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Transcript of . 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
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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
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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
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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
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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
⇒ 1 worker 10 days work = 1/32
⇒ 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
⇒ 1 worker 10 days work = 1/32
⇒ 32 workers will complete the work in 10 days.
Using statement III:
One-eighth of the work can be completed by 8
workers in 5 days.
⇒ 8 worker 5 days work = 1/8
⇒ 8 workers 1 day work = 1/5×8 = 1/40
⇒ 1 worker 1 day work = 1/320
⇒ 1 worker 10 days work = 1/32
⇒ 32 workers will complete the work in 10 days.
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.
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⇒ 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.
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88. (1)
Total number of items accepted by all the five
companies together in 2002:
Number of items accepted = Number of items
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:
Number of items accepted = Number of items
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.
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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.
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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):
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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.
124. (3) Let’s check all the options:
1) Alok, Arjun ⇒ Not definite neighbors
2) Abhinav, Abhay ⇒ Not definite if Abhinav is above
Abhay.
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3) Ajay, Alok ⇒ Definite that Ajay is above Alok.
4) Alok, Abhay ⇒ Not definite neighbors
125. (5) Let’s check all the options:
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)
Let’s arrange the question input.
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.
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133. (3)
Let’s arrange the question input.
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)
Let’s arrange the question input.
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.
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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.
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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.
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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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