1. b

There are two clues to solve this question. The first is the mandatory pair (A-B). Statement B has to come immediately after statement A as B is the response of the Labour party to the “row about the legitimacy….” mentioned in A. This leaves us with options (b) and (d). Here the difference is in the order of statements D & E. (D-E) would be the correct order as statement D introduces the event while statement E begins the description of the events that followed. Statement E also matches the tone of statement C. The paragraph has a mandatory pair (D-B) as B carries further the idea of opening of the cages, which gets introduced in D. (C-A) is another mandatory pair. Statement C compares India’s performance on social indicators with that of Asian economies. Statement A further compares India’s performance with that of poor African countries. Only options (c) and (d) have both mandatory pairs. Option (d) gets negated as the paragraph must start with C, which introduces the para. Option (c) aligns all the statements of the paragraph thereby making it coherent. Statement C clearly emerges as an introductory sentence “an episode regarding the….”. (A-E) is a mandatory pair as in statement E Akbar gives reasons to refute Hakim Ali’s argument. The only possible answer choice is (b).

6. c

Option (a) clearly refers to the opening of the story which is gruesome yet comical. Option (b) is correct as the author creats humour inspite of Brr not knowing language for very long. Refer to lines “The author isn’t about to let inexperience get in the way of a few good puns or a little witty banter” Option (c) is not a part of the story. Refer to lines “Poor Brr is not as likeable ...” Option (d) can be inferred from the lines “He picks up the art of words.......” The paragraph describes the reaction of the west towards the assassination. The tone of the passage seems to be justifying this reaction. Thus (c) is the best answer as it provides the conclusion the argument on why assassinations are not viewed with shock or horror. The theme of the paragraph is the dependence of women on their male counterparts and option (b) takes forward the idea of women deciding to own a house and free themselves. Options (a) and (d) although related describe how this change is not restricted to a particular class of women and hence cannot come directly after the paragraph given. Option (c) is inconsistent with the theme, as the focus here seems to be more on possession and not on their need for independence. The paragraph talks about a revolution and gives examples where the internet and social networking sites have been used. Option (d) takes it forward by giving yet another example on the same lines. The last line of the passage “But the idea that Germany should itself seek to adjust, through lower saving and higher consumption and investment....” clearly shows that statement A is supported by the author. ‘B’ is incorrect. From the second last paragraph, it can be inferred that a prudent Germany has invested in countries like Spain and Greece. Also these countries have been profligate to some extent. This has been responsible for their economic problems and not just German exports. ‘C’ is incorrect as the quality of Germans goods is not mentioned in the passage. Option (a) is incorrect and cannot be inferred. Germany does not believe itself to be responsible for the state of affairs in the deficit euro-area countries . Options (b), (c) and (d) can be inferred from the last paragraph of the passage.

7. c 2. c

8. b

3. b

9. d 4. d Options (a) and (b) are in the last lines of the paragraph 2 as a central reason why Brr’s boring backstory is maintained. Refer to the lines “And it costs us readers 200 pages of retreading old backstory, even while the author is telegraphing a larger plot and doing nothing about it.” Option (c) can also be inferred as the author points to it in paragraph 2 “Some new folds in the character are revealed, and it wasn’t badly done, but it more colored in the outline of his character than expanded it.” Option (d) although mentioned in the passage is not given as a reason for the novel being boring. 1 cannot be conclusively inferred as his feeling of being an outcast probably stems from a number of reasons which haven’t been explored in detail. All that we know about his past is that he was a motherless cub and had a stint in a university lab. ‘3’ can be inferred from the lines in paragraph 2”The other secondary characters are also good………the Lion’s share of print’. ‘2’ is incorrect as it is Brr’s inaction which has adverse consequences.‘4’ is implied at the end of the last paragraph “He has a tendency for vanity………………………where he not pushed around by circumstances”. This provides support for ‘4’ as an inference.

10. a

5. a

11. a

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12. c

Option (a) is true as seen from paragraph 1 and the start of paragraph 2. Option (b) can be discerned from the last lines of paragraph 2. Option (d) can be inferred from the lines in paragraph 2 “German business ruthlessly held down real wages. Unit labour costs fell by an annual average of 1.4% in 2000-08 in Germany, compared with a decline of 0.7% in America and rises of 0.8% and 0.9% in France and Britain.... “. It is clear from the subsequent paras that Germany’s ability to hold down costs is responsible for its superior economic condition inspite of the recession. But option (c) is not true. Southern Europe is considering preserving its wages and its restricted labour markets. The correct phrase should have been “lay siege to” the city. ‘Lay siege to’ means ‘to besiege’ The correct phrase should have been “fall through the cracks”. ‘fall through the cracks’ means ‘to be overlooked’ or ‘to be neglected’. Statement 1 is stated in the passage. Refer to 3rd line in paragraph 2, “while cosmology and general relativity were stuck where they had been in the 1930s”. Statement 2 is incorrect as what was thought to be extremely difficult was only theory of relativity. Statement 3 is incorrect as we cannot infer from the passage that there was nothing to be learnt. The given passage is a narrative about the author’s research and choices he made. The entire discussion about cosmology and particle physics has been done in that backdrop. Option (c) captures this. Statement 1 is correct. Refer to “but similar field theories didn’t seem to work. Indeed, the Cambridge school, in particular, held that there was no underlying field theory.” Statement 2 is incorrect as theory of relativity was ignored as its physical significance was not realized. Author is silent on available physical evidence. Statement 3 is incorrect as the passage gives no evidence to comment on whether or not Hoyle would have been a better guide. Statement B is the opening line of the paragraph. (C-A) is a mandatory pair that describes the changes or disintegration that did not affect the continent. Statement D has to follow (C-A) as the “patch work suggested by…..” is a reference to the changes in (C-A). Statement 1 has a subject verb agreement error. The Georgian windows in Bishop Terrick’s dining room looks out on to the gardens of Fulham Palace. It should be “look” as the subject of the sentence is the Georgian windows. Statement 4 is incorrect as it should be “they’re framed by clouds of wisteria”

20. a

Statement 1 should have an article “as a grown man…..”. Statement 2 has inconsistency in tenses. The correct statement should be “A storm blows up and the surrounding countryside is cloaked in rain and fog” .In statement 4 the correct preposition should be “beside the road”. Besides means ‘in addition to’ which is inappropriate here.

13. c

For questions 21 to 24: The first batsman to get out can be either Hayden or Gilchrist. Table 1 states that the first wicket falls at score 25. As Hayden himself scored 28 which is greater than 25, we can conclude that Gilchrist was the first to get out. At that time Hayden must have scored 25 – 7 = 18 runs. Similar analysis leads to the following table:

14. d

Fall of Wicket 1 2 3 4 5 6 7 8 9

Batsman to Batsman 'not Total get out (runs out'/batting Score scored) (runs scored) 25 34 42 57 62 75 82 86 99 Gilchrist (7) Ponting (8) Symonds (4) Hayden (28) Hodge (3) Hussey (20) Johnson (5) White (0) Clark (11) or Lee (18) Lee (18) / Williams (6) or Clark (11) / Williams (6) Hayden (18) Hayden (19) Hayden (23) Hussey (10) Hussey (12) Lee (5) Lee (7) Lee (11) Lee (13) or Clark (6) Williams (6) / Lee (18) or Williams (6) / Clark (11)

Partnership

15. a

25 9 8 15 5 13 7 4 13

16. c

17. a

10

110

11

21. b 22. d 23. a

Hodge lost his wicket between Hayden and Hussey. It can either be Clark or Lee. (See table). The second highest partnership was for the fourth wicket between Hayden and Hussey for 15 runs. Percentage contribution of Hayden (who got dismissed first):

18. d

5 × 100% = 33.33% 15

19. c

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24. c

Refer to the table. The maximum number of fours that could have been scored between the fall of any two wickets can be summarised as: 0 to 1st 1st – 2nd 2nd – 3rd 3rd – 4th 4th – 5th 5th – 6th 6th – 7th 7th – 8th 8th – 9th 9th – 10th : : : : : : : : : : 5 2 2 3 0 3 1 1 2 2 fours fours fours fours fours fours four four fours fours

28. d

From Statement A: Some of the cases possible according to Statement A are given below (left to right in the decreasing order of ages): 1. Steven Frank Wayne Lampard Rooney Gerrard 2. Steven Frank Wayne Rooney Lampard Gerrard 3. Steven Frank Rooney Wayne Gerrard Lampard Clearly, Statement A is insufficient to find who is the youngest among the 6 brothers. From Statement B: The following are the only two possibilities for the correct order of the decreasing ages (left to right) for the brothers which can be derived from Statement B: 1. Steven ______ Gerrard Wayne _______ _______ 2. Steven Gerrard ______ _______ Wayne ______ The exact place for Frank, Lampard and Rooney cannot be decided in the above two arrangements.

So in all 21 fours could have been scored. The rest of the score is by virtue of singles. For questions 25 to 27: We can make the following figure based on the information given in the question.
(50) W itho ut W ork Experie nce G AR (60) LAT 15 EC A 21 36 (50) W ith W o rk E xpe rience G AR EC A 14 24

Combining Statement A and Statement B: The only possible order (left to right in the decreasing order of ages) is: Steven Gerrard Frank Rooney Wayne Lampard So Lampard is the youngest. Hence, the question can be answered by using both the statements together, but cannot be answered by using either statement alone. For questions 29 to 31:
HP North Asia East Europe West Europe North Am erica South Am erica East Africa West Africa South Asia Australia Total 4960 285 1920 4260 150 225 180 450 650 Com paq 1440 630 780 540 1260 484 1170 4050 832 IBM 1120 330 1140 720 990 517 900 1350 364 7431 Sony 480 255 2160 480 600 128 2250 3150 754

10

G AR (40) BAT 0

EC A y 14 – y

14

G AR 20 – x x

EC A 6

26

GAR – Good Academic Record ECA – Extra Curricular Activities 25. d From the figure it can be inferred that the answer must be ‘20 – x’. As we don’t know the exact value of x the answer cannot be determined. From the figure it can be inferred that the answer must be the maximum possible value of ‘y’. As the number of students who have ECA and who have also written BAT but have no Work Experience is ‘14 – y’, we can say that the maximum possible value of ‘y’ must be 14. Answer = 21 (No Work Experience) + 14 (Work Experience) = 35

26. c

27. b

29. a 30. c Total sales in South Asia = 9000 units Total sales in West Africa = 4500 units Total sales in East Europe = 1500 units Hence, the required ratio is 6 : 3 : 1.

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31. c

Statement I: Total sales of Laptops in South America will be the sum of sales of HP, Compaq, IBM and Sony which is equal to 3000 units. So statement I is correct. Statement II: Sales of IBM in East Africa = 517 units Sales of HP in West Africa = 180 units Thus it is not 25% more. So statement II is incorrect. Statement III: Sales of Compaq in North America = 540 units Sales of Compaq in South America = 1260 units Ratio = 3 : 7. So statement III is correct.

For questions 35 to 37: Total marks obtained by S1, S2, S3, S4 and S5 are 148, 125, 121, 144 and 127 respectively. As Anup obtained 50 marks in Chemistry, he can be disguised either as S1 or S2. Rohan can either be disguised as S1 or S2 and accordingly Himanshu must be disguised as either S4 or S3. Based on the given data, we arrive at the following cases.

Case I S1 S2 S4 S5
35. d

Case II Anup Rohan Sudip Vishal

Case III Rohan Anup Vishal Sudip

Case IV Rohan Anup Sudip Vishal

Anup Rohan Vishal Sudip

S3 Himanshu Himanshu
32. d If James arrived before Lars then following possibilities exist (left to right in the order of arrival): 1. James, Lars, ______, _______ 2. James, ______, Lars, _______ 3. James, ______, ______, Lars (this possibility gives no arrangement) 4. ______, James, Lars, ______ 5. ______, James, _______, Lars 6. ______, _______, James, Lars (this possibility gives no arrangement) The only arrangements possible for the order of their arrival are: I. James, Lars, Cliff, Dave (from 1) II. James, Lars, Dave, Cliff (from 1) III. James, Cliff, Lars, Dave (from 2) IV. Cliff, James, Lars, Dave (from 4) V. Cliff, James, Dave, Lars (from 5) From the above 5 possible arrangements it is evident that Dave could never be the 2nd friend to arrive. 33. a From Statement A: Let y = |x – 1.5| + |x – 2.5| + |x – 3.5|. ‘y’ attains the minimum value at x = 2.5 and the minimum value is 2. Hence the equation |x – 1.5| + |x – 2.5| + |x – 3.5| = 2 is satisfied by only one real value of ‘x’. So the question can be answered by using Statement A alone. From Statement B: Let y = |x – 5| + |x – 10| + |x – 15| + |x – 20| ‘y’ attains the minimum value at x = 10, 11, 12, 13, 14, 15 i.e. all natural numbers from 10 to 15 both inclusive. Hence, a unique value of ‘x’ cannot be determined. So the question cannot be answered by Statement B alone. 34. d Case 1: Kurt and Cobain lie. This is not possible as then according to Kurt’s statement Cobain must speak the truth. Case 2: Cobain and Morrison lie. This is possible. Case 3: Kurt and Jim lie. This is possible. 38. b

Himanshu Himanshu

It is clear from the above table that one of Himanshu, Vishal or Sudip is disguised as S3. Let us consider all the options one by one. (i) If statement I is true then Sudip must be disguised as S5 and his total score in all three subjects must be 127. Subsequently from Case I and Case III, Himanshu can be disguised as either S3 or S4. Hence, Himanshu’s total score will be either 121 or 144. Therefore we cannot say that statement II is definitely true. (ii) If statement II is true then Himanshu, as his score cannot be the lowest, must be disguised as S4. Now, Sudip must be disguised as either S3 or S5. Hence, Sudip’s score in chemistry is either 43 or 36. Therefore we cannot say that statement I is definitely true. (iii) If statement I is false then Sudip must be disguised as either S3 or S4. Now, if Sudip has been disguised as S3 then Himanshu must be S4 (Case IV) and his total score (144) will be more than that of Sudip (121). But if Sudip has been disguised as S4 then Himanshu must be S3 (Case II) and his total score (121) will be less than that of Sudip (144). Therefore we cannot say that statement II is definitely false.

36. d

37. a

Himanshu obtained the highest marks in Mathematics. If we assume this to be true, then he must be S4. In both Case III and Case IV, Rohan is disguised as S1, who obtained the highest marks (148) in all the three subjects combined. The following arrangement is possible SRTPQV There are exactly 3 people between R and V.

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39. c

From Statement A: o c

41. d

H o ckey(1 30 ) a

Fo otba ll(15 0) b

C ricket(14 0)

Clearly, it is not possible to find out the number of members who play only Hockey as there are many unknown factors. From Statement B:
H o ckey a d c

Days of the year on which Kamla fasted: 1, 3, 6, 10, 15, 21, 28…. Days of the year on which Bimla fasted: 1, 4, 8, 13, 19, 26, 34…. It can be analysed that if Kamla fasted on the Kth day, Bimla must have fasted on the (K – 2)th day. Hence for both of them to fast on the same day, Kamla must fast on some (K – 2)th day and Kth day as well. This is impossible as in the first series no two consecutive terms will have a difference of 2 after the first two days. Distance of origin (0, 0) from the line 3y – 4x – 15 = 0:
3(0) − 4(0) − 15 = 15 = 3 units 5

42. c

Fo otba ll b

3 +4
2

2

C ricket

If we assume a + b + c + d = 5x, where x is an unknown variable then c = 3x Hence, a + b + d = 2x However, we cannot find the answer. Combining Statement A and Statement B:
H o ckey a o c

Let the new lines drawn parallel to 3y – 4x – 15 = 0 be L1 and L2. Distance of L1 from origin = 3 + 3 = 6 units Distance of L2 from origin = 3 – 3 = 0 units The circle x2 + y2 = 25 has a radius of 5 units. Hence line segment of L1 lying inside the circle will be of zero length (L1 does not cut the circle). Chord cut by L2 will be diameter = 10 units 43. c We can see that the difference between the divisor and the respective remainder is the same in each division i.e. 2–1=4–3=6–5=8–7=1 Hence the general form of such numbers will be LCM(2, 4, 6 and 8).K – 1 = 24K – 1, where ‘K’ is any natural number. Hence the numbers are 23, 23 + 24, 23 + 2 × 24, ......, 23 + 40 × 24 A total of 41 such numbers are there between 0 and 1000. 44. a

Fo otba ll b

C ricket

Now, d = 0 So a + b = 2x By using the summation of sets we get, 130 + 150 + 140 – c – (a + c) – (b + c) + c = 300 120 = a + b + 2c = 2x + 6x = 8x ∴ x = 15 ∴ a + b = 30 Hence, number of members who play only Hockey = 130 – a – c = 130 – 45 – a The above cannot be determined as we do not know the value of ‘a’. 40. a Let a, b, c, d and e denote the room numbers of A, B, C, D and E respectively. It is given that: e–c=a–e ⇒ 2e = a + c So a, e and c are in A.P.
D 1 01 1 02 1 03 1 04 1 05 Since E’s room is not next to D’s room, E must be in room number 102.

a15 + a16 + a17 + LL + a50
Sum = a15 1 + a + a2 + LL a35
 a36 − 1   = a15   where a ≠ 1  a −1   

{

}

Since a is the root of equation x5 – 1 = 0,

a5 − 1 = 0 ⇒ a5 = 1
 5 7  × a − 1  a So, Sum = a15  =1 a −1    

( )

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45. b

Let’s assume that ‘I’ denotes the integral part and ‘F’ denotes the fraction part of 5 + 19

(

)

n

.

Extend BA and CD to meet at E. ∆EBC becomes an equilateral triangle. So BE = 4 cm In ∆ADE, ∠ADE = 30° tan30° = ⇒ AE = AE 1 = 2 3 2 cm 3
2 4 3 −2 cm = 3 3

⇒ I + F = 5n + n C1.5n −1. 19 + n C2 .5n − 2.19 + n C3 .5n − 3.19. 19 + ........

Now 5 − 19

(

)

n

is a proper fraction as 5 – 19 < 1

(

)

Let’s assume that F′ = 5 − 19

(

)

n

.

∴ AB = BE − AE = 4 −

⇒ F' = 5n − n C1.5n −1. 19 + n C2 .5n − 2.19 − n C3 .5n − 3.19 19 + .....

Hence,
I + F + F′ = 2 5n +n C2 .5n − 2.19 + n C4 .5n − 4.192 + ......  

48. a

Discount to Saral = 20 + 25 –

20 × 25 = 40% 100

Evidently, F + F′ = 1 as the sum of two proper fractions is always less than 2 and greater than 0. Also the right hand side is an integer and hence F + F′ should also be an integer. Therefore I + 1 is an even number and I is an odd number. Alternative method: Putting n = 1, we get 5+ 19 whose integral part is 9. Putting n = 2, we get 25 + 19 + 10 19 whose integral part is 25 + 19 + 43 which is again an odd number. Now, through the options it can be judged that the greatest integer must always be an odd number. 46. a Let A involves 30 units, B involves 15 units and C involves 20 units of work. One welder does 3 units of A and 1 unit of C in a day. One blacksmith does 1 unit of B and 2 units of C in a day. Two welders would take 5 days to complete A. They finish A at the end of day 5. On day 6 they finish 2 units of C. Three blacksmiths finish B at the end of day 6. After days 6, two welders and three blacksmiths together would take 18/8 days to finish the rest of C. So in all it takes 33/4 days to complete all the jobs. 47. b

40 × 40 = 64% 100 Let a, b, c be the number of units sold at 20%, 40% and 64% respectively.
Discount to Himanshu = 40 + 40 −

⇒

a × 20 b × 40 c × 64 (a + b + c)50 + + = 100 100 100 100

⇒

7c b 3a − − =0 50 10 10

⇒ 7c − 5b − 15a = 0 So ‘c’ has to be a multiple of 5. Values of a, b, c satisfying the equation are (2, 1, 5); (1, 4, 5); (4, 2, 10); (3, 5, 10); (2, 8, 10). Hence the possible values for x are 8, 10, 16, 18 and 20.
49. d
D F

C 4 G 1 B

A

2

E

3

Area of ∆DEC = Area of ∆DBC ……..( ∆ 's between same parallel line and same base)

=

1 × Area of parallelogram ABCD 2
1 × Area of parallelogram EBCF 2

∠BAD = 90°
E 6 0° 3 2 c m 0° D 1 50 ° 6 0° 6 0° C 4 cm A

Area of ∆EFG =

=

1 3 × × Area of parallelogram ABCD 2 5
1 × area(ABCD) 5 ∆DEC = 2 = ∆EFG 1 × 3 × area(ABCD) 3 2 5

So,

B

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50. a

Case 1: All four balls in the same box: 1 way Since all the boxes are identical. Case 2: Two balls in 1 box and the remaining two in another box:
C2 = 3 ways 2! It is divided by 2! because the boxes are identical.
4

(6A + 37) =A 7 Hence, 7A = 6A + 37 or A = 37.
Average = 54. a Let the %age of honey and milk in the two solutions be (a%, b%) and (c%, d%) respectively. According to the question:
(a – 10) (b – 16) = (10 – c) (16 – d)

Case 3: Two balls in a box and one ball in each of the remaining two boxes: 4C = 6 ways 2 Note: We need not select a box for the remaining two balls as they will go one each in the remaining two boxes in one way only. Case 4: Three balls in one box and one remaining ball in another box: 4C = 4 ways 3 Since there is only one way of selecting the other box as the boxes are identical. Total ways = 1 + 3 + 6 + 4 = 14 51. a All numbers are of the form 4K or 4K +1 or 4K + 2 or 4K + 3. The coins are available in the denominations of 4, 8, 13 and 18. Since we have infinite 4’s (a) We can make every number of the form 4K. (b) We can make every number of the form 4K + 1 starting with 13. (c) We can make every number of the form 4K + 2 starting with 18. (d) We can make every number of the form 4K + 3 starting with (18 + 13) i.e. 31. So among these the largest number that is not possible is a number of the form 4K + 3 less than 31, which is 27. So the sum of the digits is 9. 52. a Let the roots of the equation x3 – Ax2 + Bx – C = 0 be

(It does not make a difference whether a is greater than c or whether b is greater than d, because taking –1 common from both the numerator and the denominator will nullify the effect.) Solving, we get, 16a – ad – 160 + 10d = 10b – 160 – bc + 16c So, ad – bc = 16a – 10b – 16c + 10d ...(1) Similarly from the second and third proportions we can say that ad – bc = 12a – 12b – 12c + 12d. ...(2) and ad – bc = xa – 16b – xc + 16d From (1) and (2), we get
(a – c) 1 = (d – b) 2

...(3)

Also from (2) and (3), we get
(a – c) 4 = (d – b) (12 – x)

Hence,

4 1 = (12 – x) 2

8 = 12 – x and x = 4. 55. a {x} = x – [x] or {x} + [x] = x The given equation
5[x] + 3{x} = 6 + x ⇒ 2[x] + 3([x] + {x}) = 6 + x reduces to 2[x] + 3x = 6 + x or 2[x] + 2x = 6 or [x] + x = 3

α, β, γ respectively. So the roots of x3 + Px2 + Qx – 18 = 0 will be α + 2, β + 2, γ + 2.

(α + 2)(β + 2)( γ + 2) = 18
⇒ 4 (α + β + γ ) + 2 (αβ + βγ + γα ) + αβγ + 8 = 18
⇒ 4A + 2B + C = 10

Since 3 and [x] are both integers, in the above equation x must also be an integer. ⇒ [x] = x Hence, 2x = 3 or x =

53. c

Total age of the 6 people on 1st January 2000 = 6A Total sum of ages (including child’s) of the family after 5 years = 6(A + 5) = 6A + 30 Total sum of ages (including child’s) of the family after 6 years = 6A + 30 + 7 = 6A + 37

3 2 This contradicts what we assumed above. So no real value of x satisfies the given equation.

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56. c

The first black marble should be placed between the first and the second white marble. After that a black marble should be placed after every two white marbles. Thus the arrangement of the marbles (W = white and B = black), starting from the left-most white marble, will look like: WBWWBWWBWWB… In the given arrangement it can be observed that, starting from the left, for every ‘2k’ white marbles there are ‘k’ black marbles. So if there are 50 white marbles, 25 black marbles are needed to ensure that each marble has atleast one neighbour of different colour. log5log4log3(x2 – 11x + 1) = 0 log4log3(x2 – 11x + 1) = 1 log3(x2 – 11x + 1) = 4 x2 – 11x + 1 = 34 = 81 x2 – 11x – 80 = 0 ⇒ x2 – 16x + 5x – 80 = 0 ⇒ x = 16 or - 5 or x = 16 (given x > 0)

60. a

Let the hexagon be ABCDEF. The rectangle formed by joining the mid-points P, Q, R and S is shown below:
B P C Q

A
60°

N M
60°

D

S a

R 2a E

F

57. a

Let FM ⊥ SR and AN ⊥ PS AS = SF =

AF = a units 2

In right angled ∆SMF :

58. a

Using prime factorization: 2700 = 22 . 33 . 52 = 2 × 2 × 3 × 3 × 3 × 5 × 5 × 1 (8 integers) Clearly if 2700 has to be written as product of 8 distinct integers, some of those have to be negative. 2700 = 3 × {2 × (–2) × 3 × (– 3) × 5 × (– 5) × 1 × (– 1)} 3 can be multiplied to any one out of the 8 integers inside the bracket except 1 and – 1 otherwise we’ll have two 3’s or two – 3’s. Hence, 6 ways are possible.

a 2 Hence, SR = 2a + 2SM = 3a Similarly in right angled ∆ANS :
SM = SF cos60° = NS = AS sin 60° = Hence, PS = 2NS =
3a 2
3a

2 Area of rectangle PQRS = PS × SR = 3 3a

Area of hexagon PQRS = 6

3 4a2 = 6 3a2 4

( )

59. a

Since Plant is running faster he’ll meet Page exactly once in his each lap (A to B or B to A) except when they meet at one of the end points. The time taken by Plant to cover 50 meters is same as the time taken by Page to cover 20 metres (as the ratio of their speeds is 5 : 2). After this they will meet at A and this will be their 5th meeting. After Plant completes 10 laps they will be at their respective original points and would have met 9 times. Similarly after another 10 Laps for Plant they will be at their respective original points and would have met 18 times. (Page will be at A and Plant will be at B.) The 23rd meeting will be at A again (23 = 18 + 5) as explained above.

2 2 ∴ Ratio = 3 3a : 6 3a = 1: 2

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