Mock CAT – 5

Answers and Explanations

1 4 2 5 3 2 4 1 5 2 6 2 7 5 8 5 9 3 10 2

11 1 12 2 13 1 14 2 15 5 16 4 17 3 18 3 19 4 20 4

21 3 22 4 23 2 24 5 25 1 26 3 27 2 28 3 29 2 30 4

31 4 32 1 33 3 34 3 35 4 36 2 37 4 38 3 39 1 40 3

41 2 42 4 43 3 44 5 45 2 46 3 47 1 48 2 49 4 50 3

51 2 52 1 53 3 54 2 55 4 56 5 57 1 58 2 59 3 60 2

61 1 62 1 63 2 64 5 65 2 66 2 67 4 68 4 69 2 70 5

71 1 72 3 73 4 74 5 75 3

MY PERFORMANCE
Total Time Taken Total Correct Incorrect Net
Questions (Min) Attempts Attempts Attempts Score
Logical Reasoning based
Section I 25
Data Interpretation
Language Comprehension
Section II 25
and English Usage
Quantitative Ability Section III 25

TOTAL 75 150

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MCT-0009/08
005 1
1. 4 Total weight (in ‘000 tonnes) of the crops produced in the For questions 6 and 7:
states of H, U, P and T = 18 + 11 + 14 + 13 = 56. Using the method described in Q5, the candidates deduced the correct
Total weight (in ‘000 tonnes) of the crops W, P, B, M produced answer options for all the questions from the lists of correct answer
= 18 + 25 + 9 + 17 = 69. options given by the three institutes. Based on these answer options,
following table can be drawn.
Assuming that the excess weight (in ‘000 tonnes) is all that of
crop P which will be equal to 69 – 56 = 13.
Correct Answer
Maximum weight (in ‘000 tonnes) of crop P produced in the Options (As
state G = 13 Rohit Megha Amar Feroz Richa
deduced by the
candidates)
13
Required percentage = × 100 = 81.25% 1 a √ √ √ √ √
16
2 b × √ – √ –
3 a × √ √ × √
2. 5 The possible combinations of states where the crop PO and
O can be grown are (W, A, G), (W, T, K), (W, M, K), 4 c √ √ × × ×
(MP, A, P), (T, M, MP) and (U, K, MP). 5 e × √ √ √ √
So, H is definitely not the state where the crop PO is grown. 6 d √ √ √ √ √
7 d × √ √ √ √
For questions 3 and 4: 8 c × √ √ – √
Minimum weight (in’000 tonnes) of the crop amongst the ten states is
produced in the state W, i.e. 7. 9 a √ √ √ √ √
10 c × – √ √ –
Lets say ‘x’ number of crops are produced in equal quantities in the √ √ √ √
11 a ×
given states and the value of ‘x’ is maximum possible.
12 Not sure × × × × ×
For this the aggregate weight of these ‘x’ crops produced in India 13 c × √ √ √ √
should not be more than 70.
14 d √ √ √ – √
So, we need to pick those crops which have been produced in least 15 e × √ √ × ×
quantity.
6. 2 Total marks of the 5 candidates, as per the above table:
Aggregate weight of the crops R, M, B, J and PO produced = 13 + 17
Rohit = (12 × 2 – 2 × 2 – 1 × 4) + (12 × 2 – 2 × 2 – 1 × 4) +
+ 9 + 11 + 16 = 66.
(12 × 2 – 2 × 2 – 1 × 4) = 46
If we take one more crop, then the total weight will cross 70, therefore Megha = (12 × 5) + (12 × 4) + (12 × 4 – 1 × 2) = 154
maximum possible value of ‘x’ is 5. Amar = (12 × 3 – 1 × 2) + (12 × 5) + (12 × 3 – 2 × 2) = 126
Feroz = (12 × 3 – 2 × 2) + (12 × 4) + (12 × 2 – 2 × 2) = 100
3. 2 Therefore, there are three crops namely W, P and O that are Richa = (12 × 3 – 2 × 1) + (12 × 4) + (12 × 3 – 2 × 2) = 114
not produced in equal quantities in the given ten states. Hence, Megha, Amar and Richa expected to become eligible
for the second round of selection.

4. 1 The aggregate weight (in ‘000 tonnes) of the crops R, M, B, J New scores
Except for the correct answer option of the question number
 66 
and PO produced in the state MP =   = 6.6 12, the correct answer options of all the other questions in
 10  the “SLRI” list match with the answer options in the list of
So, maximum possible weight (in ‘000 tonnes) of the crop W correct answer options deduced by the candidates based on
produced in the state MP = 9 – 6.6 = 2.4. the lists published by the three institutes. The total marks
obtained by the five candidates as per the “SLRI” list are:
5. 2 In Section - 3, the correct answer options for question numbers Rohit = 46 + 16 = 62
11, 13, 14 and 15 are a, c, d and e respectively. As no two Megha = 154 + 0 = 154
institutes had published the same answer option as the correct Amar = 126 + 14 = 140
answer option for question number 12 hence any answer Feroz = 100 + 14 = 114
options marked for this question have to be considered Richa = 114 + 0 = 114
incorrect. The answer options marked by Megha for question Hence, Megha and Amar were eligible for the second round
numbers 11, 13, 14 and 15 are correct and that marked for of selection but Richa may or may not have been eligible.
question number 12 is incorrect. This way, Mehga obtained
the highest total marks in Section - 3 and her marks are 7. 5 We need not calculate Rohit’s total marks as it is too low to
calculated to be 12 × 4 – 2 × 1 = 46. Hence, option (2) is make him eligible for the second round of selection.
correct. After deleting the question numbers 3, 8 and 15, the total
marks of the five candidates, as per the “SLRI” list are :
Megha = (12 × 4) + (12 × 3) + (12 × 3 – 1 × 2) = 118
Amar = (12 × 2 – 2 × 1) + (12 × 4) + (12 × 3 – 2 × 1) = 104
Feroz = (12 × 3 – 2 × 1) + (12 × 4) + (12 × 3) = 118
Richa = 12 × 2 – 2 × 1 + 12 × 3 + 12 × 3 – 2 × 1 = 92
Hence Megha, Feroz and Amar now become eligible for the
second round of selection. When all the 15 question were
evaluated, only Megha and Amar were definitely eligible for
second round. When only 12 questions are evaluated, Megha
and Amar still remain eligible but we are not sure about Feroz‘s
or Richa’s eligibility. Hence option (5).

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8. 5 Institutes A, B and C had matched the correct answer options Using statement B:
for 12, 13 and 11 questions respectively, with the correct Given that and angular distance covered by the line segment
answer options published by the “SLRI”. Hence institute B AB in every 10 seconds is 640°.
was rated as the best institute. Therefore, the angular distance covered by the line segment
AB in 30 seconds = 640° × 3 = 1920° = 2 × 900° + 120°.
9. 3 Since, W was not the third product that was launched, Therefore, we can not conclude that at time t = 30 seconds
therefore it was not launched in May or March. the line segment AB will lie in which of the four quadrants.
Also, X and W were launched in months that have names Hence, statement B alone is not sufficient to answer the
starting from the same alphabet; therefore X and W are question.
launched in January and June not necessarily in that order. Hence, option (1) is the correct choice.

Using Statement A: 12. 2 Using statement A:

Given that Y was not the second or the last product that was
launched, which means that Y was not launched in the months Y X
of March and August. A C B
So, Y was launched in May.
10x
It is given that Z was not launched in the last month, i.e.
August, so we can conclude that Z was launched in the 5x
month of March.
Hence, statement A alone is sufficient to answer the question.
A. In the worst case of Y is standing very near to B.
Using Statement B: First meeting will happen face to face near B.
Given that K was not the third or the last product that was As ratio of speed 4:5, Y will overtake X somewhere
launched, which means that K was not launched in the months between A and C hence in their second meeting they will
of May and August. not meet face to face.
So, K was launched in March. B. If Y is standing near C. They will between C and B.
We also, know that Z was not launched in August, so we can Lets take total distance = 90 metres and speed
conclude that Z was launched in the month of May. X = 5 m/s
Hence, statement B alone is sufficient to answer the question. Y = 4 m/s
Therefore, option (3) is the correct choice. Y X

10. 2 Let the number of dogs, cats and rabbits be ‘d’, ‘c’ and ‘r’ A C B
Initial Position
respectively.
It is also given that d + c + r = 12 and .
First meetings C B
The possible sets of values of ‘d’, ‘c’ and ‘r’ not necessarily in
that particular order are (2, 4, 6); (2, 5, 5); (3, 4, 5); (3, 3, 6)
A 20m 25m
and (4, 4, 4).
45m
Using Statement A:
The number of cats and dogs with Carol could be (5, 5) or 5
By the time Y reaches B X will cover 45 × = 56.25m
(3, 3) or (4, 4). 4
The number of rabbits when the number of dogs is 5, 3 and 4 ⇒ 2nd meeting will happen face to face.
is 2, 6 and 4 respectively. Hence, statement A alone is not sufficient to answer the
It is also given that the number of rabbits is not more than the question.
number of dogs, so the number of rabbits could be either
2 or 4. Using statement B:
Hence, statement A alone is not sufficient to answer the Let us assume Total distance = 90 metres and Speed of X = 4
question. m/s and Speed of Y = 1 m/s
Using Statement B: Y X
The number of cats and dogs with Carol could be (5, 2) or
(6, 3) in this particular order. A C B
30m 60m
The number of rabbits with Carol when there are 5 cats is
90m
5 and the number of rabbits with Carol when there are 6 cats
is 3. I. Y is standing at C when Y will reach B and reverse the
Since the number of rabbits with Carol is less than the number direction X has covered on 15 metres.
of cats, therefore the number of rabbits with Carol is 3. ⇒ Y will catch X from the back.
Hence, statement B alone is sufficient to answer the question. II. Similar when Y is standing near A. Y will catch X from the
back.
11. 1 Using statement A:
Given that the angular distance covered by the line segment Hence, statement B alone is sufficient to answer the question.
AB in every 5 seconds is 615°.
Therefore, the angular distance covered by the line segment
AB in 30 seconds = 615° × 6 = 3690° = 4 × 900° + 90°.
Therefore, we can conclude that at time t = 30 seconds the
line segment AB will lie in the second quadrant.
Hence, statement A alone is sufficient to answer the question.

005 3
For questions 13 and 14: 3x y 6y z x 25
Let the marks obtained by Aseem, Ayaan and Samantha in all the Since, ≥ and ≥ ⇒ ≥
25 10 25 10 z 72
mentioned six subjects be ‘x’, ‘y’ and ‘z’. Therefore, the marks obtained by Aseem in Science is at least

13. 1 Given that z : y : x = 1 : 2 : 5 9 25 1

× × 24 × 25 = 37
z = k, y = 2k and x = 5k 50 72 2
y
Marks obtained by Ayaan in Social Studies = 10% of y = 16. 4 Marks lost by Ayaan in English due to reason Q
10
36 1 24
Marks obtained by Samantha in Social Studies = × × y
100 4 100
z
= 20% of z = 36 1 24
5 Therefore, × × × y = 45
100 4 100
4x 125 × 50
Marks obtained by Aseem in Commerce = 16% of x = ⇒y=
25 3
Marks lost by Samantha in Maths due to reason
 y  z 1 1 z
  T= × × =
Required percentage =  10  × 100 =  5y  × 100 5 5 5 125
z
 + 4x   10z + 8x 
 5 25  y 5
Also, ≥
z 12
 10k  Maximum possible value of ‘z’ = 5000
=  × 100 = 20%
 10k + 40k  Maximum possible number of marks lost by Samantha in
Maths due to reason T is 40.
14. 2 Marks obtained by Aseem, Ayaan and Samantha in English
For questions 17 to 21:
x 6y 4z The total number of different days of the week in January and February
is , and respectively..
10 25 25 2008 are as follows:

x x Mondays: 8 (4 each in January and February)

Marks lost by Aseem in English = 40% of =
10 25 Tuesdays: 9 (5 in January and 4 in February)
Similarly, marks lost by Ayaan and Samantha in English is Wednesdays: 9 (5 in January and 4 in February)
Thursdays: 9 (5 in January and 4 in February)
3y 4z Fridays: 9 (4 in January and 5 in February)
and respectively..
50 125 Saturdays: 8 (4 each in January and February)
Sundays: 8 (4 each in January and February)
x 3y 4z
As per the question, = = ⇒ 10x = 15y = 8z Total number of days in January and February 2008 = 60
25 50 125 Total number of days on which he drove the car of brand Honda
Marks lost by Ayaan in Science due to reason R = 25 % of 60 = 15.
Total number of days on which he drove the car of brand BMW
 36 1 1  9y
= × × × y = = 15 % of 60 = 9.
 100 4 10  1000 Total number of days on which he drove the car of brand Hyundai
Required Percentage = = 10 % of 60 = 6.
Given that HO > ME > CH > BM > HY > FE
 9 y   9 10 
 × × 100  =  × × 100  = 0.6% So, we get that 15 > ME > CH > 9 > 6 > FE.
 1000 x   1000 15  We also know that ME + CH + FE = 60 – (15 + 9 + 6) = 30.
From the table given in the question set, we can know the brands of
For questions 15 and 16: cars that were driven by Mr. Alfonso in the months of January and
February 2008.
3x
Marks obtained by Aseem in Maths = Mondays: Honda and Hyundai
25 Tuesdays: Chevrolet and Honda
Wednesdays: Ferrari and BMW
y
Marks obtained by Ayaan in Social Studies = Thursdays: Chevrolet, BMW, Hyundai
10 Fridays: Ferrari, Hyundai, Mercedes
6y Saturdays: Chevrolet, Mercedes, Honda
Marks obtained by Ayaan in English = Sundays: Ferrari, Mercedes, BMW
25

z 17. 3 Given that the number of days on which Mr. Alfonso drove the
Marks obtained by Samantha in Hindi = car of brand in January 2008 is the maximum.
10
Now, the car of brand Honda is only driven on either of the
3x y 6y z three days of any week, i.e. Monday, Tuesdays and
Now as per the information given ≥ and ≥ Saturdays.
25 10 25 10
Total number of Mondays, Tuesdays and Saturdays in January
2008 = 4 + 5 + 4 = 13.
6z Total number of days in January and February 2008 on which
15. 5 Given that = 144, therefore z = 24 × 25. he drove the car of brand Honda is equal to 15.
25
Required Answer = 15 – 13 = 2
9x
Number of marks obtained by Aseem in Science =
50

4 005
18. 3 Given that the car of brand Ferrari is driven by Mr. Alfonso 24. 5 Let us first form a group that has exactly six friends.
only once on each of the possible days of any week. Since, in one group there are six friends, therefore in each of
From this it can be concluded that number of days in January the two other groups there are two friends.
and February 2008 on which he drove the car of brand Ferrari Also, we know that D, E and F are in different groups, B and
is 3, i.e. once each on Wednesday, Friday and a Sunday. J are in the same group and so are C, I and H.
Now, we already know that 15 > ME > CH > 9 > 6 > FE and This means that in a group which has six friends, the following
ME + CH + FE = 30 five friends namely B, J, H, I and C must be there.
Also, FE = 3 as per the question Since, C cannot be in group Y and in group X there is A along
The only possible value of ME and CH that satisfies the above with one out of D, E and F, we can conclude that it is only
given constraints is 14 and 13 respectively. group Z that can have six friends.
Required Answer = 13 The sixth friend in group Z could be either E or F as D cannot
be in group Z.
19. 4 The only possible value of ME, CH and FE can be So, neither A nor G nor D cannot be in the group that has six
(14, 13, and 3), (14, 12, and 4), (14, 11, 5), (13, 12, and 5) friends.
So, from the above information we can say that option (4)
cannot be true. 25. 1 Since there are exactly five friends in group Z, so it can be
Hence, option (4) is the correct choice. concluded that H, I and C are definitely in group Z.
Since A is in group X, and D, E and F are in different groups,
For questions 20 and 21: therefore B and J who are in the same group are in group Y.
Given that Mr. Alfonso drove the car of brands Mercedes only in So, G is also in group Z.
February 2008. The fifth friend in group Z could be either E or F.
This is possible only if he drove the car of brand Mercedes on every The third friend in group Y and second friend in group X could
Friday, Saturday and Sunday in February 2008. be either D or E or F.
This also means that the total number of days on which he drove the Therefore, J is definitely in group Y.
car of brand Mercedes = 5 + 4 + 4 = 13
Hence, the only feasible combination is 15 > 13 > 12 > 9 > 6 > 5. 26. 3 The correct sequence of the passage is 1, 9, 8, 3, 10, 7, 5, 6,
4, 2. The passage is a critique of American foreign policy. In
20. 4 Required Difference = 13 – 12 = 1. the first paragraph it talks of the need for American Government
to introspect its action in the aftermath of September 11
21. 3 Mr. Alfonso will drive the car of brand Mercedes on all the attacks. The 9th paragraph is an extension of this where it
Fridays, Saturdays and Sundays in February. highlights the importance of introspection to reduce the repitition
So, on all the Wednesdays and Sundays he can drive only the of any further atrocities. It conveys the thoughts of President
car of brands Ferrari and BMW. Bush, the 8th paragraph extends the argument further and
Also, the total number of days on which he drives the car of underlines the reason for the Arab hatred towards US.
brands Ferrari and BMW is 9 + 5 = 14. Paragraph 3 is a continuation of the idea illustrated in the
Out of these 14 days, 13 will be either Wednesdays or previous paragraph where it talks of the resentment over
Sundays, therefore he cannot drive the car of brand Ferrari specific US policies in the Muslim countries and even in the
more than once on a Friday. W estern-oriented sectors in the Middle-East. Paragraph
10 cites an example to illustrate the resentment talked about in
22. 4 From additional information (4) we know that C is not in group paragraph 3. In Paragraphs 7, 5, 6 the author makes attempts
Y and D is not in group Z. to clarify that the hatred in the heart of the people of victimised
Given that C and D are in the same group, which means that countries is not for the US citizens but for US official policies.
both C and D are in group X. Paragraph 4 gives the opinion on the US official policy from an
A, C, D, H and I are in group X. external source, Passage 2 substantiates the opinion.
Since, we need to find the friend who cannot be in the group
that has exactly three friends; it means that group X will only 27. 2 The correct sequence of the passage is 1, 9, 8, 3, 10, 7, 5, 6,
have five friends. 4, 2.
From additional information (1) we know that B and J are in
the same group. 28. 3 The passage analyses the September 11 attacks on US in the
Also from additional information (2), D, E and F are in different light of its official policy. The last paragraph talks of the negative
groups. perceptions the outside world has about US foreign policy.
So, it is clear that in a group which has exactly three friends; The continuing paragraph must be the one which comments
B and J will be accompanied by either E or F. on September 11 attacks. (3) rightfully talks of how harshly
So, G will be in the group that has exactly two friends along the attacks were condemned throughout the world but with
with either E or F. certain reservations. (1) does not talk about the September 11
attacks but the reaction towards America after its attack on
23. 2 Given that F is in group X and G is not in group Y Afghanistan. (2), (4) and (5) talks of an entirely different
Since D, E and F are in different groups and D is not in group context.
Z, therefore D is in group Y and E is in group Z.
Since, C is not in group Y and C, I and H are in the same group, 29. 2 Structuralism believes that all elements of human culture are
therefore C, I and H are either in group X or group Z. parts of a system of signs - in a way this is understanding the
Also, G is either in group X or Z. unity permeating the diversity. 1 is opposite of what
Since each group has at least two friends and B and J are in structuralism seeks to do. 3 is part of structuralism’s method
the same group, therefore in group Y, there are exactly three but not the primary purpose. We cannot conclude that
friends namely B, J and D. structuralism seeks to eliminate things-as in 4. Even 5 is one
Therefore, out of the pairs mentioned the following four pairs of the purposes of structuralism but not the primary purpose.
of friends cannot be in the same group; (J, F); (D, G); (B, E)
and (D, I).

005 5
30. 4 A was not the main aim of structuralists. It is something which when evasion happens it will be forced to raise income tax
was discovered or observed by structuralists in their process rates. Statement (iv) about the vicious circle that is created
of research. B cannot be affirmed as data is insufficient. can be clearly concluded from the passage and hence it is a
C cannot be inferred. Saussure work was useful to downstream argument.
structuralists. But his absence may not have led to the
premature death of structuralism. 39. 1 (cbda). Statement (ii) about companies founded by groups
being more likely to succeed than companies founded by
31. 4 The last sentence of para 4 leads to option 4 as the correct individuals can be concluded from the passage. Statement (i)
choice. (1) is not true. (2) is a part answer. (3) and (5) cannot supports the passage by highlighting the need for pursuing
be concluded. strategies in different business areas and is therefore a lateral
argument. Statement (iii) about ventures in technology space
32. 1 Refer to the line in the first paragraph “Adelson adamantly failing is not really relevant to the argument. Statement (iv) is
advocates for understanding “Turkish” and “German” less as an upstream argument because the passage about an
historically loaded labels and more as the names of two living individual being unlikely to have the expertise of several people
cultures that exist essentially inside one another” which leads from different backgrounds can be inferred from it.
to option 1.
40. 3 Definite article ‘The’ is not used before fire or Summer. If a
33. 3 Option 3 is the correct representation of the primary purpose message is being sent, we don’t use ‘send a word’. We simply
as Adelson wants to show that the cultures touch. 1 is too use ‘Send word’. It is not ‘give an ear’ to, rather just
narrow. 2 confines itself to literature. 4 is correct but it is not ‘give ear to’.
the primary purpose. Instead it is a reason leading to the
primary purpose. 5 is correct but part of the whole. It is part of 41. 2 In A, it should be “lines of evidence”. In C, it should be ‘from’ -
option 3. ‘of’ is redundant. In D, similar is redundant - as twins
are similar.
34. 3 Novum refers to the game of dice - so Game is closest. Trope
is a figure of speech. Hence option 3. 42. 4 In A, it should be ‘shortcomings’. In B, there is no need of ‘the’
before ‘Geography’. In E ‘between’ should be used instead of
35. 4 Refer to the lines in para 5 “Focusing on this key issue, Adelson ‘among’.
emphasizes the common ground that Germans and Turks
share. In contexts like these, according to Adelson, Germans 43. 3 ‘Augur’ means to bode well ‘auger’ is a tool. ‘Oral’ is something
and Turks produce “touching tales”, and thereby reveal their verbal, ‘aural’ means sound, ‘bail’ is for surety and ‘bale’ means
overwhelming similarities”, which reveal that both carry similar a stack of cotton. ‘Cede’ means to give in and ‘seed’ is
painful impressions about the past. something that you sow. The article ‘a’ is used here to
generalize, ‘the’ is used for specifics.
36. 2 (bacd) Statement (i) is a downstream argument because the
conclusion that a candidate hired for a job should have a 44. 5 To ‘sell’ means to make a sale, ‘cell’ is a small room. ‘Grates’ is
personality that matches the job is an inference that can be correct because it means something that irritates, ‘greats’
drawn on the basis of the passage, which states that an means something great. ‘Gneiss’ means a kind of stone ‘nice’
interview is essential to check whether an applicant has a is something good. Horses ‘neigh’, the sound produced by
personality that matches the requirement of the job or not. them is a neigh. Nay means ‘no’. The subject here is ‘patterns
Statement (ii) is an upstream argument because the conclusion ‘use a plural with it.
that an interview is an essential part of a hiring program can
be drawn on the basis of the example of a faulty hiring without 45. 2 Pare means to cut, pear is a fruit, groan is a sound and grown
a formal interview. Statement (iii) is a lateral argument because means grown up, pause means break and paws are the foot
it supports the idea of a personality match with a particular of any animal. Aircraft is always used in the same form and
type of job being very important. Statement (iv) about does not take an ‘s’. A ’hangar’ is a parking space for aircraft;
professional hiring agencies is not really relevant to the hanger means something for hanging your clothes.
argument.

37. 4 (adbc). Statement (i) is an upstream argument because if 46. 3 The magician is discussing the debate related to the structuring
television satellites are necessarily heavily insured, then it of the act of magic and statements that declare that the
can be concluded that operating mishaps would lead to a audience are watching surreal acts. So, option 3 is appropriate.
corresponding surge in insurance claims. Statement (iii) is a Option 1 is one sided as it mentions celebration. In option 2
downstream argument because it can be concluded that if ‘gruntling’ means to put one in good humour which cannot be
satellites experience frequent mishaps and this leads to an used here. ‘Impounding’ means to confine or seize; it is
increase in claims, then insurance premiums for satellite completely inappropriate here. ‘Floundering’ means to move in
insurance would go up. Statement (iv) underscores the a clumsy manner. This word does not fit in the context here.
importance of television satellites and supports the fact why
they are heavily insured. Statement (ii) about the kind of market 47. 1 The magician is appalled by the fact that the magicians are the
the manufacturers operate in is irrelevant to the passage. ones who are deliberately hindering the feelings of wonder
and astonishment among their audiences. He believes, instead
38. 3 (dacb). Since the passage is about income taxes, the issue of of that, magicians should be promoting this feeling of wonder.
collection of corporate taxes is irrelevant to the argument. In the given context, ‘aggrandise’ is incorrect because it means
Statement (ii) is an upstream argument, because if income tax to enlarge or extend. ‘Permeate’ means to penetrate or pass
collections are important to the government, then it can be through. It is not entirely appropriate because it does not
concluded that tax evasion would force the government to contain the element of ‘deliberately hindering’ that is present in
raise income tax rates. Statement (iii) supports the passage sabotage. ‘Corroborate’ and ‘ratify’, both mean to confirm.
because if, in fixing income tax rates, the government does These do not fit with the magician’s point of view that is
not allow for revenues that will be lost through evasion then presented in this sentence.

6 005
48. 2 ‘Pecuniary’ pertains to monetary. This is inappropriate because 53. 3 Using Statement A:
the magician is referring to the advantage or the positive D C
effect of creating the feeling of amazement and wonder.
‘Underlying’ means basic. This cannot be chosen because it P 4
refers to the fundamental value that the feeling of wonder 2 4
may bring. However, the magician is trying to convey that Q
there are positive effects of these feelings of amazement. 3
Just like the magician felt when he was a kid and watched x
magic for the first time. ‘Perimetric’ means of the border and
‘esoteric’ means that which is understood by a select few. A B
Neither of these two pertain to the context of the given As length of perpendicular from P on AD = 2 cm and on BC
paragraph. Only ‘therapeutic’ reinstates the positive effect of = 4 cm
promoting the feeling of wonder. ⇒ Length of AB = 6 cm
As length of perpendicular from Q on AB = x and on BC = 3 cm
49. 4 The ‘real’ world is not magical or surreal or sorcerous or
theurgic. Also, the magician does not express his point of ⇒ Q is the mid point of BD
view regarding the ‘execrable’ world. He consistently urges As length of perpendicular from Q on CD = 4 cm
to promote the feeling of wonder in a world that is ‘prosaic’. ⇒ Length of side BC = 8 cm
Hence, statement A alone is sufficient to answer the question.
50. 3 ‘Paradigm’ is most appropriate because it explains that the Using Statement B:
magician’s assumptions or principles regarding the
presentation of magic are very different from the ones’ in his D 2 J L 4 C
childhood days. ‘Affliction’ or distress is completely out of 1
context here. ‘Quandary’ refers to a dilemma which is not K
2 P 3
suitable because the magician is mentioning his fundamental
approach of presenting magic.
M
‘Ambitions’ may seem correct because the magician admits Q 4
his ambition in the last line; however, it does not fit in with the
sentence in question because the magician is comparing his
fundamental belief and approach to magic with respect to A B
two time frames-the present and his childhood. ‘Desire’ means ∆DJP, ∆DLQ and ∆DCB all are similarly to each other.
to wish or long for, which does not fit here.
DL DJ 2
=
QL PJ 1
= (as ∆ DJP : ∆DLQ)
51. 2 Using statement A:
⇒ DL = 3 × 2 = 6 cm
P and Q are prime numbers and each of them completely
⇒ DC = DL + CL = 6 + 4 = 10 cm
divides (P + Q).
This can happen only when P = Q.
So, for multiple values of P and Q, the remainder will be and
DJ DC 2
= = ( as ∆DJP : ∆DCB)
PJ BC 1
different.
Hence, statement A alone is not sufficient to answer the 10
question. ⇒ BC = = 5 cm
2
Using statement B: ⇒ Area of the rectangle = 10 cm × 5 cm = 50 cm
Any perfect square when divided by 3 leaves a remainder of Hence, statement B alone is sufficient to answer the question.
0 or 1. Here it is given that P and Q are not divisible by 3. Hence, option (3) is the correct choice.
Hence each P6, P4 and Q will leave a remainder of 1 when
divided by 3 leaving the net remainder as 0. 54. 2 Using statement A:
Hence statement B is sufficient to answer the question. Given that the quantity of water in the second beaker is 50%
Option (2) is the correct choice. (by volume) of the total solution in it.
So, the quantity of milk and water in the second beaker is
52. 1 Using statement A: 1 liter each.
If the roots are real and equal, then we can write, This information is not sufficient to find the total volume of the
solution left in the first beaker.
⇒ Q2 – 4 PR = 0
Hence, statement A alone is not sufficient to answer the
Here, there are multiple values which will satisfy the above question.
equation.
But if (P + Q + R) is equal to the smallest possible non negative Using statement B:
integer, then the values of Q, P and R that satisfy the condition Given that the quantity of milk left in the first beaker is
are Q = –2, P = R = 1. 40%(by volume) of the original quantity of milk in the that
Hence, the sum of the roots of the given equation = –2. beaker.
Hence, statement A alone is sufficient to answer the question. Therefore, 60%(by volume) of the original quantity of milk is in
the second beaker.
Using statement B: 60%(by volume) of the original quantity of milk is equivalent to
R 2 liters of solution, therefore 100%(by volume) of the original
is the product of the roots which is equal to 2. 100 10
P quantity of milk will be equivalent to 2 × = liters.
The roots of the given equation can be (1, 2) or (–1, –2). 60 3
Therefore, the sum of the roots of the given equation can be Therefore, the volume of the solution left in the first beaker
–3 or 3. 10 4
= − 2 = liters.
Hence, statement B alone is not sufficient to answer the 3 3
question. Hence, statement B alone is sufficient to answer the question.
Hence, option (1) is the correct choice. Hence, option (2) is the correct choice.

005 7
55. 4 Let the price at which the shopkeeper bought each note book
1 1
be equal to ‘c’. ⇒ area(∆ TPS) = × a × h4 = × 2 3 × 3 3 = 9cm 2
2 2
 14 − c   300 − 25c 
  × 100 = p and   × 100 = q Case III:
 c   25c 
T
14 − c p h1
As per the information given in the question = =2
12 − c q P Q
h2
⇒ c = 10
 130 − 100  S
The shopkeeper will make a profit of   × 100 = 30% R h3
 100 
in his transaction with Vijay
h4
Hence, option (4) is the correct choice.

56. 5 Two cases are possible: 1 1

area(∆TSR) – area(∆TPQ) = × a × (h2 − h1) = a2 = 2cm2
2 2
Case I:
⇒ a = 2 cm
h3
1 1
h4 Now, area( ∆TQR) = × a × h3 = × 2 × h3 = 3 cm2
2 2
T
⇒ h3 = 3cm and h4 = (h3 + a) = 5 cm
h1
Q
P 1 1
h2 ⇒ area(∆TPS) = × a × h4 = × 2 × 5 = 5 cm2
2 2
2 2 2
Hence, area of ∆TPS can be 1 cm or 9 cm or 5 cm .
S R Option (5) is the only correct option.
a

1 1 57. 1 To make a valid pair a boy must be paired with a girl. First
area(∆TSR) – area(∆TPQ) = × a × (h2 − h1) = a2 = 2cm2 choose a girl partner for each of the three boys.
2 2
⇒ a = 2 cm The number of ways choosing 3 girls is = 7C3
The number of ways the chosen three girls can be paired
1 1
Now, area( ∆TQR) = × a × h3 = × 2 × h3 = 3 cm2 with three boys is = 3! × 7C3
2 2
The number of ways in which the remaining four girls can be
⇒ h3 = 3cm and h4 = (h3 − a) = 1 cm
 4 C2 
⇒ area(∆ TPS) =
1 1
× a × h4 = × 2 × 1 = 1 cm2 paired among themselves  

2 2  2! 

Case II:  4 C2 
Hence, total number of ways = (3! × 7 C3 ) ×   = 630
 2! 
 

h4
58. 2 To find the common roots of the two equations, we subtract
P Q the first equation from the second and get (P – Q)(x + 1) = 9
h1
9 9
T ⇒x+1= or x = –1 ...(i)
a P–Q (P – Q)
h2 For x to be an integer in equation (i), (P – Q) should be 9 or 3
or 1. For (P – Q) = 1, 3 and 9, x = 8, 2 and 0 respectively. We
need to check at which of these values of x, both the quadratic
S R
equations are satisfied and also conform with the
h3 corresponding values of P and Q.

1 1 Case I: (P – Q) = 1 and x = 8
area(∆TPQ) + area( ∆TSR) = × a × (h1 + h2 ) = a2 = 6cm 2 Putting x = 8 and P = Q + 1, in the two quadratic equations, we
2 2
64 55
⇒ a = 2 3 cm get P = and Q = . As P and Q are integers, this case is
9 9
1 rejected.
Now, area( ∆TQR) = × a × h3 = 3cm 2
2
Case II: (P – Q) = 3 and x = 2
⇒ h3 = 3 cm and h4 = (a + h3 ) = (2 3 + 3) = 3 3 cm Putting x = 2 and P = 3 + Q in the two quadratic equations,
we get P = 2 and Q = –1. As P = 2, Q = –1 and x = 2 satisfy the
two quadratic equations, this case is acceptable.

8 005
 Case III: (P – Q) = 9 and x = 0 63. 2 According to the question, the mathematical version of the
Putting x = 0 and P = 9 + Q in the two quadratic equations, we statements will be as follows:
get P = 8 and Q = –1. As P = 8, Q = –1 and x = 2 satisfy the
 1  1  2  1 
two qudratic equations, this case is also acceptable.   +   –   =  
 a   b   c   10a 
Hence, 2 values of (P – Q) are possible. Hence, option (2) is 1 2 1
the correct option. ⇒ 1 +  – a =
b c 10
(2b – c)a 9
59. 3 The prime factorization of N must be 2a3b7c where each a, b, ⇒ = ...(i)
and c is either 0, 1, or 2 and at least one of a, b, and c is bc 10
positive. It follows that the sum of all such N is given by Please note that in equation (i), ‘bc’ will always be the multiples
= (20 + 21 + 22 )(30 + 31 + 32 )(70 + 71 + 72 ) – 1 of 10 which are obtained by multiplying any two of the natural
= (1+ 2 + 4)(1+ 3 + 9) (1+ 7 + 49) – 1 numbers belonging to the set (1, 2, 3….9). They are 10, 20, 30
= 7 × 13 × 57 – 1 and 40 only. Now the solution for (b, c) cannot be (5, 6) or
= 5186. (6,5) as it will never satisfy the equation (i) for any value of
Hence, option (3) is the correct choice ‘a’.

60. 2 Two-digit numbers representing the year (yy), should have Therefore bc must be 10, 20 or 40. The only solution for (a, b,
the maximum number of factors so that we can express ‘yy’ c) are (6, 4, 5) or (3, 5, 4).
as a product of two numbers (‘dd’ and ‘mm’), maximum number Therefore, a + b = 2c. Hence option (2) is the correct choice.
of times.
But we have constrains that 'dd' ≤ 31and 'mm' ≤ 12. 64. 5 If (x + p) (x + 10) + q and (x + m) (x + n) + r are equal, then we
can write
One would observe that 60, 72, 84, 90 and 96 are those two
(x + p) (x + 10) + q = (x + m)(x + n) + r
digit numbers which have a maximum number of factors, i.e.
⇒ (x + p)(x + 10) + 1 = (x + m)(x + n) .....(Since q – r = 1)
12.
⇒ x2 + (10 + p) x + 1 + 10 p = x2 + (m + n) x + mn
Out of these numbers, 60 has 7 possible cases for (dd, mm)
⇒ m + n = (10 + p) and 1 + 10p = mn ...(ii)
which are (30, 2), (20, 3), (15, 4), (5, 12), (12, 5), (10, 6) and
Solving the above two equations, we get
(6, 10) which is the maximum. In all of these cases, ‘yy’ = 60.
⇒ (m – 10)(n – 10) = 1 ...(i)
Hence, option (2) is the correct choice.
If m and n are both integers, then solution for (m, n) can only
61. 1 W e have to find two digit numbers ‘yy’ such that they are
be (11, 11) and (9, 9)
prime numbers greater than 31. There are 14 prime numbers
From (ii), we get the corresponding values of p as 12 and 8
greater than 31 and less than 100.
There are 6 more cases: Hence, option (5) is the correct choice.
a. The second multiples of 37, 41, 43 and 49 will give us 4
65. 2 Since there are 10 percent fewer B than A,
cases.
b. The second multiple of 29 and 31 will give us two more ∴ A : B = 10 : 9
cases since if ‘yy’ = 58 or 62, in those year the month of
10
February will have only 28 days. Number of fishes of type 144 × = 160
9
So, the total number of cases = 14 + 4 + 2 = 20. Since there are 25 percent more A than C,
Option (1) is the correct choice. A:C=5:4
or
62. 1 Let x be the number of MP3 players that cost $200 each, y the
4
number of MP3 players that cost $400 each, and z the profit. Number of fishes of type C = 160 × = 128
The profit is ‘25x + 30y’. The constraints which should be 5
considered are 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
i) x + y ≤ 300 ⇒ y ≤ – x + 300
100
1 pond, the pond must have 432 × = 540 fishes.
ii) 200x + 400y ≤ 8000 ⇒ y ≤ – x + 200 80
2 Hence option (2) is the correct choice
iii) x, y ≥ 0
66. 2 Probability of not getting a blue ball in two consecutive draws
The number of each type of MP3 player that will maximize the
( ) 81
profit will be the coordinates of one of the vertices of the 2
= P(B) =
region satisfying all three of the constraints. These co-ordinates 100
are (0, 0), (0, 200), (200, 100) and (300, 0) shown as follows. 9 1
⇒ P(B) = ⇒ P(B) =
X 10 10
⇒ There is only one blue ball in the box.
(0,300) 2
49  7 
Similarly, as (P(G) ) =
2
=
100  10 
(0,200) (200,100)
(400,0)
Y ⇒ There are 7 Green balls.
(0,0) (300,0) ⇒There are 10 – (7+1) = 2 Red Balls.
There are 6 different ways of getting 3 different balls of 3
The maximum profit is $8000 which is attained when 200 of different colours viz. RGB, RBG, BGR, BRG, GBR, and GRB.
the $200 MP3 players and 100 of the $400 MP3 players are The probability corresponding to each of the six cases is
sold. Hence option (1) is the correct choice. 2 7 1 7
× × =
10 10 10 500

7 21
⇒The required probability = 6 × =
500 250
005 9
67. 4 Let the co-ordinates of points P and Q be (4a, a) and (3b, b)
respectively. As R(10, 3) is the midpoint of PQ, we must have: 72. 3
4a + 3b = 20
a+b=6
Solving the above equations we get a = 2 and b = 4. Hence,
the co-ordinates of points P and Q be (8, 2) and (12, 4)
respectively and the length of line segment PQ is 2 5 units.
Hence (1) is the correct option.
Lets denote the five shaded regions by a, b, c, d and e. As all
68. 4 the vertical lines are parallel to each other, the areas of a, b, c,
T d and e must be the same as the areas of the regions marked
O by (1,1), (1, 2), (1, 3), (1, 4) and (1, 5) respectively. As all the
horizonal lines are equally spaced, we must have:
S
1
area{(1,1)+(1,2)+(1,3)+(1,4)+(1,5)} = × (area of ABCD)
4
A R M B = area (a + b + c + d + e)
Let radius of the circle be r = 10 cm and AM = BM = R. AB is 1
tangent to the circle hence AT = AM = BM = R cm. From the Hence, the required ratio is . Hence, option (3) is the
4
right triangle ATB,
correct option.
BT = (AB2 − AT 2 ) = R 3 cm
Also, we have BM2 = BS × BT
73. 4 The value of M is 10 as ten distinct numbers ‘abc’ are possible:
⇒ R2 = (R 3 – 2r) × R 3 ⇒ R = 10 3 cm 235, 236, 237, 256, 257, 267, 356, 357, 367 and 567 and the
This gives BS = BT – ST = 30 – 20 = 10 cm. value of N is 6 as only six distinct numbers ‘pq’ are possible:
10, 80, 90, 81, 91 and 98 Hence (M – N) = 4.
69. 2 Rates in 0600 to 1200 hours time slot = $125 per minute Hence (4) is the correct option.

70000
So, number of viewers in this time slot = 125 ×
300 74. 5 A
= 29166 x x
Hence option (2) is the correct choice
D F
r
70. 5 W e need to maximize the number of advertisements in the O r
times slot 0600 to 1200 hours x–2
x+2
Suppose x and y are the number of one-minute advertisements
shown in 0600 to 1200 hours and 1200 to 1800 hours time r
slots respectively.
B E C
Then we have the following constraints: x+2 x–2
300x + 125y ≤ 10000 ...(i)
Let, Side AC = (2x – 2) cm
y 1 Side CB = (2x) cm
≥ ...(ii) Side BA = (2x + 2) cm, where x is a natural number greater
x+y 4
than 1.
One needs to maximise x with the given constraints.
Substituting different values of x in (i) and (ii) we find that We know that Area of ∆ABC =
x = 29 and y = 10 is the optimum solution. Area (∆OBC + ∆OAC + ∆OAB)
Therefore, (x + y) = 39.
1
The correct choice is option (5).
= r {(2x − 2) + (2x) + 2x + 2}
2
71. 1 Since 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 27 × 34 × 5 × 7, the =(3rx)
digits 5 and 7 must be present in the required number. Also, semiperimeter

The products of 2’s and 3’s can be done in several ways (2x − 2) + (2x) + (2x + 2)
s= = 3x
such as 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2 × 4 × 4 × 2
4 × 9 × 9 = 6 × 6 × 6 × 6 × 8 = 2 × 8 × 8 × 9 × 9. This cannot be
done with fewer than 5 digits since 94 = 6561 < 27 × 34 Area of ∆ABC = 3x(x + 2)(x)(x − 2) = 3rx
= 10368.
(x + 2)(x)(x − 2)
⇒ =r
To do this with 5 digits where at least one digit is a 2 requires 3x
the product 2 × 8 × 8 × 9 × 9. Adding in the 5 and 7 digits so that
the digits form the least possible integer gives the answer ⇒ (x 2 − 4) = 3r 2
2578899. ⇒ x2 = 3r2 + 4 ... (i)
For x to be a natural number (3r 2 + 4) has to be a perfect
square
Only option (5) does not make x a natural number
Hence, option (5) is the correct choice.

10 005
75. 3 If a particular year, which is not a leap year, starts with Monday,
it will also end with Monday. For minimum number of days,
let’s consider a leap year N and assume its first day is
Tuesday.

Let Gautam start his visit on the last day of the year (N – 1).

Year First day Last Day

N–1 Monday Monday
N Tuesday Wednesday
N+1 Thursday Thursday
N+2 Friday Friday
N+3 Saturday Saturday
N+4 Sunday Monday

So, on the first day of year (N + 4), Gautam will meet the given
condition.
So, required number of days = 365 × 4 + 3 = 1463
Option (3) is the correct choice

005 11