a multiple of E but the last term which is(*1)44& 1

!o the digit sum of 4444is 1.

6ow the divisibility rule for 3- E- 2<... is the sum of the digits should be divisible by 3- E- 2<respectively. In each case the digit sum is either multiple of 3 or E.

!o for any value of n W 1- the given e$pression is not divisible by3n

<. 1D1A5=2D2A5=3D3A5....2B12D2B12A5 & Ans: 013!"11D1A5&1 2AF1⇒

1D1A5=2D2A5&1=&' ⇒ 3AF11D1A5=2D2A5=3D3A5&1==1;&23 ⇒ AF1...............................................1D1A5=2D2A5=3D3A5=........=2B12D2B12A5&2B13AF1



&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&1. (ne day /esha started 3B min late from home and reached her office 'B min late while driving2'O slower than her usual speed. How much time in min does eesha usually take to reach her officefrom homens: 4e know that !peed is inversely proportional to time

4hile she drives 2'O slower means she drove at 34D!5

4e know that & ! $ T

4hen speed became 34D!5 then Time taken should be 43DT5

i.e- !he has taken 43DT5 F T e$tra to cover the distance.

/$tra Time &T3& 2B min Das 2B min late due to slow driving5

ctual time T & B Jinutes

2. In 2BB3 there are 2; days in +ebruary and 3' days in a year in 2BB there are 2E days in+ebruary and 3 days in the year. If the date march 11 2BB3 is Tuesday- then which one of thefollowing would the date march 11 2BB would be

ns: If 11F3F2BB3 is Tuesday- Then 11F3 F 2BB is Thursday

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

35 How many positive integers less than 'BB can be formed using the numbers 1-2-3-and ' fordigits- each digit being used only once.ns: !ingle digit numbers & ouble digit numbers & $ 3 & 12Three digit numbers & 3 $ 3 $ 2 $ 1 & 1;Total & 3

5 circular swimming pool is surrounded by a concrete wall feet wide.if the area of the wall is11C2' of the area of the pool- then the radius of the pool in feet is

"et the radius of the pool be r. Then area of the wall and pool & π(r+4)2

rea of the pool &π(r)2

rea of the wall &π(r+4)2*π(r)2

Livenπ(r+4)2*π(r)2&1125(πr2)

r2+8r+16*r2=1125r2

11r2*200r*400=0!olving r & 2B

'5 survey of n people in the town of badaville found that 'BO of them prefer brand . nothersurvey of 1BB people in the town of chottaville found that BO prefer brand .In total ''O of allthe people surveyed together prefer 8rand .4hat is the total number of people surveyed!ol: 'BO Dn5 = BO D1BB 5 & ''O Dn = 1BB5!olving we get n & 1BB

5 In the simple subtraction problem below some single digits are replaced by letters .+ined thevalue of <='= '' F18;<FFFFFFFFFF



<!ol: 1' F < & ; !o & ;1B = D F15 F ; & !o & 31B = D'F15 F 8 & < !o 8 & <DF15 F 1 & !o & ;< = ' = & ' = B = 1 & 2B

<5 Two full tanks one shaped like the cylinder and the other like a cone contain liquid fuel thecylindrical tank held 'BB lts more then the conical tank fter 2BB lts of fuel is pumped out fromeach tank the cylindrical tank now contains twice the amount of fuel in the canonical tank Howmany lts of fuel did the cylindrical tank have when it was fullns: "et the cylindrical tank capacity $ = 'BB then the conical tank capacity & $fter 2BB lts pumped out- then remaining fuel with the tanks & $ =3BB- $ F 2BBLiven that first term is doubt the second.

x+300x*200=21

!olving we get $ & <BB

ylindrical tank capacity & 12BB lts

;. shop sells chocolates It is used to sell chocolates for Rs.2 each but there were no sales at thatprice.4hen it reduced the price all the chocolates sold out enabling the shopkeeper to realiGe Rs1.EB from the chocolates alone If the new price was not less than half the original price quotedHow many chocolates were sold!ol: 1EB & 2 X ' X 1< X E<6ow now chocolate price should be greater than 1 and less than 2. !o 2 $ ' $ 1< & 1<B!o Total chocolates sold & E< and 6ew chocolate price & Rs.1.<

E5 /esha bought two varities of rice costing 'BRs per kg and B Rs per kg and mi$ed them in some

ratio.Then she sold that mi$ture at <B Rs per kg making a profit of 2B O 4hat was the ratio of themic$ture!ol: !elling price of the mi$ture & <B and profit & 2BO

ost price of the mi$ture &70×100120=70×56

8y applying alligation rule:

!o ratio &60*1753:1753*50& 1 : '

1B. !tar question: If fD15& and fD$=y5&fD$5=fDy5=<$y=-then fD25=fD'5&!ol: "et $ &1 and y & 1fD1 = 15 & fD15 = fD15 = < $ 1 $ 1 = fD25 & 1E⇒

"et $ &2 and y & 2fD2 = 25 & 1E = 1E = < $ 2 $ 2 = fD5 & <B⇒

"et $ & 1 and y & fD 1 = 5 & = <B = 2; = & 1B

fD25 = fD'5 & 12'&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&1. In a staircase- there ar 1B steps. child is attempting to climb the staircase. /ach time she can



either make 1 step or 2 steps. In how many different ways can she climb the staricasea5 1Bb5 21c5 3d5 6one of thesens: d

,se fibonnacci series- with starting two terms as 1- 2. !o ne$t terms are 3- '- ;- 13- 21- 3- ''- ;E

2. boy buys 1; sharpners- D8rownCwhite5 for Rs.1BB. +or every white sharpener- he pays onerupee more than the brown sharpener. 4hat is the cost of white sharpener and how much did hebuya5 '- 13b5 '- 1Bc5 - 1Bd5 6one of thesens: ssume that he bought b- brown sharpeners and w- white sharpeners and the cost of brown

sharpener is $ and white sharpener is $ = 1!o wD$=15 = b$ & 1BBw = b & 1;b & 1; F w!ubstituting in equation 1- we get wD$=15 = D1; Fw5$ & 1BB so w = 1; $ & 1BBTake option 1: If white sharpners are 13- $ & D1BB F 135 C1; & .;33(ption 2- If white sharpeners are 1B- $ & D1BB F 1B5C1; & ' !o white sharpeners cost is .(ption 3 !atisfies this condition.

3. "etters of alphabets no from 1 to 2 are consecutively with 1 assigned to and 2 to Y. 8y 2<thletter we mean - 2;th 8. In general 2m=n- m and n negative inteGers is same as the lettersnumbered n."et ? & - strange country military general sends this secret message according ot the followingcodification scheme. In codifying a sentence- the 1st time a letter occurs it is replaced by the pthletter from it. 2nd time if occurred it is replaced by ?V2 letter from it. 3rd time it occurred it isreplaced by pV3 letter from it. 4hat is the code word for 88TI"a5 LH66Y((Rb5 LHPZY(HRc5 LHHLY(LRd5 LH"PY(IRns:

should be coded as 1= & L Dit occurred for first time58 should be coded as 2= & H Dit occurred for first time58 !hould be coded as 2 = 3 & 3; F 2 & 12 & " Dit occurred for second time5(ption is correct

. (f a set of 3B numbers- average of 1st 1B numbers is equal to average of last 2B numbers. Thesum of last 2B numbers isa5 2 $ sum of last 1B numbersb5 2 $ sum of 1st 1B numbersc5 sum of 1st 1B numbersd5 annot be determined

ns: 8"et average of first 1B numbers is a. Then sum & 1Baverage of last 1B nmbers also a. Then their sum & 2Ba



+rom the options 8 correct

'. In how many ways a team of 11 must be selected a team ' men and 11 women such that the teammust comprise of not more than 3 men.a5 1''b5 22'

c5 2'd5 123ns: 8Ja$imum 3 men can be played which means there can be B- 1- 2- 3 men in the team.

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

. The wages of 2 men and 1 women amount to 11BB per day. Half the number of men and 3<women has same money. The daily wages paid to each man isa5 3<'b5 BB

c5 3'Bd5 32'ns: 2m = 1w & 11BB12m = 3< w & 11BB!olving we get 12 m & 21w!ubstituting in the first equation we get- 2w = 1 w & 11BB w & 2BB⇒

J & 3'B

<. number when successively divided by '- 3- 2 gives remainder B- 2- 1 respectively in that order. 4hat will be the remainder when the same number is divided successively by 2- 3- ' in that order

a5 - 3- 2b5 1- B-c5 2- 1- 3d5 - 1- 2ns: 8

use this simple technique.[ D1 $ 35 = 2\ & '[D' $ '5 = B\ & 2'

?rocedure:

"et the number & 66ow 6 & 'PP & 3" = 2

" & 2J = 1P & 3D2J = 15 = 2 & J = '6 & 'DJ = '5 & 3B J = 2'



+or J & B we get the least number as 2'. 6ow when 2' is divided by 2- we get 12 as quotient and 1as remainder. 4hen 12 is divided by 3 we get as quotient- and B as remainder. 4hen is dividedby ' we gt as remainder.

;. a-b-c-d-e are distinct numbers. if D<'Fa5D<'Fb5D<'Fc5D<'Fd5D<'Fe5&22EE then a=b=c=d& Hint:22EE is divisible by 11.

22EE &11×11×19×1×1=11×*11×19×*1×1=Two of the terms in the given e$pression should equal to 1. s all the digits are distinct- two of theterms should be negative.(ne possible solution & D<' F 5D<' F '5D<' F ;5D<' F <5D<' F <5Then a = b = c = d = e & = ' = ; = < = < & 3'8ut as the sum of only terms was asked- we have to subtract one term.!o given answer can be one of 2E2- 3B- 2<B- 2;2- 2;B

E. If V8 means raised to the power of 8- in which of the following choices must ? be greaterthan 0

a5 B.EV?&B.EV0b5 B.EV?&B.E2V0c5 B.EV?WB.EVq(ption is wrong as ? & 0

(ption 8 is wrong asPQ=Log0.92Log0.9=0.79139

(ption is also wrong asaP>aQthen ?W0 if a W 1

1B. 2 gears one with 12 teeth and other one with 1 teeth are engaged with each other. (ne teeth insmaller and one tooth in bigger are marked and initially those 2 marked teeth are in contact witheach other. fter how many rotations of the smaller gear with the marked teeth in the other gear will

again come into contact for the first timea5<b5 12c5 ata insufficientd5 ;orrect (ption : ssume the distance between the teeth is 1 cm. Then the circumference of first gear is 12 cm andthe second is 1 cm.

6ow "J D12- 15 & ;. !o to cover ; cm- the first gear has to rotate8412& < rounds Dthe second

gear rotates ; C 1 & rounds as it is bigger5

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&1. hmed- 8abu- hitra- avid and /esha each choose a large different number. hmed says- #Jy number is not the largest and not the smallest%. 8abu says- #Jy number is not the largest andnot the smallest%. hitra says- #Jy number is the largest%. avid says- # Jy number is thesmallest%. /esha says- # Jy number is not the smallest%. /$actly one of the five children is lying.The others are telling the truth. 4ho has the largest numbera5 /eshab5 avidc5 hitrad5 8abu

ns: "argest FW 8 /



+ TC+ TC+ TC+ TC+

8 TC+ + TC+ TC+ TC+

+ + T + +

TC+ TC+ TC+ + TC+

/ TC+ TC+ TC+ TC+ T

+rom the above table- If we assume that has the largest then and both are lying. !imilarly if

we find the truthfullness of the remaining people- it is clear that / has the largest and lied. D(nly

one + in the last column5

2. In the equation = 8 = = = / & +L where +L is the two digit number whose value is 1B+ =

L and letters - 8 - - - /- + and L each represent different digits. If +L is as large as possible.

4hat is the value of L

a5 b5 2

c5 1

d5 3

ns: 8

+L is as large as possible and all the < numbers should be different.

8y trial and /rror method-

E = ; = < = = ' & 3']' is getting repeated twice.

E = ; = < = = & 3] is getting repeated

E = ; = < = ' = & 33]3 repeatsE = ; = = ' = & 32

6one of the numbers repeat in the above case and 32 is the ma$imum number +L can have. The

value of L is 2.

3. farmer has a rose garden. /very day he either plucks < or or 2 or 23 roses. The rose plants

are intelligent and when the farmer plucks these numbers of roses- the ne$t day 3< or 3 or E or 1;

new roses bloom in the garden respectively. (n Jonday- he counts 1;E roses in the garden. He

plucks the roses as per his plan on consecutive days and the new roses bloom as per intelligence of

the plants mentioned above. fter some days which of the following can be the number of roses inthe garden

a5

b5 <

c5 3B

d5 3<

ns:

If he plucks 23- then only 1; grows the ne$t day. This means total roses get decreases by '. !o after

n days assume the number of roses got decreased 1;' where n & 3<- then roses left.

. 4hat is the value of D'>;;;;;;;'>2=3;5CV2

a5 ;;;;;;;3



b5 ;;;;;;;

c5 ;;;;;;;;

d5 3

ns:

"et $ &

(x+1)×(2x*3)×(x*2)+(x*6)x2

(x2*x*2)×(2x*3)+(x*6)x2

2x3*2x2*4x*3x2+3x+6+x*6x2

2x3*5x2x2=2x*5

!ubstituting the value of $ in 2$ F '- we get ;;;;;;;3

. +or which of the following #n% is the number 2V< =2V2B';=2V2n is a perfect square

a5 2B12b5 21BB

c5 2B11

d5 2B2B

ns:

2V< =2V2B';=2V2n &K2

2V< =2V2B';=2V2n &(237)2+22058+(2n)2

4e try to write this e$pression as(a+b)2=a2+2ab+b2

6ow a &237- 2ab &22058and b &2n

!ubstituting the value of a in 2ab- we get b & 2B2B

'. RaN writes a number. He sees that the number of two digits e$ceeds four times the sum of its

digit by 3. If the number is increased by 1;- the result is the same as the number formed by

reversing the digit. +ind the number

a5 3'

b5 '<

c5 2d5 E

ns:

Loing by the options- 3' & ;D5 = 3.

. 4eight of J- and I is <. !um of and I is greater than J. I is BO less than . 4hat is

Ms weight.

ns: 1B

J = = I & < F F F D15

D = I5 F J & F F F D25

I & 410 'I & 2 I & 2C' F F F D35⇒ ⇒



dding D15 and D25 we get 2 = 2I & 12B

!ubstituting the value of I in the above equation-

2D+2(2D5)=120 1 & BB⇒

& 3BBC< & 2.;⇒

<. +ather is ' times faster than son. +ather completes a work in B days before son. If both of them

work together- when will the work get complete

a. ; days

b. ; 1C3 days

c. 1B days

d. 2B days

ns: 8

s efficiency is inversely proportional to days- If +ather : sonMs efficiency is ' : 1- then ays taken

by them should be 1 : '. ssume- the days taken by them are k- 'k.

Liven that father takes B days less. !o 'k F k & B k & 1B⇒

+ather takes 1B days to complete the work. Total work is 1B $ ' & 'B units.

If both of them work together- they complete ' = 1 units a day. Cday. To complete 'B units- they

take 'BC & ; 1C3 days.

;. beaker contains 1;B liters of alcohol. (n 1st day- B l of alcohol is taken out and replaced by

water. 2nd day- B l of mi$ture iss taken out and replaced by water and the process continues day

after day. 4hat will be the quantity of alcohol in beaker after 3 days

ns: '3.3

,se the formula-

+inal lcohol & Initial lcohol×(1*ReplacementquantityFinalVolume)n

+inal lcohol &180(1*60180)3&180×(23)3=53.3

E. If fDfDn55 = fDn5 & 2n=3- fDB5 & 1 then fD2B125 &

ns: 2B13

f DfDB55 = fDB5 & 2DB5 = 3 fD15 & 3F1 & 2- fD15 & 2⇒

fDfD155 = fD15 & 2D15 = 3 fD25 & 'F2 & 3- fD25 & 3⇒

fDfD255 = fD25 & 2D25 = 3 fD35 & <F3 & - fD35 & ⇒

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

fD2B125 & 2B13

1B. 4hat will be in the ne$t series1- <- ;- E- '- '<- 33- ...

ns: 3



1 & 1

< & 1 $ <

; & 1 $ < = 1

E & < $ < = 1

'B & < $ < = 1

' & ; $ <'< & ; $ < = 1

33 & E $ <

6e$t term should be E $ < = 1 & 3

11. In a 3 $ 3 grid- comprising E tiles can be painted in red or blue. 4hen tile is rotated by 1;B

degrees- there is no difference which can be spotted. How many such possibilities are there

a. 1

b. 32

c. d. 2'

ns: 8

This grid even rotated 1;B degrees the relative positions of the tiles do not change. !o we paint tile

number 1Ms with red or blue Donly one color should be used5 - 2Ms with red or blue.....tile ' red or

blue. Then total possibilities are25

& 32

apptitude zooming

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