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1 | P a g e
Answer Key for IBPS RRB Officer Scale 1
Prelims 2017 Model Question Paper
1. 5 11. 5 21. 4 31. 3 41. 3 51. 2 61. 5 71. 3
2. 4 12. 5 22. 3 32. 1 42. 1 52. 4 62. 4 72. 1
3. 1 13. 1 23. 4 33. 5 43. 3 53. 1 63. 5 73. 2
4. 1 14. 5 24. 2 34. 1 44. 1 54. 1 64. 1 74. 4
5. 3 15. 4 25. 3 35. 1 45. 2 55. 1 65. 3 75. 4
6. 1 16. 4 26. 4 36. 3 46. 3 56. 2 66. 3 76. 5
7. 1 17. 5 27. 5 37. 4 47. 1 57. 5 67. 1 77. 1
8. 2 18. 5 28. 1 38. 2 48. 2 58. 1 68. 2 78. 5
9. 1 19. 5 29. 1 39. 3 49. 3 59. 4 69. 5 79. 2
10. 2 20. 5 30. 3 40. 1 50. 3 60. 1 70. 4 80. 1
1. Let the cost of 1 kg of Apple and nuts be x and y respectively.
According to question,
⇒ 1800 = 12 × x
⇒ x = Rs. 150
Also,
⇒ 5 × y = 20 × x
⇒ y = (20 × 150)/5
⇒ y = Rs. 600
Thus cost of 1 Kg of nuts = Rs. 600
Let us suppose the quarterly income of Ritu be ‘I’. Thus,
2 | P a g e
⇒ I = 68 × 600
⇒ I = Rs. 40800
We know that,
⇒ Annual income = 4 × Quarterly income
∴ Annual income of Ritu
= 4 × 40800
= Rs. 163200
Hence the annual salary of Ritu is Rs. 1,63,200
2. We know that, average = Sum of all quantities/Number of quantities
∴ Average marks = Sum of marks in all subjects/Number of subjects
∵ Average of 4 subjects is 60%,
60 = Sum of marks of 4 subjects/4
⇒ Sum of marks of 4 subjects = 60 × 4 = 240
Let the number of remaining subjects be x
∵ Average of x subjects is 72%,
72 = Sum of marks in x subjects/x
⇒ Sum of marks in x subjects = 72x
Total number of subjects = x + 4
∴ Total marks obtained in (x + 4) subjects = 240 + 72x
∵ Average of (x + 4) subjects is 64%,
3 | P a g e
64 =240+72𝑥
𝑥+4
⇒ 64 × (x + 4) = 240 + 72x
⇒ 64x + 256 = 240 + 72x
⇒ 8x = 16
⇒ x = 2
∴ Number of remaining subjects = 2
⇒ Total no. of subjects = 4 + 2 = 6
3. Let the investment made by Hariprasad and Madhusudan be 2m and 3m
respectively.
Then we are given that
⇒ (2m + 10000) : 3m = 3 : 2
⇒ 2 × (2m + 10000) = 3 × 3m
⇒ 4m + 20000 = 9m
⇒ 20000 = 9m – 4m
⇒ 5m = 20000
⇒ m = 20000/5
⇒ m = 4000
Hence, original investment made by Hariprasad is
= 2m
= 2 × 4000
4 | P a g e
= 8000
4. Compound Interest = Principal × [(1 +𝑅
100)𝑇
− 1]
For Jill, T = 1, R = 20, Principal = 12000
⇒ Compound Interest for Jill = 12000 × [(1 +20
100) − 1] = 2400
In case of Jack, interest is compounded half yearly.
For Jack, there will be two time periods of six months each and rate of interest per six
months will be 10%.
T = 2, R = 10, Principal = 12000
⇒ Compound Interest for Jack = 12000 × [(1 +10
100)2
− 1] = 12000 × 0.21 = 2520
∴ Jack will earn Rs. 120 more as interest.
5. Suppose wholesaler sells to retailer at T% profit.
Then, in a series of successive selling from manufacturer to customer, the effective
selling price will be given by
Price to customer = Manufacturer cost × (1 + 20/100) × (1 + T/100) × (1 + 10/100)
⇒ 155 = 100 × 1.2 × (1 + 0.01T) × 1.1
⇒ 1 + 0.01T = 1.1742
⇒ T = 17.42
∴ Wholesaler sells to retailer at 17.42% profit.
5 | P a g e
6. Ginny bought a scooter at Rs. 30000, and sold it to Honey at a loss of 20%.
Loss made by Ginny = 20% of Rs. 30000 = Rs. 30000 × (20/100) = Rs. 6000
Later, Ginny bought the scooter back from Honey at Rs. 28000.
Net Loss = 10000
To compensate for losses, Ginny must sell it now at a profit of Rs. 10000.
We know, Profit percentage = (Profit/Cost price) × 100
∴ Profit at which Ginny should sell the scooter = (10000/28000) × 100 = 35.71%
7. Let the initial investments of Vinay and Atul be Rs. 4x and Rs. 5x respectively.
If, Vinay had invested an additional amount of Rs. 29,000; investment of Vinay and
Atul would be Rs. (4x + 29000) and Rs. 5x respectively.
Thus, according to question
⇒4𝑥+29000
5𝑥=
9
4
⇒ 16x + 116000 = 45x
⇒ 29x = 116000
⇒ x = 4000
Thus, initial investments of Vinay = Rs. (4 × 4000) = Rs. 16,000.
8. Machines P and Q can together produce 3,60,000 m of cloth in 10 hours
∴ Machines P and Q together produce 36,000 m of cloth in 1 hours
6 | P a g e
Machine Q alone can produce 3,60,000 m of cloth in 18 hours
∴ Machine Q alone can produce 20,000 m of cloth in 1 hour
So, Machine P can produce 16,000 m (36,000 – 20,000) of cloth in 1 hour
∴ In 10 hours, machine P can produce = 16,000 × 10
Hence, in 10 hours machine P can produce = 1,60,000 m of cloth
9. P has a capacity twice of Q.
Let us suppose that capacities of tanks P and Q are 2T liters and T liters, respectively.
When tank P is filled at the rate of 2 liters per minute, it takes 20 minutes to fill more as
compared to the time taken when tank Q is filled at 3 liters per minute.
⇒ Time taken to fill tank P at 2 liters per minute = 2T/2 = T minutes
⇒ Time taken to fill tank Q at 3 liters per minute = T/3 minutes
T = 20 + (T/3)
⇒ 2T/3 = 20
⇒ T = 30
∴ Capacity of tank Q is 30 liters.
10. Assume that he walks for t hours.
Total journey time 8.5 hours.
∴ He rides his scooter for (8.5 – t) hours.
He rides a scooter at 12 km/hr.
7 | P a g e
We know that, Distance = Speed × Time
∴ Distance travelled by scooter = 12 × (8.5 – t) km
He walks the remaining distance at 4.5 km/hr
∴ Distance travelled by walking = 4.5 × t km
Total distance travelled = Distance travelled in scooter + Distance travelled by walking
Total distance he has travelled = 12 × (8.5 – t) km. + 4.5 × t km.
Rehman has to travel total journey of 72 km.
∴ Total distance travelled = 72 km.
So, 72 km = 12 × (8.5 – t) km. + 4.5 × t km
⇒ 72 = 102 – 12t + 4.5t
⇒ 30 = 7.5t
⇒ t = 4 hours
∴ Distance travelled by scooter = 12 × (8.5 – t) km.
= 12 × (8.5 – 4)
= 12 × 4.5
= 54 km
11. As per the problem statement,
Length of train B = 400 m
Since, length of train A is twice the length of train B,
⇒ Length of train A = 2 × 400 m = 800 m
8 | P a g e
When a train crosses a pole, it covers a distance equal to its length (as length of pole can
be neglected when compared to length of train).
Both trains take same time to cross the pole, and we know that
Time=Distance/Speed
∴ (Distance covered by train A)/(Speed of train A) = (Distance covered by train
B)/(Speed of train B)
⇒ Speed of train A = [800 m × 60 km/hr]/400 m = 120 km/hr
Now, the trains are moving in the same direction and faster train will overtake the
slower train. In doing so, the faster train will cover a distance that is equal to the sum of
lengths of both trains.
Distance covered = Length of train A + Length of train B = 800 m + 400 m = 1200 m
The relative speed of faster train A, with respect to, slower train B will be given by
Speed of train A w.r.t. train B = Speed of train A – Speed of train B = 120km/hr – 60
km/hr = 60 km/hr
This speed when converted to the units of m/sec, will be [60 × (5/18)] m/sec = (50/3)
m/sec
We know,Time=Distance/Speed
Time taken =1200𝑚
(50
3)𝑚/𝑠𝑒𝑐
= 72 seconds
∴ Time taken by train A to overtake train B = 72 seconds
12. Let length of rectangle = x units and breadth = y units
∴ side of triangle = y units
9 | P a g e
Since their perimeters are equal,
⇒ 2x + 2y = 3y
⇒ 2x = y …………(i)
∴ Area of rectangle = xy
Area of triangle = y2√3/4
Required ratio =𝑥𝑦
𝑦2√3
4
=4𝑥
𝑦√3=
4𝑥
2𝑥√3=
2
√3
13. If three balls are picked, there can be balls of exactly two colours in the following
cases:
i) Two blue balls and one yellow ball
This can be done in 7C2 × 8C1 ways, i.e. 21 × 8 = 168 ways.
ii) Two blue balls and one green ball
This can be done in 7C2 × 5C1 ways, i.e. 21 × 5 = 105 ways.
iii) Two yellow balls and one blue ball
This can be done in 8C2 × 7C1 ways, i.e. 28 × 7 = 196 ways.
iv) Two yellow balls and one green ball
This can be done in 8C2 × 5C1 ways, i.e. 28 × 5 = 140 ways.
v) Two green balls and one blue ball
This can be done in 5C2 × 7C1 ways, i.e. 10 × 7 = 70 ways.
vi) Two green balls and one yellow ball
This can be done in 5C2 × 8C1 ways, i.e. 10 × 8 = 80 ways.
10 | P a g e
∴ Total number of ways in which it can be done = 168 + 105 + 196 + 140 + 70 + 80 =
759.
14. The pattern of the given number series is as following:
→ 3 × 12 = 3,
→ 3 × 22 = 12,
→ 3 × 32 = 27,
→ 3 × 42 = 48,
→ 3 × 52 = 75,
→ 3 × 62 = 108,
→ 3 × 72 = 147
Hence, the required term in place of question mark in the given number series is 147
15. The pattern of the given number series is as follows:
Every third term is overall discount of previous two terms successive discount,
→ 28 = 20 + 10 - (20 × 10)/100 = 30 - 2 = 28
→ 28 = 30 + 10 - (30 × 10)/100 = 40 - 3 = 37
Hence, the required number in place of question mark in the given number series would
be 37
16. The pattern of the given number series is as:
11 | P a g e
→ 16,
→ 16 × 1 – 2 = 14,
→ 14 × 2 – 4 = 24,
→ 24 × 3 – 6 = 66,
→ 66 × 4 – 8 = 256,
→ 256 × 5 – 10 = 1270,
→ 1270 × 6 – 12 = 7608
Hence, the required answer is 7608,
17. The pattern of given series is:
→ 144,
→ 173 = 144 + 29,
→ 140 = 173 – 33,
→ 169 = 140 + 29,
→ 136 = 169 – 33,
Following this pattern, the next term will be:
→ ? = 136 + 29,
→ ? = 165
18. The pattern of given series is:
→ 321 – 33 = 288
12 | P a g e
→ 465 – 321 = 144 [288/2 = 144]
→ 537 – 465 = 72 [144/2 = 72]
→ 573 – 537 = 36 [72/2 = 36]
→ 590 – 573 = 17* [36/2 = 18]
→ 600 – 590 = 10* [18/2 = 9]
If we replace 590 by the number 591, then the following changes take place:
591 – 573 = 18 & 600 – 591 = 9
∴ The difference between adjacent terms is half of the difference between previous pair.
Therefore, 590 is the wrong term.
19. Amount of first alloy = 24 kg
∴ Amount of A = 5/12 × 24 = 10 kg
⇒ Amount of B = (24 - 10) = 14 kg
Amount of second alloy = 7 kg
∴ Amount of A = 4 kg
⇒ Amount of B = 3 kg
Let the amount of metal B mixed be x kg
∴ Total amount of A = 10 + 4 = 14 kg
⇒ Total amount of B = 14 + 3 + x = (17 + x) kg
⇒ Total weight = A + B = 14 + 17 + x = (x + 31) kg
Percentage of A in mixture = 40%
13 | P a g e
∴ [14/(x + 31)] × 100 = 40
⇒ (14 × 100)/40 = x + 31
⇒ 35 = x + 31
⇒ x = 4 kg
∴ Amount of metal B mixed = 4 kg
20. Total no. of people = 9
Total no. of people to be selected = 3
∴ No. of ways of selecting 3 people out of 9 = 9C3
There are 5 females and no. of ways of selecting 3 out of them = 5C3
∴ Probability of selecting all the females for quiz = 5C3/ 9C3 = 5/42
Comprehension Start:
21. ∴ Total production of all the products is given in the table below
Year
Production
of P (in
tonne)
Production
of Q (in
tonne)
Production
of R (in
tonne)
Production
of S (in
tonne)
Total
production(in
tonne)
2005 45 95 75 115 45 + 95 + 75
+ 115 = 330
2006 25 40 95 155 315
2007 40 105 102 165 412
14 | P a g e
2008 35 60 62 140 297
2009 75 40 135 85 335
2010 55 70 120 95 340
∴ Total production (of all products) is second highest in the year 2010
22. Total production and average production is calculated in the table below
Year
Production
of P (in
tonne)
Production
of Q (in
tonne)
Production
of R (in
tonne)
Production
of S (in
tonne)
2005 45 95 75 115
2006 25 40 95 155
2007 40 105 102 165
2008 35 60 62 140
2009 75 40 135 85
2010 55 70 120 95
Total
production in
all the year
275 410 589 755
Average
production(total
production ÷ 6)
275/6 =
45.83
410/6 =
68.33
589/6 =
98.16
755/6 =
125.83
∵ Stability of the production = Averageproduction
Maximumproduction−Minimumproduction
15 | P a g e
∴ Stability of the production of P =45.83
75−25=
45.83
50= 0.9166
∴ Stability of the production of Q =68.33
105−40=
68.33
65= 1.051
∴ Stability of the production of R =98.16
135−62=
98.16
73= 1.344
∴ Stability of the production of S =125.83
165−85=
125.83
80= 1.572
∴ Product R is the second most stable
23. Since P, Q, R and S are sold at price of Rs. 9, 4, 13 and 3, respectively
during 2005-10
Year
Production
of P (in
tonne)
Production
of Q (in
tonne)
Production
of R (in
tonne)
Production
of S (in
tonne)
Total revenue
2005 45 95 75 115 (45 × 9)+ (95 × 4) + (75 × 13) + (115 ×
3) = Rs.2105
2006 25 40 95 155 (25 × 9)+ (40 × 4) + (95 × 13) + (155
× 3) = Rs.2085
2007 40 105 102 165 (40 × 9)+ (105 × 4) + (102 × 13) +
(165 × 3) = Rs.2601
2008 35 60 62 140 (35 × 9)+ (60 × 4) + (62 × 13) + (140
× 3) = Rs.1781
2009 75 40 135 85 (75 × 9)+ (40 × 4) + (135 × 13) + (85
× 3) = Rs.2845
2010 55 70 120 95 (55 × 9)+ (70 × 4) + (120 × 13) + (95
× 3) = Rs.2620
16 | P a g e
∴ Total revenue of all the products is highest in the year 2009
24. Since revenue of P, Q, R and S for the entire period (2005-10) is calculated
based on the price of Rs. 9, 4, 13 and 3, respectively
Total production and average production is calculated in the table below
Year
Production
of P (in
tonne)
Production
of Q (in
tonne)
Production
of R (in
tonne)
Production
of S (in
tonne)
2005 45 95 75 115
2006 25 40 95 155
2007 40 105 102 165
2008 35 60 62 140
2009 75 40 135 85
2010 55 70 120 95
Total
production in
all the year
275 410 589 755
Revenue 275 × 9 =
Rs. 2475
410 × 4 =
Rs.1640
589 × 13 =
Rs. 7657
755 × 3 =
Rs. 2265
∴ Product Q fetches the lowest revenue
25. For option a:
17 | P a g e
Year
Production
of P (in
tonne)
Production
of Q (in
tonne)
Production
of R (in
tonne)
Production
of S (in
tonne)
2005 45 95 75 115
2006 25 40 95 155
2007 40 105 102 165
2008 35 60 62 140
2009 75 40 135 85
2010 55 70 120 95
Total
production in
all the year
275 410 589 755
Revenue 275 × 9 =
Rs. 2475
410 × 4 =
Rs.1640
589 × 13 =
Rs. 7657
755 × 3 =
Rs. 2265
From above table Product R fetches highest revenue across all products
∴ Option a is false
For option b:
Sum of revenue of P, Q and S in the year 2009 = (75 × 9) + (40 × 4) + (85 × 3) = Rs.
1090
∴ Revenue of R in the year 2009 = 135 × 13 = Rs. 1755
∴ Option b is false since sum of revenue of P, Q and S is less than the revenue of R in
2009
18 | P a g e
For option c:
∴ Sum of revenue of P and Q in the year 2008 = (35 × 9) + (60 × 4) = Rs. 555
∴ Revenue of S in the year 2008 = 140 × 3 = Rs. 420
∴ Option c is true
Comprehension End
Comprehension Start:
26. ∵ Total number of students in year 2001 = 1800
∵ Total number of students in year 2002 = 2200
∴ Total number of students studying all disciplines during these two years = 1800 +
2200 = 4000
∴ Number of students studying Engineering in the years 2001 and 2002 = 15% of 1800 +
14% of 2200 = 270 + 308 = 578
∴ Required percentage =578
4000× 100 = 14.45
27. ∵ Total number of students in year 2001 = 1800
∵ Total number of students in year 2002 = 2200
∴ Number of students studying Medicine in years 2001 = 9% of 1800 = 162
∴ Number of students studying Arts in years 2002 = 11% of 2200 = 242
∴ Required ratio = 162 : 242 = 81 : 121
19 | P a g e
28. ∵ Total number of students in year 2001 = 1800
∵ Total number of students in year 2002 = 2200
∴ Number of students studying Commerce and Science together in years 2001 = (28 +
24)% of 1800 = 52% of 1800 = 936
∴ Number of students studying Commerce and Science together in years 2002 = (30 +
25)% of 2200 = 55% of 2200 = 1210
∴ Required percentage =936
1210× 100 = 77.35 ≈ 77
29. ∵ Total number of students in year 2001 = 1800
∵ Total number of students in year 2002 = 2200
∴ Number of students studying Management in years 2001 = 8% of 1800 = 144
∴ Number of students studying Management in years 2002 = 10% of 2200 = 220
∴ Percentage increase =220−144
144× 100 = 52.77 ≈ 52% increase
30. ∵ Total number of students in year 2001 = 1800
∵ Total number of students in year 2002 = 2200
∴ Number of students studying Agriculture in years 2001 = 4% of 1800 = 72
∴ Number of students studying Agriculture in years 2002 = 3% of 2200 = 66
∴ Require ratio = 72 : 66 = 12 : 11
31. We know that, total number of boys = 189
Now, given that 2/3rd of boys participated in group song.
Hence 1/3rd of boys participated in solo song and dance in the ratio of 4:5.
20 | P a g e
⇒ One-third of total boys = 1/3 × 189
= 63
Let the number of boys participated in dance be D then,
∴ Number of boys participated in solo song = 63 – D
Hence, 63−𝐷
𝐷=
4
5
⇒ 315 – 5D = 4D
⇒ 9D = 315
Hence, Number of boys participated in dance, D = 35
Now, given that out of total girls 20% of them participated in dance. Thus,
Number of girls participated in dance =20
100× 315
= 63
Hence, the required difference = 63 – 35
= 28
32. Hence 1/3rd of boys participated in solo song and dance in the ratio of 4:5.
⇒ One-third of total boys = 1/3 × 189
= 63
⇒ Number of boys participated in solo =4
4+5× 63
= 28
Given that 80% of girls participated in solo, group song and drama in ratio 2 : 3 : 4.
⇒ 80% of total girls =80
100× 315
21 | P a g e
= 252
⇒ Number of girls participated in solo =2
2+3+4× 252
= 56
Hence, the required ratio = 28/56
= 1 : 2
33. Given that 80% of girls participated in solo, group song and drama in ratio
2 : 3 : 4.
⇒ 80% of total girls = 80/100 × 315
= 252
∴ Number of girls participated in group song and drama
=3+4
2+3+4× 252
= 196
34. Given that 80% of girls participated in solo, group song and drama in ratio
2 : 3 : 4.
⇒ 80% of total girls =80
100× 315
= 252
⇒ Number of girls participated in solo =2
2+3+4× 252
= 56
Total number of participants = 504
∴ Required percentage = 56/504 × 100
22 | P a g e
= 11.11%
= 11% (approx.)
35. Given that total number of students are 504 in which girls to boys ratio is 5
: 3.
Let the total number of girls be X then total number of boys will be (504 - X).
∴𝑋
504−𝑋=
5
3
3X = 2520 – 5X
⇒ 8X = 2520
⇒ Total number of girls, X = 315
And, total number of boys = 504 – 315
= 189
Now, given that 2/3rd of boys participated in group song.
Hence 1/3rd of boys participated in solo song and dance in the ratio of 4:5.
⇒ One-third of total boys = 1/3 × 189
= 63
Let the number of boys participated in dance be D then,
∴ Number of boys participated in solo song = 63 – D
Hence, 63−𝐷
𝐷=
4
5
⇒ 315 – 5D = 4D
23 | P a g e
⇒ 9D = 315
Hence, Number of boys participated in dance, D = 35
∴ Required percentage = 35/189 × 100
= 18.518%
~ 19 %
Comprehension End
36. (9.11 % of 936) – (12.5 % of 498)
= (936 ×9.11
100) − (498 ×
12.5
100)
= 85.26 – 62.25
= 23.01
37. Given, 0.1×0.1×0.1+0.02×0.02×0.02
0.2×0.2×0.2+0.04×0.04×0.04
Multiplying and dividing by 1000000, we get
=1000+(2×2×2)
8000+(4×4×4)
= 1008/8064
= 1/8
38. follow BODMAS rule to solve this question, as per the order given below,
Step-1-Parts of an equation enclosed in 'Brackets' must be solved first,
24 | P a g e
Step-2-Any mathematical 'Of' or 'Exponent' must be solved next,
Step-3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are
calculated,
Step-4-Last but not least, the parts of the equation that contain 'Addition' and
'Subtraction' should be calculated.
The given expression is:
12 × 35%𝑜𝑓1000 − √1176493
÷ 7 = 𝑎 + 32
⇒ 12 ×35
100× 1000 − 49 ÷ 7 = 𝑎 + 32
⇒ 12 × 350 - 7 = a + 32
⇒ 4200 - 7 = a + 32
⇒ 4193 = a + 32
⇒ a = 4193 – 32
⇒ a = 4161
39. Follow BODMAS rule to solve this question, as per the order given below,
Step-1- Parts of an equation enclosed in 'Brackets' must be solved first, and in the
bracket, the BODMAS rule must be followed,
Step-2 - Any mathematical 'Of' or 'Exponent' must be solved next,
Step-3 - Next, the parts of the equation that contain 'Division' and 'Multiplication' are
calculated,
Step-4 - Last but not least, the parts of the equation that contain 'Addition' and
'Subtraction' should be calculated.
25 | P a g e
The given expression is,
(200 – 102) ÷ 5 – 2000% of 15 ÷ 300% of 10
= (200 – 100) ÷ 5 – 2000% of 15 ÷ 300% of 10
= 100 ÷ 5 – (2000/100) × 15 ÷ (300/100) × 10
= 20 – 300 ÷ 30
= 20 – 10
= 10
40. Follow BODMAS rule to solve this question, as per the order given below,
Step-1- Parts of an equation enclosed in 'Brackets' must be solved first, and in the
bracket, the BODMAS rule must be followed,
Step-2- Any mathematical 'Of' or 'Exponent' must be solved next,
Step-3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are
calculated,
Step-4-Last but not least, the parts of the equation that contain 'Addition' and
'Subtraction' should be calculated
We have the expression:
⇒ (80 × 0.40)3 ÷ (40 × 1.6) × (128)3 = (2)? + 7
⇒ 323 ÷ 64 × (128)3 = (2)? + 7
(We know that 64 = 32 × 2)
⇒ (323 ÷ 32 × 2) × (128)3 = (2)? + 7
⇒ 512 × (128)3 = (2)? + 7
26 | P a g e
(We know that 512 = 29 and 128 = 27)
⇒ 29 × 27×3 = (2)? + 7
⇒ 230 = (2)? + 7
⇒ 27+23 = (2)? + 7
⇒ ? = 23
41. i) There is a group of seven persons A, M, C, P, E, F, and R.
ii) There are four males, three females, two married couples in the group.
iii) The seven persons are seated in a row on the bench and their professions are:
Teacher, Mechanic, Carpenter, Dentist, POLICE, Painter and Guitarist.
Here ‘+’ → Male; ‘ - ‘ → Female
1st to 7th are ranks according to intelligence as 1st for most intelligent and 7th for least
intelligent.
Now given,
1. P is an Painter; he is sitting on the leftmost corner.
2. The Guitarist is sitting on the rightmost corner of the bench.
3. The least intelligent of the group is sitting on the immediate right of P, followed by the
most intelligent.
4. The Dentist is the most intelligent and the Mechanic is the least intelligent of the
group.
27 | P a g e
5. On the bench, followed by P, there are three females sitting in succession and the
Dentist is a female.
6. Thus all other including P are males because given in group three are females and
four are males.
7. The Carpenter is married to C. C is the second most intelligent of the group, followed
on the bench by her husband.
8. Thus C is a female & 2nd most intelligent and thus C must be sitting at 4th from the
right because only three females are there in the group. Carpenter is C’s husband which
is immediate right to C.
9. Teacher is married to the Mechanic, who is the least intelligent of the group.
10. Thus Mechanic and Teacher are also married and after placing Carpenter only 1
place is left for Teacher so he will be automatically placed.
28 | P a g e
11. M is not married and another person, the Dentist, is the most intelligent.
12. Thus M is neither Carpenter nor Teacher nor Dentist thus M is clearly a Guitarist
and is bachelor.
13. The Guitarist is more intelligent than the Painter, who is more intelligent than only
one person, F.
14. So F is least intelligent. Thus F is a Mechanic. So Painter is at 6th position and
Guitarist at 5th position.
15. Neither A nor R is a female.
16. Thus A and R are males and are either Carpenter or Teacher because only this is left
for them. Thus E is the Dentist because only she is left.
17. There are as many more intelligent persons than the Teacher as there are less
intelligent. Thus Teacher is 4th intelligent and Carpenter is 3rd intelligent because only
this intelligent position left for M.
18. Thus after filling this only C is left so she is a POLICE.
Hence the final arrangement is as follows:
29 | P a g e
Clearly, F is sitting immediate right of P.
42. i) There is a group of seven persons A, M, C, P, E, F, and R.
ii) There are four males, three females, two married couples in the group.
iii) The seven persons are seated in a row on the bench and their professions are:
Teacher, Mechanic, Carpenter, Dentist, POLICE, Painter and Guitarist.
Here ‘+’ → Male; ‘ - ‘ → Female
1st to 7th are ranks according to intelligence as 1st for most intelligent and 7th for least
intelligent.
Now given,
1. P is an Painter; he is sitting on the leftmost corner.
2. The Guitarist is sitting on the rightmost corner of the bench.
3. The least intelligent of the group is sitting on the immediate right of P, followed by the
most intelligent.
4. The Dentist is the most intelligent and the Mechanic is the least intelligent of the
group.
30 | P a g e
5. On the bench, followed by P, there are three females sitting in succession and the
Dentist is a female.
6. Thus all other including P are males because given in group three are females and
four are males.
7. The Carpenter is married to C. C is the second most intelligent of the group, followed
on the bench by her husband.
8. Thus C is a female & 2nd most intelligent and thus C must be sitting at 4th from the
right because only three females are there in the group. Carpenter is C’s husband which
is immediate right to C.
9. Teacher is married to the Mechanic, who is the least intelligent of the group.
10. Thus Mechanic and Teacher are also married and after placing Carpenter only 1
place is left for Teacher so he will be automatically placed.
31 | P a g e
11. M is not married and another person, the Dentist, is the most intelligent.
12. Thus M is neither Carpenter nor Teacher nor Dentist thus M is clearly a Guitarist
and is bachelor.
13. The Guitarist is more intelligent than the Painter, who is more intelligent than only
one person, F.
14. So F is least intelligent. Thus F is a Mechanic. So Painter is at 6th position and
Guitarist at 5th position.
15. Neither A nor R is a female.
16. Thus A and R are males and are either Carpenter or Teacher because only this is left
for them. Thus E is the Dentist because only she is left.
17. There are as many more intelligent persons than the Teacher as there are less
intelligent. Thus Teacher is 4th intelligent and Carpenter is 3rd intelligent because only
this intelligent position left for M.
18. Thus after filling this only C is left so she is a POLICE.
Hence the final arrangement is as follows:
32 | P a g e
As we can see increasing order of intelligence is FME.
43. i) There is a group of seven persons A, M, C, P, E, F, and R.
ii) There are four males, three females, two married couples in the group.
iii) The seven persons are seated in a row on the bench and their professions are:
Teacher, Mechanic, Carpenter, Dentist, POLICE, Painter and Guitarist.
Here ‘+’ → Male; ‘ - ‘ → Female
1st to 7th are ranks according to intelligence as 1st for most intelligent and 7th for least
intelligent.
Now given,
1. P is an Painter; he is sitting on the leftmost corner.
2. The Guitarist is sitting on the rightmost corner of the bench.
3. The least intelligent of the group is sitting on the immediate right of P, followed by the
most intelligent.
4. The Dentist is the most intelligent and the Mechanic is the least intelligent of the
group.
33 | P a g e
5. On the bench, followed by P, there are three females sitting in succession and the
Dentist is a female.
6. Thus all other including P are males because given in group three are females and
four are males.
7. The Carpenter is married to C. C is the second most intelligent of the group, followed
on the bench by her husband.
8. Thus C is a female & 2nd most intelligent and thus C must be sitting at 4th from the
right because only three females are there in the group. Carpenter is C’s husband which
is immediate right to C.
9. Teacher is married to the Mechanic, who is the least intelligent of the group.
10. Thus Mechanic and Teacher are also married and after placing Carpenter only 1
place is left for Teacher so he will be automatically placed.
34 | P a g e
11. M is not married and another person, the Dentist, is the most intelligent.
12. Thus M is neither Carpenter nor Teacher nor Dentist thus M is clearly a Guitarist
and is bachelor.
13. The Guitarist is more intelligent than the Painter, who is more intelligent than only
one person, F.
14. So F is least intelligent. Thus F is a Mechanic. So Painter is at 6th position and
Guitarist at 5th position.
15. Neither A nor R is a female.
16. Thus A and R are males and are either Carpenter or Teacher because only this is left
for them. Thus E is the Dentist because only she is left.
17. There are as many more intelligent persons than the Teacher as there are less
intelligent. Thus Teacher is 4th intelligent and Carpenter is 3rd intelligent because only
this intelligent position left for M.
18. Thus after filling this only C is left so she is a POLICE.
Hence the final arrangement is as follows:
35 | P a g e
As the row follows that A or R can be a teacher.
44. i) There is a group of seven persons A, M, C, P, E, F, and R.
ii) There are four males, three females, two married couples in the group.
iii) The seven persons are seated in a row on the bench and their professions are:
Teacher, Mechanic, Carpenter, Dentist, POLICE, Painter and Guitarist.
Here ‘+’ → Male; ‘ - ‘ → Female
1st to 7th are ranks according to intelligence as 1st for most intelligent and 7th for least
intelligent.
Now given,
1. P is an Painter; he is sitting on the leftmost corner.
2. The Guitarist is sitting on the rightmost corner of the bench.
3. The least intelligent of the group is sitting on the immediate right of P, followed by the
most intelligent.
4. The Dentist is the most intelligent and the Mechanic is the least intelligent of the
group.
36 | P a g e
5. On the bench, followed by P, there are three females sitting in succession and the
Dentist is a female.
6. Thus all other including P are males because given in group three are females and
four are males.
7. The Carpenter is married to C. C is the second most intelligent of the group, followed
on the bench by her husband.
8. Thus C is a female & 2nd most intelligent and thus C must be sitting at 4th from the
right because only three females are there in the group. Carpenter is C’s husband which
is immediate right to C.
9. Teacher is married to the Mechanic, who is the least intelligent of the group.
10. Thus Mechanic and Teacher are also married and after placing Carpenter only 1
place is left for Teacher so he will be automatically placed.
37 | P a g e
11. M is not married and another person, the Dentist, is the most intelligent.
12. Thus M is neither Carpenter nor Teacher nor Dentist thus M is clearly a Guitarist
and is bachelor.
13. The Guitarist is more intelligent than the Painter, who is more intelligent than only
one person, F.
14. So F is least intelligent. Thus F is a Mechanic. So Painter is at 6th position and
Guitarist at 5th position.
15. Neither A nor R is a female.
16. Thus A and R are males and are either Carpenter or Teacher because only this is left
for them. Thus E is the Dentist because only she is left.
17. There are as many more intelligent persons than the Teacher as there are less
intelligent. Thus Teacher is 4th intelligent and Carpenter is 3rd intelligent because only
this intelligent position left for M.
18. Thus after filling this only C is left so she is a POLICE.
Hence the final arrangement is as follows:
38 | P a g e
As we can see above arrangement F is least intelligence.
45. i) There is a group of seven persons A, M, C, P, E, F, and R.
ii) There are four males, three females, two married couples in the group.
iii) The seven persons are seated in a row on the bench and their professions are:
Teacher, Mechanic, Carpenter, Dentist, POLICE, Painter and Guitarist.
Here ‘+’ → Male; ‘ - ‘ → Female
1st to 7th are ranks according to intelligence as 1st for most intelligent and 7th for least
intelligent.
Now given,
1. P is an Painter; he is sitting on the leftmost corner.
2. The Guitarist is sitting on the rightmost corner of the bench.
3. The least intelligent of the group is sitting on the immediate right of P, followed by the
most intelligent.
4. The Dentist is the most intelligent and the Mechanic is the least intelligent of the
group.
39 | P a g e
5. On the bench, followed by P, there are three females sitting in succession and the
Dentist is a female.
6. Thus all other including P are males because given in group three are females and
four are males.
7. The Carpenter is married to C. C is the second most intelligent of the group, followed
on the bench by her husband.
8. Thus C is a female & 2nd most intelligent and thus C must be sitting at 4th from the
right because only three females are there in the group. Carpenter is C’s husband which
is immediate right to C.
9. Teacher is married to the Mechanic, who is the least intelligent of the group.
10. Thus Mechanic and Teacher are also married and after placing Carpenter only 1
place is left for Teacher so he will be automatically placed.
40 | P a g e
11. M is not married and another person, the Dentist, is the most intelligent.
12. Thus M is neither Carpenter nor Teacher nor Dentist thus M is clearly a Guitarist
and is bachelor.
13. The Guitarist is more intelligent than the Painter, who is more intelligent than only
one person, F.
14. So F is least intelligent. Thus F is a Mechanic. So Painter is at 6th position and
Guitarist at 5th position.
15. Neither A nor R is a female.
16. Thus A and R are males and are either Carpenter or Teacher because only this is left
for them. Thus E is the Dentist because only she is left.
17. There are as many more intelligent persons than the Teacher as there are less
intelligent. Thus Teacher is 4th intelligent and Carpenter is 3rd intelligent because only
this intelligent position left for M.
18. Thus after filling this only C is left so she is a POLICE.
Hence the final arrangement is as follows:
41 | P a g e
As F, C, A and R married.
46. Given in the question,
Each of them is of different weight:
Weight of Lenovo is twice of Samsung.
⇒ Lenovo = 2 × Samsung
Samsung weights 8 and half times of Apple.
⇒ Samsung = 17/2 of Apple
Apple weights half of MI.
⇒ Apple = MI/2
MI weights half of Nokia.
⇒ MI = Nokia/2
The weight of Nokia is less than Lenovo but more than Apple.
⇒ Lenovo > Nokia > Apple
42 | P a g e
Solving this we get,
⇒ Lenovo = 17 Apple
⇒ Samsung = 8.5 Apple
⇒ Nokia = 4 Apple
⇒ MI = 2 Apple
Lenovo > Samsung > Nokia > MI > Apple.
So the lightest stone is Apple.
47. Given in the question,
Each of them is of different weight:
Weight of Lenovo is twice of Samsung.
⇒ Lenovo = 2 × Samsung
Samsung weights 8 and half times of Apple.
⇒ Samsung = 17/2 of Apple
Apple weights half of MI.
⇒ Apple = MI/2
MI weights half of Nokia.
⇒ MI = Nokia/2
The weight of Nokia is less than Lenovo but more than Apple.
⇒ Lenovo > Nokia > Apple
Solving this we get,
43 | P a g e
⇒ Lenovo = 17 Apple
⇒ Samsung = 8.5 Apple
⇒ Nokia = 4 Apple
⇒ MI = 2 Apple
Lenovo > Samsung > Nokia > MI > Apple.
So the heaviest is Lenovo.
48. Given in the question,
Each of them is of different weight:
Weight of Lenovo is twice of Samsung.
⇒ Lenovo = 2 × Samsung
Samsung weights 8 and half times of Apple.
⇒ Samsung = 17/2 of Apple
Apple weights half of MI.
⇒ Apple = MI/2
MI weights half of Nokia.
⇒ MI = Nokia/2
The weight of Nokia is less than Lenovo but more than Apple.
⇒ Lenovo > Nokia > Apple
Solving this we get,
⇒ Lenovo = 17 Nokia / 4
44 | P a g e
⇒ Samsung = 17 Nokia / 8
⇒ Apple = Nokia / 4
⇒ MI = Nokia / 2
Lenovo > Samsung > Nokia > MI > Apple.
Here, Nokia is heavier than Apple and MI.
49. Given words: THEN GONE EXAM HOUR GATE
After interchanging: NEHT ENOG MAXE RUOH ETAG
Arranging in dictionary order: ENOT ETAG MAXE NEHT RUOH
Hence, the third word is MAXE i.e. EXAM.
50. Given words: THEN GONE EXAM HOUR GATE
New words after changing: UGFM HNOD FWBL INVQ HZUD
Arranging in dictionary order: FWBL HNOD HZUD INVQ UGFM
4th word is INVQ, three consonants are there in this word.
51. Given statement: Y ≤ U ≤ Z ≤ X; S > Z ≥ F ≥ U; Z = Q; W < Z
On combining: Y ≤ U ≤ F ≤ Z = Q ≤ X; S > Z > W
I. X ≥ Y → True (Y ≤ U ≤ Z ≤ X → X ≥ Y)
II. Z being equal to U is a possibility → True (U ≤ Z, as there is no strict sign so
possibility is true)
45 | P a g e
III. Z > U → Not definitely true, so false (as U ≤ Z → U < Z or U = Z)
IV. W ≤ S → False (S > Z > W → S > W)
Hence, only conclusion I and II are true.
52. Given statement: B > L > O ≥ V; C < V ≤ X ≤ O; L < N < M > B;
On combining: L < N < M > B > L > O ≥ X ≥ V > C
Conclusions:
I. V > M → False (as M > B > L > O ≥ X ≥ V → M > V)
II. L = N → False (as L < N)
III. O > N → False (as N < M > B > L > O → No relation Between O and N)
IV. M is the largest → True (L < N < M > B > L > O ≥ X ≥ V > C, implies M is the largest)
Hence, Only IV is true.
53. Given statement: C < D ≤ F = E ≤ B; G < D ≤ H ≤ K ≤ B; C < N > H
On combining: H < N > C < D ≤ F = E ≤ B; G < D ≤ H ≤ K ≤ B
Conclusion:
I. K being equal to G is not a possibility → True (G < D ≤ H ≤ K, as K > G)
II. B being greater to N is not a possibility → False (N > C < D ≤ F = E ≤ B there is no
definite relation between B and N, so it is possible)
III. B > G → True (as G < D ≤ H ≤ K ≤ B → B > G)
46 | P a g e
IV. F being greater to H is not a possibility → True (H < N > C < D ≤ F → it may be
possible)
Thus, conclusion III and IV are true.
54. Given statement: N > O > P > E; C ≤ I < P; K = I; L > P
On combining: N > O > P > E; N > O > P < L; N > O > P > I ≥ C
Conclusions:
I. I < O → True (as O > P > I → O > I)
II. P > L → False (P < L)
III. I > P → False (as I < P)
IV. O > N → False (N > O)
Conclusion I are true
55. Given statement: G > P > Z > C; R ≥ K ≥ P > M ≥ Z; D < Z < T
On combining: R ≥ K ≥ P > M ≥ Z > D; Z > C; G > Z < T
Conclusion:
I. R ≥ Z → False (as R ≥ K ≥ P > M ≥ Z → R > Z)
II. G > T → False (as G > Z < T → clear relation between Z and T cannot be determined)
III. K > M → True (K ≥ P > M → K > M
IV. Z > K → False (as K ≥ P > M ≥ Z → K > Z)
Hence, conclusion I and III are true.
47 | P a g e
Comprehension Start:
56. 10 Student: A, B, C, D, E, P, Q, R, S and T.
Day: Monday to Friday
Time slot: 8:45 AM and 12:45 PM
1) S has an exam on Tuesday at 08:45 AM.
2) B have exam day is immediately before S.
3) S does not have exam on any of the days before Q.
4) Only three people have exam between Q and E.
Day Time Student
Monday 8:45 AM Q
12:45 PM B
Tuesday 8:45 AM S
12:45 PM
Wednesday 8:45 AM E
12:45 PM
Thursday 8:45 AM
12:45 PM
Friday 8:45 AM
12:45 PM
48 | P a g e
5) D does not has exam on any day of days after E.
6) There is one student who has exam at 08:45 AM immediately before T.
7) R does not has exam at 12:45 PM
(Hence T exam on 12:45 PM and R exam at 08:45 PM)
8) The number of people who have exam between Q and D is same as the number of
people who have exam between C and R.
Case – 1 Case – 2
Day Time Student Student
Monday 8:45 AM Q Q
12:45 PM B B
Tuesday 8:45 AM S S
12:45 PM D D
Wednesday 8:45 AM E E
12:45 PM C
Thursday 8:45 AM R
12:45 PM
Friday 8:45 AM R
12:45 PM C
9) Only 2 people have exam between P and T.
10) P does not have any exam on any of the day after R.
49 | P a g e
(Here case – 2 will gets eliminated)
Case – 1
Day Time Student
Monday 8:45 AM Q
12:45 PM B
Tuesday 8:45 AM S
12:45 PM D
Wednesday 8:45 AM E
12:45 PM C
Thursday 8:45 AM P
12:45 PM A
Friday 8:45 AM R
12:45 PM T
Above combination will be final combination.
Hence, A exam on Thursday at 12:45 PM exam slot.
57. 10 Student: A, B, C, D, E, P, Q, R, S and T.
Day: Monday to Friday
Time slot: 8:45 AM and 12:45 PM
1) S has an exam on Tuesday at 08:45 AM.
50 | P a g e
2) B have exam day is immediately before S.
3) S does not have exam on any of the days before Q.
4) Only three people have exam between Q and E.
Day Time Student
Monday 8:45 AM Q
12:45 PM B
Tuesday 8:45 AM S
12:45 PM
Wednesday 8:45 AM E
12:45 PM
Thursday 8:45 AM
12:45 PM
Friday 8:45 AM
12:45 PM
5) D does not has exam on any day of days after E.
6) There is one student who has exam at 08:45 AM immediately before T.
7) R does not has exam at 12:45 PM
(Hence T exam on 12:45 PM and R exam at 08:45 PM)
8) The number of people who have exam between Q and D is same as the number of
people who have exam between C and R.
51 | P a g e
Case – 1 Case – 2
Day Time Student Student
Monday 8:45 AM Q Q
12:45 PM B B
Tuesday 8:45 AM S S
12:45 PM D D
Wednesday 8:45 AM E E
12:45 PM C
Thursday 8:45 AM R
12:45 PM
Friday 8:45 AM R
12:45 PM C
9) Only 2 people have exam between P and T.
10) P does not have any exam on any of the day after R.
(Here case – 2 will gets eliminated)
Case – 1
Day Time Student
Monday 8:45 AM Q
12:45 PM B
52 | P a g e
Tuesday 8:45 AM S
12:45 PM D
Wednesday 8:45 AM E
12:45 PM C
Thursday 8:45 AM P
12:45 PM A
Friday 8:45 AM R
12:45 PM T
Above combination will be final combination.
There are four people between A and S exam, which is not mentioned in the options,
Thus, answer is none of above.
58. 10 Student: A, B, C, D, E, P, Q, R, S and T.
Day: Monday to Friday
Time slot: 8:45 AM and 12:45 PM
1) S has an exam on Tuesday at 08:45 AM.
2) B have exam day is immediately before S.
3) S does not have exam on any of the days before Q.
4) Only three people have exam between Q and E.
53 | P a g e
Day Time Student
Monday 8:45 AM Q
12:45 PM B
Tuesday 8:45 AM S
12:45 PM
Wednesday 8:45 AM E
12:45 PM
Thursday 8:45 AM
12:45 PM
Friday 8:45 AM
12:45 PM
5) D does not has exam on any day of days after E.
6) There is one student who has exam at 08:45 AM immediately before T.
7) R does not has exam at 12:45 PM
(Hence T exam on 12:45 PM and R exam at 08:45 PM)
8) The number of people who have exam between Q and D is same as the number of
people who have exam between C and R.
Case – 1 Case – 2
Day Time Student Student
54 | P a g e
Monday 8:45 AM Q Q
12:45 PM B B
Tuesday 8:45 AM S S
12:45 PM D D
Wednesday 8:45 AM E E
12:45 PM C
Thursday 8:45 AM R
12:45 PM
Friday 8:45 AM R
12:45 PM C
9) Only 2 people have exam between P and T.
10) P does not have any exam on any of the day after R.
(Here case – 2 will gets eliminated)
Case – 1
Day Time Student
Monday 8:45 AM Q
12:45 PM B
Tuesday 8:45 AM S
12:45 PM D
55 | P a g e
Wednesday 8:45 AM E
12:45 PM C
Thursday 8:45 AM P
12:45 PM A
Friday 8:45 AM R
12:45 PM T
Above combination will be final combination.
Hence, R and T exam is on Friday.
59. 10 Student: A, B, C, D, E, P, Q, R, S and T.
Day: Monday to Friday
Time slot: 8:45 AM and 12:45 PM
1) S has an exam on Tuesday at 08:45 AM.
2) B have exam day is immediately before S.
3) S does not have exam on any of the days before Q.
4) Only three people have exam between Q and E.
Day Time Student
Monday 8:45 AM Q
12:45 PM B
56 | P a g e
Tuesday 8:45 AM S
12:45 PM
Wednesday 8:45 AM E
12:45 PM
Thursday 8:45 AM
12:45 PM
Friday 8:45 AM
12:45 PM
5) D does not has exam on any day of days after E.
6) There is one student who has exam at 08:45 AM immediately before T.
7) R does not has exam at 12:45 PM
(Hence T exam on 12:45 PM and R exam at 08:45 PM)
8) The number of people who have exam between Q and D is same as the number of
people who have exam between C and R.
Case – 1 Case – 2
Day Time Student Student
Monday 8:45 AM Q Q
12:45 PM B B
Tuesday 8:45 AM S S
57 | P a g e
12:45 PM D D
Wednesday 8:45 AM E E
12:45 PM C
Thursday 8:45 AM R
12:45 PM
Friday 8:45 AM R
12:45 PM C
9) Only 2 people have exam between P and T.
10) P does not have any exam on any of the day after R.
(Here case – 2 will gets eliminated)
Case – 1
Day Time Student
Monday 8:45 AM Q
12:45 PM B
Tuesday 8:45 AM S
12:45 PM D
Wednesday 8:45 AM E
12:45 PM C
Thursday 8:45 AM P
58 | P a g e
12:45 PM A
Friday 8:45 AM R
12:45 PM T
Above combination will be final combination.
1) S exam on Tuesday ⇒ True
2) D exam on Tuesday ⇒ True
3) R exam on Friday ⇒ True
4) P exam on Monday ⇒ False
Hence, “P exam on Monday” is false statement.
60. 10 Student: A, B, C, D, E, P, Q, R, S and T.
Day: Monday to Friday
Time slot: 8:45 AM and 12:45 PM
1) S has an exam on Tuesday at 08:45 AM.
2) B have exam day is immediately before S.
3) S does not have exam on any of the days before Q.
4) Only three people have exam between Q and E.
Day Time Student
Monday 8:45 AM Q
59 | P a g e
12:45 PM B
Tuesday 8:45 AM S
12:45 PM
Wednesday 8:45 AM E
12:45 PM
Thursday 8:45 AM
12:45 PM
Friday 8:45 AM
12:45 PM
5) D does not has exam on any day of days after E.
6) There is one student who has exam at 08:45 AM immediately before T.
7) R does not has exam at 12:45 PM
(Hence T exam on 12:45 PM and R exam at 08:45 PM)
8) The number of people who have exam between Q and D is same as the number of
people who have exam between C and R.
Case – 1 Case – 2
Day Time Student Student
Monday 8:45 AM Q Q
12:45 PM B B
60 | P a g e
Tuesday 8:45 AM S S
12:45 PM D D
Wednesday 8:45 AM E E
12:45 PM C
Thursday 8:45 AM R
12:45 PM
Friday 8:45 AM R
12:45 PM C
9) Only 2 people have exam between P and T.
10) P does not have any exam on any of the day after R.
(Here case – 2 will gets eliminated)
Case – 1
Day Time Student
Monday 8:45 AM Q
12:45 PM B
Tuesday 8:45 AM S
12:45 PM D
Wednesday 8:45 AM E
12:45 PM C
61 | P a g e
Thursday 8:45 AM P
12:45 PM A
Friday 8:45 AM R
12:45 PM T
Above combination will be final combination.
Hence, Q, S, E, P and R are student whose exam slot is 8:45 AM.
Comprehension End.
Comprehension Start:
61. People: A, B, C, D, E, F, G and H.
Floor: 1 to 8
Language: Italian, Polish, French, Chinese, German, Danish, Swedish and Thai
1) F lives an odd numbered floor above the floor numbered four.
2) Only one person lives between F and the one who knows French.
3) The number of people living above F’s floor is same as the number of people living
between F and D. 4) Only three people live between D and the one who knows Danish.
5) C lives on one of the odd numbered floors above the one who knows Danish.
6) Only two people live between C and the one who knows Italian.
Case – 1
Case – 2
Floor Name Language Name Language
8
62 | P a g e
7 C French F
6 Italian
5 F Danish D French
4 Italian
3 French C
2
1 D Danish
7) Only three people live between G and A.
8) The one who knows Swedish lives immediately above G, G knows neither Danish nor
Italian.
Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
4 Italian Swedish Italian
3 Swedish C French
2 G A A
63 | P a g e
1 D Danish D
9) The one who knows Thai lives immediately above the one who knows Polish, but not
on the topmost floor.
10) Only one person lives between the one who knows Thai and H.
Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
4 H Italian Swedish H Italian
3 Swedish C Thai French
2 G Thai A Polish A Thai
1 D Polish H Danish D Polish
11) Only one person lives between B and the one who knows German.
(Here case - 2 will gets eliminated)
12) E does not know Swedish.
Floor Name Language
64 | P a g e
8 B Chinese
7 C Swedish
6 G German
5 F Danish
4 H Italian
3 E French
2 A Thai
1 D Polish
Above combination will be final combination.
Hence, E who know French is staying in 3rd floor.
62. People: A, B, C, D, E, F, G and H.
Floor: 1 to 8
Language: Italian, Polish, French, Chinese, German, Danish, Swedish and Thai
1) F lives an odd numbered floor above the floor numbered four.
2) Only one person lives between F and the one who knows French.
3) The number of people living above F’s floor is same as the number of people living
between F and D. 4) Only three people live between D and the one who knows Danish.
5) C lives on one of the odd numbered floors above the one who knows Danish.
6) Only two people live between C and the one who knows Italian.
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Case – 1
Case – 2
Floor Name Language Name Language
8
7 C French F
6 Italian
5 F Danish D French
4 Italian
3 French C
2
1 D Danish
7) Only three people live between G and A.
8) The one who knows Swedish lives immediately above G, G knows neither Danish nor
Italian.
Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
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4 Italian Swedish Italian
3 Swedish C French
2 G A A
1 D Danish D
9) The one who knows Thai lives immediately above the one who knows Polish, but not
on the topmost floor.
10) Only one person lives between the one who knows Thai and H.
Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
4 H Italian Swedish H Italian
3 Swedish C Thai French
2 G Thai A Polish A Thai
1 D Polish H Danish D Polish
11) Only one person lives between B and the one who knows German.
(Here case - 2 will gets eliminated)
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12) E does not know Swedish.
Floor Name Language
8 B Chinese
7 C Swedish
6 G German
5 F Danish
4 H Italian
3 E French
2 A Thai
1 D Polish
Above combination will be final combination.
As C is staying in 7th floor and person who knows Thai stay at 2nd floor.
Therefore, there are four 4 people in between person who knows Thai and C.
63. People: A, B, C, D, E, F, G and H.
Floor: 1 to 8
Language: Italian, Polish, French, Chinese, German, Danish, Swedish and Thai
1) F lives an odd numbered floor above the floor numbered four.
2) Only one person lives between F and the one who knows French.
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3) The number of people living above F’s floor is same as the number of people living
between F and D. 4) Only three people live between D and the one who knows Danish.
5) C lives on one of the odd numbered floors above the one who knows Danish.
6) Only two people live between C and the one who knows Italian.
Case – 1
Case – 2
Floor Name Language Name Language
8
7 C French F
6 Italian
5 F Danish D French
4 Italian
3 French C
2
1 D Danish
7) Only three people live between G and A.
8) The one who knows Swedish lives immediately above G, G knows neither Danish nor
Italian.
Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
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7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
4 Italian Swedish Italian
3 Swedish C French
2 G A A
1 D Danish D
9) The one who knows Thai lives immediately above the one who knows Polish, but not
on the topmost floor.
10) Only one person lives between the one who knows Thai and H.
Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
4 H Italian Swedish H Italian
3 Swedish C Thai French
2 G Thai A Polish A Thai
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1 D Polish H Danish D Polish
11) Only one person lives between B and the one who knows German.
(Here case - 2 will gets eliminated)
12) E does not know Swedish.
Floor Name Language
8 B Chinese
7 C Swedish
6 G German
5 F Danish
4 H Italian
3 E French
2 A Thai
1 D Polish
Above combination will be final combination.
Hence, There are 0 person who live between C and G.
64. People: A, B, C, D, E, F, G and H.
Floor: 1 to 8
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Language: Italian, Polish, French, Chinese, German, Danish, Swedish and Thai
1) F lives an odd numbered floor above the floor numbered four.
2) Only one person lives between F and the one who knows French.
3) The number of people living above F’s floor is same as the number of people living
between F and D. 4) Only three people live between D and the one who knows Danish.
5) C lives on one of the odd numbered floors above the one who knows Danish.
6) Only two people live between C and the one who knows Italian.
Case – 1
Case – 2
Floor Name Language Name Language
8
7 C French F
6 Italian
5 F Danish D French
4 Italian
3 French C
2
1 D Danish
7) Only three people live between G and A.
8) The one who knows Swedish lives immediately above G, G knows neither Danish nor
Italian.
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Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
4 Italian Swedish Italian
3 Swedish C French
2 G A A
1 D Danish D
9) The one who knows Thai lives immediately above the one who knows Polish, but not
on the topmost floor.
10) Only one person lives between the one who knows Thai and H.
Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
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4 H Italian Swedish H Italian
3 Swedish C Thai French
2 G Thai A Polish A Thai
1 D Polish H Danish D Polish
11) Only one person lives between B and the one who knows German.
(Here case - 2 will gets eliminated)
12) E does not know Swedish.
Floor Name Language
8 B Chinese
7 C Swedish
6 G German
5 F Danish
4 H Italian
3 E French
2 A Thai
1 D Polish
Above combination will be final combination.
1) A – 3 – Thai ⇒ False (as A – 2 – Thai )
2) C – 7 – Swedish ⇒ True
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3) G – 6 – German⇒ True
4) H – 4 – Italian ⇒ True
Hence, A – 2 – Thai is false combination.
65. People: A, B, C, D, E, F, G and H.
Floor: 1 to 8
Language: Italian, Polish, French, Chinese, German, Danish, Swedish and Thai
1) F lives an odd numbered floor above the floor numbered four.
2) Only one person lives between F and the one who knows French.
3) The number of people living above F’s floor is same as the number of people living
between F and D. 4) Only three people live between D and the one who knows Danish.
5) C lives on one of the odd numbered floors above the one who knows Danish.
6) Only two people live between C and the one who knows Italian.
Case – 1
Case – 2
Floor Name Language Name Language
8
7 C French F
6 Italian
5 F Danish D French
4 Italian
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3 French C
2
1 D Danish
7) Only three people live between G and A.
8) The one who knows Swedish lives immediately above G, G knows neither Danish nor
Italian.
Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
4 Italian Swedish Italian
3 Swedish C French
2 G A A
1 D Danish D
9) The one who knows Thai lives immediately above the one who knows Polish, but not
on the topmost floor.
10) Only one person lives between the one who knows Thai and H.
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Case – 1
Case – 2 Case – 3
Floor Name Language Name Language Name Language
8
7 C French F Swedish C Swedish
6 A G Italian G
5 F Danish D French F Danish
4 H Italian Swedish H Italian
3 Swedish C Thai French
2 G Thai A Polish A Thai
1 D Polish H Danish D Polish
11) Only one person lives between B and the one who knows German.
(Here case - 2 will gets eliminated)
12) E does not know Swedish.
Floor Name Language
8 B Chinese
7 C Swedish
6 G German
5 F Danish
4 H Italian
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3 E French
2 A Thai
1 D Polish
Above combination will be final combination.
1) A live Immediate below to person who know Thai. ⇒ False
2) B live Immediate below to person who know French. ⇒ False
3) G live Immediate below to person who know Swedish. ⇒ True
4) H Live immediate above A. ⇒ False
5) None of these
Comprehension End.
Comprehension Start:
66. Here, in a joint family of seven persons L, M, N, O, P, Q and R two are
married couples.
1) ‘R’ is a housewife and her husband is a lawyer.
2) ‘N’ is the wife of ‘M’.
3) ‘L’ is an engineer and is the granddaughter of ‘R’.
4) ‘O’ is the father-in-law of ‘N’, a doctor, and father of ‘P’, a professor.
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‘Q’ is L’s brother and M’s son:
Therefore, Q is the grandson of O.
67. Here, in a joint family of seven persons L, M, N, O, P, Q and R two are
married couples.
1) ‘R’ is a housewife and her husband is a lawyer.
79 | P a g e
2) ‘N’ is the wife of ‘M’.
3) ‘L’ is an engineer and is the granddaughter of ‘R’.
4) ‘O’ is the father-in-law of ‘N’, a doctor, and father of ‘P’, a professor.
‘Q’ is L’s brother and M’s son:
Therefore, O is father of M.
Comprehension End.
68. Given,
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Number 6 4 3 8 2 1 7 9
Position I II III IV V VI VII VIII
Now, arranging the given numbers in ascending order:-
Number 1 2 3 4 6 7 8 9
Position I II III IV V VI VII VIII
Only 3 and 9 don’t change there position.
Hence, answer is two.
69. Analyzing information given, we get the following diagram,
Thus ‘finance’ is coded as ‘ma’.
70. Analyzing information given, we get the following diagram,
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Thus ‘supply’ is coded as either ‘jo’ or ‘ta’.
71. Given,
Number 6 4 3 8 2 1 7 9
Position I II III IV V VI VII VIII
Now, arranging the given numbers in ascending order:-
Number 1 2 3 4 6 7 8 9
Position I II III IV V VI VII VIII
Only 3 and 9 don’t change there position.
Hence, answer is two.
72. Numbers:
432 851 719 643 287
Interchanging the positions of 1st and 3rd digits:
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234 158 917 346 782
Arranging in descending order:
917 782 346 234 158
⇒ 2nd highest number = 782
⇒ Middle digit of 2nd highest number = 8
73. Note: Here right turn means moving in clockwise direction and left turn
means moving in anticlockwise direction.
From statement 1: If a person walks 5m towards north from point L, and takes two
consecutive right turns each after walking 5m, he would reach point K, which is 9m
away from point S.
Here point S can be in any direction of point K. So we can’t find the answer
Hence statement 1 is not sufficient to answer the question.
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From statement 2: Point D is 3m towards east of point L and 5m towards the west of
point S
Clearly point L is to the west of point S.
Hence statement 2 alone is sufficient to answer the question.
74. The least possible Venn diagram for the given statements is as follows,
Conclusions:
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I. Some Jacket being Pants is a possibility → It’s true.
II. Some Jacket are clothes → True as some Jacket are Blazer and all Blazer are clothes,
therefore some Jacket are clothes.
III. Some Blazer being Cap is a possibility→ It’s true.
IV. All Jacket are Blazer → Not true as it is given that some Jacket are Blazer.
Hence, conclusion (I), (II) and (III) follow.
75. Let’s draw a least possible Venn diagram using given statements:
Conclusions:
I. No CD is a picnic: It is possible but not definite.
II. Some picnics are definitely not CD’s: It is possible but not definite.
So, none of the conclusions follow.
76. From Statement I:
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No saucers are clips. Some clips are bonds. Some trays are tumblers. No tumblers are
knives
i) Some knives are trays → False (Possible but not definitely true).
ii) Some trays are saucers is a possibility → True
iii) Some knives are clips is a possibility → True
iv) Some knives are tumblers → False (Possible but not definitely true).
Therefore, Statement I will not result in the given conclusions.
From Statement II:
All saucers are clips. All clips are trays. No trays are tumblers. All tumblers are knives
86 | P a g e
i) Some knives are trays→ False (Possible but not definitely true).
ii) Some trays are saucers is a possibility → False (Possible but not definitely true).
iii) Some knives are clips is a possibility → False (Possible but not definitely true).
iv) Some knives are tumblers→ True
Therefore, Statement II will not result in the given conclusions.
From Statement III:
Some saucers are clips. No clips are bonds. No trays are tumblers. Some tumblers are
knives
87 | P a g e
i) Some knives are trays → False (Possible but not definitely true).
ii) Some trays are saucers is a possibility → True
iii) Some knives are clips is a possibility → True
iv) Some knives are tumblers → True
Therefore, Statement III will not result in the given conclusions.
From Statement IV
No saucers are clips. Clips are not trays. Trays are not tumblers. All tumblers are knives
88 | P a g e
i) Some knives are trays → False (Possible but not definitely true).
ii) Some trays are saucers is a possibility → True
iii) Some knives are clips is a possibility → True
iv) Some knives are tumblers → True
Therefore, Statement IV will not result in the given conclusions.
From Statement V
All saucers are clips. Some clips are bonds. Some trays are tumblers. All tumblers are
knives
i) Some knives are trays → True
ii) Some trays are saucers is a possibility → True
iii) Some knives are clips is a possibility → True
iv) Some knives are tumblers → True
Therefore, Statement V will result in the given conclusions.
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77. There are books on eight subjects English, Physics, Sociology, Chemistry,
Mathematics, French, German and History which are placed in the shelves. One
subject book must in one shelf only.
1) The books on History should be in a shelf opposite the shelf containing French books.
2) The books on Physics and those on Chemistry should be on shelves opposite each
other while books on Physics are placed to the immediate right of French books.
90 | P a g e
3) The books on French, German and English should be in three shelves placed side by
side.
So we get following 2 possibilities,
If the shelf containing the books on Physics is placed between to the shelves containing
French and Sociology books and there is only one shelf between the shelf containing
history books and sociology books, then the shelf would contain mathematics books.
91 | P a g e
78. There are books on eight subjects English, Physics, Sociology, Chemistry,
Mathematics, French, German and History which are placed in the shelves. One
subject book must in one shelf only.
1) The books on History should be in a shelf opposite the shelf containing French books.
2) The books on Physics and those on Chemistry should be on shelves opposite each
other while books on Physics are placed to the immediate right of French books.
3) The books on French, German and English should be in three shelves placed side by
side.
92 | P a g e
So we get following 2 possibilities,
If Mathematics books are placed to the immediate right of Physics books, we get
following possibilities,
Then the shelf with Sociology books are placed third to left or fourth to left with respect
to the English books.
93 | P a g e
79. There are books on eight subjects English, Physics, Sociology, Chemistry,
Mathematics, French, German and History which are placed in the shelves. One
subject book must in one shelf only.
1) The books on History should be in a shelf opposite the shelf containing French books.
2) The books on Physics and those on Chemistry should be on shelves opposite each
other while books on Physics are placed to the immediate right of French books.
3) The books on French, German and English should be in three shelves placed side by
side.
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So we get following 2 possibilities,
If the books on Sociology is placed near the shelf containing physics books, we get
following possibilities,
Then the shelf containing Mathematics books will be placed on the immediate left with
respect to the shelf containing History books.
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80. There are books on eight subjects English, Physics, Sociology, Chemistry,
Mathematics, French, German and History which are placed in the shelves. One
subject book must in one shelf only.
1) The books on History should be in a shelf opposite the shelf containing French books.
2) The books on Physics and those on Chemistry should be on shelves opposite each
other while books on Physics are placed to the immediate right of French books.
3) The books on French, German and English should be in three shelves placed side by
side.
96 | P a g e
So we get following 2 possibilities,
If the books on German are opposite the shelf of Mathematics books, we get following
possibilities,
Then English and Sociology shelves are opposite to each other.