Crack · 2016-08-04 · 22. The lengths of the sides of triangle ABC are 60 80, and 100 with Ð =...

SECTION I

This section contains 25 questions

Directions (Q. 1-6) : Answer the questions independently of each other.

1. The domain of y x= log 5 is

(1) x > 5 (2) ( , )0 ¥ (3) ( , ) { }0 1¥ - (4) [ , )5 ¥

(5) None of these

2. Let P = ´ ´ ´2 3 5 73 5 7 , Q = ´ ´2 3 54 6 8, R = ´ ´ ´2 3 5 74 3 2 2 and S P Q R= ´ ´ . Find the number of

factors of S which are multiple of P Q, and R simultaneously.

(1) 1152 (2) 1440 (3) 1520 (4) 12960

(5) None of these

3. In an isosceles triangle, R is the radius of its circumscribed circle and r is the radius of its inscribed circle.Find the distance between the centre of these two circles.

(1) 2R R r( )- (2) R R r( )- 2 (3) R Rr- 2 (4) R r- 3

(5) None of these

4. Suppose a b c, , are the sides of a triangle and a b c a b c a b2 2( ) ( )+ - + + - + + - =c a b c P2 ( ) , then

(1) P abc£ 3 (2) P a b c£ + +3 ( ) (3) P abc£6 3

4(4) P abc³

3

4

(5) P a b c£ + +3 3 ( )

5. The functions f x y( , ) satisfy the following three conditions for every non-negative integer x y, .

(i) f y y( , )0 1= + (ii) f x f x( , ) ( , )+ =1 0 1 (iii) f x y f x f x y( , ) ( , ( , ))+ + = +1 1 1

Find the value of f ( , )4 2008 .

(1) 42008 (2) ( )2 32011+ (3) ( )2 32008

- (4) 2 322

2

...

-Total 2s are 2011

(5) [(( ) ) ... ]2 32 2-

Total 2s are 2011

Common Admission Test

Crack

6. Determine all real numbers x which satisfy the inequality

3 11

2- - + >x x

(1) 0 13

2£ < -x (2) 0 1

39

8< < -x (3) 4 £ £ ¥x (4) - £ < -1 1

31

8x

(5) None of these

Directions (Q. 7 and 8) : Answer the questions on the basis of the information given below.

An 8 8´ chess board is decomposed into ‘ ’m non-overlapping or disjoint rectangles along the lines betweensquares such that each rectangle consists of as many black squares as white squares and each rectangle has adifferent number of squares.

7. Find the greatest possible value of m.(1) 7 (2) 8 (3) 9 (4) 6(5) None of these

8. Find the maximum possible number of total squares in a rectangle which corresponds to the greatest valueof m.(1) 16 (2) 18 (3) 20 (4) 22(5) None of these

Directions (Q. 9-25) : Answer the questions independently of each other.

9. A function f is defined for all the natural numbers and satisfies f ( )1 231= and

f f f f n n f n( ) ( ) ( ) ( ) ( )1 2 3 2+ + + =K for all n > 1. Find the value of f ( ).21

(1) 441 (2) 121 (3) 21 (4) 11(5) 1

10. 2 tractors, 3 camels and 4 bulls can plough a piece of farm in 6 h. 3 tractors, 4 camels and 2 bulls can plough

the same piece of farm in 3 h. 4 tractors, 2 camels and 3 bulls can plough the same piece of farm in 2 h. Then

a tractor, a camel and a bull can plough the piece of land in

(1) 8 h (2) 9 h (3) 12 h (4) Data inconsistent(5) None of these

11. Celeb communications is an SMS (Short Messaging Services) dedicated mobile phone company, which

offers special coupons/vouchers on every festive season at a very discounted price to all its customers. It

launches a special SMS coupon on the grand festival of Deewali worth Re 1. According to the scheme a

subscriber availing this coupon, can send total m SMSes in n successive days ( )n > 1 . On the first day of its

activation, he can send one SMS and 1/11 of the remaining ( )m - 1 SMSes. On the second day, two SMSes

and 1/11 of the remaining SMSes; and so on. On the nth and last day,the remaining n SMSes can be sent. If

he does not utilise the quota of any particular day it gets automatically vanished. Also if he tries to send

more SMSes than that of stipulated quota of any day, he will be prohibited to do so. What is the value of

( )m n+ ?

(1) 11! (2) 12! (3) 99 (4) 110

(5) None of these

12. A triangle with sides a b, and c is given. Denoted by s is the semi-perimeter, that is sa b c

=+ +

2. Construct a

triangle with sides s a s b s c- - -, , . This process is repeated until a triangle can no longer be constructed

with the side lengths given. For which original triangles can this process be repeated indefinitely?

(1) Right angled (2) Scalene (3) Obtuse angle (4) Equilateral

(5) None of these

4 Crack CAT 1 Common Admission Test·

13. Let P P P P1 2 1993 0, , K = be distinct points in the x y- plane with the following properties :

(i) both coordinates of Pi are integers, for i = 1 2 1993, , , ;K

(ii) there is no point other than Pi and Pi + 1 on the line segment joining Pi with Pi + 1 whose coordinates areboth integers, for i = 0, 1 ..., 1992.

For some i i, 0 1992£ £ , there exists a point Q with coordinates ( , )q qx y on the line segment joining Pi

with Pi + 1 such that both 2q x and 2q y are(1) even integers (2) odd integers(3) irrational numbers (4) negative even integers(5) None of these

14. Let a a an1 2, , ,K be a sequence of integers with values between 2 and 1995 such that

(i) Any two of the a si are relatively prime.(ii) Eacha i is either a prime or a product of primes. Determine the smallest possible values of n to make sure

that the sequence will contain a prime number.(1) 42 (2) 7 (3) 12 (4) 24(5) 14

15. Find the minimum positive integer n such that there exists a function from the set Z of all integers to{ , , ,1 2 K n} with the property that f x f y( ) ( )¹ whenever| | { , , }x y- Î 5 7 12 .(1) 6 (2) 3 (3) 2 (4) 5(5) 4

16. Let a b c, , be the lengths of the sides of a triangle and

Let p a b c b c a c a b= + - + + - + + -

and q a b c= + + , then

(1) p a< (2) q p= 2 (3) p q> (4) p q³

(5) p q£

17. Given S = +

+

+

+ +

+ +11

11

3

1

11

3

1

6

K

1

11

3

1

6

1

1993006+ + + +K

where the denominators contain partial sums of the sequence of reciprocals of triangular number

( ,( )

ie Kn n

=+ 1

2for n = 1 2 1996, , , )K . Then, which of the following is correct?

(1) S > 9999 (2) S > 1111 (3) S > 1001 (4) S > 1996(5) None of these

18. Find an integer n, where100 1997£ £n , such that2 2n

n

+is also an integer.

(1) 256 (2) 234 (3) 946 (4) 998(5) 729

19. For any positive integers m and n, ( ) ( )36 36m n m n+ + can be equal to

(1) 232 (2) 264 (3) 236 (4) 272

(5) None of these

20. Determine the number of factors of the greatest integer n with the property that n is divisible by all positive

integers that are less than n3 .

(1) 24 (2) 18 (3) 10 (4) 42(5) 6

21. D ABC is right angled at A. The circle with centre A and radius AB cuts BC and AC

internally at D and E respectively. If BD = 20 and DC = 16, determine AC.

(1) 1023 (2) 6 26 (3) 1032

(4) 5 119 (5) Data is insufficient

5Crack CAT 1 Common Admission Test·

E

DCB

A

22. The lengths of the sides of triangle ABC are 60 80, and 100 with Ð = °A 90 . The line AD

divides triangle ABC into two triangles of equal perimeter. Calculate the length of AD.

(1) 2796(2) 40(3) 2880(4) Can’t be determined(5) None of the above

23. There are ten prizes, five A’s, three B’s and two C’s, placed in identical sealed envelopes for the top tencontestants in a mathematics contest. The prizes are awarded by allowing winners to select an envelope atrandom from those remaining. When the eighth contestant goes to select a prize, what is the probability thatthe remaining three prizes are one A, one B and one C?

(1)1

4(2)

1

3(3)

1

2(4)

1

6(5) None of these

24. Between 6 am and 7 am when do the two hands of a clock coincide?(1) 6 : 23 : 00 (2) 6 : 32 : 00 (3) 6 : 35 : 17 (4) 6 : 32 : 43(5) 6 : 48 : 51

25. Find the number of ways of putting five distinct rings on four fingers of the left hand. (Ignore different sizesof rings and the fingers).(1) 1250 (2) 6720 (3) 5260 (4) 3250(5) 1260

SECTION II


Directions (Q. 26-28) : Answer the questions on the basis of the following charts.

The pie charts show the details of the Bollywood people who received the nomination for filmfare award of theyear 2009 as on 20-09-2009.

• Total number of nominees for this year’s filmfare award = 250• Genres, categories and ages all are exclusively divided into various divisions.


D

BA

C

ND (8%)

OTH (4%)

MH (6%)

BR (10%)PB(10%)

SI (4%)

WB(12%)

UP(46%)

Bollywood nomineesby location

Sci-fi(6%)

Musical(14%)Horror

(18%)Animation

(4%)

Comedy(28%) Romantic

(30%)

Bollywoodnomineesgenre-wise

30%

28%

12-14

14-16

Morethan 16 0-2

2-4

4-6

6-8

8-10

10-12

2% 10%

6% 6% 6%

4%

8%

Bollywoodnomineesby their experience (inyears)

0

5

10

15

20

30

35

Bollywood Nominees by their age

25

0-5 5-10 10-15 15-20 20-25 25-30 30-35 over 35

Pe

rce

nta

ge

of

No

min

ee

s

26. If the 20% of the nominees in Comedy genre belonge to UP, then how many nominees of UP are notnominated for Romantic genre?(1) 101 (2) 111 (3) 110 (4) 100(5) None of these

27. If the 50% of Animation, 20% of Horror, 40% of Musical, 60% of Sci-fi and 30% Comedy genre nomineesbelong to the group of 10-12 years experience, then how many nominees in the Romantic genre belong tothe group of 10-12 years experience?(1) 28 (2) 18 (3) 11 (4) 12(5) Can’t be determined

28. If 50% of the nominees from the age group of 25-30 years are nominated for the Horror, Musical, Romanticand Comedy genres, then how much percentage of Sci-fi genre’s nominee belongs to the age group of 20-30years?(1) 10 (2) 30 (3) 60 (4) Data insufficient(5) None of these

Directions (Q. 29-31) : Answer the questions on the basis of the information given below.

Just before the parliamentary election 5 national level parties—BJP, Congress, SP, BSP, JDU, sought thepermission from Election Commission to hold the rally in 3 different towns—Lucknow, Barabanki andSitapur. No two political parties can hold the election rally in the same town on the same day.Rallies can be held within five days period from Monday to Friday. No political party will hold the rally in any

town for more than one day. Every political party is given a battalion of paramilitary forces—B1,B2 ,B3 ,B4 and

B5 in such a way that B1 can protect BJP rally and JDU rally. B2 can protect only BJP and Congress rallies. B3

can protect only Congress and SP rallies. B4 can protect only SP and BSP rallies. B5 can protect only BSP and

JDU rallies. No political party will hold the rally without protection by a battalion of paramilitary forces.

Below is the table gives partial information about the rallies during the given period.

DayLucknow Barabanki Sitapur

Pol. Party Battalion Pol. Party Battalion Pol. Party Battalion

Monday SP B5

Tuesday B3 B1

Wednesday B2 SP

Thrusday B4 JDU

Friday JDU Congress

29. B1 and B5 protected which political parties in Lucknow rally?

(1) BJP and BSP (2) BJP and JDU (3) JDU and BSP (4) Can’t be determined(5) None of these

30. In Barabanki, SP and Congress rallies are protected respectively by(1) B3 and B2 (2) B and B4 2 (3) B and B4 3 (4) Data insufficient(5) None of these

31. In Sitapur, B1 and B5 protected the rallies of

(1) BJP and JDU respectively (2) JDU and BSP respectively(3) BJP and BSP respectively (4) Can’t be determined(5) None of these

Directions (Q. 32 and 33) : Answer the questions based on the information given below.

On, Tw, Th, Fr, Fv, Sx, Sn, Et, Nn and Tn are the ten people belong to two different families. On, Fv, Sx, andNn are two couples whereas Tw, Th, Fr, Sn, Et and Tn are their children. Each couple has three children. Theyare sitting in a row. No two members of the same family sit together. Tn sits in between Fr and Et. While Sx hastwo sons. Et has two sisters while Tn has two brothers. On is the father of Th and Tn while Nn is the mother ofSn and Et. The male heads of the two families sit on either ends of the row while the female heads of the twofamilies sit together exactly at the centre of the row.


32. Which of the following statements is definitely false?(1) Tw is son of Fv whose daughter is Sn.(2) Th and Tw are the brothers of Tn.(3) Sx is the wife of On and Tn is her daughter.(4) Fr and Sn are the daughters of Fv whose wife is Nn.(5) None of the above

33. In how many ways can these ten members sit in the row?(1) 4 (2) 8 (3) 2 (4) 6(5) None of these

Directions (Q. 34-38) : Answer the questions on the basis of the information given below.

There are six TV groups—Star, Zee TV, NDTV, UTV, Sony Entertainment, Cartoon Network, each of themhave 10 channels. These channels are supposed to be shown in a very planned manner from Monday toSaturday as given below.

• Every day (from Monday to Saturday) exactly 10 channels are being shown.

• Not more than four channels of a single TV group will be shown in any day.

• On Monday 3 channels of NDTV and one channel of Star group is being shown.

• On Tuesday, Wednesday and Thursday, no three channels of any group are being shown.

• On Wednesday, no any channel of UTV group is being shown but 4 channels of Sony Entertainmentare being shown.

• Every day atleast one channel of Star group and NDTV is being shown.

• Four channels of Cartoon Network are being shown on Thursday but not even a single channel of ZeeTV being shown on Thursday.

• Among the six days (Monday to Saturday) only on the three days, including Monday and Friday,channels of all the six groups are being shown.

• Total number of channels of UTV, Sony Entertainment and Cartoon network shown on each ofTuesday and Friday is equal to 6.

• Sum of the number of channels of Star group being shown on Monday, Tuesday and Wednesday andthe sum of number of channels of Zee TV being shown on Wednesday, Friday and Saturday is 6.

34. What is the difference between the number of channels of Star group being shown on Friday to the numberof channels of Sony Entertainment being shown on Monday?(1) 2 (2) 3 (3) 1 (4) Can’t be determined

(5) None of these

35. What is the difference between the number of channels of Sony Entertainment being shown on Wednesdayto the sum of the number of channels of Cartoon Network being shown on Tuesday and Friday?(1) 1 (2) 3 (3) 2 (4) Can’t be determined

(5) None of these

36. What is the sum of number of channels of Star group being shown on Wednesday and number of channelsof Zee TV being shown on Monday?(1) 7 (2) 6 (3) 9 (4) 8

(5) None of these

37. Which of the following is definitely true?(1) Number of channels of Star group being shown on Monday is 4.(2) Not more than 1 channel of Cartoon Network can be shown on Friday.(3) Number of channels of UTV being shown on Tuesday and Friday are same.(4) Total number of Zee TV channels being shown on Friday is more than the total number of Star group

channels being shown on Tuesday.(5) Sum of number of Star group channels being shown on Wednesday and Thursday is 1 more than the

sum of number of Sony Entertainment channels being shown on Tuesday and Friday.


38. If the number of UTV channels being shown on Thursday is greater than the number of Sony Entertainmentchannels being shown on the same day, then which among the following is definitely true?(1) Sum of number of Sony Entertainment channels being shown on Tuesday, Wednesday, Thursday and

Friday is 8.(2) Number of NDTV channels being shown on each of Tuesday, Wednesday, Thursday and Friday is same.(3) Number of UTV channels on Tuesday and number of Cartoon Network’s channels on Friday are equal

and 3 in each case.(4) Sum of number of channels of Star group being shown on Monday, Tuesday and Wednesday is 1 more

than the sum of number of channels of UTV, Sony Entertainment and Cartoon Network being shown onWednesday.

(5) None of the above

Directions (Q. 39-42) : Answer the questions of the basis of the information given below.

Six trading partners—A, B, C, D, E and F formed a close group. They send and receive the orders andconsignments within the group only. The following bar graphs show the details regarding the number ofconsignments sent by these partners, within the group during the last financial year. The first bar diagramgives the details of total number of consignments received by A from his five trading partners in last financialyear and the second bar graph gives the details of the total consignments sent by these five trading partnerswithin the group (ie, excluding A) in the same financial year.

A consignment is always sent and received in the same financial year. Total consignments received by A in thelast financial year is 3000 and assume the total consignment sent by five trading partners within the group(except A) is Q.

39. What is the least possible value of Q?(1) 3500 (2) 3800 (3) 4000 (4) 4200(5) None of these

40. Who has sent the least percentage of his consignments to A?(1) E (2) B (3) C (4) D(5) None of these

Additional information for questions 41 and 42 : It is also known that at least 25% of the total consignmentssent by B are to A and at most 40% of the consignments sent by E are to A.

41. What is the max. possible number of persons who have sent more than 80% of their consignments to A?(1) 2 (2) 1 (3) 4 (4) 3(5) None of these

42. Which of the following can be the value of K if K % of the consignments sent by D are to A?(1) 40 (2) 50 (3) 90 (4) 30(5) Can’t be determined

Directions (Q. 43-45) : Answer the questions based on the following information.

After the Parliamentary election 2009, in India, every State rushed to Delhi in order to grab maximum numberof chairs in various ministries.A ministry (in Indian parliament) is run by the council of ministers of India consists of three differentcategories which are as follows :Union Cabinet Minister : The Union Cabinet Ministers of India are in-charge of a ministry.


0%

10%

20%

30%

B C D E F

10

25

20

15

30

0%

10%

20%

30%

B C D E F

10

20

16

2524

Minister of State (Independent charge) : A minister of State (Independent charge) is a junior minister in thecouncil of ministers in the Central Govt. who have independent charge of a ministry.Minister of State (MoS) : A junior minister, supervised by the cabinet minister, usually task specificresponsibility in that ministry.Let Union cabinet Minister, Min. of State (Independent charge) and Minister of State, be denoted by UCM, MSIand MoS respectively. Let us consider, for a while, that in a single ministry there are three ministers—UCM,MSI and MoS. Various candidates wanted to become the minister so they will be selected by the PrimeMinister.• Every MP has given a choice of ministry he/she wanted.• No MP can apply for more than one ministry.• The MP who secured maximum votes (voted by his/her constituency) in the list of applicants for a

particular ministry, will be appointed as UCM. Similarly, the MPs who secured second highest andthird highest votes will be appointed as MSI and MoS respectively.

• These ministers belong to various States of India.

Name of the States

UP MP Bihar JK J&K UK MH RJ KN TN

UCM 26 22 17 13 9 8 6 4 4 3

MSI 28 20 16 20 6 10 2 6 1 3

MoS 15 15 11 16 18 6 8 12 6 5

Aggregate Points 99 90 36

The above table shows the chairs (or berths) grabbed by the top 10 States in the Minister of Councils. Aggregatepoints are awarded based on the number of UCM, MSI and MoS chairs. Each chair carries a certain positiveintegral points.The ranking of the States is based on the following order of priority(1) Aggregate points (Higher points, higher rank)

(2) The number of UCM chairs (more chairs, higher rank)

(3) The number of MSI chairs (more chairs, higher rank)

(4) The alphabetical order of states name (consider the first letter of Abbreviated Names)

• Rank 1 is the highest and rank 10 is the lowest of all. Also• After the revelation that few newly anointed ministers have cheated in concerned election, eight

ministers have been sacked and all of the sacked ministers belong to the states mentioned above.• In a particular ministry, only one minister was sacked.• At least 3 UCM and 3 MSI chairs have been forfeited.• Only the MPs who are holding the UCM chairs and MSI chairs were found guilty of manoeuvring

(or manoeuvering) of votes in the concerned elections.• KN (Karnataka) gained at least 1 UCM chair after the revelation of manoeuvring in votes.

43. In case of forfeiture of a UCM chair, all the UCM chairs and MSI chairs are being taken up by these 10 listedstates only. What could be the maximum possible number of points that a state other than these 10 listedstates may have after the revelation of manoeuvring in votes?(1) 33 (2) 35 (3) 29 (4) 37(5) 28

44. What is the highest possible rank MH may attain, after the revelation of manoeuvring in votes?(1) 2 (2) 4 (3) 5 (4) 6(5) 3

45. If at least 1 chair was forfeited from each state and the total number of chairs obtained by each stateincreased by at most 1 chair,then which of the following is definitely FALSE?(1) A minimum of four states retained their original ranks after the revelation of scam (as mentioned in the

question).(2) Maximum change in difference of points between any two states after and before the revelation of scam is 6.(3) These 10 listed states can only be the states who obtained the chairs in Union Ministry (or minister of

councils).(4) Both (1) and (3)(5) None of the above


Directions (Q. 46-50) : Answer the questions based on the following information.

As a chief guest for inauguration when I have been invited by India’s first ever multi car showroom, I came to

know that there were 4 brands of cars—Honda, Maruti, Tata and GM. Also every brand was represented by 4

different class of cars—class B, class C, class D and class E. The following facts about these cars were also

known.

• There were only 21 cars.

• There were at least two cars of D class from Tata brand.

• None of the brands include more than three cars from any class.

• The number of cars from the class B was exactly half the number of cars from each of the other classes.

• There was no car from the class B from Honda and apart from this, all brands, including Honda,represented at least one car from each class.

• Had there been one car less from Tata brand, then the number of cars from Maruti would have beentwice that from each of the other brands.

46. Which of the following combination is not possible?

(1) One car from the class D of Honda brand and one car from the class D of Maruti brand.

(2) Three cars from the class E of Maruti brand and one car from the class C of the Honda brand.

(3) Two cars from the class D of Maruti brand and one car from the class C of Honda brand.

(4) Two cars from class D of Maruti brand and two cars from the class C of Honda brand.

(5) All of the above

47. If there was only one car from class D of Maruti brand, which of the following is definitely not true about the

cars?

(1) There are three cars from the class E of Maruti brand.

(2) There are two cars from the class C of Maruti brand.

(3) There is one car from the class E of Honda brand.

(4) There is one car from the class C of Honda brand.


48. What can be said about the following two statements regarding the number of cars from the class E of Maruti

brand?

Statement A The number is atleast two.

Statement B The number is atmost three.

(1) Statement A is necessarily true, while statement B is not necessarily true.

(2) Statement B is necessarily true, while statement A is not necessarily true.

(3) Both statement A and statement B are necessarily true.

(4) Neither statement A nor statement B is necessarily true.

(5) None of the above

49. Which of the following cannot be determined uniquely from the given information?

(1) Number of cars from the class E of Honda company (or brand).

(2) Number of cars from the class C of Tata company.

(3) Number of cars from the class C of GM company.

(4) Number of cars from the class B of Honda company.


50. What is the number of cars from class D of Tata company?

(1) 1 (2) 2

(3) 3 (4) 4

(5) None of these


SECTION III


Directions (Q. 51-55) : Sentences given in each question, when properly sequenced form a coherentparagraph. Each sentence is labelled with a letter. Choose the most logical order of sentences from among thefive given choices to construct a coherent paragraph.

51. A. I like to think that Popeye had a positive effect on my upbringing. At least I preferred him as a role modelover Dillinger and other prominent citizens of the Depression.

B. He was good, he was powerful, and he was funny . ... I wrote Segar a fan letter in 1934 and received areply.

C. He was as basic and down to earth as mud. “I yam what I am and that's all that I yam” is a piece ofPhilosophy that stands besides Aristotle’s ‘Know thyself’....I loved Popeye”.

D. Popeye, to me, was the first super-hero. Not only did he possess super-strength but he exhibited superhonesty.

E. It was a drawing of Popeye and Wimpy with the inscription, “To me fren Mort Walker.’’ It has hung onmy wall for over 70 years filling the room with its friendly glow... Who knows what influences take a partin making a man what he is?

(1) DCBEA (2) DCABE (3) EDCBA (4) ACDEB(5) ABDCE

52. A. For Geoffrey Canada,the non-profit’s longtime CEO, the imperative was clear.B. So, in 2002, it changed its name and sharpened its focus.C. Now, simply called the Harlem Children's Zone (HCZ), the agency linked its original mission to a very

concrete statement of the impact it intended to have : namely, that 3000 children, ages 0 to 18, living inthe zone should have demographic and achievement profiles consistent with those found in an averageUS middle-class community.

D. To help the greatest possible number of kids lead healthy lives, stay in school, and grow up to becomeindependent, productive adults, Rheedlen would have to step up its performance.

E. Despite Rheedlen's many good programs, however , the prospects for Harlem's children appeared to begetting worse, not better.

(1) EADBC (2) CBDAE (3) ABCDE (4) BCDEA(5) BADCE

53. A. I take your point, and in fact I agree that there are many examples of what you are talking about.B. I certainly remember early on, after 9/11,what the impact was on the political culture here.C. The question is how we interpret such an historical moment, and from our perspective, we certainly see

more evidence of damage than we see of people being able to wend their way through it without beingaffected.

D. This is particularly important if we think about our own times, and what the War on Terrorism hasgenerated in terms of its impact.

E. This was a very serious thing, and we cannot be cavalier about the impact of this kind of dynamic.(1) CAEBD (2) AEBDC (3) BEDAC (4) DCBEA(5) DBEAC

54. A. The aim is to help children think more creatively and expand their vocabulary.B. A London-based enterprise, Philosophy Shop, is recruiting university graduates to teach philosophy in

12 primary schools in London and 10 elsewhere in the UK.C. A survey of 105 10 -year olds demonstrated significant improvements in tests of verbal, numerical and

spatial abilities at the end of sixteen months of lessons compared to a control group.D. In 2002 psychologists in Scotland studied the benefits of teaching philosophy to school children.E. However, the Campaign for Real Education say schools should concentrate on teaching the 3 Rs in view

of the number of youngsters who leave school without a fundamental grasp of these basics.(1) BEADC (2) DECBA (3) BADCE (4) EDCBA(5) ACBDE


55. A. The rise in debt was the flipside of jobs being lost to the East.B. Eventually, the credit bubble burst, As an economic strategy, It made little sense, even for the Burberrys

of this world.C. Free trade driven by cost cutting feeds and nourishes credit bubbles. It does not benefit the workers, but

it has failed corporations too.D. After seven years of debt-fuelled growth, stock markets are now lower than they were in 2000.E. As more and more companies fled the West in search of cheaper production bases, the central banks

were obliged to keep interest rates low, to stimulate economic growth.(1) EACBD (2) EABDC (3) CDABE (4) EACDB(5) BACDE

Directions (Q. 56-60) : There are two blanks in each of the following sentences. From the pairs of words givenbelow, choose the pair that fills the blanks most appropriately.

56. Not only is much of India’s unfolding economic success not reaching many of its poor citizens but itspolitical ..... at home and abroad seem to make very little difference to their ..... lives.(1) turmoils, abject (2) successes, embittered (3) advancement, bettered (4) aspirations, battered(5) will, daily

57. India and Pakistan should abandon the practice of arresting fishermen who ..... cross into their territorialwaters and instead ..... a mechanism for the informal repatriation of these innocents.(1) wantonly, create (2) mistakenly, discover(3) inadvertently, install (4) deliberately, constitute(5) desperately, install

58. The special purpose vehicle for infrastructure projects has been proposed to impart a ..... to such core sectorprojects as also to ensure the completion of such projects which remained ..... on account of paucity offunds.(1) breakthrough, shelved (2) rick, passive(3) thrust, stalled (4) push, inconclusive(5) force, dormant

59. Finally, even at the high levels that the stock market has reached these days, the question is as to whetherthe stock market mechanism in India is good enough ..... of the nation’s economic health is far from .....(1) augury, unanswered (2) thermometer, decided(3) indicator, ambiguous (4) barometer, settled(5) healer, stock market

60. State hospitality extended to a visiting foreign dignitary is often used symbolically to convey ..... messagesand very often pomp and ceremony serve to ..... sharp differences.(1) sharp, hide (2) loud, comouflage (3) important, accentuate (4) subtle, mark(5) hidden, hide

Directions (Q. 61-75) : Refer to the passages below and answer the questions that follow.

PASSAGE 1

Investigators of the Bermuda Triangle have long noted the existence of another mystery area in the world’soceans, southeast of Japan, between Japan and the Bonin Islands, specifically between two Jima and MarcusIsland, with a record and reputation indicative of special danger to ships and planes. Whether the shipshave been lost from underwater volcanoes or sudden tidal waves, this area, often called the Devil’s Sea,enjoys at least officially an even more sinister reputation than the Bermuda Triangle in that the Japaneseauthorities have proclaimed it a danger zone. This action came out often an investigation carried out by aJapanese surface craft in 1955.The Devil’s Sea had long been dreaded by fishermen, who believed it was inhabited by devils, demons andmonsters who seized the ships of the unwary. Aircraft and boats had disappeared in the area over a periodof many years, but during the time when Japan was at peace, nine modern ships disappeared in the period


of 1950 to 1954, with crews totaling several hundred persons, in circumstances characteristic (extensiveair-sea searches, lack of wreckage or oil slicks) of the happenings in the Bermuda Triangle.The Bermuda Triangle and the Devil’s Sea share a striking coincidence. The Bermuda Triangle includes,almost at its western terminus, longitude 80° west, a line where true north and magnetic north becomealigned with no compass variation to be calculated. And this same 80°W changes its designation when itpasses the poles, becoming 150°E. From the North Pole south, it continues on, passing east of Japan, andcrosses the middle of the Devils’s Sea. At this point in the centre of the Devils’s Sea, a compass needle willalso point to true north and magnetic north at the same time, just as it does at the western border of theBermuda Triangle on the other side of the world.The unexplained losses in the Japanese equivalent of the Bermuda Triangle were instrumental in inspiringa government-sponsored investigation of the area, which took place in 1955. This expedition with scientiststaking data as their ship, the Kaiyo Maru No. 5, cruised the Devil’s Sea, ended on a rather spectacular note –the survey ship suddenly vanished with its crew and the investigating scientists!The presence of one or more other areas of disappearance in the world’s oceans has led to some unusualspeculations. Theories concerning antigravity warps have been advanced, presupposing areas where thelaws of gravity and normal magnetic attraction no longer function in ways with which we are familiar.Ralph Barker, author of Great Mysteries of the Air, noting that new developments in physics point to theevidence of the existence of anti-gravitational particles of matter’ suggests ‘the presence of anti-gravitationalor “contra-terrene” matter, of a nature completely contrary to those known on this planet ... of appallingexplosive character when [it] comes into proximity of matter as we know it ... embedded in localized areasof the earth....’ He offers the possibility that this matter may have arrived from space and become embeddedunder the earth’s crust, sometimes under the land but more often under the sea.Consideration of this theory offers a possible explanation of electronic and magnetic malfunction withingiven areas but would not, however, explain the many losses of ships and planes within sight of land. Oneremembers, in this connection, reports of other areas of magnetic anomaly over other bodies of waterthroughout the world where the pull of something under the water is stronger than that of the NorthMagnetic Pole.A more detailed study of the Bermuda Triangle and other suspect areas was made by Ivan Sanderson anddiscussed in his article “The Twelve Devil’s Graveyards Around the World,” written for Saga magazine. Inplotting ship and plane disappearances throughout the world, Sanderson and his associates first found thatthe majority of these mysterious losses occurred in six areas, all of them having more or less the samelozenge shape and coincidentally located between latitudes 30° and 40° north and south of the Equator andincluding the Bermuda Triangle and the Devil’s Sea.Developing his theory further, Ivan Sanderson established a network of twelve ‘anomalies’ atseventy-two-degree intervals around the world, centered more precisely at 36° north and south latitude,making five in the Northern Hemisphere, five in the Southern and including the North and South poles.The reason for the Bermuda Triangle being the most celebrated, he suggests, is that it is the most travelledwhile the others, although located in less travelled areas, also give considerable evidence, of magnetic spacetime anomalies.The majority of these active areas lie due east of continental land masses where warm ocean currents goingnorth collide with cold currents going south. In addition to this current collision, these areas also representthe nodal points where the surface ocean currents turn one way and the subsurface currents turn in anotherdirection. The great subsurface tidal currents sweeping tangentially, and influenced by differenttemperatures, set up magnetic vortices, affecting radio communication, magnetism – perhaps even gravity –and eventually, in special conditions, causing air and surface craft to vanish – sailing or flying off into adifferent point in time and space. An interesting sidelight on the erratic behaviour of these areas isunderlined by Sanderson in describing the astonishing ‘early arrivals’ of carefully clocked-in air flightswhere planes have arrived so far ahead of schedule that the only possible explanation would be that theyhad a tail wind behind them blowing, for example, at 500 miles per hour. Such incidents may be the resultof unrecorded winds but they seem to occur most frequently within the Bermuda Triangle and other vortexareas, as if these particular planes had encountered the anomaly but had skirted or been propelled safelythrough the ‘hole in the sky’ that had cost so many travellers their lives.


61. The “Japanese equivalent of the Bermuda Triangle”refers to(1) The Kaiyo Maru No. 5 (2) The Devil’s Sea(3) An area of the ocean near Japan (4) Devil’s Grave yard(5) Sinister island

62. According to the passage, the theory that anti-gravity warps are responsible for disappearance in oceanswould be termed as(1) inconclusive (2) imaginative (3) speculative (4) baseless(5) plausible

63. According to Ralph Barker, when contra-terrene matter comes in contact with matter, it has(1) a vacuuming effect (2) a paralysing effect(3) a draining effect (4) an explosive effect(5) a wavy effect

64. Which of the following represents an area that is suggested to be an anomaly by Ivan Sanderson?(1) 36° north latitude (2) The equator(3) The South Pole (4) Asian gulf(5) None of these

65. Which of the following is not said to be a feature of the “active” areas?(1) Tangential subsurface tidal currents(2) Antigravity warps(3) Magnetic vortices(4) Collison points for tidal currents of different temperatures(5) None of the above

66. One may infer from the passage that the “early arrivals” in the Bermuda triangle can be attributed to(1) unrecorded winds (2) vortices(3) holes in the sky (4) space-time warps(5) poler friction

67. An appropriate title for the passage would be(1) The Bermuda Triangle — some theories (2) Geographic anomalies and their causes(3) Anti-matter and Space-Time Warps (4) The Bermuda Triangle’s disappearances(5) Shocks of the nature

PASSAGE 2

The people of the Skagit love the Alpine scenery where their river is born. They climb the mountain peaks,hunt in the hills, and fish in the icy streams – and mourn the passing of the day when the river almostoverran its banks with steelhead and salmon.“When I was a lad,” old-timer Rudy Clark told me, “I used to have to whip the horse to make him cross. Hewas terrified, trying to walk through the fish. But when we reached the other side, the spokes would havethrown enough salmon up into the wagon to last the family for days. But what fish are left today,well,theIndians get ‘em”.The controversy surrounding fish and Indians is only one of many on Washington State's SkagitRiver,which flows from the high wilderness of the Cascade Range into Puget Sound and the Pacific. Itsyear-round ice-ridden tributary valleys are deep and cold. Yet, where its glacial waters reach the seascarcely a hundred miles away, the Skagit delta is one of the most fertile gardens of the globe.Five million pounds of vegetable seeds were harvested here in 1976, including two-thirds of the bestspinach, beet, and cabbage seeds in the country. Lush dairy lands share the delta with loamy fields sownwith tulips, irises, daffodils, and strawberries. From the mountains that slope down from the feet of Skagitglaciers come hundreds of millions of board feet of prime fir and cedar and hemlock.Yet the Skagit has been proposed for federal protection under the Wild and Scenic Rivers Act. Thequestions, then,if not the answers, are clear. How does one declare a river scenic or recreational when the


people of its floodplain, with their farms and homes and highways, need more dams for flood protection?When it already turns turbines at five major hydroelectric dams, and a higher one is on the drawing board?How does one elect to preserve even more of the river's forests as wilderness when those already reservedare coveted by the timber industry? When a thousand square miles of its mountains and lakes have recentlybecome national park, national recreation area, or federal wilderness? When the secrets of its fishing andboating and climbing and camping have leaked out to a world hungry for recreation?And add to that controversial coming of nuclear power to the Skagit. And the spectre of a horde of foreignhomesteaders, from such places as California, Connecticut,and Canada—anxious to buy, along with Skagitdrizzle and fog, some peace and quiet to calm the soul.Big Beaver Valley is a forested cathedral, its columns giant cedars, its nave leads the eye upward towardssuch altars as ice-mantled Mount Shuksan, and the white cone of volcanic Mount Baker, its crater smokingas if with incense.If Seattle City Light is to obtain another 270000 kilowatts of peaking power for what is already one of themost electrified cities in the world, it must flood part of Big Beaver Valley by raising Ross Dam another 121feet, adding to its already very high 400-foot waterhead.Paul Kraabel and Tim Hill, councilmen of the city of Seattle are trying to decide whether they really wantmore power for Seattle at such expense. “Our City,” Tim told me, “owns the electric company and is madeup of people who constantly use more power. But also of people who love these mountains and valleys.Besides, if the Federal Power Commission and City Council vote to raise Ross Dam, the flooding wouldswallow up seven square miles of recreation land in Canada.That could provoke a lot of anti-Americanism.”I knew. An irate Canadian had already told me: “We made a bad bargain with City Light ten years ago.Traded miles of the Canadian Skagit for a pitiable payoff of $34,556 a year.”But that was before the environmental revolution of the late 1960's. In 1974-75, Norman Pearson, thenBritish Columbia's deputy minister of lands, tried to reverse the agreement.“The entire thing,” he told me, “is tangled up in politics. The best hope for the Skagit is for your FederalPower Commission to heed the environmentalists and other agencies, kill the high dam project, and get ourtwo national governments off the hook.”However that international wound may fester, City Light is proposing to construct Copper Creek Dam onthe main stem of the river entirely within Washington State. Hundreds of Skagit floodplain farmers wouldapplaud.“In 1975, on this river, we lost more than 100 cattle in a flood,” said Bob Hulbert, a supervisor of the SkagitConservation District. “No loss of human life , thank God. But it was bad enough.” “The Copper Creek Damwouldn't really help all that much, though. Two Skagit tributaries are the real problems, the Sauk and theSuiattle. And they’d be kept free of dams if they became scenic rivers.”“If we can't get a dam on the Sauk, we’ll have to build a major diversion channel for flood water, and raiseand strengthen our dikes. Either way, it’s going to cost a lot of money.”

68. According to the passage, the Skagit River arises in the(1) Alpine Mountains (2) Glacier Park(3) Cascade Range (4) Newhalem Mountain(5) Skagit delta

69. The underlined word means a land that(1) is infertile (2) has silt (3) is rich in nutrients (4) lies alongside rivers(5) is yellow

70. The Skagit is said to provide an abundance of all of the following except(1) vegetables (2) timber (3) electricity (4) ski slopes(5) timber and electricity both

71. The description of Big Beaver Valley is intended to(1) arouse a sense of awe (2) evoke nostalgia(3) create an atmosphere of piety (4) evoke a feeling of serenity(5) expose the political ideology


72. Which of the following types of wood is not said to be among those found in the Skagit?(1) Cedar (2) Teak(3) Hemlock (4) Fir(5) Fir and Teak

73. By posing various questions related to federal protection of the river, the author seeks to suggest that thefederal protection of the river is(1) not necessarily desirable as there are many practical aspects of the issue to be considered.(2) a task that poses more questions than its answer.(3) a task that will be based on difficult choices.(4) a desirable but unlikely outcome as there is too much already invested in the area.(5) a technical issue which requires flood water specialist to look into the matter.

74. The basic problem with raising Ross Dam is that(1) it would antagonise Canadians. (2) it would result in flooding.(3) it would result in loss of farmland. (4) it harms the irrigation system.(5) All of these

75. Which of the following can be inferred from the passage?(1) The construction of the Copper Creek Dam is being resisted by American farmers.(2) The Sauk is situated in Canada.(3) Dams are not built on rivers declared to be scenic.(4) The flooding of the Skagit can only be stopped by damming the tributaries.(5) Constructing dams requires public mandate.


SECTION I

1. (3) yx

x= =loglog

log5

5

and x > 0

\ x ( , ) { }0 1¥ -

or x Î È ¥( , ) ( , )0 1 1

Hence, choice (3) is correct.

2. (2) P = ´ ´ ´2 3 5 73 5 7

Q = ´ ´2 3 54 6 8

R = ´ ´ ´2 3 5 74 3 2 2

S P Q R= ´ ´ = ´ ´ ´2 3 5 711 14 17 3

LCM of P Q R, , = ´ ´ ´2 3 5 74 6 8 2

\ S = ´ ´ ´ ´ ´ ´ ´( ) ( )2 3 5 7 2 3 5 74 6 8 2 7 8 9

\ Required number of factors of

S = + + + +( ) ( ) ( ) ( )7 1 8 1 9 1 1 1

= ´ ´ ´8 9 10 2

= 1440


3. (2)

Let the triangle be ABC having

two sides equal as AB AC= .

Let the incentre be I and the

circumcentre be O. Let the

distance IO be d; d is positive if

A and BC are on different sides

of O [fig (i)] otherwise d is

negative if A and BC are on the

same side of O [fig(ii)].

Let ÐOAB be q,

Then,r

R d+= sin q

andr d

Rr d R

+= Þ + =cos ( ) cos2 2q q

[ ]Q Ð = ÐDOC DAC2Þ r d R+ = -( sin )1 2 2 q

[ cos sin ]Q 2 1 2 2q q= -

Þ r d Rr

R d+ = -

+

æ

èçç

ö

ø÷÷

é

ë

êê

ù

û

úú

1 2

2

Þ R r d r Rrd d R d2 2 3 22+ + + - + - + =Rd R Rr2 3 22 0

Þ ( )2 22 2 2 3 2R r R r d r Rrd d R d- + + + -

+ - + =Rd R Rr2 3 22 0

Þ d d R d r R d3 2 2 2+ + -

- - + + + =R R r Rrd R r Rr3 2 2 22 2 2 0

Þ d d R r R d R r2 2( ) ( )+ + - + +

+ + + =2 0Rr d R r( )

Þ ( ) ( )d R r d R Rr+ + - + =2 2 2 0

Þ ( ) [ ( )]d R r d R R r+ + - - =2 2 0

If d R r+ + = 0 Þ d R r= - +( )

Since, OI OA< , so d R r¹ - +( )

\ d R R r2 2 0- - =( )

Þ d R R r2 2= -( )

Þ d R R r= -( )2

Hence, choice (2) is correct

4. (1) Consider, arbitrarily, an equilateral triangle with

a b c= = = 1

Then, a b c a b c a b2 2( ) ( )+ - + + - + + - =c a b c P2( )

3 = P

But 3 3abc =

Hence, choice (1) seems to be true.

Further consider a right triangle with

a b c= = =3 4 5, ,

Then, P = 168, but 3 180abc =

Again only choice (1) conforms to the relation betweenP and a b c, , .

5. (4) f n f f n( , ) ( , ( , ))1 0 1 1= - [using function (iii)]

Þ f n f n( , ) ( , )1 1 1 1= - +

[using function (i)]

Þ f n f n( , ) ( , )1 1 2 2= - +

Þ f n f n( , ) ( , )1 1 3 3= - +

……………………….

……………………….

……………………….

Þ f n f n( , ) ( , )1 1 0= +

Þ f n f n( , ) ( , )1 0 1= + [using function (ii)]

Þ f n n n( , ) ( )1 1 1 2= + + = + [using function (i)]

qR

dO

I

r

DCB

A

r Iq

dD

OR

CB

A

Fig. (i) Fig. (ii)

q q

OR

CD

IB

A

90°

20°

Fig. (iii)

Answers with Explanations

Crack CAT 1

Crack CAT 1 Common Admission Test·Similarly,

f n f f n( , ) ( , ( , ))2 1 2 1= - [using function (iii)]

Þ f n f n( , ) ( , )2 2 1 2= - + [using function (i)]

\ f n f n( , ) ( , )2 2 0 2= +

Þ f n f n( , ) ( , )2 1 1 2= + [using function (ii)]

Þ f n n( , )2 2 3= + [using function (i)]

Similarly,

f n f f n( , ) ( , ( , ))3 2 3 1= - [using function (iii)]

f n f n( , ) ( , )3 2 3 1 3= - + [ ( , ) ]Q f n n2 2 3= +

Let U f nn = +( , )3 3

\ f n f n( , ) ( , ) ( )3 3 2 3 1 3 3+ = - + +

Þ f n f n( , ) [ ( , ) ]3 3 2 3 1 3+ = - +

Þ U Un n= -2 1( )

Also, U f0 3 0 3= +( , ) [ ( , ) ]QU f nn = +3 3

Þ U f0 2 1 3= +( , ) [ ( , ) ( , )]Q f x f x+ =1 0 1

Þ U0 2 1 3 3 8= ´ + + =( ) [ ( , ) ]Q f n n2 2 3= +

\ Unn= +2 3

Þ f n n( , )3 3 2 3+ = +

Þ f n n( , )3 2 33= -+

Now, f n f f n( ) [ , ( )]4, 4,= -3 1

Þ f n f n( ) ( , )4, = -- +2 34 1 3

[ ( , ) ]Q f n n3 2 33= -+

\ f f( , ) ( , )4 0 3 1= [ ( , ) ( , )]Q f x f x+ =1 0 1

= -2 34 [ ( , ) ]f n n3 2 33= -+

= 13

To observe the pattern we would like to compute twomore terms.

f f( , ) ( , )4 1 2 34 0 3= -+ [ ( , ) ]( , )Q f n f n4 2 34 1 3= -- +

Þ f ( , ) ( )4 1 2 32 3 34

= -- + [ ( , ) ]Q f 4 0 2 34= -

f ( , )4 1 2 3 2 32 24 22

= - = -

Similarly, f f( , ) ( , )4 2 2 34 1 3= -+

[ ( , ) ]( , )Q f n f n4 2 34 1 3= -- +

Þ f ( , ) ( )4 2 2 32 3 32 4

= -- + [ ( , ) ]Q f 4 1 2 324

= -

Þ f ( , )4 2 2 322 4

= -

Þ f ( , )4 2 2 32222

= -

\ f ( , ) [ ]..

.

4 2008 2 322

2

= -Total 2011 twos

6. (4) ( ) ( )3 11

2- - + >x x

Squaring both sides, we get

( ) ( ) ( ) ( )3 1 2 3 11

4- + + - - + >x x x x

Þ - - + >-

2 3 115

4( ) ( )x x

Þ ( ) ( )3 115

8- + <x x

Þ - - - <( )x x2 2 315

8

Þ - - + <( )x 1 415

82

Þ - - + <( )x 1 4225

642

Þ - - <-

( )x 131

642

Þ ( )x - >131

642

Þ | |x - >131

8

Þ 131

81

31

8+

æ

èçç

ö

ø÷÷ < < -

æ

èçç

ö

ø÷÷x

However, it is clear that if x > 1, then

( ) ( )3 1 0- - + <x x , which is inadmissible.

Further we need to understand that the expressionsunder the root signs are not negative, thus

- £ £1 3xQ3 0 3

1 0 1

- ³ Þ £

+ ³ Þ ³ -

ìíî

üýþ

x x

x xand

Therefore, possible values of x are expressed asbelow.

- £ < -1 131

8x


7. (1) Our requirement is that the number of black andwhite squares be equal is equivalent to requiring thatthe each rectangle has an even number of squares,which is as follows.

2 4 6 8 10 12 14 16

8

72 64+ + + + + + + = >1 2444444 3444444

Therefore, m < 8, so max( ) .m = 7Hence, choice (1) is correct.

8. (3) There are 5 probable divisions of 64 squares into 7unequal even numbers as follows

2 4 6 8 10 12 22+ + + + + +2 4 6 8 10 14 20+ + + + + +2 4 6 8 10 16 18+ + + + + +2 4 6 8 12 14 18+ + + + + +

2 4 6 10 12 14 16+ + + + + +

But since 22 2 11= ´ . There can’t be 11 squares in asingle row as atmost 8 squares are possible in any row,so first combination is ruled out. Thus there are onlyfour cases remaining.Hence, maximum 20 squares can be found in anysuch rectangle.Hence, choice (3) is correct.

9. (5) By definition,f f n n f n( ) ( ) ( ) ( )1 2 2+ + + =K …(i)

and f f f n n f n( ) ( ) ( ) ( ) ( )1 2 1 1 12+ + + - = - -K …(ii)

Subtracting Eq. (ii) from Eq. (i), we get

f n n f n n f n( ) ( ) ( ) ( )= - - -2 21 1

Þ n f n f n n f n2 21 1( ) ( ) ( ) ( )- = - -

Þ ( ) ( ) ( ) ( )n f n n f n2 21 1 1- = - -

Þ f nn

nf n( ) ( )=

-

+

æ

èçç

ö

ø÷÷ -

1

11

19

Crack CAT 1 Common Admission Test·

\ f ( )2120

22

19

21

18

20

17

19= ´ ´ ´ ´K ´ ´ ´ ´

3

5

2

4

1

31f ( )

Þ f f( ) ( )212

22 211=

´´

Þ f ( )211

11 21231=

´´

Þ f ( )21 1=


10. (4) Let a tractor can plough along in T hours, a camel canplough alone in C hours and a bull can plough alone inB hours.

Then,2 3 4 1

6T C B+ + = …(i)

3 4 2 1

3T C B+ + = …(ii)

4 2 3 1

2T C B+ + = …(iii)

Adding all the above equations, we get

91 1 1

1T C B

+ +æ

èç

ö

ø÷ =

Þ1 1 1 1

9T C B+ + = …(iv)

Now, subtracting Eq. (i) from Eq. (ii), we get1 1 2 1

6T C B+ - =

But from Eq. (iv),1 1 1

9

1

T C B+ = -

Þ1

9

1 2 1

6-

æ

èç

ö

ø÷ - =

B B

Þ1

9

1

6

3- =

B

Þ B = -54 , which is impossible.

Hence, data is inconsistent.


11. (4) Let ar = number of SMSes still not sent at the start of

the rth day. (Please note that ar includes only thoseSMSes which are officially available to be sent on therth day and afterward, it does not include any SMSwhich was available at our disposal in previous day (s)but has not been sent).

\ a m1 = and a nn = ( )1 £ £r n \ an + =1 0

So, a a r a rr r r+ = - - -1

1

11( ) ( )

Þ a a rr r+ = -1

10

11( )

Þ 10 11 101a a rr r- =+

Þ 10 11 101a a rr r= ++

Þ 10 100 11 1101a r a rr r+ = ++

Þ 10 100 11 110 110 1101a r a rr r+ = + + -+

Þ 10 100 11 110 1 1101a r a rr r+ = + + -+ ( )

Þ 10 100 10 110 11 110 11a r a rr r+ - = + ++( ) ( )

- -110 10 110( )

Þ 10 100 10 110 11 1a r ar r+ - = +( )+ + -110 1 11 110( ) ( )r

Þ 10 10 110 11 10 1 1101( ) [ ( ) ]a r a rr r+ - = + + -+

Þ a r a rr r+ + + - = + -1 10 1 11010

1110 110( ) ( )

\ a a2 110 2 11010

1110 1 110+ - = + -( ) [ ( ) ]

Similarly, a a3 210 3 11010

1110 2 110+ - = + -( ) [ ( ) ]

=æ

èç

ö

ø÷ + -

10

1110 110

2

1[ ]a

Similarly, a a4 310 4 11010

1110 3 110+ - = + -( ) [ ( ) ]

=æ

èç

ö

ø÷ + -

10

1110 110

3

1[ ]a

\ a r ar

r

+ - =æ

èç

ö

ø÷ + -

-

10 11010

1110 110

1

1( )

Since, a m an1 1 0= =+, for every 1 1£ £ +r n

a n an

n

+ + + - =æ

èç

ö

ø÷ + -1 110 1 110

10

1110 110( ) ( )

Þ 10 1 11010

111001( ) ( )n a

n

+ - =æ

èç

ö

ø÷ - [ ]an + =1 0

\ 11 10 10 1001n nn m( ) ( )- = -- [ ]a m1 =

Þ mn n

n= +

--

10010 11

10 1

( )

Since, m is an integer, therefore( )n n

n

--

10 11

10 1must be

an integral number.Again, since 11 and 10 are coprime, so 10 1n - mustdivide n - 10.But 10 101n n- > - so n = 10 and m = 100


• Below is the visual explanation of the question isgiven

DayKnown

no. of theSMS sent

1/11 of theremainingno. of SMS

sent

TotalSMS

sent perday/by

that day

RemainingSMSs forthe next

day

1 1 9 10/10 90

2 2 8 10/20 80

3 3 7 10/30 70

4 4 6 10/40 60

5 5 5 10/50 50

6 6 4 10/60 40

7 7 3 10/70 30

8 8 2 10/80 20

9 9 1 10/90 10

10 10 0 10/100 0

12. (4) Semiperimeter sa b c

=+ +( )

2. Therefore, sides of

another triangle are( )

,( )

,- + + - +a b c a b c

2 2( )a b c+ -

2. We may ignore the factor 1/2, since clearly

20

Crack CAT 1 Common Admission Test·a triangle with sides x y z, , can be constructed iff atriangle with sides 2 2 2x y z, , can be constructed.The advantage of considering the process asgenerating ( ), ( )- + + - +a b c a b c , ( )a b c+ - froma b c, , is that the sum of the sides remains unchangedat a b c+ + , so we can focus on just one of the threesides. Thus we are looking at the sequencea a b c a, ( ) ,+ + - 2 a b c a b c+ + - - + +2 ( ), ... . Letd a b c= - -2 . We show that the process generates thesequence a, a d a d a d- + -, , ,3 a d a d+ -5 11, , a d+ 21 ,... . Let the nth term be a a dn

n+ -( )1 . We claim that

a an n n+ = + -1 2 1( ) . This is an easy induction, for we

have a a dnn+ - +

+( )1 11 = + +a b c - + -2 1( ( ) )a a dn

n

and hence ( )- ++1 1

1n

na d = - - -d a dnn2 1( ) , and hence

a an nn

+ = + -1 2 1( ) . But this shows that an is

unbounded. Hence, if d is non-zero, then the processultimately generates a negative number. Thus anecessary condition for the process to generatetriangles indefinitely is that 2a b c= + . Similarly,2b c a= + is a necessary condition. But these twoequations imply (subtracting) a b= and hence a c= .So, a necessary condition is that the triangle isequilateral. But this is obviously also sufficient.Hence, choice (4) is correct.

13. (2) We consider the mid point of each side. We say that avertex ( , )x y is pure if x and y have the same parity andimpure if x and y have opposite parity. Since, the totalnumber of vertices is odd, there must be two adjacentpure vertices P and Q or two adjacent impure verticesP and Q. But in either case the mid point of P and Qeither has both coordinates integers, which we aretold does not happen, or as both coordinates of theform an integer plus half, which therefore must occur.Hence, choice (2) is correct.

14. (5) We can exhibit a sequence with 13 terms which doesnot contain a prime: 2 101× = 202, 3 97× = 291,5 89× = 445, 7 83× = 581, 11 79× = 869, 13 73× = 949,17 71× = 1207, 19 67× = 1273, 23 61× = 1403,29 59× = 1711, 31 53× = 1643, 37 47× = 1739,

41 43× = 1763. So, certainly n ³ 14.

If there is a sequence with n ³ 14 not containing anyprimes, then since there are only 13 primes notexceeding 41, at least one member of the sequencemust have at least two prime factors exceeding 41.Hence, it must be at least 43.47 = 2021 which exceeds1995. So, n = 14 is impossible.Hence, choice (5) is correct.

15. (5) Each pair of 0, 5, 12 differ by 5, 7 or 12, sof f f( ), ( ), ( )0 5 12 must all be different, so n ³ 3.We can exhibit an f with n = 4. Define f m( ) = 1 form = 1 3 5 7 9 11, , , , , (mod 24), f m( ) = 2 for m = 2, 4, 6, 8,10, 12 (mod 24), f m( ) = 3 for m = 13, 15, 17, 19, 21, 23(mod 24), f m( ) = 4 for m = 14, 16, 18, 20, 22,0 (mod 24). Hence, n = 4.Hence, choice (5) is correct.

16. (5) Let A b c a B c a b2 2= + - = + -, , C a b c2 = + - . Then,

A B c2 2 2+ = . Also, A B= iff a b= . We have

( )A B- ³2 0, with equality iff A B= . Hence,

A B AB2 2 2+ ³ and so 2 2 2 2( ) ( )A B A B+ ³ + or

4 2c A B³ +( ) or 2 c A B³ + , with equality iff A B= .

Adding the two similar relations we get the desiredinequality, with equality iff the triangle is equilateral.Hence, choice (5) is correct.

17. (4) Let T nn n

n = + + + =+

1 21

2K

( )

Then, ST T T

nn

= + + +1 1 1

1 2

K

Therefore,1

21 1

1T m mm

= -+

æ

èçç

ö

ø÷÷

Hence,S

nn

21

1

1= -

+( )

So,1 1 1

2S

n

n

=+ /

Hence,1 1 1

1 2S S Sn

+ + +K

= ++ + + +æ

èç

ö

ø÷

1996

2

1 1 2 1 3 1 1996

2

/ / ... /

Now,

11

2

1

3

1

4

1

5

1

6

1

7

1

8

1

9

1

16+ + +

æ

èç

ö

ø÷ + + + +

æ

èç

ö

ø÷ + + +

æ

èç

ö

ø÷ +K

1

17

1

32

1

33

1

64+ +

æ

èç

ö

ø÷ + + +

æ

èç

ö

ø÷ +K ...

1

65

1

128

1

129

1

256+ +

æ

èç

ö

ø÷ + + +

æ

èç

ö

ø÷K ... +

1

257

1

512

1

513

1

1024+ +

æ

èç

ö

ø÷ + + +

æ

èç

ö

ø÷K ... >

11

2

1

2

1

2+ + + +K = 6. So,

1 1 1

1 2S S Sn

+ + +K

= + = + =1996

2

6

2998 3 1001


18. (3) We have 2 11p - = mod p for any prime p, so if we can

find h in {1, 2, ..., p - 2} for which2 2h = - mod p, then

2 2k = - mod p for any h k= mod p. Thus we find that

2 2k = - mod 5 for k = 3 mod 4, and2 2k = - mod 11 for

k = 6 mod 10. So, we might, then hope that 5 11× = 3mod 4 and 2 13 6× = mod 10. Unfortunately, it doesnot! But we try searching for more examples.The simplest would be to look at pq. Suppose first thatp and q are both odd, so that pq is odd. If k h= modp -1, then we need h to be odd (otherwise pq wouldhave to be even). So, the first step is to get a list ofprimes p with 2 2h = - mod p for some odd h p< . We

always have 2 11p - = mod p, so we sometimes have

2 11 2( )/p - = - mod p and hence 2 21 2( )/p + = - mod p. Ifp + 1

2is to be odd, then p = 1 mod 4. So, searching

such primes we find 3 mod 5, 7 mod 13, 15 mod 29, 19mod 37, 27 mod 53, 31 mod 61. We require pq to lie inthe range 100-1997, so we check5 29× (not = 3 mod 4),5 37× (not =3 mod 4),5 53× (not = 3 mod 4),5 61× (not = 3mod 4),13 29× (not = 7 mod 12),13 37× (not = 7 mod 12),13 53× (not = 7 mod 12),13 61× (not = 7 mod 12),29 37×(not = 15 mod 28), 29 53× (not = 15 mod 28), 29 61×(not = 15 mod 28), 37 53× (not = 19 mod 36). So, thatdoes not advance matters much!

2p will not work (at least with hp

=+( )1

2) because we

cannot have 21

2 1p

p

p=

+

-

( )

mod. So, we try looking at

2pq. This requires that p and q = 3 mod 4. So,

21

Crack CAT 1 Common Admission Test·searching for suitable p we find 6 mod 11, 10 mod 19,22 mod 43, 30 mod 59, 34 mod 67, 42 mod 83. So, welook at 2 11 43× × = 946, which works. Hence, the onlysuch number is 946.Hence, choice (3) is correct.Alternatively : Note that 2 divides 2 2n + for all n.

Also, 11 divides 2 2n + if and only if n º 6 (mod 10),

and 43 divides 2 2n + if and only if n º 8 (mod 14).

Since, n = = × ×946 2 11 43 satisfies both congruences,n divides 2 2n + .

Please note that, one can prove that there are infinitely

many nsuch that2 2

2

n +is an integer. Also, any such n

is even, since by a theorem ofW . Sierpinski we cannot

have2 1

2

n +as an integer unless n = 1.

Alternatively : The first step would be to choosen p= 2 , for some prime number p. Unfortunately, thiscannot work by Fermat’s little theorem. So, let us trysetting n pq= 2 , with p q, different prime numbers. We

need2 12 1pq

pq

- +as an integer and so must have

-æ

èçç

ö

ø÷÷ =

-æ

èçç

ö

ø÷÷ =

2 21

p q.

Also, using Fermat’s little theorem,2 12 1q

p

- +and

2 12 1p

q

- +. A small verification shows that q = 3 5 7, , are

not good choices, so let us try q = 11. In this case we

obtain p = 43, and so it suffices to show that2 12 1pq

pq

- +

is an integer for q = 11 and p = 43. This is immediate,since the hard work has already been completed : we

have shown that it suffices to have2 12 1q

p

- +and

2 12 1p

q

- +as integers and

-æ

èçç

ö

ø÷÷ =

-æ

èçç

ö

ø÷÷ =

2 21

p qin order to

have2 12 1pq

pq

- +an integer. But as one can easily check,

all these conditions are satisfied and the number2 11 43 946× × = is a valid answer.

Proving that it is unique is harder. The easiest way is to use acomputer to search (approx 5 min to write a Maple program orsimilar and a few seconds to run it).

19. (5) Take m n£ and the smallest possible m. Now,

( )36m n+ and ( )m n+ 36 must each be powers of 2, asper the given choices. Hence, 4 divides n and 4 divides

m. So,m n

2 2and is a smaller solution with

mm

2< .

Contradiction.Hence, choice (5) is correct.

20. (1) Let N be a positive integer satisfying the condition and

let n be the largest integer not exceeding its cube root.

If n = 7, then 3 4 5 7 420× × × = must divide N. But N

cannot exceed 8 1 5113 - = , so the largest such N

is 420.

If n ³ 8, then 3 8 5 7 840× × × = divides N, so N > =729 93.

Hence, 9 divides N, and hence 3 840 2520× = divides

N. But we show that no N > 2000 can satisfy the

condition.

Note that 2 1 3 3( )x x- > for any x > 4 . Hence, [ ]xx3

3

2>

for x > 4 . So, certainly if N > 2000, we have nN3

2> .

Now, let pk be the highest power of k which does not

exceed n. Then, pn

kk > . Hence, p p p

n N2 3 5

3

30 60> > . But

since N > 2000, we have 7 11< < n and hence

p p p2 3 5 7 11, , , , are all £ n. But 77 p p p N2 3 5 > , so N

cannot satisfy the condition. Therefore n = 420.

Now, 420 2 3 5 72= × × × , therefore number of factors of

420 2 1 1 1 1 1 1 1= + + + + =( ) ( )( )( ) 24.


21. (2) Drop a perpendicular from centre A on BC, it willbisect the chord BD.

\ BF FD= = 10

Now, since Ð = ÐBAC AFC

and so Ð = ÐFAC ABC

\ D ABC is similar to DFAC

\AC

BC

FC

AC=

Þ AC BC FC2 = ´

AC2 36 26= ´

AC = 6 26


22. (3) Let AB = 80, then BC = 60.

We know that

AB BD AD BC BD CD+ + = + +

80 60+ + = +BD AD BD

+ -( )AC AD

Þ AD = 40

\ CD = 60 (Q AC = 100)

22

E

DCB

A

F 16

20

10 10

90°x y

xy

A

B CF 2610

F

B CE

A

80

60

D

Crack CAT 1 Common Admission Test·Now, drop the twoperpendiculars DE on BC andDF on AB, as shown in thediagram.

Since, FD BC||

\AF

BF

AD

CD=

ÞAF

BF= =

40

60

2

3

Þ AF BF= =32 48and

Similarly, DE AB||

\EC

EB

DC

DA= = =

60

40

3

2

Þ EC BE= =36 24and

But BD BE ED= +2 2

BD BE BF= +2 2 ( )Q DE BF=

BD = +24 482 2 BD = 2880


23. (1) There are 103C groups of three letters to be chosen.

There are 5 ways to choose in A, 3 ways to choose a Band 2 ways to choose a C.That is there are 30 ways( )5 3 2´ ´ to choose A B, or C.

The required probability is5 3 2 5 3 2

120103

´ ´=

´ ´

C

=1

4


24. (4) Time =30

55h = ´

30

5560 min

= 32 min 43 s

\Required time = 6 : 32 : 43


25. (2) 4 5 6 7 8 6720´ ´ ´ ´ =


SECTION II

Solutions (Q. 26-28)

26. (1) Total nominees for Comedy genre =´

=28 250

10070

20% of 70 14=

Total nominees from UP =´

=46 250

100115

\Total nominees from UP who are not nominated forComedy genre = - =115 14 101Hence, choice (1) is correct.

27. (4) 50% of Animation genre nominees = 520% of Horror genre nominess = 940% of Musical genre nominees = 1460% of Sci-fi genre nominees = 930% of Comedy genre nominees = 21Total = 58Total nominees with 10-12 years experience

=´

=28 250

10070

\12 70 58( )= - nominees in the Romantic genre belongto the group of 10-12 years experience.Hence, choice (4) is correct.

28. (5) Total nominees of age group of25 30250 20

10050- =

´=

\25 nominees belong to Horror, Musical, Romanticand Comedy genre and rest 25 nominees belong toAnimation and Sci-fi genre.But we know that in Animation and Sci-fi altogetherthere are 10% (25 nominees) of the total 250.

So the remaining 25 nominees from the age group of25-30 will be nominated for the 25 nominations ofAnimation and Sci-fi.Hence, all the nominations of Sci-fi genre belong tothe age group of 25-30 years.

So, the answer should be 100%.


Solutions (Q. 29-31)In the context of Lucknow

B2 is already engaged on Wednesday (in Barabanki).

Also, B2 is not entitled to protect JDU. Therefore,

B2 will be protecting some rally on Monday in

Lucknow.

In the context of Sitapur

B4 can protect either SP rally or BSP rally. But B4 is

already engaged on Thursday, so B4 has to protect SP

rally on Wednesdey in Sitapur.

Again in Sitapur, on Thursday there will be either

B2 or B3 on duty and consequently the rally can be

held either by BJP, or Congress or SP.

But since SP and Congress are holding their rallies on

Wednesday and Friday respectively, so BJP is the only

party which will hold rally on Thursday and hence

the protection will be provided by B2 (not by B3).

Therefore,

Congress B¬® 3

BJP B¬® 2

SP B¬® 4

JDU B¬® 1

BSP B¬® 5

In the context of BarabankiB3 can’t be there on Tuesday as B3 is already engaged

in Lucknow.

B3 can’t be there on Wednesday as B2 is there.

B3 can’t be there on Thursday as JDU (ie, B / B1 5) is

there.

B3 can’t be there on Friday as B3 is already engaged in

Sitapur.

23

CB

D

A

E

CB

F D

A

Crack CAT 1 Common Admission Test·Therefore, B3 will be protecting SP on Monday.

Now, since Congress rally is protected by either B2 or

B3, but B3 is already engaged with SP rally, so only B2

is there to protect Congress.

Similarly, BJP rally can be protected by B1 as B2 is

already engaged with Congress rally. Further, since B1

cannot be engaged on Tuesday in Barabanki, so B1 will

protect BJP rally on Friday.

Now, we know that since JDU rally can be protected

only by either B1 or B5. But B1 is already engaged with

BJP, so JDU will be protected by B5.

Hence, the remaining combination of party and

battalion is BSP and B4 will fit in on Tuesday.

Therefore,

SP B¬® 3

BSP B¬® 4

Congress B¬® 2

JDU B¬® 5

BJP B¬® 1

In the context of LucknowSince, B1 is already engaged on Friday with BJP rally

(in Barabanki), so JDU can be protected by B5.

Hence, B1 will be there on Wednesday.

Since, B5 protects JDU, so BSP will be protected by B4.

Since, B4 protects BSP, so SP will be protected by B3.

Since, B3 protects SP, so Congress will be protected by

B2.

Hence, B1 will protect the remaining party-BJP on

Wednesday.

Rally Day

Lucknow Barabanki Sitapur

Rally ofthe Party

Battali

onNumber

Rallyof

theParty

BattalionNumber

Rallyof theParty

BattalionNumber

Monday Congress B2 SP B3 BSP B5

Tuesday SP B3 BSP B4 JDU B1

Wednesday BJP B1 Congress

B2 SP B4

Thursday BSP B4 JDU B5 BJP B2

Friday JDU B5 BJP B1 Congress

B3

29. Choice (2)

30. Choice (1)

31. Choice (2)

Solutions (Q. 32 and 33)

32. Choice (1)

33. Choice (2)


Case I

StarZeeTV

NDTV UTV SonyCartoon

Total

Monday 1 2 3 2 1 1 10

Tuesday 1 2 1 2 2 2 10

Wednesday 4 1 1 0 4 0 10

Thursday 2 0 1 2 1 4 10

Friday 1 2 1 1 2 3 10

Saturday 1 3 3 3 0 0 10

Total 10 10 10 10 10 10

Case II

StarZeeTV


Total

Monday 1 2 3 2 1 1 10

Tuesday 1 2 1 2 2 2 10

Wednesday 4 1 1 0 4 0 10

Thursday 2 0 1 1 2 4 10

Friday 1 2 1 2 1 3 10

Saturday 1 3 3 3 0 0 10

Total 10 10 10 10 10 10

StarZeeTV


Total

Monday 1 2 3 1 2 1 10

Tuesday 1 2 1 2 2 2 10

Wednesday 4 1 1 0 4 0 10

Thursday 2 0 1 2 1 4 10

Friday 1 2 1 2 1 3 10

Saturday 1 3 3 3 0 0 10

Total 10 10 10 10 10 10

34. (4) Number of Star channels on Friday = 1Number of Sony channels on Monday = 1 or 2\Difference = 0 or 1Hence, choice (4) is correct.

35. (1) Number of Sony channels on Wednesday = 4

Number of Cartoon channels on Tuesday and Friday

= + =2 3 5\Required difference = - =5 4 1Hence, choice (1) is correct.

36. (2) Number of Star channels on Wednesday = 4Number of Zee TV channels on Monday = 2\Required answer = + =4 2 6Hence, choice (2) is correct.

37. (4) Zee TV channels on Friday are greater than Star Pluschannels on Tuesday.So, only statement 4 is true.Hence, choice (4) is correct.

38. Choice (2)

24

Crack CAT 1 Common Admission Test·Solutions (Q. 39-42)

39. Choice (2)

40. Choice (1)

41. Choice (4)

42. Choice (2)


Hint : UCM, MSI and MoS carry 3, 2 and 1 points respectively.

43. Choice (1)

44. Choice (3)

45. Choice (5)


The given data can be represented in the table form as :

CompanyClass

TotalB C D E

Honda 0 4

Maruti 1 8

Tata 1 1 2 1 5

GM 1 1 1 1 4

Total 3 6 6 6

46. Choice (1) 47. Choice (2) 48. Choice (3)

49. Choice (1) 50. Choice (2)

SECTION III


51. (1) BE is a very strong link. EA is also a possible link.Hence, choice (1) is correct.

52. (1) AD is a very clear link. However the para (paragraph)can’t start with sentence B.Hence, choice (1) is correct.

53. (5) DBE is a very important link. Also, AC link talks aboutmore number of such historical moments or events.Hence, choice (5) is correct.

54. (3) BA is a very obvious link. Besides DCE is also asignificant link.Hence, choice (3) is correct.

55. (2) Finding a link EAB is very easy.Hence, choice (2) is correct.


56. Choice (4)

57. Choice (3)

58. Choice (3)

59. Choice (4)

60. Choice (4)


61. Choice (2). In the first para, the author refers to expedition

in the devil’s sea as being inspired by “unexplained

losses in the Japanese equivalent ….”.

62. Choice (3). Refer to para 5, first line –“…some unusual

speculations.”

63. Choice (4). Refer to para 5, “… of appalling explosive

character when…..”.

64. Choice (3). Refer to para 8—The areas are located between

36° north and 36° south latitude. Also, refer last para of

the sentence “…. including the north and the south

poles”.

Confusion might arise from the fact that the sentence stated “moreprecisely at 36° north..”again ; note that this refers to the area of 36°north to 36° south where 10 anomalies are located. It does not pointthe exact location of an anomaly.

65. Choice (2). Refer to last para choice (2) is not mentioned.

66. Choice (4). Refer last sentence—the idea of unrecordedwinds is not taken to apply to the planes crossing theBermuda triangle. Rather they are attributed withmysterious causes associated with anomalies - in thecontext of the passage the cause mentioned is a timewarp.

67. Choice (2).The passage does not restrict itself only to bethe Bermuda Triangle-it looks at different such areas andhypotheses or possible reasons.

68. Choice (3). Refer to para 3-“… flows from the highwildness of the Cascade Range……..”.

69. Choice (2). Local soil refers to a type of soil rich in clay,sand slit.

70. Choice (4). Here, choice (4) is not mentioned in thepassage.

71. Choice (1). The valley and its features have beencompared to a cathedral and its characteristics to conveythe sense of awe and grandeur associated with cathedral.

72. Choice (2). Refer to para 4.(2) is not mentioned .

73. Choice (3). Refer to para 5 and para 6. The questionsreveal that there are too many difficult choices to be madewith no clear answer.

74. Choice (1). The dam would in fact, prevent flooding.

75. Choice (3). Choice (1) does not follow as the Americanformers are all for it. No information relating to (2) hasbeen provided . As for (4), raising and strengthening dikesis an option to constructing dams. Only (3) is valid as caninferred from statement that “they’d be kept free of damsif they became scenic rivers”.

25

Crack · 2016-08-04 · 22. The lengths of the sides of triangle ABC are 60 80, and 100 with Ð =...

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