answer key for ibps rrb officer scale 1 prelims 2017 model

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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,

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⇒ 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%,



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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



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= 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.



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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



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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.



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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



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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



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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


iii) Two yellow balls and one blue ball


iv) Two yellow balls and one green ball


v) Two green balls and one blue ball


vi) Two green balls and one yellow ball




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∴ 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:



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→ 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



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→ 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%



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∴ [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



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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



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∴ 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



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∴ 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:



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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



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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


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



∴ 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



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∴ 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



∴ 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



∴ 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.



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⇒ 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.


= 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



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= 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



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= 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.


= 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



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⇒ 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,



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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.



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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



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(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.



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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.



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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.







intelligent.

Now given,




most intelligent.


group.



30 | P a g e




four are males.











31 | P a g e



and is bachelor.


one person, F.













32 | P a g e

As we can see increasing order of intelligence is FME.







intelligent.

Now given,




most intelligent.


group.



33 | P a g e




four are males.











34 | P a g e



and is bachelor.


one person, F.













35 | P a g e

As the row follows that A or R can be a teacher.







intelligent.

Now given,




most intelligent.


group.



36 | P a g e




four are males.











37 | P a g e



and is bachelor.


one person, F.













38 | P a g e

As we can see above arrangement F is least intelligence.







intelligent.

Now given,




most intelligent.


group.



39 | P a g e




four are males.











40 | P a g e



and is bachelor.


one person, F.













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.








⇒ Apple = MI/2


⇒ MI = Nokia/2






43 | P a g e

⇒ Lenovo = 17 Apple

⇒ Samsung = 8.5 Apple

⇒ Nokia = 4 Apple

⇒ MI = 2 Apple


So the heaviest is Lenovo.








⇒ Apple = MI/2


⇒ MI = Nokia/2




⇒ Lenovo = 17 Nokia / 4



44 | P a g e

⇒ Samsung = 17 Nokia / 8

⇒ Apple = Nokia / 4

⇒ MI = Nokia / 2


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


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.







50 | 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









51 | 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


12:45 PM C

Thursday 8:45 AM R

12:45 PM

Friday 8:45 AM R

12:45 PM C




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


There are four people between A and S exam, which is not mentioned in the options,

Thus, answer is none of above.










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











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


12:45 PM C

Thursday 8:45 AM R

12:45 PM

Friday 8:45 AM R

12:45 PM C




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


Hence, R and T exam is on Friday.








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









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


12:45 PM C

Thursday 8:45 AM R

12:45 PM

Friday 8:45 AM R

12:45 PM C




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


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.








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









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


12:45 PM C

Thursday 8:45 AM R

12:45 PM

Friday 8:45 AM R

12:45 PM C




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


Hence, Q, S, E, P and R are student whose exam slot is 8:45 AM.

Comprehension End.


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


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



8


6 A G Italian G


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


Hence, E who know French is staying in 3rd floor.


Floor: 1 to 8










65 | P a g e

Case – 1

Case – 2


8

7 C French F

6 Italian

5 F Danish D French

4 Italian

3 French C

2

1 D Danish



Italian.

Case – 1



8


6 A G Italian G




66 | P a g e


3 Swedish C French

2 G A A

1 D Danish D




Case – 1



8


6 A G Italian G










67 | P a g e


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


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.


Floor: 1 to 8






68 | P a g e





Case – 1

Case – 2


8

7 C French F

6 Italian

5 F Danish D French

4 Italian

3 French C

2

1 D Danish



Italian.

Case – 1



8



69 | P a g e


6 A G Italian G



3 Swedish C French

2 G A A

1 D Danish D




Case – 1



8


6 A G Italian G







70 | P a g e





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


Hence, There are 0 person who live between C and G.


Floor: 1 to 8



71 | P a g e








Case – 1

Case – 2


8

7 C French F

6 Italian

5 F Danish D French

4 Italian

3 French C

2

1 D Danish



Italian.



72 | P a g e

Case – 1



8


6 A G Italian G



3 Swedish C French

2 G A A

1 D Danish D




Case – 1



8


6 A G Italian G




73 | P a g e








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


1) A – 3 – Thai ⇒ False (as A – 2 – Thai )

2) C – 7 – Swedish ⇒ True



74 | P a g e

3) G – 6 – German⇒ True

4) H – 4 – Italian ⇒ True

Hence, A – 2 – Thai is false combination.


Floor: 1 to 8








Case – 1

Case – 2


8

7 C French F

6 Italian

5 F Danish D French

4 Italian



75 | P a g e

3 French C

2

1 D Danish



Italian.

Case – 1



8


6 A G Italian G



3 Swedish C French

2 G A A

1 D Danish D






76 | P a g e

Case – 1



8


6 A G Italian G









Floor Name Language

8 B Chinese

7 C Swedish

6 G German

5 F Danish

4 H Italian



77 | P a g e

3 E French

2 A Thai

1 D Polish


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.


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.



78 | P a g e

‘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,



80 | P a g e

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


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,



81 | P a g e

Thus ‘supply’ is coded as either ‘jo’ or ‘ta’.

71. Given,

Number 6 4 3 8 2 1 7 9


Now, arranging the given numbers in ascending order:-

Number 1 2 3 4 6 7 8 9


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:



82 | P a g e

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.



83 | P a g e

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:



84 | P a g e

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:



85 | P a g e

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




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





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




Therefore, Statement V will result in the given conclusions.



89 | P a g e

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








side.



92 | P a g e


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








side.



94 | P a g e


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.



95 | P a g e








side.



96 | P a g e


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.



answer key for ibps rrb officer scale 1 prelims 2017 model

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