. 1 48. 5 49. 4 50. 2 Quantitative Aptitude Solutions (51-53) 51. (2) 2, 4, 12, 48, 240, …. The...

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BOB & BOM Manipal PO Mock Test

Solutions

ENGLISH LANGAUGE

Answer (1-10)

1. 1

2. 1

3. 3

4. 5

5. 4

6. 5

7. 5

8. 3

9. 3

10. 4

Solution (11-20):-

11. 2; Refer to 3rd para last line.

12. 4; Refer to last two paras of the passage.

13. 1; Refer to 4th para, 5th sentence.

14. 3

15. 4

16. 5;

17. 1

18. 1

19. 2

20. 4

Solution (21-25):-

21. (2) Replace ‘neither sulking nor’ with ‘either sulking

or’.

22. (4) Replace ‘goes’ with ‘go’.

23. (2) Replace ‘storing’ with ‘storage’.

24. (5)

25. (4) Replace ‘travel and perform’ with ‘travelling and

performing’.

Solution (26-30):-

26. 2

27. 5

28. 3

29. 4

30. 2

Solution (31-35):-

Correct Sequence: G F A D E C B

31. 4

32. 4

33. 2

34. 1

35. 3

Solution (36-40):-

36. (2)

37. (3)

38. (2)

39. (1)

40. (4)

Solution (41-50):-

41. 4

42. 1;

43. 2

44. 3

45. 3;

46. 4

47. 1

48. 5

49. 4

50. 2



Quantitative Aptitude Solutions (51-53)

51. (2) 2, 4, 12, 48, 240, ….

The pattern is: to arrive at a term, the previous term

is being multiplied by (n+1) where ‘n’ keeps on

increasing by 1 for every term.

2

4 = 2 × (2 + 0)

12 = 4 × (2 + 1)

48 = 12 × (2 + 2)

240 = 48 × (2 + 3)

⇒ Next term = 240 × (2 + 4) = 240 × 6 = 1440

52. (3) 2, 5, 9, 19, 37, …..

The pattern is: every number is arrived at previous

number multiplied by 2 and then alternate addition

and subtraction by 1 i.e.

2

5=2×2+1

9=5×2-1

19=9×2+1

37=19×2-1

the next term 37×2+1 = 75

53. (2)

4, -8, 16, -32, 64, ….

The pattern is: Every number is arrived at by

multiplying previous alternate number with ‘4’ as

shown below:

Hence, ‘-128’ is the correct answer.

Solutions (54-55)

54. (5)

2, 9, 28, 65, 126, 216, 344.

The pattern in the series is that the series is triangular

as shown below:

In the triangular series, the difference between

consecutive terms is written below the numbers and

then, difference between consecutive differences is

written below & this process carries on until all the

difference become equal. In the figure above there

was an error & we have corrected it.

55. (4)

10, 26, 74, 218, 654, 1946, 5834

The pattern is: to arrive at next term, the previous is

multiplied by 3 and subtracted by 4:

10

10 × 3 – 4 = 26

26 × 3 – 4 = 74

74 × 3 – 4 = 218

218 × 3 – 4 = 650 ≠ 654

650 × 3 – 4 = 1946

1946 × 3 – 4 = 5834

Here, ‘654’ was wrong.

56. (4) People: 7 men and 6 women

Condition: 5 persons to be selected, at least 3 men.

Number of ways to do that = 7C3 × 6C2 + 7C4 × 6C1 + 7C5

= 7!

3!.4!×

6!

2!.4!+

7!

3!.4!×

6!

1!.5!+

7!

2!.5!

= 7×6×5

3×2×

6×5

2+

7×6×5

3×2×

6×1

1+

7×6

2

= 35 × 15 + 35 × 6 + 21

= 21 ( 5×5 + 5×2 + 1)

= 21 (25 + 10 + 1) = 21 × 36

= 756

Clearly, there are 756 ways to do that.

57. (2)

The lawn can be represented, diagrammatically, as:

Area of roads (running parallel to length & breadth) =

Area of road parallel to length + Area of road parallel

to breadth – Area of square double counted (shaded

region)

= 4 × 55 + 4 × 35 – 4 × 4

= 4 ( 55 + 35) – 16

= 360 – 16 = 344 m2

⇒ Cost of gravelling the roads at 75 paise per square

metre =75

100× 344

= ¾ × 344 = 3 × 86 = Rs. 258

58. (2) Let the required number of hours be ‘x’

Speed for working of 1st and 2nd set = ½ : 1/3

More work, More Time ⇒ Direct Proportion

Less speed, More time ⇒ Indirect Proportion

Work = 1 : 2 ∷ 25 : x

Speed = ½ : 1

3

∴ 1 ×1

3× 𝑥 = 2 ×

1

2× 25

⇒ x = 75



59. (2) Let the principal be ‘P’

S.I. = 𝑃×𝑅×𝑇

100

C.I. = 𝑃 [(1 +𝑅

100)

𝑇− 1]

For compounded half-yearly, time becomes double &

rate becomes half.

Difference for 2 years = 𝑃 [(1 +𝑅

100)

𝑇− 1] –

𝑃×𝑅×𝑇

100

= P {[ (1 +5

100)

4− 1] – (10 × 2)/100]}

124.05 = 𝑃 {[ (21

20)

4 – 1] –

1

5}

⇒ 124.05 = 𝑃 {[194481

160000– 1 ] –

1

5}

⇒ 124.05 = 𝑃 {[34481

160000] –

1

5}

⇒ 124.05 = 𝑃{34481−32000}

160000

⇒ (124.05 × 160000)/2481 = 𝑃

⇒ P = 8,000

60. (2) Let cost of 1 liter milk be Re. 1

Milk in 1 liter mix in 1st can = ¾ liter, CP = Re. ¾

Milk in 2 liter mix in 2nd can = ½ liter, CP = Re. ½

Milk in final mix = 5/8, CP = 5/8

By the rule of allegation,

⇒ Ratio of two mixtures = 1/8 : 1/8 = 1: 1

So, quantity of mixture to be taken from each can = ½ ×

12 L = 6L each

61. (1)

By the rule of allegation, we have:

So, the ratio of 1st and 2nd quantities = 7: 14 = 1: 2

⇒ Required quantity replaced = 1/3

62. (4) A train 150 m long passes a Km stone in 15

seconds : 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛

𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛 = Time

150

𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛= 15 𝑠𝑒𝑐

⇒ Speed of Train = 150

15 = 10 m/s

First Train passes another train of the same length

travelling in opposite direction in 8 seconds: 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛 𝐼+𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛 𝐼𝐼

𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛 𝐼+𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑇𝑟𝑎𝑖𝑛 𝐼𝐼= 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛

150+150

10+𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑇𝑟𝑎𝑖𝑛 𝐼𝐼= 8

⇒300

8– 10 = 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡𝑟𝑎𝑖𝑛 𝐼𝐼

⇒ 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 2𝑛𝑑 𝑡𝑟𝑎𝑖𝑛 =55

2𝑚𝑝𝑠 𝑜𝑟 99 𝑘𝑚𝑝ℎ.

63. (4)

Two pipes A and B can fill a cistern in 12 minutes and

15 minutes respectively while a third pipe C can

empty the full tank in 6 minutes. A and B are kept

open for 5 minutes in the beginning then C is also

opened. In what time is the cistern emptied

Work done by A in 1 minute = 1

12

Work done by B in 1 minute = 1

15

Work done by both A and B in 1 minute = 1

12+

1

15=

3

20

Work done by both A and B in 5 minutes = 5 ×3

20=

15

20=

3

4

Work done by C in 1 minute = −1

6

Work done by A, B and C in 1 minute =3

20–

1

6= −

1

60

Work done in 6th minute = −1

60

Tank left = ¾ −1

60=

45 – 1

60=

44

60

Similarly, Work done in next 44 minutes = 44 ×

−1

60= −

44

60

Tank left =44

60–

44

60= 0

Now, tank is empty in 1 + 44 = 45 min after C is

started.

64. (3)

7 men can complete a work in 12 days.

In one day, the work done by 7 men = 1

12

In one day, work done by one men = 1

12×7=

1

84

Work done in 5 days = 5 ×1

12=

5

12

Left work = 1 –5

12 =

7

12

After 5 Days, Two Men Left

5 men’s 1 day work = 5 ×1

84=

5

84

Days required to complete 7

12 of work by 5 men =

7

12×

84

5 =

49/5 ≈ 9 days



65. (4)

A, B and C = 7

2:

4

3:

6

5=

7×15

30:

4×10

30:

6×6

30= 105 ∶ 40: 36.

After 4 months, A increases his share by 50%.

A: B: C = 105 × 4 + 105 ×150

100× 8 ∶ 40 × 12: 36 × 12

= 105 + 105 ×3

2× 2: 120: 108 = 35 + 35 × 3 ∶ 40 ∶ 36

A: B: C= 140: 40: 36 = 35: 10: 9

⇒ A’s share = 35

54; B’s share =

10

54, C’s share =

9

54

Total Profit at the End of One Year = Rs. 21,600

B’s share = 10

54× 21600 = Rs. 4,000

66. (4)

Let the cost price be 100

Gain% = 35%

CP for 16 articles = 16 × 100

⇒ Gain = 16 × 35 = Rs 560

SP of 15 articles = 1600 + 560 = 2160

SP of 1 article = 2160

15= 𝑅𝑠. 144

∵ Discount is 4% ⇒ IF SP is 96 then MP is 100

If SP is 144 then MP = 100

96× 144 = 𝑅𝑠 150

CP = 100, MP = Rs 150

Hence, MP is 50% over CP

67. (3)

Let the amount of bill be ‘x’

Amount after Two successive discounts of 20% =

x(1 –20

100)

2 =

16𝑥

25

⇒ Discount = 𝑥 –16𝑥

25=

9𝑥

25 ------(1)

Stand alone discount of 35% = 35x/100 = 7x/20 -------

--(2)

(1) – (2) = 22 .. Given

9x/25 –7x/20 = 22

⇒ 5x/500 = 22

⇒ x = 2200

Hence, the amount of bill is Rs. 2,200.

68. (2)

Let the present age of Naman be ‘X’ 𝑋 – 6

18= 𝐴𝑚𝑎𝑛’𝑠 𝑎𝑔𝑒 ------(1)

Aman is 2 years younger to Madan whose age is 5

years:

Aman’s age = Madan’s age (5) – 2 = 5 – 2 = 3 years

Substituting in (1)

X – 6 = 18 × 3 = 54

⇒ X = 54 + 6 = 60 years.

Hence, Naman’s age is 60 years old.

69. (2)

8796 × 223 + 8796 × 77 =?

= 8796 (223 + 77)

= 8796 × 300

= 2638800

Hence, ? = 2638800

Solutions (70-74)

70. (5)

I. 4x2= 16

⇒ x2 = 4

⇒ x = ± 2

II. y2 - 10y + 16 = 0

Using ax2 + bx + c

x = −𝑏±√𝑏2−4𝑎𝑐

2𝑎

y = 10±√102−4×1×16

2×1=

10±6

2= 8, 2

x = 2, -2 & y = 2, 8

⇒hence y ≥ x

71. (3) I. 2x2 + 40 = 18x

2x2 -18x + 40 = 0

x2 – 9x + 20 = 0

Using

x = 9±√92−4×20

2

x = 5, 4

II. y2 = 13y - 42

y2 – 13y + 42 = 0

y = 13±√132−4×42

2

Y = 7, 6

Hence, Y > X.

72. (5). I. 12y2 + 8y = 4y + 8

12y2 + 4y – 8 = 0

3y2 + y - 2 = 0

y = −1±√12+4×2×3

6

⇒ y = -1, 2/3

II.x2 -11 = 2 – 12

⇒ x2 = -10+11

⇒ x = √1 = ±1

Clearly, the relation does not exist.

73. (2). I. x - 7 = 0

⇒ x = 7

II. 3y2 - 10y + 7 = 0

Using

y = −𝑏±√𝑏2−4𝑎𝑐

2𝑎

Y = 10±√102−4×3×7

6

Y = 1, 7/3

⇒ X > Y



74. (1). I. 5x2 + 11x + 6 = 0

Using

x = −𝑏±√𝑏2−4𝑎𝑐

2𝑎

x = −11±√112−4×5×6

10

X = -1, -6/5

II. 4y2 + 10y - 66 = 0

2y2 + 5y - 33 = 0

y = −5±√52+4×2×33

4

y = 3, -11/2

⇒ Relationship between X and Y can’t be established.

75. (3)

Probability that A speaks truth = 75

100=

3

4

Probability that A speaks Lie = 1 –3

4=

1

4

Probability that B speaks truth = 80

100=

4

5

Probability that B speaks Lie = 1 –4

5=

1

5

A and B will contradict each other when one speaks

truth while other lie.

Probability of Contradiction cases =3

4×

1

5+

1

4×

4

5=

3 + 4

20=

7

20

In percentage = 7

20× 100 = 35%

Solutions (76-80)

76. (4)

We’ve to find: No. of articles sold.

I. Total profit earned was Rs. 1596

II. Cost Price per article was Rs. 632

III. Selling price per article was Rs. 765

Statement II and III: give profit per article = SP per

article – CP per article = 765 – 632 = 133

Using Statement III with I, II: We get that Number of

articles = Total profit ÷ Profit per article = 1596

÷133= 12

Hence, all statements are required to answer the

question.

77. (4). Find: The ratio in which 3 partners will distribute

the profit.

Using statement I:

Rahul got one-fourth of the profit. ⇒ We can’t tell the

share of other two partners.

Using statement II:

Rahul & Vivek contributed 75% of the total

investment. ⇒ Rahul & Vivek have 75% while Anurag

has 25% ownership. But we can’t tell what the

individual ownership of Rahul & Vivek is.

Using statement I and II together:

Rahul has ¼ of shares and, Rahul & Vivek together

have ¾ of shares ⇒ Anurag has ¼ of share

Also, Vivek’s share = ¾ - ¼ = ½

Hence, Rahul: Vivek: Anurag = ¼: ½: ¼ = 1: 2: 1

Hence, both statements are required to answer the

question.

78. (5). No. of days = 10

Find: No. of workers

Using statement I:

20% work can be completed by 8 workers in 8 days.

⇒ 8 worker 8 days work = 1/5

⇒ 8 workers 1 day work = 1/5×8 = 1/40

⇒ 1 worker 1 day work = 1/320


⇒ 32 workers will complete the work in 10 days.

Using statement II:

20 workers can complete the work in 16 days.

20 worker’s 1 day work = 1/16

⇒ 1 worker 1 day work = 1/16×20 = 1/320



Using statement III:

One-eighth of the work can be completed by 8

workers in 5 days.


⇒ 8 workers 1 day work = 1/5×8 = 1/40

⇒ 1 worker 1 day work = 1/320



Hence, the solution can be found using any one of the

three statements.

79. (4). Find: Time to fill.

Using statement I:

A is 50% more efficient than B.

We don’t have total time to fill the tank or the

individual time taken by A or B. So, not complete

information.

Using statement II:

A alone takes 16 hours to fill the tank. We don’t have

total time to fill the tank or the individual time taken

by B. So, not complete information.

Using statement I and II together:

A is 50% more efficient than B.

A alone takes 16 hours to fill the tank.



⇒ Time taken by 𝐴 ×150

100 = Time taken by B

⇒ Time by B = 16 ×3

2= 24 hours

so, 1/total time =1

16+

1

24=

5

48

⇒ Total Time =48

5 hrs.

Hence, both the statements are required to find the

solution.

80. (3)Find: Distance between two stations.

Using statement I:

The speed of mail train is 12 kmph more than that of

an express train.

We don’t have time to cover the distance by the trains.

So, not complete information.

Using statement II:

A mail train takes 40 min less than an express train to

cover the distance.

We don’t have speed to cover the distance by the

trains. So, not complete information.

Using statement I and II:

Difference in speed of both train = 12 kmph

Difference in time by both train = 40

Let total distance be ‘D’

Let time by Express train be TE & by Mail be Tm.

⇒ TE - Tm = 40 min or 2/3 hrs ----(1)

Also, Speed of mail Train – Speed of Express Train =

12 kmph

D ÷ Tm – D ÷ Te = 12 kmph

⇒ D (1/Tm – 1/TE) = 12 kmph

⇒ D ( TE - TM)/TM.TE = 12

⇒ D ( 2/3)/TM.TE = 12

⇒ D/ TM.TE = 18

∵ we don’t know the value of TM.TE so the distance

can’t be found. Hence, data even in both statements

(I) and (II) together are not sufficient to answer the

question.

Solutions (81-85)

81. (2)The given information can be summarized as:

States Total Cars Diesel Car Petrol car

State-1 98 42 56

State-2 196 70 126

State-3 224 140 84

State-4 182 91 91

Difference between the number of diesel engine cars in

State-2 and the number of petrol engine cars in State-4 =

70 – 91 = 21

82. (1)Number of petrol engine cars in State-3 = 84

Number of diesel engine cars in State-1 = 42

The number of petrol engine cars in State-3 is percent

more than the number of diesel engine cars in State-1

= 84 – 42

42× 100 =

42

42× 100 = 100% 𝑚𝑜𝑟𝑒

83. (4)

Diesel Engine Cars In State-3 = 140

⇒ 25% are AC

⇒ 75% are non-AC: 75 ×140

100 = 105

Hence, Non-ac diesel car in State-3 are 105.

84. (5)

Total number of cars in State-3 = 224

Number of petrol engine cars in State-2 = 126

⇒ Difference = 224 – 126 = 98

85. (2)

Average number of petrol engine cars = [𝑃𝑒𝑡𝑟𝑜𝑙 𝑐𝑎𝑟 𝑖𝑛 (𝑆𝑡𝑎𝑡𝑒1 + 𝑆𝑡𝑎𝑡𝑒2 + 𝑆𝑡𝑎𝑡𝑒 3 + 𝑆𝑡𝑎𝑡𝑒4)]

4

[56+126+84+91]

4=

357

4= 89.25

So, average number of Petrol cars in all the states is 89.25

Solutions (86-90)

86. (3)

Company B, year 2003, Rejected item = 2600

Company B, year 2003, Manufactured item = 152000

Percentage of items rejected out of total items

manufactured by Company B in the year 2003 = 2600

152000× 100 = 1.71 ~ 1.71%

87. (4)

Number of items accepted = Number of items

manufactured – Number of items rejected.

Items accepted by A in 2004 = 156 – 2.2 = 153.8 or

153800

Sold items by A in 2004 = 145000

⇒ Unsold items by A in 2004 = Accepted – Sold = 153800

– 145000

= 8,800

⇒ Unsold items out of accepted by A in 2004 = 8,800.



88. (1)

Total number of items accepted by all the five

companies together in 2002:


manufactured – Number of items rejected

A: 164 – 1.7 = 162.3

B: 115 – 1.1 = 113.9

C: 172 – 2.9 = 169.1

D: 169 – 1.9 = 167.1

E: 96 – 0.8 = 95.2

⇒ Total accepted items = 162.3 + 113.9 + 169.1 +

167.1 + 95.2 = 707.6 or 707600

89. (5)

Average number of items rejected by Company D for

all the given years

2001: 1.5

2002: 1.9

2003: 2.3

2004: 2.1

2005: 2.0

2006: 2.4

⇒ Total = 1.5 + 1.9 + 2.3 + 2.1 + 2 + 2.4 = 12.2 or

12200

Average number of items rejected by Company D for

all the given years = 12200

6 = 2033 ~ 2030.

90. (3)

Total number of items accepted by all the five

companies together in 2006:


manufactured – Number of items rejected

A: 175 – 2.8 = 172.2

B: 168 – 2.2 = 165.8

C: 180 – 2.4 = 177.6

D: 171 – 2.4 = 168.6

E: 105 – 0.8 = 104.2

⇒ Total accepted items = 172.2+ 165.8+ 177.6+

168.6+ 104.2= 788.4 or 788400

Solutions (91-95)

91. (3)

Given information can be summarized as:

Days Rahul Gita Naveen Monday 250 130 360 Tuesday 180 200 260 Wednesday 460 420 120 Thursday 320 400 150 Total 1210 1150 890

Gita’s average earning over all the days together

= Total earning ÷ 4

= 1210 ÷ 4

= 302

92. (1)

Total amount earned by Rahul and Naveen on

Tuesday and Thursday

Tuesday = 180 + 260 = 440

Thursday = 320 + 150 = 470

⇒ Total earning = 440 + 470 = 910

93. (3)

Gita donated her earnings of Wednesday to Naveen.

Gita’s income on Wednesday = 420

Naveen’s Income on Wednesday = 120

Here, Naveen’s new income = 120 + Geeta’s income = 120

+ 420 = 540

94. (3)

Difference between Rahul’s earning on Monday and

Gita’s earning on Tuesday

Rahul’s earning on Monday = 250

Gita’s earning on Tuesday = 200

⇒ Difference = 250 – 200 = 50

95. (3)

Ratio of Naveen’s earning on Monday, Wednesday and

Thursday

Earning on Monday = 360

Earning on Wednesday = 120

Earning on Thursday = 150

⇒ Ratio = 360: 120: 150 = 36: 12: 15

= 12: 4: 5

Solutions (96-100)

96. (3)

The difference in number of candidates appeared

from Mumbai:

Mumbai Difference 2001 35145 2002 17264 35145 – 17264 = 17881 2003 24800 24800 – 17264 = 7536 2004 28316 3516 2005 36503 8287 2006 29129 7374 2007 32438 3309 ⇒ Minimum

Clearly, the difference between appeared candidates is

minimum in 2007.



97. (1)

The number of candidates qualified from Chennai

Centre → Year ↓

Chennai % Qualified Number of Qualified

2001 37346 9 3361 2002 48932 12 5872 2003 51406 10 5140 2004 52315 8 4185 2005 55492 13 7214 2006 57365 11 6310 2007 58492 14 8189 ⇒

Maximum

The numbers of candidates qualified from Chennai are

maximum in year 2007.

98. (4)

Candidates qualified from Delhi in 2002 = 58248 ×28

100 = 16,310

Candidates qualified from Delhi in 2006 = 59216 ×20

100 = 11,843

Total number of candidates qualified from Delhi in

2002 and 2006 together = 16310 + 11843 = 28,153

99. (2)

Candidates appearing from Kolkata in 2004 = 71253

Qualified percentage = 19%

⇒ Candidates qualified in the competitive

examination = 71253 ×19

100 = 13,538 ≡ 13,540

100. (5)

Candidates from Hyderabad

Year 2001 2002

Appeared 51124 50248

Qualified% 17 21

Qualified 8,691 10,552

⇒ Difference in qualified in 2001 to 2002 = 10552 – 8691

= 1861 ~ 1860.



Reasoning Ability

Solutions: (101 – 105)

101. (2)Given statements are: M ≥ E > I; E = P < T

Conclusions

I. M ≥ T (False as M ≥ E and E < T hence relation

between M and T can’t be established)

II. T > I (True because I < E < T)

Hence only conclusion II follows

102. (5) Given statements are: A < K = D ≥ J; K < F; A ≥ G

On combining we get G ≤ A < K = D ≥ J; K< F

Conclusions

I. F > J (True because F > K ≥ J)

II. G < D (True because G ≤ A < K = D)

Hence both the conclusions are true

103. (2)Given statements are: Z = Q > V ≤ S; Q < N ≤ U

On combining: U ≥ N > Q > V ≤ S; Q = Z

Conclusions:

I. U ≥ V (False because U > V)

II. N > V (Clearly true) Hence only conclusion II follows.

104. (4) Given statement is: P > J ≥ X < K = Q Conclusions:

I. X < P(True because P > J and J≥X)

II. Q < X(false because Q > X) Hence neither conclusion I nor II follows

105. (2) Given statements are: D > N = O ≤ W; N ≥ Q; W ≤

H

On combining: D > N = O ≤ W ≤ H;N ≥ Q

Conclusions:

I. D ≥ Q (False because D > N ≥ Q it means D > Q)

II. H ≥ N (True)

Hence only conclusion II follows.

Solutions (106-108)

106. (3) The Coding can be illustrated as:

Here, the code for ‘Can be’ = Laj Kaj

And, for ‘it’ doesn’t exist in the question hence, it must

be a new code.

Out of all given options only option d) fits the criteria.

107. (4)

Here, Aj Maj = The Good.

And, ‘Raj’ doesn’t exist in the code so it must be a new

word.

Out of all given options only option c) fits the criteria.

108. (2)

Here, ‘Laj’ is clearly either the code for ‘Can’ or ‘Be.

Solutions (109-113)

109. (3)

110. (2)

111. (4)

.

112. (5):



113. (3)

Solutions (114-117)

114. (3)

Couples: Mr. & Mrs. Aggarwal, Mr. & Mrs.

Bhatia, Mr. & Mrs. Chodhary, Mr. & Mrs.

Dhawan

People: 8

For the sake of simplicity following symbols have

been used:

Man → Square, Woman → Circle, Aggarwal →A,

Bhatia → B, Chodhary → C and Dhawan → D

115. (2) Clearly, Mr. Chodhary is sitting across Mrs.

Aggarwal.

116. (3) Clearly, three people are sitting between Mrs.

Bhatia & Mrs. Chodhary.

117. (4) Mr. Dhawan is sitting 3rd to the left of Mr.

Aggarwal.

Solutions (118-120)

118. (4) People: X, Y, Z, A, B, P, Q,R and U

Generations: 2

Clearly, X, A, P and Q/R are male but since we don’t

know the gender of Y.

So the number of males is either 4 or 5.

119. (3)Clearly, Z’s husband is either Q or R so it can’t be

determined.

120. (2) P’s brother is Z’s husband. And, U is Z’s mother.

So, U is P’s brother’s mother-in-law.

Solutions (121-125)

People: Arjun, Ajay, Aman, Abhay, Abhinav and Alok

Subjects: English, Biology, Civics, Accounts, Hindi and

History

Cartoons: , Power-puff Girls, Tom & Jerry, Duck

Tales, Scooby Doo and Dexter

121. (3)

122. (4) Clearly, out of given options, Aman is adjacent to

English studying person i.e. Ajay in 2nd possibility.

123. (3) Let’s check all the options:

1) Arjun ⇒ We don’t know the sure position

2) Civics ⇒ We don’t know the sure position

3) Popeye ⇒ We know the sure position

4) Abhay ⇒ We don’t know the sure position

Clearly, Popeye is odd-one out.


1) Alok, Arjun ⇒ Not definite neighbors

2) Abhinav, Abhay ⇒ Not definite if Abhinav is above

Abhay.



3) Ajay, Alok ⇒ Definite that Ajay is above Alok.

4) Alok, Abhay ⇒ Not definite neighbors


1) Ajay ⇒ Can study Civics in possibility 1st.

2) Arjun ⇒ Can study Civics in both the possibilities.

3) Abhay ⇒ can study Civics in both the possibilities.

4) Aman ⇒ Can study Civics in possibility 2nd.

Hence, none of these is the answer.

Solutions (126-129)

People: A, B, C, D, E and F ⇒ Row 1 → Facing North

(or Upward)

P, Q, R, S, T and U ⇒ Row 2 → Facing South (or

Downward)

126. (1)

127. (3)Clearly, B is sitting diagonally opposite to P.

128. (4) In possibility 1, 3 people are sitting between S

and R while in 2nd possibility 2 people are sitting

between S and R.

So, we can’t determine that.

129. (4) Let’s check all the statements:

a) E is third to the right of A ⇒ This is true in

possibility 1.

b) Q faces E ⇒ This is not true in any of the possibility.

c) A is diagonally opposite to R ⇒ This is true in

possibility 1.

d) U is facing B ⇒ This is true in possibility 2.

130. (3) Using statement I:

I. A is taller than C. ⇒ A > C.

We can’t tell who is the tallest. Hence, statement I isn’t

sufficient alone.

Using statement II:

II. B is taller than C and D. ⇒ B > C, D.

We can’t tell who is the tallest. Hence, statement II

isn’t sufficient alone.

On combining I and II:

A is taller than C; B is taller than C and D.

A > C; B > C,D

We can’t tell who is the tallest. Hence, statement I and

II aren’t sufficient even together.

Solutions (131-135)

131. (5) We can see in the last step, all numbers are

arranged in increasing order of increasing digit sum

as number with lowest sum 710(7+1+0 = 8) is at 1st

place and number with highest digit sum 689(6+8+9=

23) is at the last place, from left to right.

In step 1, number with lowest digit sum i.e. 710 comes

at 1st place. And, 316 with 2nd lowest digit sum is

automatically filled.

In step 2, number with 3rd lowest digit sum i.e. 245 is

exchanging position with 3rd position number.

In step 3, number with 4th lowest digit sum i.e. 436 is

exchanging position with 4th position number.

Similarly, the process is repeated till all are arranged

in ascending manner by digit sum.

Let’s arrange the question input.

Input: 655, 436, 764, 799, 977, 572, 333

Digit sum: 655 = 16, 436 = 13, 764 = 17, 799=25, 977=23,

572=14, 333=9

Step I: 333,655, 436, 764, 799, 977, 572,

Step II: 333, 436, 655, 764, 799, 977, 572

Step III: 333, 436, 572, 764, 799, 977, 655

Step IV: 333, 436, 572, 655, 799, 977, 764

Step V: 333, 436, 572, 655, 764, 977, 799

[here 977 and 799 have automatically arranged after

arranging 764]

Clearly, ‘333, 436, 572, 655, 977, 764, 799’ is

132. (1)


Input: 544, 653, 325, 688, 461, 231, 857

Digit sum: 544=13; 653=14; 325=10, 688=22, 461=11,

231=6, 857=20

Step I: 231, 544, 653, 325, 688, 461, 857

Step II: 231, 325, 653, 544, 688, 461, 857

Step III: 231, 325, 461, 544, 688, 653 , 857[Here 544 has

been automatically filled]

Step IV: 231, 325, 461, 544, 653, 688, 857

Step V: 231, 325, 461, 544, 653, 857 , 688

Clearly, 5 steps are required to complete the given input.



133. (3)


Input: 236, 522, 824, 765, 622, 463, 358

Digit sum: 236=11, 522=9, 824=14, 765=18, 622=10,

463=13, 358=16

Step I: 522, 236, 824, 765, 622, 463, 358

Step II: 522, 622, 824, 765, 236 , 463, 358

Step III: 522, 622, 236, 765, 824, 463, 358

Clearly, 3rd step is ‘522, 622, 236, 765, 824, 463, 358’

134. (2)


Step II: 620, 415, 344, 537, 787, 634, 977

Digit sum: 620=8, 415=10, 344=11, 537=15, 787=22,

634=13, 977=23

Here 620, 415 and 344 are automatically arranged.

Step III: 620, 415, 344, 634, 787, 537 , 977

Step IV: 620, 415, 344, 634, 537, 787 , 977

[Here 787 and 977 have been automatically arranged].

Hence, step 4th is 620, 415, 344, 634, 537, 787, 977

135. (1) Let’s arrange the question input.

Input: 473, 442, 735, 542, 367, 234, 549

Digit sum: 473=14, 442=10, 735=15, 542=11, 367=16,

234=9, 549=18

Step I: 234, 442, 735, 542, 367, 473, 549 [Here 442 has

been automatically arranged]

Step II: 234, 442, 542, 735 , 367, 473, 549

Step III: 234, 442, 542, 473 , 367, 735, 549

Step IV: 234, 442, 542, 473 , 735, 367 , 549 [ Here 367 and

549 have automatically been arranged].

Step IV is the last step.

Hence, last step is 234, 442, 542, 473, 735, 367, 549

136. (4)

Using statement I:

I. The month begins on Monday. ⇒ We don’t know the

number of days. Hence, we can’t tell the number of

Sunday just with the 1st day being a Monday.

Hence, statement I isn’t sufficient alone.

Using statement II:

II. The month ends on Wednesday. ⇒ We don’t know

the number of days. Hence, we can’t tell the number of

Sunday just with the last day being a Monday.

hence, statement II isn’t sufficient alone.

On combining I and II:

1st day = Monday, last day = Wednesday.

There can be either 28 or 29 or 30 or 31 days.

Last day = Wednesday and this condition is only getting fulfilled when there are 31 days in a month. Hence, no. of days = 30 where 7th,14th,21st and 28th day is a Sunday. So, number of Sundays = 4. Hence, data in both statements (I) and (II) together are necessary to answer the question. 137. (3)

Using statement I: I. I counted 132 pages from the beginning of this book. ⇒ We can’t tell the total number of pages in the given book because there can be ‘n’ more pages in the end. Hence, statement I isn’t sufficient alone. Using statement II: II. My wife counted 138 pages starting from the end of the same book. . ⇒ We can’t tell the total number of pages in the given book because there can be ‘n’ more pages in the beginning. Hence, statement II isn’t sufficient alone. On combining I and II: I counted 132 pages from the beginning; My wife counted 138 pages starting from the end. . ⇒ We can’t tell the total number of pages in the given book because we don’t know they both stopped counting at the same page. Therefore, Data even in both statements (I) and (II) together are not sufficient to answer the question.

Solutions (138-142)

138. (2)

139. (4) Clearly, A is the grandfather of E.

140. (2) Clearly, there are three male members in the

family.

141. (1) Clearly A is a doctor.

142. (3) Clearly AD are a couple.

143. (5) Given: 4 subjects[Physics, Chemistry,

Mathematics and Biology] taught for 1 hr. each from

8.00 am.



Solutions (144-146) 144. (5) Government isn’t allowed to invest in NALCO due

to restriction imposed by NPCIL. In the passage it is written that, “government had to turn down a Rs 12,000 crore investment proposal by the National Aluminium Company (Nalco) to become a silent partner with the Nuclear Power Corporation of India (NPCIL)” ⇒ That NPCIL didn’t force government to become silent partner & forbid investment in NalCo. In fact on reading the passage, we can see that it’s the effect of restrictive law i.e. Un-amended Nuclear Act that is stopping the government from investing in itself (NALCO). Hence, the given statement is definitely false.

145. (1) Currently PSUs that are non-subsidiary of Department of Atomic Energy can invest in nuclear energy sector. In the passage it is written that, “If passed, under the

new and expanded scope of the law, public sector units that are not subsidiaries of the Department of Atomic Energy would be able to invest in the nuclear energy sector.” hence, non-subsidiary PSUs will be able to invest in Nuclear Energy Sector, if only the amendment is passed. ⇒ Currently they can’t invest as law hasn’t yet been amended. Hence, the given statement is definitely false.

146. (1) Restrictive laws don’t include the new &

upcoming amendment to Atomic Energy Act, 1962. In the beginning of the passage, it’s shown that after the new amendment is passed, the non-subsidiary PSUs will be able to invest in Nuclear Sector, which means the amendment is empowering them. Whereas in the end of the passage, it’s written that the restrictive law isn’t allowing the government to invest in itself. So, the new & upcoming amendment isn’t at all restrictive as it allows them to invest rather than restricting them. So, the given statement is definitely false.

147. (3) The Government should force US to lessen the

production of Atomic items to empower Indian

Nuclear position:

As we can see in the passage, there is no mention of

US and it’s Atomic Items so inference can’t be drawn

as data is inadequate to make this statement true or

false.



148. (4) Number of people = 7

1. X is taller than Y. ⇒ X > Y

2. W is taller than A, B and C but not as tall as Z. ⇒ Z >

W > A,B, C

3. Z is taller than Y ⇒ Z > Y

⇒ On combining all the three:

X > Y < Z > A, B, C

Here, either X or Z can be the tallest so we cannot

determine the answer.

149. (4)

The Government has appealed to all citizens to use

potable water judiciously as there is an acute shortage

in supply. Excessive use may lead to huge scarcity in

future months.

1. People may ignore the appeal and continue using

water as per their consideration. => This is not

implied. Has it been so then government wouldn’t

have made the request in the first place. Whenever we

make a request, it is assumed that the other person

will pay heed to our request.

2. Government may be able to tap those who do not

respond to the appeal. =>it’s also not implied as

government has not specified any penalties or such

which could be borne by those not abiding to the

appeal.

3. Government may be able to put in place alternate

sources of water in the event of a crisis situation. =>

It’s definitely not implied. Has it been the case then

government wouldn’t have made the appeal. Instead

it is requesting people to save water for crisis.

4. Large number of people may positively respond to

the Government’s appeal and help tide over the crisis.

=> It’s definitely implied. Whenever we appeal

something to someone we assume that they will pay

heed and respond to our appeal. Government must

have been thinking the same.

5. Only poor are going to suffer from this shortage of

water supply. => Not implied. There is no information

about the class of victims of water shortage in the

passage.

Hence, only 4th implies.

150. (4) Statement I: The prices of petrol and diesel in

the domestic market have remained unchanged for

the past few months.

Statement II: The crude oil prices in the international

market have gown down substantially in the last few

months.

We can see the prices of oil have gone down in the

international market but that is not the case in the

national market.

We can imagine hypothetically that Prices have gone

down in World because there must have been an

increase in supply or decrease in demand or any other

factor like that which derives the prices down.

Similarly, the prices of oil in the country must have

remained constant despite downward trend in the

world market because government must have

increased tax(excise duty) on oil or any other factor

like that which derives the prices upward.

So, we can say that both the statements are the effect

of independent causes. Their causes have no

relationship among each other.



GENERAL AWARENESS

151. (3)

152. (2)

153. (2)

154. (3)

155. (3)

156. (5)

157. (3)

158. (2)

159. (4)

160. (1)

161. (4)

162. (4)

163. (3)

164. (1)

165. (2)

166. (2)

167. (3)

168. (1)

169. (5)

170. (4)

171. (3)

172. (1)

173. (4)

174. (3)

175. (4)

176. (3)

177. (1)

178. (3)

179. (3)

180. (3)

181. (2)

182. (3)

183. (3)

184. (4)

185. (2)

186. (1)

187. (4)

188. (3)

189. (4)

190. (4)

191. (2)

192. (1)

193. (3)

194. (2)

195. (3)

196. (2)

197. (3)

198. (4)

199. (4)

200. (2)




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