**Ans . **

(2) satirical

**Explanation :**The writer is using satire to mildly tease the French winemaker. (1), (3) and (4) are rather extreme choices.

**Ans . **

(1) Follow the labeling strategy of the English-speaking countries

**Explanation :**Refer to the part some areas … have now produced a generation of growers using the varietal names on their labels. The writer says that (1) is probably the only option left for French winemakers.

**Ans . **

(2) ‘education’ that consumers have derived from wine labels from English speaking countries.

**Explanation :**Refer to the part it is on every wine label … the name of the grape from which the wine is made … acquired a basic lexicon. (2) well describes that the French winemakers are scared of this trend.

**Ans . **

(4) Long-term surveys in southern France showed that the incidence of coronary heart disease was significantly lower in red wine drinkers than in those who did not drink red wine.

**Explanation :**Option (4) is the most substantiated reason to support Dr. Renaud’s findings. The development in (4) would support Dr. Renaud's findings that fat-derived cholesterols can be dispersed by the tannins in wine.

**Ans . **

(3) Consumers are able to appreciate better quality wines.

**Explanation :**Option (1), (2) and (4) are stated in the 4th paragraph. (3) is unlikely. A consumer may still not be enough of a connoisseur to discriminate wine tastes.

**Ans . **

(3) It was afraid that if India refused to pay, Britain’s war efforts would be jeopardized.

**Explanation :**Refer to the part India would resist payment, and paralyze the war effort. (3) is clearly the answer.

**Ans . **

(3) That the British were a small ethnic group.

**Explanation :**Refer to the part it reminded the British vividly. (3) is clearly the answer. (1) was an outcome, not a cause. (2) is a minor factor. (4) is far-sighted.

**Ans . **

(2) The decreasing returns from imperial loot and increasing costs of conquest.

**Explanation :**(1), (3) and (4) are stated in the third paragraph. (2) is not a reason for the emergence of the 'white man's burden'. It is a consequence, not a cause.

**Ans . **

(1) The British claim to a civilizing mission directed at ensuring the good of the natives.

**Explanation :**Refer to the part it was supposedly for the good of the conquered. (1) entirely captures the meaning of the 'white man's burden'.

**Ans . **

(4) To illustrate how erosion of the financial basis of an empire supports the granting of independence to an empire’s constituents.

**Explanation :**Refer to the last line of the first paragraph, the second paragraph and the last line of the passage. They amply support (4) as the answer. (1) does not touch on the financial implications. White man’s burden is a single aspect of the passage, not the main idea, so (2) is not right. (3) can be ruled out straightaway.

**Ans . **

(3) MNCs are not the only group of actors in genetically-modified food research.

**Explanation :**Refer to the part much of biotechnology research is also funded by governments. (3) is clearly the answer.

**Ans . **

(3) Germany and France.

**Explanation :**Refer to the part anti-GM campaign has been quite effective in Europe. (3) is clearly the answer.

**Ans . **

(2) forcing application of stronger herbicides to kill weeds which have become resistant to weak herbicides.

**Explanation :**Refer to the part use of ever-stronger herbicides which are poisonous. The last line specifically supports (2) as the answer and not (1) which is discussed in a different context. The passage has no intention of keeping competing plants standing at all, let alone keeping them weed-free, so (3) is wrong.

**Ans . **

(4) addresses the concerns of rich and poor countries.

**Explanation :**Refer to the part much of biotechnology research is also funded by governments in both developing and developed countries. (4) is the answer. (1), (2) and (3) are disputed in the passage.

**Ans . **

(1) Indian media generally covers a subject of scientific importance when its mass application is likely.

**Explanation :**Refer to the part GM controversy will soon hit the headlines in India … use the protato in its midday meal program for schools. (1) can be inferred. (2) is, of course, wrong. (3) is doubtful. (4) is also not true.

**Ans . **

(1) Large parties consisting of casual acquaintances and strangers.

**Explanation :**The last sentence of the 2nd paragraph states these large gatherings which continues as they in the 3rd paragraph. (1) is clearly the answer.

**Ans . **

(3) lamenting the drying up of our real social life.

**Explanation :**The passage begins with description of social life and towards the last few paragraphs, moves on to show drying up of our social life. …(3) is clearly the answer. (2) and (4) are rather extreme observations. (1) is also a blunt statement, whereas the passage does have a subtle tone.

**Ans . **

(2) people possess qualities like wonder and interest.

**Explanation :**Refer to the part Interest, wonder … the need of the first two must not be underrated. (2) is clearly the answer.

**Ans . **

(1) recognize

**Explanation :**Discriminate means to recognize passionate attitude, distinguish is too technical a word to fit the requirement. (2) and (4) are irrelevant.

**Ans . **

(4) Achieving a high degree of sociability does not stop the poor from hating the rich.

**Explanation :**The correct ans. is (4) as can be seen by the first line of the second last para. If you read the previous para also you’ll find that what the author is actually saying is that the so called social life is not as per the real definitions. (1). is not right as the author is nowhere showing that the crowds in poor Calcutta can turn violent anytime. He is just giving a couple of instances to prove his point. We can’t generalize like this. (2) is the opposite of what the author is trying to show. (3) again is a generalization.

**Ans . **

(1) Apparently they did not think it necessary to experiment.

**Explanation :**Refer to the part it remains a fact that the Greeks…never seem to have realized the importance of experiment. (1) is clearly the answer. The Greek preference for geometry is not mentioned in the passage, so (2) and (4) are out. (3) is a superficial answer.

**Ans . **

(3) physical phenomena conformed to mathematical laws.

**Explanation :**Refer to the part physical processes of nature would prove to be unfolding themselves according to rigorous mathematical laws. (3) is clearly the answer. (1) is not true. (2) is also refuted and (4) is irrelevant.

**Ans . **

(2) married physics with mathematics.

**Explanation :**Refer to the part account be taken of his joint contributions to mathematics and physics. (2) is clearly the answer. (1), (3) and (4) are specific aspects.

**Ans . **

(4) New knowledge about natural phenomena builds on existing knowledge.

**Explanation :**4 Refer to the part extension of the validity. The writer states that Einstein's special principle is an extension of the validity of the classical Newtonian principle. This being the concluding sentence makes (4) the best answer. (1) and (2) are not correct observations. (3) sounds plausible but it is actually a vague observation.

**Ans . **

(3) there are limits to which experimentation can be used to understand some physical phenomena.

**Explanation :**The correct answer is (3) If you read the 6th line of last para it’s given that the principle’s assertion was that “absolute velocity must ever escape all experimental detection.” Which means that sometimes we can’t experiment. This is very similar to (3). Ans. choice (1) is a fact and not an “implication”. (2). Is again a fact and in (4). The word “meaningless” is too strong and this choice is a generalization from a specific point. Generalizations need not be correct.

**Ans . **

(2) Don’t rush to your goal; the journey is what enriches you.

**Explanation :**Refer to the part better if it lasts for years …wealthy with all you have gained on the way. (2) is clearly the answer. (3) is far-fetched. (1) is an isolated observation. (4) is totally incorrect.

**Ans . **

(1) You can gain knowledge as well as sensual experience.

**Explanation :**Refer to the part as many sensual perfumes as you can … to gather stores of knowledge. (1) is clearly the answer. (2), (3) and (4) are short-sighted observations.

**Ans . **

(4) life’s distant goal.

**Explanation :**Refer to the part Keep Ithaka always in your mind. Arriving there is what you are destined for. (4) is undoubtedly the answer.

**Ans . **

(3) Intra-personal obstacles that hinder one’s journey.

**Explanation :**Refer to the part you bring them along inside your soul. (3) is undoubtedly the answer.

**Ans . **

(2) Exhorting

**Explanation :**Refer to the part Ithaka gave you the marvelous journey, without her you would not have set out. The poem has a tone of encouragement and promise. (2) is clearly the answer. (1), (3) and (4) are ridiculous choices.

**Ans . **

(2) B

**Explanation :**Running … consists has singular subject-verb agreement. Again, more than it costs is the right diction.

**Ans . **

(3) C

**Explanation :**B and D have inappropriate temporal references. A is also wrong as products did not lead to the heightened focus. C is the answer as the second and third part of the sentence when put together is complete by itself.

**Ans . **

(1) A

**Explanation :**Improper use as in “falling back” and “explanations” rule out B and C. Fall back on is the right prepositional phrase and thus A is right.

**Ans . **

(4) D

**Explanation :**is regarded should go together. Valuable in itself is the right expression. Not only as …but also as has parallel construction.

**Ans . **

(2) B

**Explanation :**it would be ideal expresses a satisfactory proposition. Reflection should precede action, and thought should facilitate behavior.

**Ans . **

(3) ADBCE

**Explanation :**ADB is a clear sequence. So is CE. A has a suitable opening with A few months ago. The invitation and the response follow in DB. she in E has a clear reference to One senior in C.

**Ans . **

(1) CABDE

**Explanation :**CA gives the sequence of action. BD follows with reaction. The outcome is in E. CA outlines the consecutive bids. BD gives Mr. Conway's statements. Moreover in D adds to B.

**Ans . **

(4) CEBDA

**Explanation :**C is the best beginning to the paragraph. C spells out the misnomer. E makes a statement on terror that is justified though B and in D as Besides. The humanitarian context of D is given in A.

**Ans . **

(1) EACBD

**Explanation :**The “these types are rare” of D should follow B. AC also is mandatory as “these cases” of C is an explanation of A. Also D looks like the logical ending and E the logical beginning. Hence the correct ans. is (1)

**Ans . **

(4) CEBDA

**Explanation :**C is the best beginning to the paragraph. C spells out the misnomer. E makes a statement on terror that is justified though B and in D as Besides. The humanitarian context of D is given in A.

**Ans . **

(2) CEABD

**Explanation :**CE gives the problem. A gives the solution. BD gives the Dvorak angle. Pay attention to the openers, To avoid this answers the problem. Similarly, D presents a contrast with Yet.

**Ans . **

(4) It was sheer luck that brought a bundle of boy-scouts to where I was lying wounded.

**Explanation :**bundle of boy-scouts is incorrect usage. The correct usage is a group of boy-scouts.

**Ans . **

(1) He is distinct about what is right and what is wrong.

**Explanation :**He is clear about what is would have been a better expression. The correct usage is “clear” about certain things.

**Ans . **

(1) Everyone appreciated the headmaster’s implication in raising flood relief in the village.

**Explanation :**appreciated the headmaster’s gesture of raising is the correct expression, implication implies negativity.

**Ans . **

(3) Ranchi will play the host to the next national film festival.

**Explanation :**Ranchi will play the host to is incorrect. The correct sentence should be ‘Ranchi will host’ the next national film festival.

**Ans . **

(2) Farmers of all sort attended the rally.

**Farmers of “all sorts” is the correct expression.**

**Ans . **

(2) conceded, offload

**Explanation :**conceded and offload are the most appropriate pair of words to fit here. announced do not go with formally, so (3) is out. Nor does ratified, so (4) is out. Acquire does not go logically with purchasers, so (1) is out.

**Ans . **

(3) identification, complicated

**Explanation :**If you have friends outside college, they tend to mask adjustment problems with college colleagues. treatment cannot be compounded, so (1) is out. If signals are masked, nothing is facilitated, so (2) is out. For similar reasons, helped in (4) cannot fill the second blank. Identification and complicated is thus the right pair.

**Ans . **

(1) different, discrete

**Explanation :**In the first blank the confusion could be between “different” and “distinct”. However once you know that certain regions of Spain are unique, only then can you call them distinct, not before. Which is why the first blank can’t be distinct. So the first blank should be different. Now between (1) and (4) the correct answer is (1) because discrete means distinct and so we are carrying forward the thought of difference between regions and then in the regions themselves.

**Ans . **

(1) resent, replacing

**Explanation :**resent and replacing is the most appropriate pair of words to fit here. welcome cannot go with the implication in unhappy so (3) is out. Resist is too extreme to fit in a teacher's situation, so (2) is out. are in (4) also indicate a compulsive situation which is not evident in the sentence, so (4) is out.

**Ans . **

(4)withholding, fostering

**Explanation :**Negative reinforcements foster negative behavior. (1), (2) and (3) are easily ruled out as giving, bestowing or conferring rewards cannot possibly encourage negative behaviour. Withholding and fostering thus presents the right situation here.

Statement B: The success rate of moving from written test to interview stage for females was better in 2002 than in 2003.

Choose (1) if only A is true

Choose (2) if only B is true

Choose (3) if both A and B are true

Choose (4) if neither A nor B is true

**Ans . **

(4) if neither A nor B is true

**Explanation :**Statement A: Success rate for males in 2003

=\( \frac{637}{60133} \)× 100 ≈ 1.06%

Success rate for females in 2003

=\( \frac{399}{40763} \)× 100 ≈ 0.98%

Hence ‘A’ is false.

Statement B: Success rate for females in 2002

=\( \frac{138}{15389} \)× 100 ≈ 0.89%

Success rate for females in 2003

=\( \frac{399}{40763} \)× 100 ≈ 0.98%

Hence ‘B’ is false.

52.Statement A: In 2002, the number of females selected for the course as a proportion of the number
of females who bought application forms, was higher than the corresponding proportion for males.

Statement B: In 2002, among those called for interview, males had a greater success rate than
females.

**Ans . **

(4) if neither A nor B is true

**Explanation :**Statement A: Females selected

=\( \frac{48}{19236} \)× 100 ≈ 0.25%

Males selected

=\( \frac{171}{61205} \)× 100 ≈ 0.28%

Hence ‘A’ is false.

Statement B: Success rate for Males

=\( \frac{17}{684} \)× 100 ≈ 25%

Success rate for Females

=\( \frac{48}{138} \)× 100 ≈ 34.8%

Hence ‘B’ is false.

53.Statement A: The percentage of absentees in the written test among females decreased from 2002
to 2003.

Statement B: The percentage of absentees in the written test among males was larger than among
females in 2003.

**Ans . **

(1) if only A is true

**Explanation :**Statement A: Female absentees in 2002 (19236 – 15389) = 3847

=\( \frac{3847}{19236} \)× 100 ≈ 20%

Female absentees in 2003 (45292 – 40763) = 4529

=\( \frac{4529}{45292} \)× 100 ≈ 10%

Hence ‘A’ is true.

Statement B: Male absentees in 2003 (63298 – 60133) = 3165

=\( \frac{63298 – 60133}{63298} \)× 100 ≈ 5%

Hence ‘B’ is false.

DIRECTIONS for Questions 54 to 57:Answer the questions on the basis of the information given below.

The length of an infant is one of the measures of his/her development in the early stages of his/her life.
The figure below shows the growth chart of four infants in the first five months of life.

**Ans . **

(2) Third month

**Explanation :**It is evident from the graph that Seeta's growth rate decreased from third month as this is the first time the slope has decreased.

55.Who grew at the fastest rate in the first two months of life?

**Ans . **

(1) Geeta

**Explanation :**Geeta grew at the fastest rate in the first two months (the slope of the line in this period is steepest for Geeta).

56.Who grew at the fastest rate in the first two months of life?

**Ans . **

(1) Geeta

**Explanation :**Geeta grew the lowest in the third month (during this period, the slope was least for Geeta).

57.Among the four infants, who grew the least in the first five months of life?

**Ans . **

(4) Shyam

**Explanation :**Seeta increased by 7 cm on 50 cm and Shyam by 7 cm on 53 cm. Hence, Shyam grew least.

DIRECTIONS for Questions 58 to 60: Answer the questions on the basis of the information given below.
The table below provides certain demographic details of 30 respondents who were part of a survey. The demographic characteristics are: gender, number of children, and age of respondents. The first number in
each cell is the number of respondents in that group. The minimum and maximum age of respondents in each group is given in brackets. For example, there are five female respondents with no children and
among these five, the youngest is 34 years old, while the oldest is 49.

**Ans . **

(4) 30%

**Explanation :**The possible combinations when the respondents are aged less than 40 years is minimum can be:

(i) No children – 1 male(aged 38) and atleast 1 female(aged 34)

(ii) 1 child – 1 male(aged 32) and atleast 1 female(aged 35)

(iii) 2 children – at least 1 male(aged 21) and at least 1 female(aged 37)

(iv) 3 children – 2 males(aged 32 and 33) and 1 female(aged 27)

i.e. there is at least 9 such respondents.

Required percentage = \( \frac{9}{30} \)× 100 = 30%

59.Given the information above, the percentage of respondents older than 35 can be at most

**Ans . **

(3) 76.67%

**Explanation :**The possible combinations when the respondents are aged more than 35 years is maximum can be:

(i) No children – 1 male(aged 38) and atmost 4 females

(ii) 1 children – 0 male and at most 7 female

(iii) 2 children – at most 7 males and 3 females

(iv) 3 children – 0 male and 1 female(aged 40)

i.e. there can be at most 23 such respondents.

Required percentage = \( \frac{23}{30} \) × 100 = 76.67%

60.The percentage of respondents that fall into the 35 to 40 years age group (both inclusive) is at least

**Ans . **

(3) 13.33%

**Explanation :**The possible combinations when the respondents are aged between 35 and 40 years(both inclusive) is minimum can be:

(i) No children – 1 male(aged 38) and 0 female

(ii) 1 children – 0 male and at least 1 female(aged 35)

(iii) 2 children – 0 males and at least 1 female (aged 37)

(iv) 3 children – 0 male and 1 female(aged 40)

i.e. there can be at least 4 such respondents.

Required percentage = \( \frac{4}{30} \) × 100 = 13.33%

DIRECTIONS for Questions 61 to 63: Answer the questions on the basis of the information given below.
Spam that enters our electronic mailboxes can be classified under several spam heads. The following table
shows the distribution of such spam worldwide over time. The total number of spam emails received during
December 2002 was larger than the number received in June 2003. The total number of spam emails
received during September 2002 was larger than the number received in March 2003. The figures in the
table represent the percentage of all spam emails received during that period, falling into those respective
categories.

**Ans . **

(3) Products

**Explanation :**In case of Products, percentage of spam emails is increasing but at decreasing rate, from Sep 2002 to Dec 2002 products increased by \( \frac{7 - 3}{3} \) ≈ 133% and in

Mar 2003 about \( \frac{7 - 4}{7} \) ≈ 43% and in Jun 2003 ( \( \frac{11- 10}{10} \) ≈ 10% )

62.In the health category, the number of spam emails received in December 2002 as compared to
June 2003.

**Ans . **

(1) was larger

**Explanation :**Since percentage of spam is Dec 2002 is higher than June 2003, and the number of total e-mails received is higher, hence number received in Dec 2002 is higher.

63.In the financial category, the number of spam emails received in September 2002 as compared to
March 2003.

**Ans . **

(4) cannot be determined

**Explanation :**Cannot be determined as in Sept 2002 percentage is lower as compared to March 2003, however the total number of emails received in Sept 2003 is higher than that in March 2002. Thus we cannot say anything.

DIRECTIONS for Questions 64 to 66: Answer the questions on the basis of the information given below.
One of the functions of the Reserve Bank of India is to mobilize funds for the Government of India by issuing
securities. The following table shows details of funds mobilized during the period July 2002 - July 2003.
Notice that on each date there were two rounds of issues, each with a different maturity.

**Ans . **

(2) 1

**Explanation :**It happened only once i.e; on 17-Jul-02

65.Which of the following is true?

**Ans . **

(3) On at least one occasion, the second round issue having lower maturity received a higher number of competitive bids.

**Explanation :**From the table we can see that for issue dated, 04 June-03, the 2nd round issue has a lower maturity and the competitive bids received are higher.

66.Which of the following statements is NOT true?

**Ans . **

(4) The value of non-competitive bids accepted in the first round is always greater than that in the second round.

**Explanation :**On 07-Nov 02, the value of non-competitive bids in the 2nd round is greater than that of 1st round. So option (4) is not true.

DIRECTIONS for Questions 67 to 69: Answer the questions on the basis of the information given below.
Each point in the graph below shows the profit and turnover data for a company. Each company belongs
to one of the three industries: textile, cement and steel.

**Ans . **

(2) 7

**Explanation :**Here the scale of the profit axis is exactly 10% of the scale of turnover axis. Just draw a diagonal line from bottom left point to top right point. All companies lying above this line have profit in excess of 10% of turnover. From the graph, there are 7 companies, has the profit 10% of turnover.

68.For how many steel companies with a turnover of more than 2000 is the profit less than 300?

**Ans . **

(3) 2

**Explanation :**From the graph there are 2 steel companies with a turnover of more than 2000 and profit less than 300.

69.An investor wants to buy stock of only steel or cement companies with a turnover more than
1000 and profit exceeding 10% of turnover. How many choices are available to the investor?

**Ans . **

(2) 5

**Explanation :**From the graph there are 5 companies.

DIRECTIONS for Questions 70 to 72: Answer the questions on the basis of the information given below.
Details of the top 20 MBA schools in the US as ranked by US News and World Report, 1997 are given
below.

**Ans . **

(4) University of California - Berkeley

**Explanation :**By looking up the table, in University of California - Berkeley median starting salary is $70,000 and annual tuition fee is $18,788.

71.In terms of staring salary and tuition fee, how many schools are uniformly better (higher median
starting salary AND lower tuition fee) than Dartmouth College?

**Ans . **

(2) 2

**Explanation :**By looking up the table, the number of schools, uniformly better than Dartmouth College is 2, namely Stanford and New York University.

72.How many schools in the list above have single digit rankings on at least 3 of the 4 parameters
(overall ranking, ranking by academics, ranking by recruiters and ranking by placement)?

**Ans . **

(2) 8

**Explanation :**8 universities namely, Stanford, Harvard, Pennsylvania, Massachusetts, Chicago, Northwestern, Columbia and Duke university have single digit ranking on atleast 3 of the 4 parameters.

DIRECTIONS for Questions 73 to 75: Answer the questions on the basis of the information given below.
Table A below provides data about ages of children in a school. For the age given in the first column, the
second column gives the number of children not exceeding the age. For example, first entry indicates that
there are 9 children aged 4 years or less. Tables B and C provide data on the heights and weights respectively
of the same group of children in a similar format. Assuming that an older child is always taller and
weighs more than a younger child, answer the following questions.

**Ans . **

(2) 45

**Explanation :**Number of children with age ≤ 9 years = 45

Number of children with height ≤ 135 cm = 48

Therefore, the number of children of age 9 years or less whose height does not exceed 135 cm will be the

common of the two (age ≤ 9 years and height ≤ 135 cm) = minimum(45, 48) = 45

74.How many children of age more than 10 years are taller than 150 cm and do not weigh more than
48 kg?

**Ans . **

(1) 16

**Explanation :**Number of children aged more than 10 years = 100 – 60 = 40

Number of children taller than 150 cm = 100 – 75 = 25

Number of children with weight more than 48 kg = 100 – 91 = 9

These 9 children are surely included in the 25 children

taller than 150 cm and more than 10 years of age because of the assumption given.

Thus, 25 – 9 = 16 children satisfy the given condition.

75.Among the children older than 6 years but not exceeding 12 years, how many weigh more than
38 kg.?

**Ans . **

(3) 44

**Explanation :**Number of children older than 6 years but not exceeding 12 years = 77 – 22 = 55

Number of children with weights not exceeding 38 kg = 33

These 33 children includes the 22 children with age not exceeding 6 years.

Therefore, the remaining (33 – 22) = 11 comes from

the 55 children of ages older than 6 years but not exceeding 12 years.

Therefore, 55 – 11 = 44 children satisfy the given condition.

DIRECTIONS for Questions 76 to 77: Answer the questions on the basis of the information given below.
An industry comprises four firms (A, B, C, and D). Financial details of these firms and of the industry as a
whole for a particular year are given below. Profitability of a firm is defined as profit as a percentage of sales.

**Ans . **

(4) D

**Explanation :**Profitability is defined as percentage of sales. Approximately, Firm A has 20% profit, B has 16.66%, C has 20% and D has approximately 25% profit.

77.If firm A acquires firm B, approximately what percentage of the total market (total sales) will they
corner together?

**Ans . **

(3) 44

**Explanation :**\( \frac{24568 + 25468}{89570} \) × 100 ≈ 55%

DIRECTIONS for Questions 78 to 80: Answer the questions on the basis of the information given below.
A, B, C, D, E, and F are a group of friends. There are two housewives, one professor, one engineer, one
accountant and one lawyer in the group. There are only two married couples in the group. The lawyer is
married to D, who is a housewife. No woman in the group is either an engineer or an accountant. C, the
accountant, is married to F, who is a professor. A is married to a housewife. E is not a housewife.

78.Which firm has the highest profitability?

**Ans . **

(4)

79.What is E's profession?

**Ans . **

(1)

80.How many members of the group are males?

**Ans . **

(2)

DIRECTIONS for Questions 81 and 82: Answer the questions on the basis of the information given below.
The Head of a newly formed government desires to appoint five of the six elected members A, B, C, D, E
and F to portfolios of Home, Power, Defence, Telecom and Finance. F does not want any portfolio if D gets
one of the five. C wants either Home or Finance or no portfolio. B says that if D gets either Power or
Telecom, then she must get the other one. E insists on a portfolio if A gets one.

81.Which is a valid assignment?

**Ans . **

(2) C-Home, D-Power, A-Defence, B-Telecom, E-Finance.

**Explanation :**For questions 81 and 82:

If D gets portfolio, F does not or vice-versa. C wants only Home or Finance or none. If D gets Power, B must get Telecom or if D gets Telecom, then B must get Power. If A gets a portfolio, E should get the same.

81. 2 Option (1) gets eliminated because C can have either Home or Finance. Option (3) gets eliminated because F and D cannot be in the same team. Option (4) gets eliminated because C cannot have Telecom portfolio. Hence, option (2) is the correct answer.

82.If A gets Home and C gets Finance, then which is NOT a valid assignment of Defense and Telecom?

**Ans . **

(4) B-Defence, D-Telecom.

**Explanation :**B-Defence, D - Telecom because if D gets Telecom then B must get Power.

DIRECTIONS for Questions 83 to 85: Answer the questions on the basis of the information given below.
Rang Barsey Paint Company (RBPC) is in the business of manufacturing paints. RBPC buys RED, YELLOW,
WHITE, ORANGE, and PINK paints. ORANGE paint can be also produced by mixing RED and YELLOW
paints in equal proportions. Similarly, PINK paint can also be produced by mixing equal amounts of RED
and WHITE paints. Among other paints, RBPC sells CREAM paint, (formed by mixing WHITE and YELLOW
in the ratio 70:30) AVOCADO paint (formed by mixing equal amounts of ORANGE and PINK paint) and
WASHEDORANGE paint (formed by mixing equal amounts of ORANGE and WHITE paint). The following
table provides the price at which RBPC buys paints.

**Ans . **

(2) Rs. 19.75 per litre

**Explanation :**AVOCADO paint is mixture of ORANGE and PINK in equal quantities.

If ORANGE is made using RED and YELLOW, then the

cost of ORANGE would be \( \frac{20 + 25}{2} \)= 22.5 which is

greater than the cost of the ORANGE.

If we make PINK by RED and WHITE, the cost of PINK would be \( \frac{20 + 15}{2} \)= 17.5 which is less than the cost of the PINK paint.

Hence, the cost of the AVOCADO is \( \frac{22 + 17.5}{2} \)= 19.5

84. WASHEDORANGE can be manufactured by mixing

**Ans . **

(4) RED, YELLOW, and WHITE in the ratio 1:1:2.

**Explanation :**Mixing equal amounts of ORANGE and WHITE can make WASHEDORANGE, ORANGE can be made by mixing equal amounts of RED and YELLOW. So the ratio of RED, YELLOW and WHITE is 1 : 1 : 2

85. Assume that AVOCADO, CREAM and WASHEDORANGE each sells for the same price. Which of the three is the most profitable to manufacture?

**Ans . **

(2) CREAM

**Explanation :**If cost of AVOCADO paint is Rs.19.75

The cost of the CREAM is \( \frac{7 x 15 + 3 x 75}{3} \) = Rs.18

And cost of WASHEDORANGE is Rs.18.50 So CREAM is the most profitable.

DIRECTIONS for Questions 86 to 88: Answer the questions on the basis of the information given below.
Seven varsity basketball players (A, B, C, D, E, F, and G) are to be honoured at a special luncheon. The
players will be seated on the dais in a row. A and G have to leave the luncheon early and so must be seated
at the extreme right. B will receive the most valuable player's trophy and so must be in the centre to
facilitate presentation. C and D are bitter rivals and therefore must be seated as far apart as possible.

**Ans . **

(3) F

**Explanation :**From given options, F is the only possibility.

87.Which of the following pairs cannot be seated together?

**Ans . **

(4) E & A

**Explanation :**If we look at the options, D and G can sit together. C and F can sit together B and D can sit together. Hence, E and A is the only option which is not possible.

88.Which of the following pairs cannot occupy the seats on either side of B?

**Ans . **

(3) E & G

**Explanation :**E and G is the only possibility.

DIRECTIONS for Questions 89 to 92: In each question there are two statements: A and B.

Choose (1) if the question can be answered by one of the statements alone but not by the other.

Choose (2) if the question can be answered by using either statement alone.

Choose (3) if the question can be answered by using both the statements together but cannot be answered
using either statement alone.

Choose (4) if the question cannot be answered even by using both the statements A and B.

89. F and M are father and mother of S, respectively. S has four uncles and three aunts. F has two
siblings. The siblings of F and M are unmarried. How many brothers does M have?

A. F has two brothers.

B. M has five siblings.

**Ans . **

(1)

**Explanation :**S has 4 uncles and from statement A. F has two brothers. Hence, the other 2 uncles of S must be the

brothers of M. Statement B does not give any additional information.

90. A game consists of tossing a coin successively. There is an entry fee of Rs. 10 and an additional fee of Re. 1 for each toss of coin. The game is considered to have ended normally when the coin turns heads on two consecutive throws. In this case the player is paid Rs. 100. Alternatively, the player can choose to terminate the game prematurely after any of the tosses. Ram has incurred a loss of Rs. 50 by playing this game. How many times did he toss the coin?

A. The game ended normally.

B. The total number of tails obtained in the game was 138.

**Ans . **

(2)

**Explanation :**If Ram tossed the coin x number of times, then from statement A, we get the equation 10 + x – 100 = 50.

Thus, x = 140.

From statement II individually, we have x > 138.

Thus, we are sure that he has paid up more than 148. If he incurs a loss of only Rs. 50, the game has to endno other scenario, a loss of just Rs.50 and 138 tails will show up.

91. Each packet of SOAP costs Rs. 10. Inside each packet is a gift coupon labelled with one of the letters S, O, A and P. If a customer submits four such coupons that make up the word SOAP, the customer gets a free SOAP packets. Ms. X kept buying packet after packet of SOAP till she could get one set of coupons that formed the word SOAP. How many coupons with label P did she get in the above process?

A. The last label obtained by her was S and the total amount spent was Rs. 210.

B. The total number of vowels obtained was 18.

**Ans . **

(3)

**Explanation :**Since Ms. X bought 21 packets out of which there are 18 O’s and A’s in total. Since she got one S, there has to be 2 P’s which she bought. Hence, both the statements are required to answer the question.

92. If A and B run a race, then A wins by 60 seconds. If B and C run the same race, then B wins by 30 seconds. Assuming that C maintains a uniform speed what is the time taken by C to finish the race?

A. A and C run the same race and A wins by 375 metres.

B. The length of the race is 1 km.

**Ans . **

(3)

**Explanation :**Either of the statements is alone not sufficient. Using both the statements together: If A takes X seconds then B takes (x + 60) seconds to run 1000 m. Ratio of speeds of A and C = 1000 : 625 = 8 : 5 Ratio of time taken by A and C = 5 : 8 If B takes y second then C takes (y + 30) seconds to run 1000 m, then 5 (y + 30) = 8x …(i) and \( \frac{1000}{x + 60 } \)= \( \frac{1000}{y } \) …(ii)

Solving, we get the values of x and y.

Hence, both statements are required.

DIRECTIONS for Questions 93 to 94: Answer the questions on the basis of the information given below.
Some children were taking free throws at the basketball court in school during lunch break. Below are
some facts about how many baskets these children shot.
i. Ganesh shot 8 baskets less than Ashish.

ii. Dhanraj and Ramesh together shot 37 baskets.

iii. Jugraj shot 8 baskets more than Dhanraj.

iv. Ashish shot 5 baskets more than Dhanraj.

v. Ashish and Ganesh together shot 40 baskets.

**Ans . **

1.

**Explanation :**For questions 93 and 94: G + 8 = A

D + R = 37

J = D + 8

A + G = 40

Solving the above equations, we get

2G = 32, G = 16, A = 24

D = 19, J = 27, R = 18

94.Which of the following statements is true?

**Ans . **

1.

**Explanation :**D + J = 46

DIRECTIONS for Questions 95 to 97: Answer the questions on the basis of the information given below.
Five women decided to go shopping to M.G. Road, Bangalore. They arrived at the designated meeting
place in the following order: 1. Archana, 2. Chellamma, 3. Dhenuka, 4. Helen, and 5. Shahnaz. Each
woman spent at least Rs. 1000. Below are some additional facts about how much they spent during their
shopping spree.
i. The woman who spent Rs. 2234 arrived before the lady who spent Rs. 1193.
ii. One woman spent Rs. 1340 and she was not Dhenuka.
iii. One woman spent Rs. 1378 more than Chellamma.
iv. One woman spent Rs. 2517 and she was not Archana.
v. Helen spent more than Dhenuka.
vi. Shahnaz spent the largest amount and Chellamma the smallest.

For questions 95 to 97:
Four of the amounts spent by the five women are Rs.2234,
Rs.1193, Rs.1340 ad Rs.2517.

Two cases arise:

(i) The lowest amount spent is Rs.1193(by Chellamma):
Then, the fifth amount will be Rs(1193 + 1378) = Rs.2571,

which will then be the highest amount and is spent by Shahnaz.
As Archana arrived before Chellamma, so she must have
spent Rs.2234.

This implies Helen spent Rs.2517 and Dhenuka
spent Rs.1340, which is a contradiction.

Hence, this case is not possible.

(ii) The highest amount spent is Rs.2517 (by Shahnaz):
Then the fifth amount will be Rs.(2517 – 1378) = Rs.1139.
Since it is the lowest amount, it will be spent by Chellamma.

Further analysis leads to the following table:

**Ans . **

2.

96.Which of the following amounts was spent by one of them?

**Ans . **

1.

97.The woman who spent Rs. 1193 is

**Ans . **

3.

DIRECTIONS for Questions 98 to 100: Answer the questions on the basis of the information given below.
Five friends meet every morning at Sree Sagar restaurant for an idli-vada breakfast. Each consumes a
different number of idlis and vadas. The number of idlis consumed are 1, 4, 5, 6, and 8, while the number of
vadas consumed are 0, 1, 2, 4, and 6. Below are some more facts about who eats what and how much.

i. The number of vadas eaten by Ignesh is three times the number of vadas consumed by the person
who eats four idlis.

ii. Three persons, including the one who eats four vadas eat without chutney.

iii. Sandeep does not take any chutney.

iv. The one who eats one idli a day does not eat any vadas or chutney. Further, he is not Mukesh.

v. Daljit eats idli with chutney and also eats vada.

vi. Mukesh, who does not take chutney, eats half as many vadas as the person who eats twice as
many idlis as he does.

vii. Bimal eats two more idlis than Ignesh, but Ignesh eats two more vadas than Bimal.

For questions 98 to 100:

From statement (i), possible number of vadas consumed by
Ignesh is 6, being the only multiple of 3.Therefore, another
person had 4 idlis and 2 vadas.
From statement (vii), Bimal had (6 – 2) = 4 vadas.
Using these inferences and statements (ii), (iii) and (vi), we
get that Bimal, Sandeep and Mukesh do not have chutney,
while Ignesh and Daljit consume chutney.
From (iii) and (iv), Sandeep has only one idli and no vada.
From (vii), Bimal has two more idlis than Ignesh. This implies
that Bimal can have either 6 or 8 idlis. If Bimal had 6 idlis, then
Ignesh had 4 idlis. This contradicts statement (i). Therefore,
Bimal had 8 idlis and Ignesh had 6 idlis.
Mukesh has half the number of idlis as one other person and
the only number satisfying this is 4. Therefore, he must have 2
vadas.
These inferences can be summarised in the table below:

**Ans . **

1.

99.Which of the following statements is true?

**Ans . **

3.

100.Which of the following statements is true?

**Ans . **

3.

DIRECTIONS for Questions 101 and 102: Answer the questions on the basis of the information given
below.
A certain perfume is available at a duty-free shop at the Bangkok international airport. It is priced in the Thai
currency Baht but other currencies are also acceptable. In particular, the shop accepts Euro and US Dollar
at the following rates of exchange:
US Dollar 1 = 41 Bahts
Euro 1= 46 Bahts
The perfume is priced at 520 Bahts per bottle. After one bottle is purchased, subsequent bottles are
available at a discount of 30%. Three friends S, R and M together purchase three bottles of the perfume,
agreeing to share the cost equally. R pays 2 Euros. M pays 4 Euros and 27 Thai Bahts and S pays the
remaining amount in US Dollars.

For questions 101 to 102:
S, M and R in all spend 1248 Bahts.
Initially M pays 211 Bahts and R pays 92 Bahts.
Remaining is paid by S i.e; 945 Bahts
If 1248 is divided equally among S, M and R, each has to spend
415 Bahts.
Hence, M has to pay 205 Bahts which is 5 Dollars to S.
and R has to pay 324 Bahts to S.
101.How much does R owe to S in Thai Baht?

**Ans . **

4.

102.How much does M owe to S in US Dollars?

**Ans . **

3.

DIRECTIONS for Questions 103 and 104: Answer the questions on the basis of the information given
below.
New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number of projects
handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All
three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has
2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number
of projects for New Age Consultants is one less than twice the number of projects in which more than one
consultant is involved.

**Ans . **

4.

**Explanation :**Putting the value of M in either equation, we get

G + B = 17.

Hence neither of two can be uniquely determined.

104.What is the number of projects in which Medha alone is involved?

**Ans . **

2.

**Explanation :** G + B = M + 16

Also, M + B + G + 19 = (2 × 19) – 1

i.e. (G + B) = 18 – M

Thus, M + 16 = 18 – M

i.e. M = 1

DIRECTIONS for Questions 105 to 110: Answer the questions independently of each other.

**Ans . **

3.

**Explanation :**2^{x}– x – 1 = 0

⇒ 2^{x} – 1 = x

If we put x = 0, then this is satisfied and if we put

x = 1, then also this is satisfied.

Now, if we put x = 2, the equation this is not valid.

106.When the curves y = log_{10}x and y = x^{-1} are drawn in the x-y plane, how many times do they intersect for values x ≥ 1 ?

**Ans . **

2.

**Explanation :**For the curves to intersect,

log_{10}x = x^{-1}

Thus, log_{10}x = \( \frac{1}{x} \)or x^{x} = 10

This is possible for only one value of x such that

2 < x < 3.

107.Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be

**Ans . **

4.

**Explanation :**The surface area of a sphere is proportional to the square of the radius.

Thus,\( \frac{s_B}{s_A} \)= \( \frac{4}{1} \) (Surface area of B is 300% higher than A)

∴ \( \frac{r_B}{r_A} \)= \( \frac{2}{1} \)

The volume of a sphere is proportional to the cube of the radius.

Thus,\( \frac{v_B}{v_A} \)= \( \frac{8}{1} \)

Hence, V_A is \( \frac{7}{8} \)th less than B i.e.\( \frac{7}{8} \)× 100 = 87.5%.

108.Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, such that p + q + r ≠ 0?

x+ 2y – 3z = p

2x + 6y – 11z = q

x – 2y + 7z = r

**Ans . **

1.

**Explanation :**It is given that p + q + r ≠ 0 , if we consider the first option, and multiply the first equation by 5, second by

–2 and third by –1, we see that the coefficients of x,

y and z all add up-to zero.

Thus, 5p – 2q – r = 0

No other option satisfies this.

109.A leather factory produces two kinds of bags, standard and deluxe. The profit margin is Rs. 20 on a
standard bag and Rs. 30 on a deluxe bag. Every bag must be processed on machine A and on
Machine B. The processing times per bag on the two machines are as follows:

**Ans . **

(1)

**Explanation :**Let 'x' be the number of standard bags and 'y' be the number of deluxe bags. Thus, 4x + 5y ≤ 700 and 6x + 10y ≤ 1250

Among the choices, (3) and (4) do not satisfy the second equation.

Choice (2) is eliminated as, in order to maximize profits

the number of deluxe bags should be higher than the number of standard bags because the profit margin is higher in a deluxe bag.

110.The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?

**Ans . **

3.

**Explanation :**Let the 1st term be ‘a’ and common difference be ‘d’

then we have 3rd term = a + 2d

15th term = a + 14d

6th term = a + 5d

11th term = a + 10d

13th term = a + 12d

Since sum of 3rd and 15th term = sum of 6th, 11th and

13th term, therefore we have

2a + 16d = 3a + 27d

⇒ a + 11d = 0

Which is the 12th term.

DIRECTIONS for Questions 111 to 113: Answer the questions on the basis of the information given below.
A city has two perfectly circular and concentric ring roads, the outer ring road (OR) being twice as long as
the inner ring road (IR). There are also four (straight line) chord roads from E1, the east end point of OR to
N2, the north end point of IR; from N1, the north end point of OR to W2, the west end point of IR; from W1,
the west end point of OR, to S2, the south end point of IR; and from S1 the south end point of OR to E2, the
east end point of IR. Traffic moves at a constant speed of 30π km/hr on the OR road, 20π km/hr on the IR
road, and 15 5 km/hr on all the chord roads.

**Explanation :**For questions 111 to 113:

The length of IR = 2π r, that of OR = 4π r and that of the chord roads are r√5 (Pythagoras theorem) The corresponding speeds are 20π, 30π and 15√5 kmph.

Thus time taken to travel one circumference of

IR = \( \frac{r}{10} \) hr,one circumference of OR = \( \frac{r}{7.5} \) hr and

one length of the chord road = \( \frac{r}{15} \)hr

111.The ratio of the sum of the lengths of all chord roads to the length of the outer ring road is

**Ans . **

3.

**Explanation :**Sum of the length of the chord roads = 4r√5 and the

length of OR = 4π r.

Thus the required ratio = √5 : π

112.Amit wants to reach N2 from S1. It would take him 90 minutes if he goes on minor arc S1 – E1 on OR, and then on the chord road E1 – N2. What is the radius of the outer ring road in kms?

**Ans . **

3.

**Explanation :**The total time taken by the route given

= \( \frac{r}{30} \) + \( \frac{r}{15} \)= \( \frac{3}{2} \)

(i.e. 90 min.)

Thus, r = 15 km. The radius of OR = 2r = 30 kms

113.Amit wants to reach E2 from N1 using first the chord N1 – W2 and then the inner ring road. What will be his travel time in minutes on the basis of information given in the above question?

**Ans . **

4.

**Explanation :**The total time taken = \( \frac{r}{20} \) + \( \frac{r}{15} \)= \( \frac{7r}{60} \)

Since r = 15, total time taken = \( \frac{7}{4} \)hr = 105 min.

DIRECTIONS for Questions 114 to 120: Answer the questions independently of each other.

**Ans . **

3.

**Explanation :**Let the number of questions answered correctly be

'x', that of answered wrongly be 'y' and that of left

unattempted be 'z'.

Thus, x + y + z = 50 …(i)

And x– \( \frac{y}{3} \)- \( \frac{z}{6} \)= 32

The second equation can be written as,

6x – 2y – z = 192 …(ii)

Adding the two equations, we get,

7x – y = 242 or x = \( \frac{242 + y}{7} \)

Since x and y are both integers, the minimum value of y must be 3.

DIRECTIONS for Questions 114 to 120: Answer the questions independently of each other.

**Ans . **

2.

**Explanation :**Since there are 27 people, each person can have 0 upto 26 acquaintances.

If a person has zero acquaintances, then the maximum number of acquaintances any of the other persons can have is 25.

Similarly, if a person has one acquaintance, then the maximum number of acquaintances any of the other persons can have is 26.

Therefore, the number of acquaintances can be any number from 0 to 25 or from 1 to 26. This rules out options (1) and (3).

The congregation consists of 27 people whereas the number of acquaintances any person can have is 26

(either 0 to 25 or 1 to 26). This implies that there is one person who share the same number of acquaintances

as atleast one of the other persons. This contradicts option (2).

Hence, (2) is the desired option.

NOTE: If we consider the situation other wise, to satisfy condition 2, the first person must have 26 acquaintances, the second 25, third 24 and so on. If we continue, the last one should have 0 acquaintance, which is not possible.

116.Let g(x) = max(5 – x, x + 2). The smallest possible value of g(x) is

**Ans . **

4.

**Explanation :**We can see that x + 2 is an increasing function and

5 – x is a decreasing function. This system of equation

will have smallest value at the point of intersection of

the two. i.e. 5 – x = x + 2 or x = 1.5.

Thus smallest value of g(x) = 3.5

117.The function f(x) = |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at

**Ans . **

2.

**Explanation :**Case 1: If x < 2, then y = 2 – x + 2.5 – x + 3.6 – x

= 8.1 – 3x.

This will be least if x is highest i.e. just less than 2.

In this case y will be just more than 2.1

Case 2: If 2 ≤ x < 2.5 , then y = x – 2 + 2.5 – x + 3.6 – x = 4.1 – x

Again, this will be least if x is the highest i.e. just less

than 2.5. In this case y will be just more than 1.6.

Case 3: If 2.5 ≤ x < 3.6 , then y = x – 2 + x – 2.5 + 3.6

– x = x – 0.9

This will be least if x is least i.e. x = 2.5.

Case 4: If x ≥ 3.6 , then

y = x – 2 + x – 2.5 + x – 3.6 = 3x – 8.1 The minimum value of this will be at x = 3.6 and y = 2.7 Hence, the minimum value of y is attained at x = 2.5

Alternate method:

At x = 2, f(x) = 2.1

At x = 2.5, f(x) = 1.6

At x = 3.6, f(x) = 2.7

Hence, at x = 2.5, f(x) will be minimum.

118.How many even integers n, where 100 ≤ n ≤ 200 , are divisible neither by seven nor by nine?

**Ans . **

3.

**Explanation :**There are 101 integers between 100 and 200, of which

51 are even.

Between 100 and 200, there are 14 multiples of 7, of

which 7 are even.

There are 11 multiples of 9, of which 6 are even.

But there is one integer (i.e., 126) that is a multiple of

both 7 and 9 and also even.

Hence, the answer is (51 – 7 – 6 + 1) = 39.

119.A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals

**Ans . **

4.

**Explanation :**Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91. Among these only 31 and 91 are a part of the answer choices. Among these,(31)_{10}=(11111)_{2}=(1011)_{3}=(111)_{5}

Thus, all three forms have leading digit 1.

Hence the answer is 91.

120.In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?

**Ans . **

3.

**Explanation :**The ratio of the speeds of the fastest and the slowest

runners is 2 : 1. Hence they should meet at only one

point on the circumference i.e. the starting point (As

the difference in the ratio in reduced form is 1). For the

two of them to meet for the first time, the faster should

have completed one complete round over the slower

one. Since the two of them meet for the first time after

5 min, the faster one should have completed 2 rounds

(i.e. 2000 m) and the slower one should have completed

1 round. (i.e. 1000 m) in this time. Thus, the faster one

would complete the race (i.e. 4000 m) in 10 min.

DIRECTIONS for Questions 121 to 125: Each question is followed by two statements, A and B.

Answer each question using the following instructions.

Choose (1) if the question can be answered by one of the statements alone but not by the other.

Choose (2) if the question can be answered by using either statement alone.

Choose (3) if the question can be answered by using both the statements together, but cannot be answered

by using either statement alone.

Choose (4) if the question cannot be answered even by using both the statements together.

121.Is a^44 < b^11, given that a = 2 and b is an integer?

A. b is even

B. b is greater than 16

**Ans . **

(1)

**Explanation :**Since b can take any even number 2, 4, 6,…, we

cannot say anything from statement A.

Consider statement B.

If b > 16, say b = 17, then 2^44 < (16 + 1)^11. ⇒ 2^44 < (24 + 1)^11 Hence, we can answer the question using statement B alone.

122.What are the unique values of b and c in the equation 4x2 + bx + c = 0 if one of the roots of the equation is (–1/2)?

A. The second root is 1/2.

B. The ratio of c and b is 1.

**Ans . **

(2)

**Explanation :**Solution can be found using Statement A alone as we know both the roots for the equation (viz.\( \frac{1}{2} \)and - \( \frac {1}{2} \)

Also, statement B alone is sufficient.

Since ratio of c and b = 1, c = b.

Thus, the equation is 4x2 + bx + b = 0. Since x = - \( \frac {1}{2} \)is

one of the roots, substituting, we get 1 - \( \frac {b}{2} \)+ b = 0 or

b = –2.

Thus, c = –2.

123.AB is a chord of a circle. AB = 5 cm. A tangent parallel to AB touches the minor arc AB at E. What is the radius of the circle?

A. AB is not a diameter of the circle.

B. The distance between AB and the tangent at E is 5 cm.

**Ans . **

(1)

**Explanation :** We can get the answer using the second statement

only. Let the radius be r.

AC = CB = 2.5 and using statement B, CE = 5, thus OC = (r – 5).

Using Pythagoras theorem, (r – 5)^2 + (2.5)^2 = r^2

We get r = 3.125 cm.

Note: You will realize that such a circle is not possible (if r = 3.125 how can CE be 5). However we need to

check data sufficiency and not data consistency. Since we are able to find the value of r uniquely using

second statement the answer is (1).

124.Is (1/a^2 + 1/a^4 +1/a^6 +...)> (1/a + 1/a^3 +1/a^5+...)?

A. −3 ≤ a ≤ 3

B. One of the roots of the equation 4x^2–4x+1 = 0 is a

**Ans . **

(1)

**Explanation :**Both the series are infinitely diminishing series.

For the first series: First term = 1/a^2 and r = 1/a^2

For the second series: First term = 1/a^ and r = 1/a^2

The sum of the first series = (1/a^2 )/(1- 1/a^2 )= (1/a^2-1)

The sum of the second series = (1/a )/(1- 1/a^2 )= (1/a^2-1)

Now, from the first statement, the relation can be

anything (depending on whether a is positive or

negative).

But the second statement tells us, 4a^2 – 4a + 1 = 0 or

a = \( \frac{1}{2} \)\. For this value of a, the sum of second series

will always be greater than that of the first.

125.Is D, E, F are the mid points of the sides AB, BC and CA of triangle ABC respectively. What is the area of DEF in square centimeters?

A. AD = 1 cm, DF = 1 cm and perimeter of DEF = 3 cm

B. Perimeter of ABC = 6 cm, AB = 2 cm and AC = 2 cm.

**Ans . **

(2)

**Explanation :**From statement (A):

We can find the sides of triangle ABC.

Hence, area of the triangle ABC and triangle DEF can

be found.

From statement (B):

The question tells us that the area of triangle DEF will

be 1/4th the area of triangle ABC. Thus by knowing

either of the statements, we get the area of the triangle

DEF.

DIRECTIONS for Questions 126 to 150: Answer the questions independently of each other.

126.At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of
the goats at the beginning of the year and sold q% of the goats at the end of the year where p > 0
and q > 0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that
year, which of the following is true?

**Ans . **

(3)

**Explanation :**Given that Shephard had 9 dozens of goat at the end

of 1998.

∴Number of goats at the beginning of 1999 =

(1 + p%) 9 dozens

He sells off q% of this at the end of the year 1999.

∴Number of goats at the end of 1999 = (1 – q%)

(1 + p%) 9 dozens

Since every year the same process is repeated and

at the end of 2002, he has the same number as at the

end of 1998,

we can get that (1 + p%) (1 – q%) = 1

⇒ 1 + p – q – pq = 1

⇒ p – q = pq

⇒ p > q (Since p > 0 and q > 0)

127.Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90° or concave if the internal angle is 270°. If the number of convex corners in such a polygon is 25, the number of concave corners must be

**Ans . **

(3)

**Explanation :**Let the total number of angles in the polygon be x.

Therefore, the number of concave corners will be

x – 25.

∴25 × 90 + (x – 25) × 270 = (x – 2) × 180

⇒ x = 46 ⇒ x – 25 = 21.

Alternative solution:

In this kind of polygon, the number of convex angles will always be exactly 4 more than the number of concave angles.

Note: The number of vertices have to be even. Hence,

the number of concave and convex corners should add up to an even number. This is true only for the

128.The 288th term of the series a,b,b,c,c,c,d,d,d,d,e,e,e,e,e,f,f,f,f,f,f… is

**Ans . **

(4)

**Explanation :**The number of terms of the series forms the sum of

first n natural numbers i.e.

\( \frac{n(n + 2)}{2} \)

Thus the first 23 letters will account for the first

\( \frac{n(23 x 24)}{2} \)= 276 terms of the series.

The 288th term will be the 24th letter which is x.

129.Let p and q be the roots of the quadratic equation x2 − (α − 2)x − α −1 = 0 . What is the minimum possible value of p2 + q2?

**Ans . **

(4)

**Explanation :**p + q = α –2 and pq = –α – 1

(p + q)2 = p2 + q2 + 2pq,

Thus (α –2)2 = p2 + q2 + 2(–α – 1)

p2 + q2 = α2 – 4α + 4 + 2α + 2

p2 + q2 = α2 – 2α + 6

p2 + q2 = α2 – 2α + 1 + 5

p2 + q2 = (α – 1)2 + 5

Thus, minimum value of p2 + q2 is 5.

130.There are two concentric circles such that the area of the outer circle is four times the area of the inner circle. Let A, B and C be three distinct points on the perimeter of the outer circle such that AB and AC are tangents to the inner circle. If the area of the outer circle is 12 square centimeters then the area (in square centimeters) of the triangle ABC would be

**Ans . **

(3)

**Explanation :**Since the area of the outer circle is 4 times the area of

the inner circle, the radius of the outer circle should be

2 times that of the inner circle.

Since AB and AC are the tangents to the inner circle,

they should be equal. Also, BC should be a tangent to

inner circle. In other words, triangle ABC should be

equilateral.

The area of the outer circle is 12.

Hence, the area of inner circle is 3 or the radius is √(3/π)

The area of equilateral triangle \( \frac{9√3}{π} \)

131.Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?

**Ans . **

(2)

**Explanation :**(a + b + c + d)^2 = (4m + 1)^2

Thus, a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd)

= 16m^2 + 8m + 1

a^2 + b^2 + c^2 + d^2 will have the minimum value if

(ab + ac+ ad + bc + bd + cd) is the maximum.

This is possible if a = b = c = d = (m + 0.25)

[since a + b + c + d = 4m + 1]

In that case, 2(ab + ac + ad + bc + bd + cd)

= 12(m + 0.25)2 = 12m^2 + 6m + 0.75

Thus, the minimum value of a^2 + b^2 + c^2 + d^2

= (16m^2 + 8m + 1) – 2(ab + ac + ad + bc + bd + cd)

= (16m^2 + 8m + 1) – (12m^2 + 6m + 0.75)

= 4m^2 + 2m + 0.25

Since it is an integer, the actual minimum value

= 4m^2 + 2m + 1

132.Three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with center at O. Horses are tied up at P, R and S such that PO and RO are the radii of semi-circles with centers at P and R respectively, and S is the center of the circle touching the two semi-circles with diameters AO and OB. The horses tied at P and R can graze within the respective semi-circles and the horse tied at S can graze within the circle centred at S. The percentage of the area of the semi-circle with diameter AB that cannot be grazed by the horses is nearest to

**Ans . **

(2)

**Explanation :**If the radius of the field is r, then the total area of the

field = 〖πr〗^2/2

The radius of the semi-circles with centre's P and R = \( \frac{r}{2} \)

Hence, their total area = 〖πr〗^2/4

Let the radius if the circle with centre S be x.

Thus, OS = (r – x), OR = \( \frac{r}{2} \)and RS x = \( \frac{r}{2} \)+ x

Applying Pythagoras Theorem, we get

(r-x)^2 + (r/2)^2 =(r/2+x)^2

Solving this, we get x = \( \frac{r}{3} \)

Thus the area of the circle with centre s = 〖πr〗^2/9

The total area that can be grazed = 〖πr〗^2 (1/4+1/9) = \( \frac{13πr^2}{36} \)

Thus the area of the field that cannot be

grazed = 〖πr〗^2/9 - \( \frac{13πr^2}{36} \) = \( \frac{5πr^2}{36} \)

The percentage = (5/36 πr^2)/(1/2 πr^2 )× 100 = 28.

133.In the figure below, ABCDEF is a regular hexagon and ∠AOF = 90° . FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF?

**Ans . **

(1)

**Explanation :**It is very clear, that a regular hexagon can be divided

into six equilateral triangles. And triangle AOF is half

of an equilateral triangle.

Hence, the required ratio = 1 : 12.

134.How many three digit positive integers, with digits x, y and z in the hundred's, ten's and unit's place respectively, exist such that x < y, z < y and x ≠ 0 ?

**Ans . **

(3)

**Explanation :**If y = 2 (it cannot be 0 or 1), then x can take 1 value

and z can take 2 values.

Thus with y = 2, a total of 1 × 2 = 2 numbers can be

formed. With y = 3, 2 × 3 = 6 numbers can be formed.

Similarly checking for all values of y from 2 to 9 and

adding up we get the answer as 240.

135.A vertical tower OP stands at the center O of a square ABCD. Let h and b denote the length OP and AB respectively. Suppose ∠APB = 60° then the relationship between h and b can be expressed as

**Ans . **

(2)

**Explanation :**

Given∴ PQ = \( \frac{b}{2} \) × √3

Next, \( \frac{b}{2} \) , h and PQ form a right angle triangle.

\( \frac{b^2 }{4} \) + h^2 = \( \frac{3b^2}{4} \)

⇒ 2h^2 = b^2

136.In the triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with center B) is drawn. If the circle cuts AB and BC at P and Q respectively, the AP : QC is equal to

**Ans . **

(4)

**Explanation :**

Triangle ABC is a right angled triangle.

Thus \( \frac{1}{2} \) × BC × AB = \( \frac{1}{2} \)= \( \frac{1}{2} \)× BD × AC

Or, 6 × 8 = BD × 10. Thus BD = 4.8.

Therefore, BP = BQ = 4.8.

So, AP = AB – BP = 6 – 4.8 = 1.2 and CQ = BC – BQ

= 8 – 4.8 = 3.2.

Thus, AP : CQ = 1.2 : 3.2 = 3 : 8

137.In the diagram given below,

**Ans . **

(2)

**Explanation :**Using the Basic Proportionality Theorem,

\( \frac{AB}{PQ} \) = \( \frac{BD}{QD} \) and \( \frac{PQ}{CD} \) = \( \frac{BQ}{BD} \)

Multiplying the two we get, \( \frac{AB}{CD} \) = \( \frac{BQ}{QD} \)= 3 : 1.

Thus CD : PQ = BD : BQ = 4 : 3 = 1 : 0.75

138.There are 8436 steel balls, each with a radius of 1 centimeter, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is

**Ans . **

(3)

**Explanation :**Assume the number of horizontal layers in the pile be n. So Σ\( \frac{n(n+1)}{2} \)= 8436

⇒ \( \frac{1}{2} \)[Σn^2 + Σn]= 8436

\( \frac{n(n+1)(2n+1)}{12} \)= 8436

⇒ n(n + 1)[\( \frac{2n +4}{12} \)] = 8436

⇒n(n + 1)(n + 2) = 36 × 37 × 38

So n = 36

139.If the product of n positive real numbers is unity, then their sum is necessarily

**Ans . **

(3)

**Explanation :**The best way to do this is to take some value and verify.

E.g. 2,\( \frac{1}{2} \) and 1. Thus, n = 3 and the sum of the three

numbers = 3.5.

Thus, options (a), (b) and (d) get eliminated.

Alternative method:

Let the n positive numbers be a1, a2, a3, … an

a1, a2, a3, … an = 1

We know that AM ≥ GM

Hence,\( \frac{1}{n} \)(a_1 + a_2 + a_3 + a_n)≥ (a_1a_2....a_n)^1/n

140.If log_3 2, log_3 (2x – 5), log_3 (2x – 7/2) are in arithmetic progression, then the value of x is equal to

**Ans . **

(4)

**Explanation :**Using log a – log b = (a/b),(2/y-5)=(y-5/y-3.5)

where

y = 2^x

Solving we get y = 4 or 8

i.e. x = 2 or 3.

It cannot be 2 as log of negative number is not defined

(see the second expression).

141.In the figure below, AB is the chord of a circle with center O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If

**Ans . **

(1)

**Explanation :**If y = 10°, BOC = 10° (opposite equal sides)

∠OBA = 20° (external angle of ΔBOC )

∠OAB = 20 (opposite equal sides)

∠AOD = 30° (external angle of ΔAOC )

Thus k = 3

142.In the figure below, the rectangle at the corner measures 10 cm × 20 cm. The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?

**Ans . **

(3)

**Explanation :**

Let the radius be r. Thus by Pythagoras’ theorem for ΔABC we have (r – 10)^2 + (r – 20)^2 = r^2 i.e. r^2 – 60r + 500 = 0. Thus r = 10 or 50. It would be 10, if the corner of the rectangle had been lying on the inner circumference. But as per the given diagram, the radius of the circle should be 50 cm.

143.Given that −1≤ v ≤ 1, −2 ≤ u ≤ −0.5 and −2 ≤ z ≤ −0.5 and w = vz /u , then which of the following is necessarily true?

**Ans . **

(2)

**Explanation :**u is always negative. Hence, for us to have a minimum

value of \( \frac{vz}{u} \)

, vz should be positive. Also, for the least

value, the numerator has to be the maximum positive

value and the denominator has to be the smallest

negative value. In other words, vz has to be 2 and u

has to be –0.5.

Hence, the minimum value of \( \frac{vz}{u} \)= \( \frac{2}{-0.5} \)= –4.

To get the maximum value, vz has to be the smallest

negative value and u has to be the highest negative

value. Thus, vz has to be –2 and u has to be –0.5.

Hence, the maximum value of \( \frac{-2}{} -0.5\)= 4.

144.There are 6 boxes numbered 1,2,… 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

**Ans . **

(2)

**Explanation :**GRRRRR, RGRRRR, RRGRRR, RRRGRR, RRRRGR,

RRRRRG

GGRRRR, RGGRRR, RRGGRR, RRRGGR, RRRRGG

GGGRRR, RGGGRR, RRGGGR, RRRGGG

GGGGRR, RGGGGR, RRGGGG

GGGGGR, RGGGGG

GGGGGG Hence 21 ways.

145.Consider the following two curves in the x-y plane: y = x^3 + x^2 + 5 y = x^2 + x + 5 Which of following statements is true for −2 ≤ x ≤ 2 ?

**Ans . **

(4)

**Explanation :**Putting y_1 = y_2, we get

x^3 + x^2 + 5 = x^2 + x + 5Alternative method:

When we substitute two values of x in the above

curves, at x = –2, we get

y = –8 + 4 + 5 = 1

y = 4 – 2 + 5 = 7

Hence, at x = –2 the curves do not intersect.

At x = 2, y_1 = 17 and y_2 = 11

At x = –1, y1 = 5 and 2 and y2 = 5

When x = 0, y_1 = 5 and y_2 = 5

And at x = 1, y_1 = 7 and y_2 = 7

Therefore, the two curves meet thrice when x = –1, 0

and 1.

146.In a certain examination paper, there are n questions. For j = 1,2 …n, there are 2n–j students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is

**Ans . **

(1)

**Explanation :**Let us say there are only 3 questions. Then, there are,

2^3–1 = 4 students who have done 1 or more questions

wrongly, 2^3–2 = 2 students who have done 2 or more

questions wrongly and 2^3–3 = 1 student who must

have done all 3 wrongly. Thus, total number of wrong

answers = 4 + 2 + 1 = 7 = 2^3 – 1 = 2^n – 1.

In our question, the total number of wrong answers

= 4095 = 2^12 – 1. Thus n = 12.

147.If x, y, z are distinct positive real numbers the(x^2 (y+z)y^2 (x+z)z^2 (x+y))/xyz would be

**Ans . **

(3)

**Explanation :**Here x, y, z are distinct positive real number

So (x^2 (y+z)y^2 (x+z)z^2 (x+y))/xyz = \( \frac{x}{y} \)+ \( \frac{x}{z} \)+ \( \frac{y}{x} \)+ \( \frac{y}{z} \)+ \( \frac{z}{x} \)+ \( \frac{z}{y} \)

= (\( \frac{x}{y} \)+ \( \frac{y}{x} \))+ (\( \frac{y}{z} \)+ \( \frac{z}{y} \))+ (\( \frac{z}{x} \)+ \( \frac{x}{z} \))[We know that

\( \frac{a}{b} \) + \( \frac{b}{a} \)> 2if a and b are distinct numbers]

> 2 + 2 + 2 > 6

148.A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition

**Ans . **

(1)

**Explanation :**The least number of edges will be when one point is

connected to each of the other 11 points, giving a total

of 11 lines. One can move from any point to any other

point via the common point.

The maximum edges will be when a line exists between

any two points. Two points can be selected from 12

points in 12^C_2 i.e. 66 lines.

149.The number of positive integers n in the range 12 ≤ n ≤ 40 such that the product (n −1)(n − 2) 3.2.1 is not divisible by n is

**Ans . **

(2)

**Explanation :**From 12 to 40, there are 7 prime numbers, i.e., 13, 17,

19, 23, 29, 31 and 37 such that (n – 1)! is not divisible

by any of them.

150.Let T be the set of integers {3, 11, 19, 27, …, 451, 459, 467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is

**Ans . **

(4)

**Explanation :**T_n = a + (n – 1)d

⇒ 467 = 3 + (n – 1)8

⇒ n = 5929th term is 227 and 30th term is 235 and when these

two terms are added the sum is less than 470.

Hence the maximum possible values the set S can

have is 30.