ANALYSIS
ANAL YSIS
1. The state having the least maximum temperature in summer season is Sikkim, i.e., 32°C.
Ans.(4)
2. The types of soil mentioned are loamy, clay, red clay, sandy, black, rocky and hard – a total of 7 types. Ans.(3)
3. Investment in item S1 = Rs.13,620. Therefore, in pie chart (I), item S1 subtends a central angle of 13620
110467×360°=44 4.°(approx.). Adding all central angles, we get 302.4°.
Therefore, central angle of item S6 = [360° – 302.4°] = 57.6°
Therefore, relative percentage of item S6 57.6
360 100 16%
= °
°× = .
Therefore, 16% of total expenditure outlay = 110467 × 16% = Rs.17674.72 = 17,675 (approx). Ans.(2)
4. Degree measure of item H3 = 62.7°.
Degree measure of item as misread by student = 26.7°.
Therefore, difference = 36.0°. 360° represents expenditure outlay of Rs.2,21 436.
Therefore, 36° represents expenditure outlay of 36 360
2,21436 1 22,143
°
°× , = (approx.)
Ans.(4)
5. H1 + H7 = 123.5° ...(I) (given) H1 – H4 = 40.3° ...(II) (given) – + = –
Therefore, H4 + H7 = 83.2° ...(III) Now, H7 – H4 = 72.6° ...(IV) (given)
Therefore, 2H7 = 155.8°.
Therefore, H7 = 77.9° ...(V) Substitute the value in (I)
H1 + 77.9° = 123.5°.
Therefore, H1 = 123.5° – 77.9° = 45.6° ....(VI) Now, central angle of item H1 = 45.6°.
Therefore,Expenditure Outlay for H Expenditure Outlay for H
1 7
45.6 77.9 0 585
= = . (approx.). Ans.(2)
6. First we assume that the first statement of Ramu is correct. ∴ Second must be wrong and if Ramus second statement is wrong then Mohit’s second statement is correct and hence his first statement is wrong, also Ramu’s first statement is correct. ∴ Shamu’s second statement is false and hence Shamus first statement is true. Now Ramu and Shamu both did not eat the Pizza then the only person left is Mohit. Ans.(3)
7. Lower limit Rohan’s Actual roll number is 450 and upper limit is 600. Range of his roll number given by the institute can be found out as follows:
450 × 1.5 + 50 = 725 600 × 1.75 + 100 = 1150
Out of the given choices, 725 – 1150 is the most accurate range for his roll number. Ans.(2) 8. Mini’s roll number given by the institute = 14, 207
Actual roll number must be between 5001 – 10000, as 5001 × 2.25 + 200 = 11, 453 and 10000 × 2.25 + 200 = 22, 700
10. We know from the calculations done in the previous question that Virtual roll number for Actual roll number 5001 = 10, 308.
Hence, Actual roll number of Virtual roll number 10,791 must be above 5000.
Trying choice (2): 0.9{5320 x 2.25 + 200} = 10 953.
Trying choice (3). 0.9[5240 x 2.25 + 200] = 10,791. Ans.(3) For Q.11 and Q.12:
11. C is P’s grandmother and she is 77 years old. Ans.(2) 12. Relation between S and J is aunt-niece. Ans.(1)
13. Men: J, K, L Women: P, Q, R Days: M, T, W, Th, F, S Men and Women will alternate, K < R, P < L.
Since J is the host on Monday, the other two men, K and L, are the hosts on Wednesday and Friday. Since L must be a host later in the week than P, L cannot be the host on Wednesday. So, K is the host on Wednesday and L is the host on Friday. Since P must host a session earlier in the week than L, P must host the session on Thursday and R on Saturday. The schedule thus becomes
M T W T h F S
Options (1) and (4) violate the condition that Q and N must attend the same conference [N ↔ Q].
Option (3) violates the condition that K and P cannot attend the same conference.
16. Since L must be examined just before U, and K must be examined fourth, so L must not be examined third, fourth or sixth. Therefore U must not be examined first, fourth or fifth. M and S cannot be examined successively in either order.
If S is examined sixth, the possible schedules will be
1 2 3 4 5 6
L U M K T S
M L U K T S
So, M could be in the first or third time slot. Ans.(3) 17. It is possible to split a tower in the following way.
And so the 100th tower can be made into a 100 × 100 square, requiring 10,000 blocks. Ans.(2) Short-cut: Tower 1, number of blocks = 1 = 12.
Tower 2, number of blocks = 4 = 22. Tower 3, number of blocks = 9 = 32.
∴ Number of blocks needed in 100th Tower = 1002 = 10000.
18. Since Mr. Avneesh and Mr. Chandra Prakash cannot be selected for the committee, Mr. David and Mrs. Aditi must be selected for committee. Hence, Mr. Brijkishore must not be selected for the committee. Now, one remaining member will be selected from Mr. Ravi and Mrs. Rukmini. So two different committees can be formed. Ans.(3)
19. All of the tails must be converted to heads in such a way that the dragon is left with an even number of heads. The way to do this in the smallest number of strokes would be to first convert the 3 tails to 6 tails by using 3 (1 tail) strokes, then converting the 6 tails to 3 heads via 3 (2 tail) strokes. Finally, the six-headed dragon can be killed with 3 (2 head) strokes, for a total of 9 strokes. Ans.(2)
For Q.20 and Q.21:
From the condition given in the question, we can get following two equations 10 = k(4 + w ) ...(1)
11 = k(4 + w+13 ) ...(2)
By dividing equations (1) and (2) we get w = 36 and k = 1.
20. Ans.(4)
21. 14 1 36=
e
+ wj
⇒ = +14 6 w . ∴ w 8 .=∴ w = 64 kg. Ans.(4)
22. Let there be n lockers on the top row,
1 n
2n
2n + 1 3n
Therefore, the sum of four corners is 1 + n + 2n + 1 + 3n = 6n + 2
Since each player is 14 or 15 years old, minimum sum = 4 × 14 = 56 and maximum sum
= 4 × 15 = 60. But using formula, 6 × 8 + 2 = 50, 6 × 9 + 2 = 56 and 6 × 10 + 2 = 62.
Hence n = 9. So there are 3n = 3 × 9 = 27 lockers. Ans.(3) 23. On the basis of the given information, we can make
the adjacent diagram.
Q
R
T
S
X
W
Y
Z Double arrow line shows that the messages can be
sent back and forth between two computers.
For minimum number of computers, the path will be:
T→ S → X. Ans.(1)
24. Since none of the four liquids turns the reactive paper yellow, the bottles can contain pure X, pure Y, pure Z or Y mixed with Z or X mixed with Y.
Therefore options (1), (2) and (3) are ruled out.
The question itself states that only liquids that the bottle could possibly contain are X, Y, Z, X + Y, X + Z, Y + Z. Therefore the bottle cannot have X, Y and Z mixed together. Ans.(4)
25. If the mixture turns the reactive paper red, the mixture is either pure Y or Y mixed with Z. The
26. The shortest straight cut that will divide the triangular board into two equal parts will be the perpendicular drawn from B on to AC. Since ABC is an isosceles right angled triangle BD will be the median and altitude on to AC.
A 27. By converse of midpoint theorem:
∆ CQR ~ ∆ CBA, 2QR = AB and AB || QR.
⇒ 2 × CD = CE ⇒ CD = DE
Again, by converse of midpoint theorem:
|||
29. Here, in rectangle PAUB, Again, area(ABCD) = area(PQRS) – area(APB) – area(ASD) – area(CRD) – area(BQC)
= S – area(ABU) – area(ATD) – area(CDW) – area(BVC)
= S – area(ABCD) + S0 ⇒ area(ABCD) + area(ABCD) = S + S0
∴ area(ABCD) = S S+ o 2 . Ans.(1)
30. The center of the basketball and the corner of the room are on opposite ends of the space diagonal of a cube of side 8 inches (see diagram). Similarly, if we denote the radius of the softball by x, its centre and the corner of the room lie on opposite ends of the space diagonal of a cube of side x inches. By the Pythagorean theorem, the centre of the basketball is 8√3 inches from the corner of the room and the centre of the softball is x√3 inches from the corner of the room. But the distance from the centre of the basketball to the centre of the softball is the sum of their radii, namely 8 + x.
31.
A(0,2)
E(0,4) D(2 +1,4)π
B(4,0)
O(0,0) • C(2 + 1,0)π
•
If P is selected on a semi-circle with AB as diameter, then ∠APB = 90°. If P is selected from the interior of semi-circle with AB as diameter, then ∠APB will be obtuse.
Now, A(ABCDE) = A( OCDE) – A(OAB) = 4 × (2π + 1) −1
2 (2)(4) = 8π Now, l(AB) = ( )42+( )22= 20 2 5=
∴ Radius of semi-circle =2 5=
2 5
∴ Area of semi-circle = × ×1 =
2 5 5
2
π 2 π
( )
∴ Probability= = 5 2 8
5 16 π
π . Ans.(2)
32. As average distance of planets from star is ‘a’ and distance between two consecutive planets is
‘b’, the distance of the planets from the star are a – 2b, a – b, a, a + b and a + 2b.
∴ The distances travelled by the planets in 1 revolution are 2π(a – 2b), 2π (a – b), 2πa, 2π(a + b) and 2 π (a + 2b).
The average speeds of the 5 planets = 6.2x [where x = speed of slowest planet]
∴ Sum of the speeds of the 5 planets = 6.2x × 5 = 31x
∴ x + xr + xr2 + xr3 + xr4 = 31x
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33. In 10 seconds, A covers 4 × 10 = 40 m.
Initially the athletes are at L
3=51m away from each other.
The first athlete is now 40 + 51 = 91 m away from the second one.
Since the second athlete starts running at 11 m/s, the relative speed of A2 w.r.t. A1 is 11 – 4
= 7 m/s.
Therefore, A2 catches up with A1 in 91
7 =13seconds after he starts running.
After A2 catches up with A1, A3 is 51 m away from A1 and A2.
∴ Time taken =51
11= 4.64 sec. .... Qnow only A3is running T = 10 + 13 + 4.64 = 27.64 sec. Ans.(3)
34. PQRS is a rectangle. ∆DAQ is a 30 – 60 – 90 ∆.
Also, AD = BC = 6, DQ = 3, AQ = 3 3 . Again ∆DRC is a 30 – 60 – 90 ∆ . So, DC = AB = 6 3 , DR = 3 3 . RC = 6 3 × 3
2 = 9, QR = (3 3 – 3) = 3( 3 – 1).
Therefore RS = (9 – 3 3 ) = 3 3 ( 3 – 1).
Therefore area of triangle PQRS = 3( 3 – 1) ( 3 3 ) ( 3 – 1) = 18 3 (2 – 3 ) cm2. Ans.(1) 35. The big pile of apples contains the same amount of larger apples of half a rupee each (from F2),
as smaller, so average price is therefore 1 2
1 3 2
5 12
F
+HG I
KJ
= rupee. But the apples were sold for 2/5 rupee each (5 apples for 2 rupee). Or 25 60 and24
60 rupee respectively. This means that per sold apple there is a shortage of 1
60 rupee. The total shortage is 7 rupees, so the ladies together started out with 420 apples. These are worth2
5×420 168= rupees, or with equal division, 84 rupees for each. If F2 would have sold her apples herself, she would have received 105 rupees.
F2 loses 21 rupees in this deal. Ans.(1)
36. Distance Time (hrs.) Average Speed km covered
AB 1
2 5 2.5
BC 1
2 1 0 5
CD 1 12.5 12.5
DE 1
2 12.5 6.25
EF 1
2 12.5 6.25
His average speed for the whole journey =Total dis ce travelled Total time taken
tan .
=AB BC CD DE EF+ + + +
3 km per hour .
=2.5 5 12.5 6.25 6.25+ + + + = = 3
32.5
3 10.83 km per hour. Ans.(3) 37. Triangles P’OB and P’AO are congruent by SAS congruency.
Therefore ∠OP’A = ∠BP’O = 30°. In triangle P’OA tan 30° = r 10 .
Therefore r = 10
3 m. Ans.(1)
38. Triangles POB and PAO are congruent by SAS congruency.
Therefore ∠OPA = ∠BPO = 15° B
39. Since A.M. ≥ G.M, ∴ log53 log75 log97
40. The first person pays 15 cents and wins a penny. So, the booth earns 14 cents. The second person pays 15 cents and wins a nickel, so the booth earns 10 cents. Likewise, the booth earns 5 cents from the third person and loses 10 cents from the fourth person. Then the cycle repeats.
So each cycle through the game, the booth earns: 14 + 10 + 5 – 10 = 19 cents.
The problem asks how much money is made after 43 people have played the game. Because 43 is not divisible by 4, let’s look at the 40th person, the previous number divisible by 4. After the 40th person has played, the booth has been through 10 complete cycles. Having earned 19 cents per cycle, the booth has netted Rs.1.90 profit. The booth earns 14 cents of the 41st player, bringing the profit to $2.04, then 10 cents and 5 cents from the 42nd and the 43rd players, bringing the total profit to $2.19. Ans.(2)
41. The length of the tangent = (313)2−( )252=312cm=3.12m . Ans.(3) 42. Given the width of the moat = 15/√2 feet. By using the
boards of minimum length as shown in the figure (AB &
CD) we can move across the moat of width of 15/√2 .
Let the length of the boards = AB = CD = x ⇒ AC = CB = x/2. CASTLE 43. Let the two series of question booklet be A and B.
Now, according to the given condition the series can be given to the students in following ways.
First row A B A B A B
Second row A B A B A B
Or
First row B A B A B A
Second row B A B A B A
44. Let the speed of the soldier be x metre/sec. and the speed of the troop be y metre/sec.
Since, distanceαspeed, when the troop travels 5 metres the man travels 5( 1± 2 ) m.
Distance cannot be negative. So, the distance travelled by the soldier = 5(1 + √2) m. Ans.(2) 45. Area of the square = a2 unit2. In the first process: Area of each small square =a
n
2
unit2. (For convenience we assume all n squares to be of equal area.)
∴ Side of each small square= a n
2
unit= a n unit.
Radius of circular cloth pieces of maximum possible area from each of the small square
= a
In the second process: Radius of cloth piece having maximum possible area from the given square cloth piece=a
2units. Area of circular cloth piece=
F HG I
The area of cloth piece remaining after cutting the circle= − =F
−HG I
46. Wrench (x) Spanner (y) Max. Limit
Time per unit 3 hrs. 4 hrs. 190 hrs.
Raw material required 1 unit 1 unit 50 units
For Q.47 and Q.48:
As both you and your opponent seek to minimise your score at each move, both of you will move into positions with the lowest scores. ∴ Your positions in the first four moves will be as follows:
1 4 7 9
12 2 5 8
11 10 3 6
You: 1 – 12 – 2 – 4 – 7 – 5 – 2 – 10 – 3 – 5 – 8 – 6 – 3 Opponent: 6 – 3 – 10 – 2 – 5 – 7 – 4 – 7 – 5 – 3 – 5 – 8 – 9
47. At the end of second move, the required difference = (6 + 2 + 4) – (1 + 4 + 2) = 12 – 7 = 5. Ans.(1) 48. Ans.(3)
49. In one hour A will cover 60 km. Distance left is 90 km.
∴ Time required to cover 90 km is 90/140 = 38.57 min.
∴ Required time = 9 : 38.57 am. Distance from X = 60 + 38.57 = 98.57 km. Ans.(4)
50. Time taken by train A to cross B =288× 140
18
5 = 7.40 sec.
Time taken by train A to cross the train C =320× 150
18
5 = 7.68 sec. Ans.(2)
51. Ans.(3). BA, AE and ED are the keys. B, which states broadly the reasons for career change, begins the passage. A then explodes the myth that pain (a self we fear is becoming) is the only driver for change. Thereafter, E establishes the real reason we change. D then cites E as the reason for 'working identity' being a process of experimenting & learning about our possible selves. Finally, we take on an example starting with C. Hence, the order is BAEDC and the answer is option (3). In option (1), D doesn't have a link to A. In option (2), A, instead of following B, precedes it. In option (4) too, A is incorrectly separated from B and in turn separates E and D.
52. Ans.(2). BC and AE are the keys. D introduces the passage. B begins illustrating with an example.
C takes it further. A, followed by E, cites the benefits. Thus, the correct order is DBCAE and the answer is option (2). All other options have ill-connected statements.
53. Ans.(1). CA is the key. C suitably begins the passage. A takes the example further. Then comes E, which, with its 'He could be similarly strict …', connects CA to BD. Thus, CAEBD is the order and option (1) is the correct answer. In option (2), CA doesn't have a suitable link to BD, as one cannot assume that Scott would be a partner. The same goes for option (3) as regards E and CA.
56. Ans.(1). The word is incorrectly used in option (1). The correct usage is 'He is too forgiving to bear someone a grudge.' (To bear someone a grudge means to nurture a feeling of resentment against someone).
57. Ans.(3). Statements A, B and D are correct. Statement C is incorrect. Except colloquially, 'so' as an adverb of degree must not be used absolutely. In statement C, replace 'so' with 'very'.
58. Ans.(4). Statements A, B and D are correct. 'Of course' is often loosely used for 'certainly, undoubtedly'. Strictly speaking, 'of course' should be used to denote a natural or an inevitable consequence. So, statement C is incorrect.
59. Ans.(2). Only statement D is correct, as 'neither' is a distributive pronoun calling attention to the individuals forming a collection and must accordingly be followed by verbs in the singular. The object of a verb or a preposition, when it is a pronoun, should be in the objective form. In statement A, 'she' should be replaced with 'her'. 'Many' must be followed by a singular verb. In statement B, replace 'have' with 'has'. Words joined to a singular subject by 'as well as' are parenthetical.
The verb should therefore be put in the singular. In statement C, replace 'were' with 'was'.
60. Ans.(2). Only option (2) can be safely concluded from the passage. Option (1) goes beyond the scope of the passage. Options (3) and (4) are not at all suggested in the passage. Hence, option (2) is the correct answer.
61. Ans.(3). The key to ruling out options (1) and (2) lies in the word “most”. The passage doesn't indicate any assumption that a majority of people either choose work that they don't enjoy doing or want to change career directions. Option (4) is contradictory to the passage. Option (3) is a sure enough assumption for the passage. Hence, option (3) is the answer.
62. Ans.(1). Statement A has been mentioned in the second last paragraph of the passage. Statement C has been stated in the eighth paragraph of the passage.
63. Ans.(3). All the statements in this question have been wrongly inferred from the passage. They are distorted conclusions. Hence, the answer is option 3.
64. Ans.(4). Only statement B can be correctly concluded from the penultimate paragraph of the passage.
65. Ans.(1). Statements B and C are appropriate conclusions drawn from the passage. Statement B can be deduced from the second paragraph of the passage, whereas statement C from the last paragraph.
“
71. Ans.(2). Options 1, 3 and 4 can be eliminated as they do not answer the question correctly from the information given in the paragraph.
72. Ans.(1). It's a phrase in English the meaning of which is given in option 1.
73. Ans.(4). We can obtain the answer by process of elimination of wrong options. On reading the entire passage, it can be concluded that the author has written this article after a reported incident (that of PM asking CEOs to cap their salaries) and is critical about it. Hence, the answer definitely cannot be options 1, 2 and 3. It has to be a Reporter.
74. Ans.(3). On reading the entire passage, it can be concluded that the author considers the opposition "not required" i.e. unnecessary.
75. Ans.(2). The author described her style as "edgy" that means nervous.