Unproctored Mock Cat 14 Solutions

8/11/2019 Unproctored Mock Cat 14 Solutions

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Page1 Unproctored - Mock CAT 14

1 c 2 d 3 b 4 d 5 b 6 a 7 d 8 d 9 c 10 d

11 d 12 d 13 c 14 a 15 b 16 b 17 c 18 d 19 d 20 a

21 a 22 d 23 a 24 a 25 a 26 b 27 b 28 c 29 d 30 d

31 c 32 d 33 a 34 c 35 a 36 a 37 d 38 a 39 d 40 c

41 b 42 d 43 c 44 c 45 a 46 c 47 d 48 d 49 a 50 c

51 a 52 c 53 a 54 d 55 b 56 d 57 c 58 d 59 d 60 d


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1. c Option (c) is correct, Penelopes web means An endlessjob. All the other options are incorrect.

2. d Option (d) is correct, fair and square means honest andproper, all the other options are incorrect.

3. b In course of means while something is going on, (a) is incorrectas it means at a suitable time, (c) is incorrect, as it means

because. (d) means because of and (e) is grammaticallyincorrect.

4. d It is evident from the first paragraph that both options (a) and(c) contribute towards sensuality.

5. b depend on the manipulation of a limited range of forms, theauthor seems to be scornful of this facet of Handels style.Refer to the end of the 1st paragraph, which leads to option(b). Options (a) and (c) are not limitations. Option (d) is talkingabout peoples acceptance of Handels music but does notdirectly talk about the limitation in Handels music.

6. a The author ostensibly seems to be praising Handel, but in thepassage he is being critical of Handels popularity and the

limited scope of his skill. Since the tone is not extreme in thepassage - Sardonic is out. Critical suits the tone of the passage.Options (b) and (c) do not characterize the passage.

7. d A is incorrect because the correct phrase is came and went(the tense also should be the same), C is incorrect becausethe indefinite article a is missing before brief appearance.

8. d A is incorrect as the correct pronoun usage is at instead offor. C has an error of subject verb agreement, should be areinstead of is since the subject is games which is in theplural. D is incorrect as the definite article the should precedesneaker world as we are referring to a particular area ofbusiness.

9. c The lines Macondo oozes, reeks and burns even when it is

most tantalizing and entertaining. It is a place flooded with liesand liars and yet it spills over with reality. Lovers in thisnovelfront porch, clearly show how Marquez has useddreamlike or fantastic characters in an enchanted place whichis not distanced from earth or reality. Even the charactersconfirm to this fusion. This leads us to option (c). Option (a) isincorrect as the author does not talk of the narrative in thiscontext. Option (b) cannot be inferred from the passage.Option (d) becomes too specific in earthly pains. The authoris talking of reality and not just pains.

10. d A is clearly mentioned in the last two paragraphs. B isnegated by the author in the last paragraph, in the followingwords- absurdly complicated, labored and almost impossibleto read. In fact, it is none of these things. C is incorrect as

fantastic journey is implied by a quixotic expedition. Indeed,here the author mentions that the goal is not reached but thatdoes not imply a journey devoid of any goal.

11. d The author praises Marquezs style of story-telling and hisportrayal of characters and settings in the novel. In the thirdparagraph, the author also expresses admiration for thelanguage used by Marquez. Thus, it is clear from the passagethat the authors tone is laudatory. There is no indication offlattery in the authors tone, which makes adulatory incorrect.Caviling, which means to raise trivial and frivolous objections,would not fit in here. Analytical is out of context as the authoris not providing an analysis of Marquezs novel; he is justpraising his style of story-telling.

12. d Option (a) introduces a may whereas the paragraph isassertive that Raga is inseparably linked to Rasa. Option (b)leaves out Raga altogether. Option (c) is incomplete in itsdescription of Raga as also the purpose. Option (d) gives areasonably correct summary of the paragraph.

13. c Option (a) introduces a new concept foreground. Also itdoes not talk about the human hell. Option (b) leaves out the

possible purpose of Liardet. Option (d) is incorrect and readstoo much into the paragraph. Option (c) gives a correct andconcise summary of the paragraph.

14. a The crux of Para 1 lies in understanding how we havebecome what we are rather than importing our prejudices onto the past in the guise of their being eternal truths apprehendedby a supra-historical intellect . This means actively probing orstudying the factors which have led to our present state,rather than seeing the whole thing through our prejudices orfrom a third-party stance as one does while narrating historyor studying history. Option (a) hits the crux of the paragraph.Option (b) is going against what has been said. Foucault doesnot want a prejudiced approach. Seeing everything as anatural oucome is a prejudiced approach. Option (c) is too

specific to encompass what is being implied. Option (d) isvague and distorted. Foucault has not indicated any suchthing.

15. b The author is predominantly descriptive in the passage. He

describes what biopower (and also disciplinary technologies)seeks to achieve. He does not resort to arguments or counter-arguments. Foucault does argue to some extent in the firstparagraph. The author is also not exhorting anyone to doanything. Foucault has an exhorting tone, but not the author.The author does not appear to be exposing something orsomeone.

16. b Refer to the line Biopower is thus tied to the emergence ofthe discipline of statistical demography, and there begins thequantification of the phenomena of birth-rate, longevity, the

reproductive rates and fertility of a given population, its stateof health, patterns of diet and habitation.; and Biopower, onthe other hand, focuses on the body as the vehicle of specieslife. Given the nature of the phenomena with which it isconcerned it is regulatory rather than disciplinary.Options (a), (c) and (d) deal with regulatory measures which affect thepopulation as a whole. The measures are in response topopulation statistics. Option (b) is a disciplinary measure whichis not connected to a population statistic like mortality/longevityetc. Hence only option (b) is not an example of the use/ outcomeof biopower.

17. c Statement A uses the word utilizing for biopower which is

not appropriate. The philosophy of biopower is not to utilizebut to regulate to positively affect life as a whole. Also, even

disciplinary technologies can be applied to utilize an entirepopulation. B and C are correct in the illustration of thedifference between the disciplinary technologies andbiopower.

18. d Option (a) is incorrect because it does not have any impact onthe argument. It just suggests that these developments happenat an early stage in life which has already been stated in theargument.Option (b) is incorrect as it does not highlight the flaw in thereasoning. It adds a parallel dimension to the situation given inthe paragraph.Option (c) elaborates on the need for social skills and theirimpact on the childs personality. It does not impact the argumentin any other way and hence it is not the best choice.


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Option (d) is correct as the type of reasoning used in thequestion is: for Y to happen, X has to take place. If we canprove that X does not guarantee the occurrence of Y, it willhelp us to weaken the reasoning. Option (d) states that thosestudents who have an interest in music are good at adaptingto the environment and not vice-versa. This brings out theflaw in the reasoning.

19. d Mr. A is terminally ill and is going through insufferable pain. Thecourt will turn down the plea of Mr. A only when the court isconvinced that he is not terminally ill. The court is consideringthe recommendations of the group of doctors attending to Mr.A but there some doctors from the fraternity who believe thathe has not been treated properly for the disease. This canraise doubts about the deteriorating condition of Mr. A; hencethe court can reject the plea for euthanasia. Options (b) and(c) do not hold true as Mr. A is considered to be terminally illand these go against the basic premises of the case. Option(a) is beyond the scope of the argument.

20. a Option (a) is the best choice because it is within the purviewof the argument. It is clear that the government plans to takethese steps in the hope of meeting the requirement for parking

space in the city. This can be inferred from the given argumentas it is in sync with the governments intentions presentedabove.Option (b) is incorrect because it suggests that diversion oftraffic from neighbouring areas creates further problem;however, this cannot be justified from the argument.Option (c) cannot be inferred from the argument because theargument pertains to the problem of parking space in publicareas within the city and not residential areas. Hence, this isan incorrect inference.Option (d) is incorrect because it cannot be inferred from theargument. The argument does not refer to the problems relatedto following traffic rules. The behaviour of the drivers on theroad is also not discussed in the argument which is why noinference can be drawn on this basis.

21. a If a sum S can be paid using coins of denomination 5, 7 and 11then the sum S + 5N (where N is a natural number) can alsobe paid.Let us consider any sum between Re.1 and Rs.15. The sumthat cannot be paid are (by observation)1, 2, 3, 4, 6, 8, 9 and 13.Clearly, the sum (in Rs.) 5(5 1), 7(7 1), 10(5 2)11(11 1), 12(5 1 + 7 1), 14(7 2) and 15 (5 3) can bepaid.So any sum of the type 5N, 5N + 1, 5N + 2 and 5N + 4 can be

paid if N 2 .We can also see that the sum of 18 (11 1 + 7 1) can be paid(in fact, this is the first sum of the form 5N + 3 which can be

paid). So any sum of type 5N + 3 can be paid if N 3

Largest sum that can never be paid = 13.

22. d n

m aa

nlog b log b

m=

+ + + +3 9 27 n3

1 1 1 1........

log 9 log 9 log 9 log 9

1 1 3 n......

2 1 2 2= + + + +

+ + + +1 2 3 n

.....2 2 2 2

n(n 1)

4

+=

23. a

A B

CD

O

a

b

h2

h1

Let AB = b; CD = a.Let the perpendiculars from O to AB and CD be h1, h2respectively.Triangles AOB and COD are similar.Since the ratio of their areas is 1 : 4; the ratio of the proportionalsides and the heights h2, h1must be 1 : 2.

Area of the trapezium =1

2(a + b) (h1+ h2)

=1

2(3a) (3h2)

= 9 area of the triangle COD = 90 sq cm.

24. a (ab cd) (1 6)ad = = = 121bc

2

12

(bc de) 12be = = =

cd 6 6

16

(cd ef) 32cf = = =

de 2 2

1 3ad : be : cf = 12 : : = 72 : 1 : 9

6 2

Alternative Method:Let f = 1, then finding values subsequently,

1 3 1e = , d = 4, c = , b = , a = 3

2 2 3

Value of ad : be : cf = 72 : 1 : 9

25. a f(4) = 16a + 4b + cFurther, it is known that f(4) = 100.Hence, 16a + 4b + c = 100.We can observe that the value of a + b + c will be maximumwhen a = 1, b = 1 and c = 80, but it is given in the question thata, b, c are all distinct. So, a = 1, b = 2 and c = 76 will give themaximum value.So, for the maximum value, a + b + c = 1 + 2 + 76 = 79.

26. b Since there are 10 percent fewer B than A,

A :B 10 : 9 =

Number of fishes of type B =10

144 1609

=

Since there are 25 percent more A than C,A : C = 5 : 4

or

Number of fishes of type C4

=160 1285

=


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Since there are (160 + 144 + 128) = 432 fishes which are A,B, or C type, and these account for 80 percent of fishes in the

pond, the pond must have100

432 = 540 fishes.80

27. b P Q

RSM N

O

T

SPM 90 OPQ 30 = =

MPT SPT SPM 60 30 30 = = =

TPQ OPQ MPT 60 30 30 = = =

( ) ( )PTQ 180 TPQ PQT 180 30 60 90 = + = + =

PT QR 3In PTQ : sin60

PQ PQ 2 = = =

2Required ratio

3 =

28. c Out of the given options only 16 cannot be expressed as asum of 3 distinct composite numbers.21 = 4 + 8 + 9 and 23 = 6 + 8 + 9

29. d 2(x 1)log (2 x) 1

+ +

Now,1 5 1 5

x or, x2 2

+ > 1.618 or, x < 0.618.As we have to get the smallest integral value of x, it can be1(1 < 0.618).x cannot be less than1 as then 2 + x becomes negative andlog is not defined for negative numbers.

Note:

If logba < c then a < bc(if b > 1) and a > bc(if 0 < b < 1).

Alternative method:

2(x 1)log (2 x) 1

+ + 0 (as logba is defined for a > 0

1b > 0 and b 1 )

or, x >2Putting x =1 in the inequality: log

21 < 1 or, 0 < 1, which is true.

The smallest integer satisfying the inequality is 1.

30. d 8

4

4

Perimeter of one of the parts that was cut= 4 2 + 2 4 = 16 cm

31. c Ravis average score in the examination

(20 + 40 + 60 + 80 + 100)= = 60

5

As average marks for the 3 theorytests= average marks for the 2 practical tests, so each one ofthem should be equal to 60.Marks in practicalcan only be either 40 and 80or 20 and100.Subsequently marks in theory will be 20, 60 and 100 or40, 60 and 80.Maximum marks for Theory and minimum marks for Practicalare either 100 and 40 or 80 and 20(in that order).Clearly the overall weighted average will be more for the firstone.Overall weighted average for Ravi

20 + 60 + 80 + 2 (100 + 40)= = 62.86 (approx).

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32. d Let the number of people that came to the party be x.Since, exactly six people came in each of the three cars andexactly seven people in each of the remaining cars, thereforex = 7n + 18. (total number of cars = n + 3).

Also, at the party, when every person including Tarun wasseated in a group of eleven, one group fell short by fourpeople, hence x + 1 = 11m + 7 x = 11m + 6.Therefore, 7n + 18 = 11m + 6 11m7n = 12.Since, the value of x is maximum, therefore the maximumvalue of mwill be 66. (As x = 11m + 6 and x < 750)Therefore, the value of x= 11 66 + 6 = 732.Hence, the number of people who were invited by Tarun anddid not come to the party is 750732 = 18.

33. a = + = + 4 6 8 6 6N 4 6 2 2 3

( ) = + = 6 2 6 6N 2 2 3 2 733Now, 733 is a prime number, therefore the number of factors

of the number N = 7 2 = 14.

34. c (xy)z= 26= 43= 82= 641

Whenz = 1, xy = 64. The total number of factors of 64 is 7.Possible sets of solution of (x, y) are (1,64) (64,1) (2,32),(32,2) (4,16) (16,4) (8,8)Similarly,z = 2, xy = 8. The total number of factors of 8 is 4.z = 3, xy = 4. The total number of factors of 4 is 3.z = 6, xy = 2. the total number of factors of 2 is 2Hence, total number of positive integral solutions is 16.


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35. a Let Obe the center of the circle.

AOB = 2 ACB 2 30 60 . = =

OAD OBD 90 = = (as radius of a cirlce is alwaysperpendicular to the tangent at the point of contact).In quadilateral AOBD,

ADB 360 ( OAD OBD AOB) = + +

= 360(90+ 90+ 60) = 120.

=

ADB4

ACB

R

A

B

C O30

60 D

Also, since AC = CB and AD = DB, therefore the bisector ofthe angles ACB and ADB will pass through the center of thecircle.Let R be a point on the circle and lying on the angular bisector

of ACB and ADB ,

1 1RAB ( ROB) 30 15

2 2 = = =

1( DAB)

2=

RA is the angular bisector of the DAB R is the incenter of the ABD CR = 4 cm.

For question 36:

RP Q8 km 2 km

S T U

A(3v) B(v)

Let the speeds of A and B be 3vand vkm/hr.

Let the speed of the river by ukm/hr.Let R be the point where the floating cork was initially located.Let PR, RS and TQ be x, y and z kms respectively.Let U be the point where A and B cross each other.(Note: Position of U must be as shown in figure above)

Time taken by P to reach S = Time taken by the floating cork toreach S

x y y x 3v

3v u u y u

+ = =

+ ...(i)

Time taken by B to reach T = Time taken by the floating cork toreach T

y 8 z v z1

u v u u y 8

+ = = + ...(ii)

Time taken by A to reach U = Time taken by B to reach U

x y 10 z 2

3v u v u

+ + =

+ ....(iii)

Getting x in terms of yfrom (i) and z in terms of y from (ii) andputting that in (iii) we get -

3v u v u vy 10 (y 8) 1 2

u 3v u u

+ + = + +

v u 10 (v u) (v u) vy y 8 1 2

u 3v u u u

+ = + +

10v 10u 8v 10u

3v u u

=

+

10 uv10u2= 24v230 uv + 8uv10u2

24 v2= 32 uv

v 4.

u 3 = (as v 0)

36. a Relative Speed of A when A is moving upstream= 3vu.Relative speed of B when B is moving downstream= v + u.

Required ratio(3v u)

9 :7 .(v u)

= =

+

37. d If all the 11 alphabets had been different then there wouldhave been 6 x 5 or 30 terms.But (c+ d + e + f) is common in both the expressions. Whenthese common expressions are multiplied we do not get 16terms, but 6 terms less that is 10 terms.Hence the total number of terms = 306 = 24

Alternative method:

(a + b + c + d + e + f) (c + d + e + f + g)

2(a b)(c d e f) (c d e f ) (a b c d e f )g

I II III

+ + + + + + + + + + + += + +

Now, part I of this expression will give 4 2 = 8 terms

part II will give 5 3C 10= terms, and part III will give 1

6 = 6 terms.

Total number of terms = 8 + 10 + 6 = 24.

38. a A S B

U

CRD

P QM

NT

Let, AB = a units

SB = PQ2a

units3

=

2a aPM a units

3 3 = =

2a a a

MQ PQ PM units3 3 3 = = =Also, each of ASMP, MQNT and NURC are squares.

Area of shaded region2 2 2

2 2a a 2aa 3 a9 3 3

= = =

Required ratio = 2:3


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Alternate method:

If we redraw the figure,

A S B

UN

M

CR

D

P Q

T

We can see 9 identical smaller squaresRequired ratio = 6:9 = 2:3

39. d Consider 5 working days and 1 day off as one cycle for MayaSimilarly, 10 working days and 1 day off as one cycle ofBhanu.LCM(6, 11) = 66Hence the 66th working day for the office after Jan 1st wouldbe an off for both Maya and Bhanu. The 66 th working daywould be exactly at the end of 11 weeks or 77 days after Jan1st. So the day would be March 19th, which would again be a

Thursday.

40. c S1, S

2, S

3and S

4are (a + b + c), (b + c + d), (a + d + c) and

(a + b + d). As all the sums are positive, using the AM GMproperty, we get,

1/3a b c (abc)3

+ + ...(i)

1/3b c d (bcd)3

+ + ...(ii)

1/3c d a (cda)3

+ + ...(iii)

1/3a d b (adb)3

+ + ...(iv)

Multiplying (i), (ii), (iii) and (iv), we get

(a + b + c)(b + c + d)(c + d + a)(a + d + b) 34(abcd)S

1S

2S

3S

4405 .

For questions 41 to 43:From (III) and (I) we can conclude that Akash is second tallest amongthe five people as he has a car and the tallest person can not have acar.From (II) and (IV) we can conclude that one of the brothers is thesecond tallest and the other is the second shortest among the fivepeople.From (V) we can conclude that Anup is the second shortest amongthe five people as by (I) we know that the tallest person doest nothave a scooter.

From the drawn conclusions we get that (Tallest = 1, shortest = 5)

Height Vehicle Person

1

2 Car Akash

3 Ajay

4 Scooter Anup

5

41. b Akash and Anup are brothers

42. d If Arjun is the shortest among the given five people and has abus, then Anand is the tallest person and he could have either

a bicycle or a truck. The new table looks like

Height Vehicle Person

1 Truck/Bicycle Anand

2 Car Akash

3 Truck/Bicycle Ajay

4 Scooter Anup

5 Bus Arjun

43. c Statement A:

Anand is the shortest among the five people and has a truck.So, Arjun is the tallest among the five people and since Ajaydoes not have a bus, Arjun must have a bus and Ajay abicycle.So, the information given in statement (A) is sufficient.

Statement B:Ajay has a truck and Anand is the tallest, means Arjun is the

shortest. So Anand will have a bicycle and Arjun will have abus. So, the information given in statement (B) is sufficient.

Statement C:

Anand is the tallest among the five people and has a truck.So, Arjun is the shortest among the five people and has a busand Ajay has a bicycle.So, the information given in statement (C) is sufficient.

Statement D:

Arjun is not the tallest among the five people and does nothave a bus, which means Anand is the tallest person.

Anand has a bus, Arjun has a truck and Ajay has a bicycle.So, the information given in statement (D) is sufficient.So, all the four statements are independently sufficient to

determine the owner of each vehicle.

44. c Let the angles of the cyclic quadrilateral be 2x, 3x, yand2yrespectively.The following are the cases that are possible.

Case 1:

2x + 3x = 180 and y + 2y = 180The measure of the four angles is 60, 72, 120 and 108 degrees.

Case 2:

2x + 2y = 3x + y = 180The measure of the four angles is 90, 90, 135 and 45 degrees.

Using Statement I:

Given that the measure of the two angles of the quadrilateral

is the same, which means Case 2 is the only possibility. Hence,statement I alone is sufficient to answer the question.Using Statement II:

Given that the difference between the smallest and the largestangle of the quadrilateral is greater than 60 degrees, whichmeans that only Case 2 is possible.

Hence, Statement II alone is also sufficient to answer thequestion


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45. a Let P = ABCPM = (100 A + 10B + C)(100C + 10B + A)PM = 99(AC)Since, the value of PM is a two-digit number, therefore thevalue of (AC) is equal to 1.Since, C is an odd number, the possible values of A and C intheir particular order are as follows:(8, 7); (6, 5); (4, 3); (2, 1)

Using Statement I:

P > 890Since, the tens digit of P is greater than the unit s digit of P, theonly possible value of P is 897.Hence, statement I alone is sufficient to answer the question.

Using Statement II:

P < 351There are more than one value of P satisfying the above giveninequality. For example P = 221, 231 and 241.Hence, statement II alone is not sufficient to answer thequestion.

For questions 46 to 48:

The total points given by the judges J1, J2, J3, J4and J5 is 25, 20, 30, 20and 20 respectively. Please bear in the mind that judges awardeddistinct point to the candidates. The range of the possible points earnedby the five contestants is given in the following table.

Judge Samir Milind Ranvir Sahil Arjun

J1 9 6 51 or 2 or 3

or 4

1 or 2 or 3

or 4

J2 4 0 or 2 0 or 2 8 6

J31 or 2 or 4

or 511

1 or 2 or 4

or 56 7

J4 7 0 or 3 0 or 3 8 2

J5 3 5 8 0 or 4 0 or 4

46. c Minimum possible points earned by Sahil = 1 + 8 + 6 + 8 + 0= 23.

47. d Given that the judge J2and J

4gave the same points to Milind.

This is only possible if both the judges gave 0 points to Milindbecause if Milind got 1 point from the judge J

2, then Ranvir will

also get 1 point from the judge J2and this is not possible.

Given that Samir got 4 points from judge J3, which means that

Ranvir got 64 = 2 points from judge J3.

The total points earned by Ranvir = 5 + 2 + 2 + 3 + 8 = 20.

48. d The range of the total points earned by all the five contestantsis given below.Samir: 24 - 28Milind: 22 - 27

Ranvir: 14 - 23Sahil: 23 - 30Arjun: 16 - 23Therefore, both Ranvir and Arjun will definitely not earn morepoints than Sahil.

49. a As per the given information we can say thatP can be married to S, T or UQ can be married to T or UR can be married to U onlyTherefore R will be married to U, Q will be married to T and Pwill be married to S.

50. c If Yasir occupies seat number 1, then the serial number of theseat on which the remaining persons can be seated are asfollows:

Persons Numbers of the seats

Arafat 2 or 3

Rasheed 2 or 3 or 4

Ali 4 or 5

Rehman 5 or 6

Case I:Seat number 2 is unoccupied.Arafat will occupy seat number 3, Rasheed will occupy seatnumber 4 and similarly the serial number of the seat occupiedby Ali and Rehman is 5 and 6 respectively.Total number of ways in which they can be seated = 1.

Case II: Seat number 4 is unoccupied.Ali and Rehman will sit on seat number 5 and 6 respectivelyand Arafat can sit on either of the seats numbered 2 or 3.Total number of ways in which they can be seated = 2.Hence, total number of ways in which the mentioned 5 personscan be seated = 1 + 2 = 3.

51. aIncome Expenditure Savings

A 2Z 2Z20000 Rs. 20000

R Z Rs. 20000 Z20000

S 5Z 5Z20000 Rs. 20000

Using Statement I:

Given that the expenditure of S is less than thrice the expen-diture of A.5Z20000 < 3(2Z20000)Z > 40000Hence, statement I alone is sufficient to answer the question.Using Statement II:Given that twice the expenditure of A is more than 5 times the

savings of R.2(2Z20000) > 5(Z 20000)Z < 60000.Hence, statement II alone is not sufficient to answer thequestion..

52. c b2+ 2ab48a2= (b6a)(b + 8a)

Statement A:

As ax2+ bx + 4a > 0 for all values of x, we must have a > 0 and

Discriminant 0. 0 and b216a2< 0or a > 0 and (b4a)(b + 4a) < 0or (b4a) < 0 and (b + 4a) > 0

(b6a) < 0 and (b + 8a) > 0 b2+ 2ab48a2 < 0Hence (A) alone is sufficient to answer the question and theanswer is no.

Statement B:

( )2As 2ax 4b x 3 0 for only one real x, discriminant 0.+ + = =

2Discriminant ( 4b ) 4(2a)(3) 0 = =

Discriminant (b 6a) 0 = =

2 2b 2ab 48a (b 6a)(b 8a) 0 + = + =Hence (B) alone is also sufficient to answer the question andthe answer is no.


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From the two bar graphs, following data can be derived.

PeriodNumber of

Customers

Percentage of

Customers Who

went Abroad

Number of

Customers

Who Went

Abroad

January April 120 40% 48

February-May 200 25% 50

March-June 170 40% 68

April-July 180 20% 36

May-August 250 10% 25

The following representation will help in breaking the data month wise.Number of Customers Who Went Abroad:

January February March April May June July August

68

48

50

36

25

Total number of Customers:


120 250

53. a Total number of customer till August = 120 + 250 = 370.Total Number of customers who went abroad, till August= 48 + 25 = 73

The required percentage73

19.73%370

= =

54. d As March to June = 68

(January + February + July + August)= Total Customers(March to June)= 7368 = 5April to July = 36May to August = 25

Assuming that no customer went abroad in the month ofAugust, i.e. taking August = 0, we can get the minimum numberof customer that could have gone abroad in April= 3625 = 11

For the case of a maximum number of customers going abroadin the month of March, we must have:January + February = 0

March + April = 48(January + February) = 48(March)max = 48 (April)min= 4811 = 37As February to May = 50, now we get May = 2.As February to June = 68, June = 68 (37 + 11 + 2) = 18.For each month, we can now determine the number ofcustomers who went abroad:


0 0 37 11 2 18 5 0

55. b May to August = 25.For a maximum number in June, July + August = 0

May + June = 25 and,January + Feb = 5 and,March + April = 485 = 43As, February + March + April + May = 50OrFebruary + 43 + May = 50

Feb + May = 7(May)min = 7 (February)max= 75 = 2

(June)max= 25(May)min= 23The number of customers in different months is given below:


0 5 2 23 0 043

For questions 56 and 57:

The total worth of coins originally in the bag = Rs.148 + Rs.212= Rs.360Let a, b and c denote the number of coins of denomination Re.1, Rs.2and Rs.5 respectively.

Therefore, a + b + c = 100 and a + 2b + 5c = 360 and 10 a,b,c 60 Eliminating C, we get: 4a + 3b = 140.Possible values of a, b and c in that particular order are(20, 20 and 60); (17, 24 and 59); (14, 28, 58) and(11, 32 and 57)

Similarly lets assume that x, y and z denote the total number of coins

of denomination Re. 1, Rs. 2 and Rs. 5 taken out by Urvashi respectively.

Possible values of x, y and z in that particular order are(10, 4, 26); (7, 8, 25); (4, 12, 24) and (1, 16, 23).

Total number of coins of denomination Re. 1 that remain in the bagcould be either one of these 7 values namely 1, 4, 7, 10, 13, 16 and 19respectively.

56. d 13 is a possible number of coins of denomination Re.1 thatremain in the bag.

57. c If total number of coins of denomination Rs. 5 that Urvashitook out is 26, then possible number of coins of denominationRs. 2 that remain in the bag could be either one of these 4values namely 16, 20, 24 and 28.

Hence option (c) is the correct choice.


9/9



Let the consumption of High Quality and Low Quality rice by theresidents of the city X in the year 2002 be Aand Brespectively. It isgiven that A > B.

2002 2003 2004 2005 2006 2007 2008

A 1.2A 0.3A 0.15A 0.18A 0.225A 0.09A

B 0.8B 0.6B 0.84B 0.21B 0.126B 0.189B

58. d In the year 2007, the ratio of the consumption of High Qualityrice to Low Qualityrice by the residents of city Xwas morethan that in the year 2002.

59. d As A and B are positive quantities, it is very clear that theconsumption of High Qualityrice is maximum in the year 2003and the consumption of Low Quality rice is 2nd highest in2003 (it is highest in the year 2005).Total consumption of rice in the year 2003 is definitely morethan that of each of the years 2004, 2006, 2007 and 2008. Weneed to compare the data for the years 2003 and 2005.Let the total consumption of rice in the years 2003 and 2005

be x and y respectively.We know that x=1.2A + 0.8B and y = 0.15A + 0.84B.xy = 1.05A0.04B.Now, As A > B, 1.05A > 1.05B > 0.04B

1.05A > 0.04B or 1.05A0.04B > 0 xy > 0 or, x > y. The total consumption of rice in the year 2003 is the maximumin the period 2003-2008.

For question 60:

Total consumption of rice by the residents of city X in 2004= 0.3A + 0.6BTotal consumption of rice by the residents of city X in 2006= 0.18A + 0.21B

118A 21B (30A 60B)

2 + +

6A 18B A 3B

60. d Except for the year 2005, in each of the years in the period20032008, the consumption High Qualityrice is definitelymore than that of Low Qualityrice.

Unproctored Mock Cat 14 Solutions

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