# Answer Key & Explanations for Full-length test 6

Section I: DI & LR

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

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

21. 3      22. 4      23. 2

Section II: English Usage

24. 3      25. 2      26. 4      27. 4      28. 4      29. 3      30. 3      31. 1      32. 4      33. 4

34. 1      35. 4      36. 2      37. 3      38. 4      39. 4      40. 1      41. 2      42. 4      43. 1

44. 2      45. 3      46. 1      47. 4      48. 4

Section III

49. 2      50. 1      51. 3      52. 3      53. 4      54. 4      55. 3      56. 1      57. 4      58. 2

59. 2      60. 3      61. 1      62. 4      63. 2      64. 4      65. 3      66. 4      67. 4      68. 1

69. 3      70. 4

## Explanations:

1. (2)

Book Delhi – Mumbai, flight GA G8-171 on MakeATrip. 90% off on base price of Rs. 3000 will result in total price of 300 + 2290 + 50 = Rs. 2640. (Rs. 50 is service charge)

And book Mumbai – Pune, flight Jet 9W-101 on GoEasy. 50% Cash back on base fare for Jet flights means a total cost of Rs. 500 + 2031 + 50 = Rs. 2581

The total cost will be Rs. 2640 + 2581 = Rs. 5221

2. (3)

Book Delhi – Mumbai, flight IG 6E-183 on MakeATrip. Total price of 1950 + 3340 + 50 = Rs. 5340.

The flight will arrive at Bombay at 11:35 and then book connecting flight Bombay-Goa, flight IG 6E-181 on MakeATrip. With a 90% off on the base price, total cost will be 45 + 2381 + 50 = 2476

The total cost will be Rs. 5340 + 2476 = Rs. 7816

3. (3)

Delhi – Pune, Flight, KF IT-3152 on MyYatra. Total price = Rs. 2000 + 3289 + 50 = Rs. 5339

Whenever the ratio is less than 1.5, the company wastes resources. This happens in 9 months. Thus it operates under pressure in 3 months.

In each group, the middle possible score is a multiple of 3. Taking these values, and adding, we get,

3 × (E + M + G + H + P) = 1221 i.e. (E + M + G + H + P) = 407

Adding totals of groups 1 and 3; E + M + 2G + H + P = 477

Subtracting the two relation, we get, G = 70

Similarly adding groups 2 and 4: E + M + G + 2H + P = 474

Subtracting from the total of the 5 subjects, we get, H = 67

In group 3, substituting scores of G and H, we get P = 91

In group 4, substituting scores of H and P, we get E = 82

In group 5, substituting scores of P and E, we get M = 97

Now, the marks in the subjects will vary around these values (+ 1 or – 1) depending on which total is taken.

9. (3)

We cannot take the maximum total in each of the 5 groups because then the sum total of the 5 subjects will not be an integer. Remember, while adding the possible scores of each group, care should be taken such that the final total has to be a multiple of 3.

Thus, the maximum total will be 1 more than that that found by taking the middle values i.e. 408. This can be by taking the maximum total for any three groups and the middle value for the remaining two groups, say 3 × (E + M + G + H + P) = 250 + 235 + 229 + 240 + 270 = 1224.

Hence (E + M + G + H + P) = 408.

But this is not the only way, you could have other combination of marks in the various groups.

10. (3)

As found in the approximate marks of each subject, we see that the lowest score is in H and the highest score is in M. So let us try to maximize the difference between M and H.

Thus, the in the 2 groups that M appears but H does not appear, let us take the maximum possible total; in the two groups that H appears but M does not appear, let’s take the minimum possible total; and in the group that both M and H appear, let’s take the middle values. In this way, we can also ensure that the sum total of the marks of the 5 groups will be a multiple of 3.

With these chosen values, total of group 4: H + P + E = 239; total of group 5: P + E + M = 271. Subtracting the two, we have M – H = 271 – 239 = 32.

For those who are not convinced:

Adding all the 5 groups chosen total:

3 × (E + M + G + H + P) = 250 + 234 + 227 + 239 + 271 = 1221

i.e. E + M + G + H + P = 407.

Adding total of groups 2 and 5: E + 2M + G + H + P = 234 + 271 = 505

Subtracting, M = 98.

Adding totals of groups 2 and 4: E + M + G + 2H + P = 234 + 239 = 473

Subtracting from the total of 5 subjects, H = 66

Thus, the maximum difference between marks of two subject is 98 – 66 = 32.

11. (1)

Choose the total such that totals in the group having E are less and at the same time total of the 5 subjects is as high as possible (remember we are talking of averages, a higher total in other subjects would drive the marks of English lesser). Also we need to keep in mind that the total of the 5 groups should be a multiple of 3. Thus, total of 2 groups in which E appears is taken as the least and the two groups in which E does not appear as the highest. This makes it mandatory to choose the middle value for the third group in which E appears.

3 × (E + M + G + H + P) = 248 + 235 + 229 + 239 + 270 = 1221

Then (E + M + G + H + P) = 407

Adding total of groups 1 and 4: 2E + M + G + H + P = 487.

Subtracting the two, E = 80

12. (3)

If no two students have the same scores in all the 5 subjects, then their totals in all the 5 groups will also not be the same.

We could choose the total of group 1 in 3 different ways, that of group 2 in 3 different ways, that of group 3 in 3 different ways, and that of group 4 in 3 different ways. But for any particular choice of totals of the first 4 groups, the total of the fifth group can be unique. This is because the sum total of the 5 groups has to be a multiple of 3.

Thus, each of the above 3 × 3 × 3 × 3 × 1 = 81 ways of selecting the totals of the five groups would result in scores of 5 subjects such that marks scored in all the 5 subjects would not be the same.

The totals in the 5 groups being 248, 233, 227, 239, 271 respectively would result in E = 81, M = 98, G = 69, H = 66, P = 92, whereas the totals being 248, 233, 227, 240, 270 respectively would result in E = 83, M = 99, G = 68, H = 67, P = 89.

For questions 13 to 16:

Consider the net asset value of HSB is 100, that of HDF is 108.

14. (3)

Working with options:

Housing is far too small a investment in HDF to make a big difference in HDF’s net asset value.

Personal care:

In HDF, value of investment in Personal care at current prices is 20%. And a decrease in the value of this sector by 50% would imply a 50% of 20% i.e. 10% decrease in total asset value. Hence value of HDF investment will be 108 – 10.8 = 97.2

In HSB, value of investment in Personal care at current prices is 11% and a 50% decrease in this investment would imply a 11% ×50% i.e. 5.5% decrease in total asset value. Thus, value of HSB investment would be 100 – 5.5 = 94.5. Which is not same as that of HDF.

Oil and Gas:

In HDF, value of investment in Oil and Gas at current prices is 25%. And a decrease in the value of this sector by 50% would imply a 50% of 25% i.e. 12.5% decrease in total asset value. Hence value of HDF investment will be 108 – (10.8 + 2.7) = 108 – 13.5 = 94.5.

In HSB, value of investment in Oil and Gas at current prices is 11% and a 50% decrease in this investment would imply a 11% ×50% i.e. 5.5% decrease in total asset value. Thus, value of HSB investment would be 100 – 5.5 = 94.5. Which is same as that of HDF.

Thus, we see that among the number of 1’s, 2’s and 3’s, if two of them appear even number of times and one of them appear odd number of times, then the number that remains is the number which appears odd number of times. And if two of them appear odd number of times and one of them appear even number of times, then the number that remains is the number which appears even number of times.

And the above can easily be generalised also. Whenever a number is erased, the number of times it appears in the list changes state from odd to even (or from even to odd) and so does the others. Thus, if at end, only 1 remains, i.e. the number of times, 1, 2, 3 appear are odd, even, even, then at any state in the game, the number of 1’s, 2’s and 3’s will have to be odd, even, even OR even, odd, odd.

And if all three appear even number of times or odd number of times, the game would end in stale-mate. This can be verified by considering 1, 2, 3, or 1, 1, 2, 2, 3, 3 as start and then proceeding.

17. (2)

Since number of 1’s, 2’s and 3’s are even, even and odd, the last number to remain will be 3. Since the question is asking which cannot be a number remaining the answer would be 1 or 2 cannot be remaining i.e. choice (2)

18. (2)

p, q, r will be odd, even, odd for any value of n. Thus, the last number to remain will be 2.

19. (2)

Since all of them are even, the game will end in a stale-mate and the least number of numbers that can be left at the end is 2. Thos two numbers could be any of 1, 1, or 2, 2, or 3, 3.

20. (3)

In each move, the total number of numbers will reduce by 1 (since two numbers are erased and one number is written). At start there are 10 + 20 + 25 = 55 numbers. After 46 moves, the number of numbers left on the board will be 55 – 46 = 9 numbers. Thus option (2) is eliminated.

At start the number of 1’s, 2’s and 3’s are even, even and odd. Since an even number of moves are over, the number of 1’s, 2’s and 3’s would again be even, even and odd. Only option (3) satisfies this condition.

21. (3)

A leap year is found if the year is divisible by 4. But in case the year is divisible by 100, then the condition for the year to be leap is that it should be divisible by 400. Thus, each of 1700, 1800, 1900, 2100, etc, though divisible by 4 is not a leap year.

Statement I: Even if the last time the world cup was not held in a leap year, it is quite possible that this year it could be held in a leap year. E.g. Last time the world cup was held in 1900 (not a leap year) and this time it will be held in 1904 (a leap year).

Needless to say, this time the world cup could also be held in a non-leap year. E.g. last time it was held in 2007, this time it will be held in 2011.

Thus, no firm answer can be found for the questions.

Statement II: By the same using this statement independently, the question can still not be answered.

Next time the world cup will be held in 1904 means this time the world cup is held in 1900, a non-leap year. Whereas if next time the world cup is held in 2004 means this time the world cup will be held in 2000,a leap year.

Using both the statement: Last time it was not held in a leap year and next time it will be held in a leap year results only in possibilities like 1900, 1904, 1908 or 2100, 2104, 2108 or similar years. Thus, this time the world cup will surely be held in a leap year.

22. (4)

A and B are not the speakers who speak last.

Using statement I: Additionally we just know that C is not the last person to speak. The last to speak could be any of D or E as seen in the following two arrangements: B A C D E or B A C E D

Using statement II: Nothing can be said about the last person. The two examples cited above still remain a possibility and thus, the last speaker cannot be ascertained uniquely.

Using both statement: If you paid careful attention to the use of the two cases cited above for both statements independently, then you should have realised that the two cases satisfy the conditions given in both the statements simultaneously. Thus, even after using both the statements together, we cannot identify the last speaker uniquely.

23. (2)

Using statement I: Since we do not know the capacity of the basket we cannot know what does 5/6th of the basket mean.

Using Statement II: Even if we do not know the capacity of the basket, after 1 minute the basket will be 2 × 1/8th i.e. 1/4th filled. After another minute the basket will be ½ filled. And so on. Thus, we can find out after how many minutes the basket will be atleast 5/6th filled. (In-fact after 3 minutes, the basket will be filled completely.)

Section II:

24. (3)

Question Type: Tone/ Analysis

Though words and phrases like ‘ham-fisted effort’ ‘dubiously support his case’ and ‘plodding repetition’ make the author’s take on the book very clear, the answer choices here are asking you to discern the degree and nature of the author’s disappointment. (1) is too strong plus author is not dismissive of the subject, in fact states that it is especially pertinent (para 4). Author presents counterarguments to show the flaw in Callahan’s argument and not vice versa (as is stated in B). The broader approach is not ignored (as stated in (4)) but commented upon in last two paras.

25. (2)

Question Type: Implied meaning

What the author is conveying by recounting the prevalence of cheating (avalanche of stories) even in places one would expect to find honesty, is that they make one suspect that now cheating is so widespread that no one can be trusted. Thus, not just people above thirty but even those below thirty should not be trusted- which makes only the first half of statement correct. (4) may look like a reasonable alternative but the ‘sometimes’ in it makes it mean the opposite of the correct answer.

26. (4)

Question Type: Application

Note that (1) does not talk of erasing and smudging, (2) is ridiculous because if the right answer is already marked there is no scope for cheating. Note that while (3) is also possible as a means of cheating, the passage is talking of a “right answer being marked wrong because of the devices’ fault plus it does not constitute a ‘retroactive correction’.

27. (4)

Question Type: detail

Refer to para 7 and last line of para 1. Author talks about impact of technology both ways but concludes the discussion saying that cheating itself is hard to gauge. The impact of technologies is mentioned in relation to shaping our perception on cheating’s prevalence.

28. (4)

Question Type: Arrangement

A glance at the passage would show that barring the 1st and 6th, the rest of the paras are all discussing the views presented in Callahan’s The Cheating Culture and critiquing them. Also note that at the end of para one, a certain observation on the nature of cheating (which follows from the discussed example) is made which serves as a link to relate the example to the views presented in the book.

(1) is wrong because even though other examples are touched upon in the passage, it is done in the context of the books contents, not specifically to discuss various cheating related instances. (2) is wrong because passage is written with respect to Callahan’s book and not as a broader discussion on cheating. (3) wrong because had it been a case study, the subsequent passages should be exclusively discussing that specific problem/ instance.

29. (3)

Question Type: Further Application

This question is based purely on logic. If and only if p then q implies:

p happens implies q happens

q happens implies p happens

p does not happen implies q does not happen

q does not happen implies p does not happen

(1) is wrong cause it can only be pleasure or only displeasure which makes a judgement aesthetic. One can’t feel pleasure and displeasure at the same time towards the same object. Don’t get confused by “Why not?” – we should only be concerned with what is there in the passage.

(2) and (4) are the opposites – pleasure or displeasure will lead to ‘aesthetic’ judgement not ‘not aesthetic’.

(Check the 2nd para).

30. (3)

Question Type: Specific – detail

This is a specific-detail question and the answer (or its paraphrase) can be found in the middle of the 2nd para.

(1) and (2) are incorrect as there is no comparison done between judgement and belief.

(4) interchanges the words ‘belief’ and ‘judgement’

(3) is just a paraphrase of lines 6-9 of 2nd para – “And by judgement Kant… form of awareness.”

31. (1)

Question Type: Vocabulary

(2) is incorrect as ‘pleasure’ is the affect of ‘beauty’ not a synonym of ‘affect.’(2nd para 27th line)

(3) is an example of an ‘affective’ term

(4) is wrong as ‘affective’ has got nothing to do with ‘effective’

(1) – ‘affective’ means ‘Influenced by or resulting from the emotions’

32. (4)

Question Type: further application

The author has not presented his views on the matter; in fact most of the passage is about Kant’s views on the matter. So the answer has to be (4).

(2) is Kant’s view and (3) is opposite of Kant’s view.

33. (4)

Question Type: Main Idea

As you would remember from RC theory, in such questions try to eliminate answer choices that are either too narrow or too broad; in other words, specific or outside the scope.

(2) and (3), while making sense, go way beyond the passage. If this passage is a part of a bigger chapter, then one or both of them could be the answer.

Between (1) and (4) the latter is better as the passage concentrates on aesthetic judgements as Kant saw them or felt about them.

34. (1)

The correct usage is ‘to play’.

35. (4)

The correct usage is “stick with you”.

36. (2)

The correct usage is “Drink the liquid”

37. (3)

The table must have been “pushed away” to make more room.

38. (4)

The correct usage is “Open your arms”

39. (4)

Intermittent – recurrent, showing water only part of the time; Interminate – Endless

Axle – the pin or shaft on which a wheel or pair of wheels rotates; Axel – a jump performed by a skater

Isle – island; Aisle – a walkway between or along sections of seats in a theater, classroom, or the like.

Discreet – unobtrusive; Discrete – separate

Eyrie – nest of a bird of prey; Eerie – uncanny, so as to inspire superstitious fear

40. (1)

Ale – a beverage; ail – to cause pain or uneasiness

Alter – change; altar – an elevated place or structure, as a mound or platform, at which religious rites are performed

Leech – bloodsucking worm; leach – to cause (water or other liquid) to percolate through something.

Fate – fortune; fete – a day of celebration

Beach – an expanse of sand or pebbles along a shore; beech – a tree of the temperate regions

41. (2)

Mantle – cape; mantel – a construction framing the opening of a fireplace

Flea – wingless bloodsucking insect; flee – to run away

Baron – a member of nobility; barren – sterile

Braise – to cook by sautéing and simmering; braze – to make brasslike

Canvas – a closely woven heavy cloth; canvass – to solicit votes

42. (4)

Censer – a container in which incense is burned ; censor – to examine (as a publication or film) in order to suppress or delete any contents considered objectionable

Complacent – self satisfied; complaisant – inclined to please

Deviser – one who makes a plan; divisor – a number that divides another number

Weever – a kind of fish; weaver – one whose occupation is weaving.

Palette – a tablet used for mixing colors; palate – the sense of taste

43. (1)

Hanger – a frame to hang clothes; hangar – a shed or a shelter

Gild the lily – to add a feature to improve something that is already satisfactory; guild – an organization of people with common interests

Mousse – an aerosol foam used ot style hair; moose – an animal

Vain – without a real value; vane – blade of a windmill

Wail – to utter a mournful cry; wale – a streak, stripe, or ridge produced on the skin by the stroke of a rod

44. (2)

Every choice begins with ‘ultimately, therefore’ implying that the last sentence in this case is a sort of a conclusion. Remember this.

(1) is very specific which capture only one aspect of the paragraph.

(2), (3) and (4) are saying more or less the same thing. (2) is better than (3) as it carries on in present tense as is the case in the last sentence. (2) is better than (4) as ‘vision and purpose’ is much more thorough than ‘end result’.

45. (3)

From the given answer choices we know that the last sentence begins with ‘or’ which implies that the last and the second last sentences are presenting two options.

(2) is wrong cause we are concerned with foreign companies not with foreigners.

(4) is wrong as ‘labyrinth’ means a complicated irregular network of passages or paths – which does not fit.

Between (1) and (3) the latter says what (1) does and gives a reason as well and so is better.

46. (1)

(2) and (3) can be straightaway eliminated as they have no direct connection with what is being discussed.

(4) should be eliminated for the same reason; however some people might get confused because just like the passage it talks about the ‘Origin’ – remember not to get carried away by the mere presence of some word or phrase.

The way the passage is going (1) is a much better way of ending as ‘in short’ means that the author is trying to summarize or paraphrase what she/he as already talked and (1) does a good job of it.

47. (4)

If the author is saying that the answer to the problems mentioned is democratic planning. (2) does not make sense as they criticise or question democratic planning.

(1) can be further down the discussion but when compared with (4) cannot be the very next sentence as (4) tries to take the discussion forward by first asking what is democratic planning.

48. (4)

(1) makes some sense as it tends to give reasons behind the passage but can come immediately after (4) ( start a new paragraph) as (4) tends to end the passage on a sombre note. Also ‘this commercialization’ refers to what has been talked in the passage.

(2) is another way of looking at war and can come later in another paragraph.

(3) follows (2) rather than the passage.

Section III

Columns are counted from left side.

Since it is easier, let’s first assume that there is no carry from 4th column to 3rd column.

Thus, O + O ends in O. Checking possibilities, we can rule out each of 1, 2, 3, 4, 5, 6, 7, 8, 9. Thus, O = 0.

Hence, in the 2nd column R + R ends with a 0. Since R cannot be 0, it has to be 5.

Thus, for 1st column, 1 + T + T = 5 i.e. T = 2.

And since T = 2, in the fourth column we get M = 4.

There is no contradiction in terms of an alphabet representing two distinct numbers or two distinct alphabets representing the same number.

Also the options do not suggest more than one possible answer. Thus, the solution is O = 0, R = 5, T = 2 and M = 4.

The required answer is 5 + 0 + 4 + 2 = 11.

But since x is the base of a log, it cannot be negative. Also since logs of negative numbers are not defined, (1 – x) cannot be negative. Hence x = 4 is also not acceptable.

Thus neither of x = 4 or –2 is acceptable and the equation will have no solution.

52. (3)

Applying Pythagoras in right triangle AEF, AF2 = AE2 + EF2 = 48 + 16 = 64. Thus AF = 8 and radius of the circle is 4.

54. (4)

The question is NOT asking you in how many WAYS can the rook move from bottom left corner to top right corner. It is asking the different number of MOVES in which it can do so.

The minimum number of moves is 2 (in one move cover entire vertical column and in next move cover entire row) and the maximum number of moves is 14 (move 1 square at a time, any order – row-wise or column-wise or zig-zag manner – will result in same maximum number of steps).

Thus it is possible to perform the task with the number of moves being any natural number from 2 to 14 i.e. 13 different number of moves.

58. (2)

The number is divisible by 4 since last two digits are divisible by 4.

The sum of the blanks will be 1 + 2 + 3 + … + 9 = 45. And the sum of the given digits of the numbers is 90. Thus the entire number will always be divisible by 9, irrespective of how the blanks are filled in.

If you notice closely all the blanks are in the set of alternate digits, that set which includes the unit digit. Thus, sum of set of alternate digits, starting with unit digit = 17 + 45 = 62. And sum of alternate digits starting with ten’s digit = 73. The difference of these two sets = 11, which is divisible by 11. Thus, irrespective of how the blanks are filled, the number will always be divisible by 11.

Thus the number is always divisible by 4 × 9 × 11 i.e. 396.

60. (3)

We know that xn – yn is divisible by x – y. Since 2141 – 1863 = 1770 – 1492 = 278, the expression is always going to be divisible by 278. Similarly, 2141 – 1770 = 1863 – 1492 = 371 and so, the expression is also going to be divisible by 371. Since both 278 and 371 are relatively prime, the expression has to be divisible by the product 278 x 371. The product can be rewritten as 53 x 1946. Hence, of the given options, the year of birth could be 1946.

61. (1)

The line joining the centers will bisect the 60 degrees angle. Thus, there are two 30-60-90 triangles, with one side known in each.

Distance from vertex to center of smaller circle = 2 × 15 = 30

Distance from vertex to center of larger circle = 2 × 95 = 190

Thus, distance between centers = 190 – 30 = 160.

68. (1)

Each interval of 1 sec, when the dial is considered to be divided into 60 seconds is same as the interval of 1 minute when the dial is considered to be divided into 60 minutes. Thus, the minutes hand is 18 minutes ahead of the hour hand, when dial is considered to be divided into 60 minutes.

Since the hour hand travels 5 minutes (unit of distance) in 60 minutes (unit of time), it covers 1/12 minute per minute.

Option (1): The minute hand is on 36th minute mark. The hour hand has travelled 36/12 = 3 minutes more than its initial position of being on 15th minute mark (at 3:00). Thus, it is now on 18th minute mark. Thus, the minute hand is indeed 18 minutes ahead of hour hand and hence this is an answer.

The only other option to check is option (3) i.e. 6:56. Consider the hour hand: From it’s initial position of being on the 30th minute mark (at 6:00), the hour hand would have travelled 56/12 i.e. 4.6 minutes ahead i.e. the hour hand will be at 34.6 minute mark. And the minute hand is on 56th minute mark. This is not a gap of 18 minutes. Thus, this cannot be the answer.

For 69 & 70:

Rather than working on the time taken by each tap to fill in the tank, the amazingly easier way to work is to work on the efficiency of each tap. Remember days cannot be added, but efficiencies can be added.

If the efficiency of the first two taps are considered as x and x, then the efficiency of the taps 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …… will be x, x, 2x, 3x, 5x, 8x, 13x, 21x, 34x, 55x, ……

69. (3)

The time taken by the 6th and 7th tap together is same as the time taken by the 8th tap individually.

The efficiencies of the 4th and the 8th tap are in ratio 3x : 21x i.e. 1 : 7. Thus, the time taken by 4th tap and 8th tap individually will be in ratio 7 : 1. Since 4th tap takes 35 minutes, 8th tap will take 5 minutes.

70. (4)

The efficiencies of the 9th and the 10th tap are in ratio 34 : 55. Thus, the rate of flow of water through them will also be in the same ratio. Since the rate of flow through 10th tap is 550 litres per hour, that through 9th tap will be 340 litres per hour.

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