answer key for ibps rrb office assistant prelims 2016 ...€¦ · 1 | page answer key for ibps rrb...

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Answer Key for IBPS RRB Office Assistant Prelims 2016 Model Question Paper

Q1. 1 Q2. 2 Q3. 1 Q4. 3 Q5. 2 Q6. 4 Q7. 5 Q8. 2 Q9. 3 Q10. 3

Q11. 4 Q12. 5 Q13. 3 Q14. 3 Q15. 4 Q16. 5 Q17. 2 Q18. 4 Q19. 4 Q20. 4

Q21. 2 Q22. 5 Q23. 1 Q24. 5 Q25. 1 Q26. 4 Q27. 3 Q28. 3 Q29. 5 Q30. 2

Q31. 4 Q32. 2 Q33. 3 Q34. 5 Q35. 5 Q36. 3 Q37. 2 Q38. 3 Q39. 1 Q40. 1

Q41. 4 Q42. 5 Q43. 2 Q44. 4 Q45. 4 Q46. 5 Q47. 1 Q48. 2 Q49. 5 Q50. 2

Q51. 4 Q52. 5 Q53. 5 Q54. 3 Q55. 3 Q56. 5 Q57. 4 Q58. 2 Q59. 2 Q60. 2

Q61. 2 Q62. 4 Q63. 3 Q64. 5 Q65. 4 Q66. 2 Q67. 3 Q68. 4 Q69. 2 Q70. 2

Q71. 5 Q72. 4 Q73. 2 Q74. 5 Q75. 1 Q76. 3 Q77. 4 Q78. 1 Q79. 4 Q80. 2

Reasoning

1. The number in the subscript alternates between the first and second letter.

This is shown in the figure:

Hence the answer is VE7.

2. It follows the pattern as:

Hence the next term is S/30.

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3. From the given series:

Let's take respective first entity of each term, i.e., (4, 5, 7, ?, 14, 19)

4 + 1 = 5

5 + 2 = 7

7 + 3 = '?' = 10

10 + 4 = 14

14 + 5 = 19

Hence, first entity of missing term in 10.

Similarly, let's check for respective 2nd and 3rd entities of each term and relation between them:

E + 2 = G

G + 2 = I

I + 2 = K

K + 2 = M

And,

Z - 1 = Y, Y - 1 = X, X - 1 = W, W - 1 = V

Hence, the missing term is 10KW.

4. Given series follows the following pattern:-

From the above pattern we can see that, L11T will come after I7V.

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5. The pattern followed in given series is:

P 3 C, R 5 F, T 8 I, V 12 L, ?

P + 2 = R; R + 2 = T; T + 2 = V; V + 2 = X

3 + 2 = 5; 5 + 3 = 8; 8 + 4 = 12; 12 + 5 = 17

C + 3 = F; F + 3 = I; I + 3 = L; L + 3 = O

Thus X 17 O is the answer.

6. A and B, T and R, A and E.

Hence, there are three such pairs.

7. We can see same number of letters between following pairs as shown,

Hence, the answer is more than 3 pairs.

8. The pattern is as follows

Similarly

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Clearly, code for MENSTRUAL is ODPUSTWZN.

9. The pattern followed for coding KEYBOARD is as follows,

Following the same pattern,

Thus code for TOUCHPAD is FVQWEJRC.

10. According to the information given,

Rule no. First letter Last letter Action taken

i ConsonantVowel Codes are interchanged

ii ConsonantConsonantBoth are to be coded as per the code of the last letter

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iii Vowel ConsonantBoth are to be coded as ‘ ’

In the word UKVMIH, we can see that first letter is a vowel and the last letter is a consonant, hence

condition 3 is applicable here that is both first and last letter is to be coded as .

Letters L M A E J K D R Q H I U V F

Digits/Symbols Conditions 4 $ 1 2 3 % 5 @ © 6 # δ 7 9

UKVMIH is coded as δ % 7 $ # 6. After applying condition, % 7 $ # .

Hence, % 7 $ # is the answer.








In the word VMEIUF, we can see that both the first and the last letter are consonants so condition 2 is

applicable here i.e. both first and last letter are to be coded as the last letter.

From the table, the simple code for VMEIUF becomes 7$2#δ9, after condition 2 it becomes 9$2#δ9.

Hence, 9$2#δ9 is the answer.



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In the word MLEKDU, we can see that first letter is consonant and the last letter is vowel. So condition

1 is applicable here i.e. codes are interchanged.

From the table, the simple code for MLEKDU becomes $ 4 2 % 5 δ,

Now, after condition 2 is applied the code for MLEKDU becomes δ 4 2 % 5 $.

Hence, δ 4 2 % 5 $ is the answer.

13.

Rule No. Condition Action taken

(i) First letter is a vowel and the last

letter is a consonant.

Both are coded as the code for

the vowel.

(ii) Both the first and the last letters

are consonants.

Both are coded for the last

letter.

(iii) First letter is a consonant and the

last letter is a vowel. Both are to be coded as ‘ ’.

Given Digit/Symbol code:

O R T M D E I Q Z F H K A P J

© 7 6 3 9 2 1 4 # $ 5 % @ 8 δ

Given term: DKPJMO

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Here first letter is a consonant and the last letter is a vowel. Thus rule (iii) is applicable.

Thus DKPJMO will be coded as %8δ3 .

14.

Rule No. Condition Action taken

(i) First letter is a vowel and the last

letter is a consonant.

Both are coded as the code for

the vowel.

(ii) Both the first and the last letters

are consonants.

Both are coded for the last

letter.

(iii) First letter is a consonant and the

last letter is a vowel. Both are to be coded as ‘ ’.

Given Digit/Symbol code:

O R T M D E I Q Z F H K A P J

© 7 6 3 9 2 1 4 # $ 5 % @ 8 δ

Given term: IDATRJ

Here first letter is a vowel and the last letter is a consonant. Thus rule (i) is applicable.

Thus both letter ‘I’ and ‘J’ will be coded as 1.

Thus IDATRJ will be coded as 19@671.

15. Given, if Misha is shifted by 4 places towards the left, she became 12th from the right end

⇒ Misha’s earlier position from right end = 12 – 4 = 8th from right end

Given Total woman in the row = 22

Earlier position of Misha from left end = (Total woman – Her position from right) + 1 = (22 – 8) + 1 = 14 +

1 = 15th from left.

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16. Given statement is: A < B

Conclusions:

I. B ≥ D ≥ C < A (Cannot be determined as clear relation between A and B cannot be established)

II. B ≥ V > C ≥ A (Not False as B > A)

III. D ≥ B ≥ C < A (Cannot be determined as clear relation between A and B cannot be established)

IV. A ≥ C = D < B (Cannot be determined as clear relation between A and B cannot be established)

Hence, none of these conclusions give “B > A” as definitely false.

17. Given Statements:

Q = R, R > M, M ≤ O, O = P

⇒ Q = R > M ≤ O = P

Conclusions:

a) O < Q → Q > M ≤ O → Rela�onship between O and Q cannot be determined, hence false.

b) P ≥ M → M ≤ O = P → M ≤ P, hence true.

18. Given statements:

L ≤ T< N > O ≥ P

Conclusions:

I. O < T → According to statement T< N > O → Thus clear rela�onship between T and O cannot be

established, hence false.

II. P ≤ N → According to statement N > O ≥ P → N > P → Hence false.

Hence neither conclusion I nor II follows.

19. Given Statements:

B ≤ J; K < L > M; J = K; G ≥ H = B

⇒ G ≥ H = B ≤ J = K < L > M

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Conclusions:

I. M > J → J = K < L > M → J < L > M → hence rela�onship between M and J cannot be determined.

II. J > G → G ≥ H = B ≤ J → hence rela�onship between J and G cannot be determined.

Hence none of the conclusion follow.

20. Given statements: R = M ≥ L; Z ≤ W ≤ N; N ≥ L

On combining: R = M ≥ L ≤ N ≥ W ≥ Z

Conclusion:

I. W > Z → Not true as it is given that W ≥ Z therefore they can be equal as well.

II. R ≥ L → True as it is given that R = M ≥ L.

III. W = Z → Not true as it is given that W ≥ Z therefore W can be greater as well.

IV. N > Z → Not true as it is given that N ≥ W ≥ Z therefore they can be equal as well.

V. L ≥ W → Not true as it is given that L ≤ N ≥ W therefore no definite relation between L and W.

Conclusion I and III are complementary to each other.

Hence only conclusion II and either of conclusion I or III are true.

21. The least possible Venn diagram for the given statements is as follows,

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Conclusions:

I. No sweets are diet → It’s possible but not de nite, hence false.

II. No food is chocolates → It’s possible but not de nite, hence false.

III. Some sweets are diet → It’s possible but not de nite, hence false.

IV. Some sweets are food → It’s possible but not definite, hence false.

Conclusion I and III form complementary pair.

Thus, either conclusion I or III follows.


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Conclusions:

I. Some utensils are forks → It’s possible but not definite, hence false.

II. Some plates are forks → It’s possible but not de nite, hence false.

III. Some plates are spoons → It’s possible but not de nite, hence false.

IV. Some utensils are spoons → It’s possible but not de nite, hence false.

Thus none of the given conclusion follows.


Conclusions:

I. All walls are rooms → It’s possible but not de nite, hence false.

II. Some rooms are doors → It’s possible but not de nite, hence false.

III. Some rooms are walls → It’s possible but not de nite, hence false.

IV. Some floors are doors → It’s possible but not de nite, hence false.


24. The least possible Venn diagram for the given statements is,

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Conclusions:

I. Some doctors are advocates → it’s possible but not de nite, hence false.

II. All graduates are judges → it’s possible but not de nite, hence false.

III. Some doctors are graduates → it’s possible but not de nite, hence false.

IV. Some lawyers are advocates → clearly true.

Hence, only conclusion IV follows.

25. The least possible Venn diagram for the given statements is,

Conclusions:

I. All asteroids are planets → it’s possible but not de nite, hence false.

II. All asteroids are stars → it’s possible but not de nite, hence false.

III. All moon are stars → it’s possible but not de nite, hence false.

IV. Some rocks are stars → it’s possible but not de nite, hence false.


26. People: A, B, C, D, E, F, G and H

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Professions: Doctor, Engineer, Architect, Pilot, Banker, Teacher, Businessman and Politician.

1) The politician sits third to the right of G.

2) C is an immediate neighbor of G.

So we get 2 cases.

3) The architect sits second to the right of C.

So case 2 gets eliminated.

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4) B sits third to the right of H.

5) H is neither a politician nor an architect.

So we get 2 cases,

6) A and F are immediate neighbours of each other.

7) Neither A nor F is a politician.

So case 1a gets eliminated.

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8) Only one person sits between C and the teacher.

9) The doctor sits second to the right of A.

So we get 2 possible positions for doctor.

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10) The banker sits second to the left of A.

So case 1 b.1 gets eliminated.

11) Two people sit between D and the engineer.

12) D is not a politician.

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13) The pilot is not an immediate neighbor of the politician.

Therefore, A is the businessman.



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So we get 2 cases.



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So we get 2 cases,




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Therefore, F is second to the right of the politician.




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So we get 2 cases.



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So we get 2 cases,




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Therefore, doctor sits between the teacher and the engineer.



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So we get 2 cases.



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So we get 2 cases,




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Therefore, doctor sits second to the right of the businessman.




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So we get 2 cases.



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So we get 2 cases,




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Therefore, G is a pilot.

31. People facing north: P, Q, R, S, T, V and W.

Floors: 1st to 7th

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1) Q sits fourth to the left of the person living on the 6th floor.

2) Either Q or the person living on the 6th floor sits at the extreme ends of the line.

3) Only one person sits between Q and W.

4) W lives on the 3rd floor.

5) The person living on the 1st floor sits third to right of S.

6) S is not an immediate neighbor of W.

7) P and R are immediate neighbours of each other.

8) P does not live on the 6th floor.

9) Only one person lives between T and the person who lives on the 2nd floor.

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10) One who lives on the 5th floor sits third to right of the one who lives on the 7th floor.

So we get the final diagram,

Therefore, S lives on 4th floor.


Floors: 1st to 7th





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Therefore, there are 2 floors between V and P.

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Floors: 1st to 7th









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Only S is at the extreme end thus the odd one.


Floors: 1st to 7th





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Therefore, Q-6th is the only option which do not satisfy the condition of person with floor above it.

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Floors: 1st to 7th









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The one who lives on the 4th floor sits at one of the extreme ends of the line.

36. Let’s solve this according to the option given,

Taking Only I:

Clearly we cannot conclude.

Taking only II:

Clearly we cannot conclude.

Taking I & II:

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Hence we can conclude.

Hence answer is only statement I & II are required to answer the question.

37. Taking Only I:

I. If Anu turns 135° to his left, he will be facing a direction that is exactly opposite Raju.

Clearly, we only know Anu’s direction relative to Raju whose direction is not given. Hence we cannot

conclude Anu’s direction.

Taking Only II:

II. Mukesh is facing North; if he turns 90° to his left after moving 45° ACW, he will be exactly opposite to

the direction of Anu.

Since Mukesh is turning Anti Clockwise 45°, he is moving to his left. Hence he rotates, 135° left. His

direction after turning is shown by a solid line.

So Anu is facing North East. Conclusion reached.

Taking only III:

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Raju will be in the same direction of Anu if he turns 45° to his right after turning 90° to his left.

Clearly we only know Anu’s direction relative to Anu, so again we cannot conclude.

Hence, the answer is only statement II is sufficient to answer the question.

38. Students = 45

From statement 1: Sweta is five ranks below Sanjay, who is 15th from the bottom.

So, the position of Sanjay from top = 46 – 15 = 31st

Sweta is five ranks below Sanjay → So, Sweta’s rank = (31 + 5) = 36th

Hence statement 1 is sufficient to answer the question.

From statement 2: Ravina is 30th from the top and Neha is 4th from the bottom.

So, the position of Neha from top = 46 – 4 = 42.

But we have no data regarding Sweta so statement 2 isn’t sufficient alone.

From statement 3: Sweta is exactly in the middle of Ravina and Neha.

We have no data regarding position of Ravina&Neha so statement 3 isn’t sufficient alone.

On combining 2 and 3:

Statement 2 gives → Ravina is 30th from top, Neha is 42nd from top.

⇒ There are 42 – 30 + 1 = 13 people between Ravina&Neha.

Statement 3 gives → Sweta is exactly in the middle of Ravina and Neha.

⇒ Sweta’s position is 6th from both Ravina&Neha

⇒ Sweta’s rank from top = 30 + 6 = 36th

So, II and III together are sufficient.

39. From statement 1: He has only one son and twice as many daughters.

Thus, clearly Mr. X has three children one son and two daughters.

Thus statement 1 is sufficient to answer the question.

From statement 2: One of his daughters is elder than his son

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We can’t depict how many children Mr. X has with this statement.

Thus, statement II alone is not sufficient to answer the question.

Thus, data in statement I alone are sufficient to answer the question, while the data in statement II

alone are not sufficient to answer the question.

40. Statement I: Q is heavier than R and T but lighter than only S.

So S is the heaviest among them.

Statement II: R is third from the top when they are arranged in descending order of their weight and is

heavier than T and P.

Hence data in statement I alone is sufficient to answer the question, while the data in statement II alone

are not sufficient to answer the question.

Numerical Ability

41. 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,

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.

Now the given expression:

(78750 ÷ 1750) + (10 ÷ 2.5 × 120) =? × 2 × 2.5

⇒ ? × 5 = (78750/1750) + (4 × 120)

⇒ ? × 5 = 45 + 480

⇒ ? × 5 = 525

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⇒ ? = 525/5 = 105

Hence, the required number in place of question mark in the following question is 105.

42. Given expression:

⇒?=��

�.�×��.�=

��

�×��× 10 × 100

⇒?=��

��× 1000=

��×��

��

= 400

Hence, the required answer is 400.

43. Given that 11(10 – 2x) = 1

⇒ 1110 /112x = 1 (∵am – n = am/an)

⇒ 1110 = 112x

⇒ 10 = 2x (∵ If am = an then m = n)

⇒ x = 10/2 = 5

∴ x = 5

44. 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,

Step-2-Any mathematical 'Of' or 'Exponent' must be solved next,



calculated.

Now, the given expression:

3�

�+ 2

�

�+ 1

�

�=?

= 3 + 2 + 1 + ��

�+

�

�+

�

��

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= 6 +��

��= 6 +

��

��

= 6 + 1��

��= 7

��

��

45. 1.25 × 1.5 + 4% of 160 = ?

⇒ 1.25 × 1.5 + (4/100) × 160 = ?

⇒ ? = 1.875 + 6.4

⇒ ? = 8.275

46. In this type of question, we are expected to calculate Approximate value (not exact value), so we

can replace the given numbers by their nearest perfect places which makes the calculation easy.

Let, 456.675 ≈ 457, 35.7683 ≈ 36, 67.909 ≈ 68, 58.876 ≈ 59


456.675 + 35.7683 × 67.909 – 58.876 =?

⇒ ? ≈ 457 + 36 × 68 – 59

⇒ ? ≈ 457 + 2448 – 59

⇒ ? ≈ 2905 – 59

⇒ ? ≈ 2846

Since, we have approximated the value, that’s why we got 2846. But according to the options required

answer will be 2825.

47. In this type of question, we are expected to calculate Approximate value (not exact value), so we

can replace the given numbers by their nearest perfect places which makes the calculation easy.


? = 3.2 × 8.1 + 3185 ÷ 4.95

⇒ ? ≈ 3 × 8 + 3185 ÷ 5

⇒ ? ≈ 24 + 637 ≈ 661

Hence, the approximate answer is 670.

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48. We have got

121÷ ��

�×

�

�×

�

�� =?

⇒ 121 ÷ 84/200 = ?

⇒ 121 ÷ 0.42 = ?

⇒ ? = 121/0.42

⇒ ? ≈ 288

49. 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,



calculated.

The equation is,

⇒ (5/7) × 1596 + 3015 + 2150 = ? (of means ‘×’)

⇒ 1140 + 3015 + 2150 = ?

⇒ 6305 = ?, approximately, 6300

50. For approximation Problems

In this type of question, we are expected to calculate Approximate value (not exact value), so we can

replace the given numbers by their nearest perfect places which makes the calculation easy.

√339 ≈ 18 (∵ 18 × 18 = 324)

(√339 × 25) ÷ 30 = ?

⇒ (18 × 25) ÷ 30 = ?

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⇒ 450 ÷ 30 = ?

⇒ 15 = ?

51. We are denoting the positions of the numbers. Like:

The number in position 1 is 5

The number in position 2 is 6

The number in position 3 is 20 & so on.

In the following number series the pattern we observe is:

Any number = (The previous number’s position)3 + (previous number × the previous number’s position)

For example: 6 = (1)3 + (5 × 1)

[The previous number of 6 is 5 & 5’s position is 1]

Similarly, 20 = (2)3 + (6 × 2)

So, for ‘?’ we can observe:

The previous number of question mark is 20

The position of the previous number of the question mark is 3

∴ ? = (3)3 + (20 × 3)

⇒ ? = 87

For confirmation we can check if 412 can be obtained from 87 in similar fashion.

i.e., 412 = (4)3 + (87 × 4)

52. Difference between each successive number is a perfect square.

9 – 5 = 22

18 – 9 = 32

34 – 18 = 42

59 – 34 = 52

95 – 59 = 62

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Let the next number be n.

n – 95 = 72

⇒ n = 144

53. In the given series, the difference between two consecutive numbers is as follows:

→ 5 – 0 = 5

→ 18 – 5 = 13

→ 43 – 18 = 25

→ 84 – 43 = 41

→ 145 – 84 = 61

It can be observed from the values of these differences that it is increasing with every next step by an

addition of multiples of 4 starting from 8. i.e.

→ 13 – 5 = 8 = 4 × 2

→ 25 – 13 = 12 = 4 × 3

→ 41 – 25 = 16 = 4 × 4

→ 61 – 41 = 20 = 4 × 5

Hence the value of differences between the next two numbers must be

= 4 × 6

= 24

Let X is the required term in the given number series. Thus

⇒ X – 145 = 61 + 24 = 85

⇒ X = 230

Since it does not match any option thus option 5 is correct.

54. The pattern of given series can be evaluated as:

→ 4

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→ 4 × 1 + 2 = 6

Similarly,

→ 6 × 2 + 3 = 15

→ 15 × 3 + 4 = 49

→ 49 × 4 + 5 = 201

→ 201 × 5 + 6 = 1011

So, it can be seen that apart from 18 every other numbers of the series follows this pattern.

Hence, the only wrong number in the given number series is 18.

55. In the following number series the pattern we observe is:

Any number = 2 × (previous number) + 3

For example: 5 = (2 × 1) + 3

13 = (2 × 5) + 3

125 = (2 × 61) + 3

253 = (2 × 125) + 3

But, for 31 we can observe:

31 ≠ (13 × 2) + 3

& 61 ≠ (31 × 2) + 3

∴ 31 is the wrong number.

56. Let the number of pages in the book be ‘x’, then according to the question,

Number of pages read on first day = (3x/8)

Remaining pages = 1 – (3x/8) = 5x/8

Number of pages read on second day-

=�

�×

�

�� =

�

��

Now, remaining pages –

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= � ��

�+

�

��

= � ��

��

= ��

�

= x/8

According to the question,

x/8 = 40

∴ x = 320

Hence, the required number of pages in the book is 320, therefore option 5 is correct.

57. Let the cost price of 1000 gm of goods be c.

∴ cost price of 900 gm of goods = c × 1000 ÷ 900

⇒ C.P. of 900 gm of goods = 0.9 × c

Given, dealer sells his goods at C.P.

S.P. of 0.9 kg goods = Cost price of 1 kg goods = c

Profit = Selling price – Cost price

⇒ Profit = c – 0.9 × c

⇒ Profit = 0.1 × c

�� % =��

�� × 100%

⇒ �� % =�.��

�.��× 100%

⇒ �� % =��

�% = 11

�

�%

58. Simple interest = (principal × rate × time)/100

Simple interest on a sum of money is 4/9th the principal, and the number of years is equal to the rate of

interest per annum.

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Let principal be P

Then,

��

�=

��×�

��

⇒��

�= ��

On taking square root

� =��

�= 6

�

�%

59. Let us assume that Ajith travels a distance `d’ km

Also let us assume that the time taken to cover the distance `d’ km is `t’ hours

Given that he reaches in time when he travels at a speed of 48 kmph.

We know that Speed = Distance/time

∴ 48 kmph = d/t

⇒ d/t = 48 kmph ………...(1)

Similarly it is given that he reaches 15 minutes late when he travels at 36 kmph.

∴ d/(t + 15min) = 36 kmph

⇒�

��

��= 36 �� (∵`t’ is in hours hence converting 15 minutes in hours. i.e. 15 minutes = ¼ hour)

⇒��

��= 36 ��

�

��= 9 �� ……… (2)

From (1) and (2) equating the distance in both cases (∵ We know that the distance is same)

⇒ 48t = 9(4t + 1)

⇒ 48t = 36t + 9

⇒ 12t = 9

⇒ t = 9/12 = ¾ hours …….. (3)

By substituting equation (3) in (1), we get

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�

��

��= 48 ��

⇒ d = (¾) × 48 km = 36 km

∴ The distance of journey is `36 km’.

60. Let the volume of the tank be V litres

⇒ Part of tank emptied by A in minute = V/7.5

= 2V/15

Part of tank filled by pipe B in 1 minute = V/5

Part of tank emptied by C in a minute = 14 litres

Total part of tank filled in a min = 2V/15 + V/5

= (2 + 3)V/15

= V/3

Hence part of Volume changed in 1 min = ��

�14�

⇒ In 60 minutes = ��

�14�× 60

As the pipes were opened after the tank was full and was subsequently emptied in next 1 hour.

� + ��

�14� × 60 = 0

⇒ 21V = 60 × 14

⇒ V = 40 litres

61. Let the woman’s and her daughters age be a and b.

Given average age of woman and her daughter is 46 years

∴ (a+b)/2 = 46 -------------- 1

Also given their ages are in the ratio of 15 : 8

a/b = 15/8

58 | P a g e

⇒ a = 15b/8

Substituting the value of a in eq.1 we get,

��

��

�= 46

⇒ 23b/8 = 92

⇒ b = 32

62. Area of rectangle = Length × Breadth

Perimeter of rectangle = 2(Length + Breadth)

The area of a rectangle is 2891 square metres and length of the rectangle is 59 metres

∴ Length × Breadth = 2891

⇒ 59 × Breadth = 2891

⇒ Breadth = 49 metre

Perimeter of rectangle = 2(59 + 49)

= 216 metre

150% of perimeter of rectangle = (150/100) × 216

= 324 metres

63. Among the 5 courses 2 courses are mandatory.

⇒ The person is free to select only (5 – 2) = 3 courses.

Also among the 9 available courses the two mandatory courses are there.

⇒ The person is free to select from (9 – 2) = 7 courses.

⇒ The person has to select 3 courses from 7 available courses.

The number of ways this can be done = 7�� = 35.

64. Let the total weight of mixture be w.

59 | P a g e

Given, in a Mixture of milk and water the proportion of water by weight was 75%

∴ Weight of water in mixture = 0.75 × w

Weight of milk = 0.25 w

Given mixture is 60 gms

∴ Weight of water present = 0.75 × 60 = 45 gm

Now 15 gm of water was added

Total weight of mixture = 60 + 15 = 75 gms

Total weight of water = 45 + 15 = 60 gms

% of water in mixture = ��

��× 100%

⇒ % of water in mixture = 80%

65. Assume that the length of Malgudi auditorium is L.

∴ Its breadth is ¾L.

The area of rectangle is length × breadth

Area of Malgudi auditorium = L × ¾ L = ¾ L2

Given in the question, the area of the hall is 300 m2

∴ ¾ L2 = 300

⇒ L2 = 400 = 202

⇒ L = 20 meter

∴ Breadth = ¾L = ¾ × 20 = 15 meter

Difference between length & breadth is (20 – 15) meter = 5 meter

66. Let the ages of A, B and C be x, y and z

The ages of A, B and C together are 185 years.

∴ x + y + z = 185 …(1).

60 | P a g e

B is twice as old as A and C is 17 years older than A.

∴ y = 2x and z = x + 17

Put the values of y and z in equation (1)

⇒ x + 2x + x + 17 = 185

⇒ 4x = 185 - 17

⇒ 4x = 168

⇒ x = 42

Thus, y = 2 × 42 = 84 years

And z = 42 +17 = 59 years

67. Let share of B = 100m.

Then share of A = 100m + 25 % of 100m

= 100m + (25/100) × 100m

= 100m + 25m = 125m

Also, share of B is 25% less than C’s share

Hence share of B = 75 % of C’s share

100m = (75/100) × C’s share

C’s share = 100m × (100/75)

= 133.33 m.

Total share = 817

⇒ 100m + 125m + 133.33m = 817

⇒ 358.33m = 817

⇒ m = 2.28

Hence share of A = 125m

= 125 × 2.28

= 285

61 | P a g e

68. Let a and b be the length and breadth of the rectangle

We have the formula for perimeter of rectangle as

Perimeter= 2 × (L+B)

Where L and B are the length and breadth of the rectangle respectively.

⇒ 2 × (a + b) = 200

⇒ a + b = 200/2

⇒ a + b = 100 -------- [1]

We have the length of the base of triangle = (80/100) × b

⇒ Base = 0.8b

We have the formula for the area of rectangle = L × B

Where L is the length and B is the breadth

Area of given rectangle = ab

Area of right triangle =�

�× ��× ��

Area of right triangle = ½ × 0.8b × height

We are given that area of triangle = (2/3) × area of rectangle

⇒ ½ × 0.8b × height = (2/3) × a ×b ------[2]

From the above 2 equations, we cannot derive any meaningful conclusion.

Hence we can say that data is insufficient

69. ∵ Profit % = (Profit/ Cost Price) × 100

According to Question

Let Profit % = C.P. = x

⇒ x = (Profit/x) × 100 ... (i)

Also

62 | P a g e

Selling Price - Cost Price = Profit

⇒ 75 - x = Profit ... (ii)

From Equations (i) & (ii)

⇒ � =��

�× 100

⇒ x2 + 100x -7500 = 0

⇒ x2 + 150x - 50x -7500 = 0

⇒ x (x + 150) - 50 (x + 150) = 0

⇒ (x - 50) (x + 150) = 0

⇒ x = 50 or -150

As x = Cost price of something it can't be negative

⇒ x = 50

⇒ Cost Price = Profit Percentage = 50

70. We know that:

The formula for annual compound interest is

A = P × (1 + r/n)nt

Where:

A = the future value of the investment/loan, including interest

P = the principal investment amount (the initial deposit or loan amount)

r = the annual interest rate (decimal)

n = the number of times that interest is compounded per year

t = the number of years the money is invested or borrowed for

Given,

P = 7250

t = 2

r = 6%

and since the amount is compounded annually. Thus

63 | P a g e

n = 1

Hence the Amount value after 1 year

= 7250× �1 +�

��×��×�

= Rs. 8146.1

∴ The amount value after 1 year would be Rs. 8146 approximately.

71. According to the question:

3 men ≡ 6 boys

⇒ 1 man ≡ 2 boys

∴ 4 men + 4 boys ≡ 4 men + 2men = 6 men

∵ 3 men can do work in 18 days.

∴ 1 man can do a work in 18 × 3 days.

∴ 6 men can do the work in 18×3618×36 days.

∴ Required number of days

=��×�

�= 9 ��

Hence, the required number of days is 9, therefore option 5 is correct.

72. Average weight of 19 men is 73 kgs

Total weight of 19 men = 73 × 19 kg

= 1387 kg

The average weight of 38 women is 65 kgs

Total weight of 38 women = 65 × 38 kg

= 2470 kg

Total weight of 19 Men and 38 Women = Total weight of 19 men + Total weight of 38 Women

= 1387 + 2470 kg

64 | P a g e

= 3857 kg

The average weight of all the men and the women together

= Total weight of 19 men & 38 Women / Total number of persons

= 3857 / (19+38) kg

= 3857 /57 kg

= 67.666 kg

≈ 68 kg

Hence the answer is 68 kg

73. We solve both equations separately

Equation I:

p2 – 1 = 0

⇒ (p – 1) (p + 1) = 0

⇒ p = ± 1

Equation II:

�� + ��×��

��+ 3 = 0

⇒ q2 + 4q + 3 = 0

⇒ q2 + 3q + q + 3 = 0

⇒ q(q + 3) +1(q + 3) = 0

⇒ (q + 3) (q + 1) = 0

q = -1, -3

∴ p ≥ q

74. We will solve both the equations separately.

Equation I:

a2 – 11a + 24 = 0

65 | P a g e

⇒ a2 – 8a – 3a + 24 = 0

⇒ a(a – 8) – 3(a – 8) = 0

⇒ (a – 8) (a – 3) = 0

⇒ a = 3, 8

Equation II:

2b2 – 13b + 20 = 0

⇒ 2b2 – 8b – 5b + 20 = 0

⇒ 2b(b – 4) – 5(b – 4) = 0

⇒ (2b – 5) (b – 4) = 0

⇒ b = 5/2 = 2.5 or b = 4

Comparing the values of a and b, we get,

One value of a is greater than b and other is smaller, same is true for b too. So, relation cannot be

determined.

75. We will solve both the equations separately.

Equation I:

16x2 + 20x + 6 = 0

⇒ 16x2 + 12x + 8x + 6 = 0

⇒ 8x (2x + 1) + 6(2x + 1) = 0

⇒ (8x + 6) (2x + 1) = 0

⇒ x = - 6/8 = - 0.75 or x = -1/2 = - 0.5

Equation II:

10y2 + 38y + 24 = 0

⇒ 10y2 + 30y + 8y + 24 = 0

⇒ 10y(y + 3) + 8(y + 3) = 0

⇒ (10y + 8) (y + 3) = 0

66 | P a g e

⇒ y = - 8/10 = - 0.8 or y = - 3

Comparing the values of x and y, we get,

x > y

76. Number of Passed Candidates from Institute C = 1500

Number of Passed Candidates from Institute E = 2000

Total number of passed candidates from Institute C and E = 1500 + 2000

= 3500

Number of Appeared candidate from Institute A = 1500

Number of Appeared candidate from Institute F = 3500

Total Number of appeared candidate from Institute A and F = 1500 + 3500

= 5000

Number of candidates passed from institutes C and E together is percentage of the total number of

candidates appeared from institutes A and F together will be

=��

�� × 100

��

��× 100

= 70%

77. Number of Appeared candidate from Institute A = 1500

Number of Appeared candidate from Institute C = 2500

Number of Appeared candidate from Institute D = 1500


Total Number of Candidates appeared from Institute A, C, D and F = 1500 + 2500 + 1500 + 3500

= 9000

Number of Passed Candidates from Institute B = 1000


67 | P a g e

Number of Passed Candidates from Institute G = 3500

Total Number of Passed Candidates from B, E and G = 1000 + 2000 + 3500

= 6500

Difference between the number of candidates appeared from institutes A, C, D and F together and

candidates passed from institutes B, E and G together = 9000 – 6500

= 2500

78. Number of Appeared candidate from Institute B = 2000


Number of failed candidate from institute B = 2000-1000 = 1000

Number of Appeared candidate from Institute E = 3000

The respective ratio between the number of candidates who have failed from institute B and the

number of candidates who have appeared from institute E = 1000 : 3000

= 1: 3

79. Number of Appeared candidate from Institute A = 1500

Number of Passed Candidates from Institute A = 500

Difference between the appeared candidates and passed candidates = 1500 - 500

= 1000

Number of Appeared candidate from Institute B = 2000



= 1000

Number of Appeared candidate from Institute C = 2500

Number of Passed Candidates from Institute C = 1500


68 | P a g e

= 1000

Number of Appeared candidate from Institute D = 1500

Number of Passed Candidates from Institute D = 1000


= 500

Number of Appeared candidate from Institute E = 3000



= 1000


Number of Passed Candidates from Institute F = 2000


= 1500

Number of Appeared candidate from Institute G = 4000

Number of Passed Candidates from Institute G = 3500


= 500

The difference between the appeared candidates and passed candidates maximum is of Institute F and

the number is 1500.

80. Passed Candidates from Institute A = 500

Passed Candidates from Institute B = 1000

Passed Candidates from Institute C = 1500

Passed Candidates from Institute D = 1000

Passed Candidates from Institute E = 2000

Passed Candidates from Institute F = 2000

69 | P a g e

Passed Candidates from Institute G = 3500

Total Passed Candidates from All the institutes = 500 + 1000 + 1500 + 1000 + 2000 + 2000 + 3500

= 11500

Total Number of institutes = 7

Average Number of candidates passed from these institutes = Total Numbers of candidates passed /

total Number of Institutes

= 11500/7

= 1642.86 ≈ 1640

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