answer key for ibps po prelims 2017 model question paper€¦ · answer for the given blank is thus...

Answer Key for IBPS PO Prelims 2017

Model Question Paper 1. 3 2. 1 3. 4 4. 4 5. 1

6. 4 7. 3 8. 5 9. 5 10. 3

11. 1 12. 2 13. 4 14. 1 15. 3

16. 4 17. 2 18. 1 19. 4 20. 3

21. 3 22. 4 23. 4 24. 2 25. 1

26. 3 27. 1 28. 1 29. 4 30. 2

31. 2 32. 3 33. 5 34. 2 35. 4

36. 2 37. 4 38. 4 39. 2 40. 1

41. 5 42. 4 43. 3 44. 5 45. 4

46. 3 47. 4 48. 3 49. 5 50. 1

51. 1 52. 4 53. 4 54. 4 55. 3

56. 5 57. 4 58. 2 59. 1 60. 2

61. 5 62. 1 63. 3 64. 2 65. 4

66. 5 67. 5 68. 4 69. 3 70. 1

71. 2 72. 5 73. 3 74. 4 75. 1

76. 3 77. 3 78. 2 79. 3 80. 2

81. 2 82. 4 83. 5 84. 5 85. 3

86. 1 87. 2 88. 3 89. 4 90. 2

91. 5 92. 5 93. 4 94. 2 95. 5

96. 5 97. 1 98. 1 99. 4 100. 2

English Language

1. The entire passage talks about how the GST Council failed to reach a consensus on critical issues. It further says that the demonetization move has been a major cause which has led to this. It has been discussed in the passage

that due to demonetization, the production as well as the tax collection could be lower. Thus, option 2 conveys the main idea of the passage.

2. According to the first paragraph of the passage, 'While the effect of demonetization on the economy and on the states is outside the purview of the GST...'. Thus, option 5 is the best fit answer.

3. It has been stated in the first paragraph of the passage, 'With the pending legislations, including the draft model GST law, as well as the contentious issue of dual administrative control over assesses between the Centre and the States remaining unresolved, the GST Council would now meet on December 11 and 12.' This indicates that option 4 is the best fit answer.

4. The second paragraph of the passage talks about the opinions of the Finance Ministers of the States on how the demonetization move would affect the GST. From the statements of the ministers, it can be inferred that they are not very happy with the move as it would have several negative consequences which includes: a fall in production... leading to cash crunch; a fall in production; and difficulty in paying salaries of govt employees. Thus, option 4 is correct in the given context. None of the other options are either given or can be inferred from the given passage.

5. The given passage gives a report on the the GST Council that failed to reach a consensus on critical issues and how is it related to the demonetization move. The passage is a factual account and reports events as well as statements and opinions of several authoritative figures concerned with the matter. Thus, option 1 is the best fit answer and all other options are irrelevant.

6. According to the first paragraph of the given passage, some of the states, suggested an increase in the Fiscal Responsibility and Budget Management targets while others called for a relaxation of the ways and means limits. So option 4 is the correct answer. Option 1 does not give a complete answer and can be eliminated.

7. According to the second paragraph of the give passage, Kerala Finance Minister Thomas Isaac said that demonetization would result in a fall in production worth Rs.2.5 lakh crore, adding that the cash crunch would also impact state revenue. Thus, option 3 is the best fit answer.

8. According to the passage,“Tax collection can be 50 per cent lower this month as there is no business in Delhi markets,” Manish Sisodia, Deputy Chief

Minister of Delhi tweeted on Saturday. “If so, it would be hard to pay salaries of govt employees… same concern is being expressed by many state FMs in the GST council in the presence of Finance Minister.” So option 5 is the right answer.

9. Options 1, 2 and 3 all means to elude or escape someone or something. But in context of the given sentence we need a word that would show that the West African region could not avoid the outbreak of Ebola virus. Option 4 means to be irritated and is out of context here. Option 5 means to come face to face with an hostile problem which forces one to take actions against it. Hence it is the best fit answer for the given blank.

10. The given blank needs a word that would define the efforts that have been taken so far as the word 'challenges' has been mentioned. It is clear from the sentence that the various agencies are making quite an effort but various reasons are interfering in the processes. Thus options 1 and 2 are negated as they would imply that the agencies are not making any effort. Option 4 cannot be used here as it would imply that the various agencies are competing against one another to curb Ebola. Option 5 would imply that the actions taken are without purpose or motive. Option 3 is the best fit answer as it means striving to achieve or attain something in the face of difficulty or resistance and that is what the agencies are exactly trying to do.

11. The given blank needs a word that would signify how the EVD outbreak has been restrained at the given point of time withing a specific zone. The best fit answer for the given blank is thus option 1 which means control or restrain.

12. For the given blank we need a word that would explain the lowering of agricultural activities through outbreak of diseases. Thus option 2 which means to make smaller or lesser in amount is the best fit answer for the given blank.

13. We have already established that countries with diseases would see a lowering in their agricultural output. From the paragraph it is evident that these countries are some of the poorest countries in the world which often suffer from famine. Thus they as it is do not have adequate food security and it is a vicious cycle which consistently recurs or makes an occurrence. Thus the best fit word for the given blank is option 4 which means 'long lasting'.

14. Public health is a matter of concern for all countries and their governments as a good healthy public would contribute more to the country's growth. Hence they need to be prepared for serious or unexpected attacks on it. Hence option 1 is the best fit answer for the given blank which means a serious, unexpected, and often dangerous situation requiring immediate action.

15. Option 5 means to give out secrets and option 2 has already been used in the sentence with the blank, making them redundant in the given context. Option 4 means scattered and is not linked to the given context. Option 1 means to make place for something extra. Option 3 means united or with various parts or aspects linked or coordinated which is the exact approach needed for the governments to successfully tackle problems. Hence option 3 is the correct answer as it is in conjunction with the word 'holisticapproach' .

16. There are certain nouns which are plural like ‘glasses’, 'scissors' that take a plural verb. Instead of ‘it’ in the sentence it should be ‘they’. Thus the error is in the fourth part of the sentence.

17. The error is in option 2 as the subject say 'Neither' meaning 'no one' which is singular so verb has to be singular 'is' to agree with it.

18. The error lies in part 1 of the sentence where the preposition has been used incorrectly. With the verb 'aim' 'to' is a wrong preposition. We should always use the preposition 'at' after 'aim'. Thus the sentence should read: "She aims at doing well...". Thus option 1 is the correct answer.

19. Since 'cattle' signifies a plural group, it has no separate plural form.The general rule of adding –s at the end of the word is not applicable always. Thus the error is in the fourth part of the sentence.

20. The error is in option 3, Whenever we use connectors like 'as well as, together with, along with etc.' the verb follows the first subject. Here, the first subject is singular-- 'boy'. Hence to agree with it the verb too should be singular hence it should be 'shops' not 'shop.

21. The error lies is the usage of the word ‘which’ in the sentence. The correct word should be ‘who’. The term ‘which’ is generally used to refer to nonliving things. Here, the pronoun is being used for people, hence, ‘who’ should be used instead.

22. ‘Each other’ is used when there are two subjects. E.g. Jack and Jill loved each other. But if there are many subjects then ‘one another’ is used. Hence, the error is in part 4.

23. The error lies in part 4 of the sentence as the word ‘aplomb’ which means ‘great self-confidence in demanding situations’ has been used incorrectly here. It should be used followed by the preposition ‘with’ instead of ‘in’. Thus option 4 is the correct answer.

24. Here, the use of the word ‘comparatively’ suggests that there is a comparison made. In this case the use of ‘comparative degree’ will make it superfluous.

Thus, ‘fine’ must be used instead of ‘finer’. Hence, the error is in option 2.

25. The given sentence is okay, but it is what we know as "dangling modifier’. A dangling modifier occurs when the phrase makes no logical sense with the subject it seems to be modifying.

E.g. Being extremely hot, the glass-blower must be very careful when handling the molten glass.

It is not clear in this sentence that what/who is ‘hot’, ‘the glass-blower’or ‘the molten glass.’ As it is logical to consider ‘the molten glass’ to be hot we can correct the sentence as:

It being extremely hot, the glass-blower must be very careful when handling the molten glass.

In the given question we cannot clearly conclude what is rainy, as the subject is missing and grammatically the subject is assumed to be ‘the child’. Thus, ’it’ must precede ‘being a rainy day..’ to make it grammatically correct 'It being a rainy day the child could not go out'.The error lies in option 1.

26. D should be the first sentence as it introduces the topic of food wastage in an ironical way. F should follow as it states the quantitative value of food wastage. The third sentence B provides monetary implications of the same. The fourth sentence is A because it states the implications of the food wastage. E and C follow in the same order as they provide solutions to the problem of food wastage. E is a generic solution whereas C mentions specific measures needed to be implemented by the government.

The correct sequence is DFBAEC.

The second sentence is F.



The third sentence is B.



The fourth sentence is A.



The fifth sentence is E.



The last sentence is C.

Reasoning Ability

31. The information given can be represented as follows,

Hence code ‘fu’ is for the word ‘animal’ in the given code. 32. The information given can be represented as follows,

Hence code ‘pa’ is for the word ‘largest’ in the given code.


Thus code for ‘live’ is either ‘ru’ or ‘ka’.


Here code for ‘pacific’ is ‘la’;

Code for ‘ocean’ is ‘za’;

Code for ‘is’ is either ‘mu’ or ‘si’

And code for ‘beautiful’ is some new code not present in the given

information.

Thus the only possible code can be ‘tp mu la za’ where ‘tp’ represents

‘beautiful’.


Hence code ‘ru’ is either for word ‘live’ or for word ‘in’.

36. Seven family members: P, U, Q, T, C, G and R

1) U is mother of Q and likes classics “Gone with the wind”.

2) U has 2 children, a girl and a boy.

3) Q likes to watch “Star Wars” like her husband.

That means Q is a female.

4) T is son in law of U.

5) P is married to R and likes the movie “Transformer”.

6) C likes same movie as his father R.

That means C is a boy.

7) There are two couples, each having a child.

8) U is a grandmother to G.

9) C and G are first cousins.

10) G is R’s niece and both like to watch “Battlestar Galactica”.

That means G is a girl and all G, R and C likes to watch “Battlestar Galactica”.

So we get the final family tree as follows,

Thus C likes to watch “Battlestar Galactica”.







Thus Q and R are children of U.
















Thus P is U’s daughter in law.

39. Given statements: A > B ≥ C ≥ D; M > E; D > E = F ≥ G; H = G ≥ I

On combining: A > B ≥ C ≥ D > E = F ≥ G = H ≥ I; M > E

Conclusions:

I. F ≥ I → True (as F ≥ G = H ≥ I → F ≥ I)

II. M > G → True (as E = F ≥ G and M > E → M > E = F ≥ G → M > G)

III. C ≥ F → False (as C ≥ D > E = F → C > F)

Therefore, only conclusion I and II follows.

40. Directions: In the following question assuming the given statements

to be true, find which of the conclusion among given three conclusions is

/are definitely true and then give your answers accordingly.

Statements:

Q = P ≥ R; T < U; A = T; A > E; T ≥ S > R

Conclusions:

I. R ≤ U

II. S = E

III. R ≤ A

41. Given statements: M ≤ N ≤ O < P ≥ Q; R = S; P < R; U ≤ M; T < M

On combining: M ≤ N ≤ O < P ≥ Q; P < R = S; U ≤ M > T

Conclusions:

I. M < P → True (as M ≤ N ≤ O < P → M < P)

II. S > Q → True (as P ≥ Q and P < R = S → S = R > P ≥ Q → S > Q)

III. T < O → True (as M ≤ N ≤ O and M > T → T < M ≤ N ≤ O → T < O)

Therefore, all the conclusions follow.

42. According to the question, we have 7 individuals – Alan, Barton, Dana,

Kyle, Peter, Remus and Sarah.

Considering the information provided in the question:

1. Sarah weights more than Barton and Alan. Hence Sarah > (Alan, Barton). 2. Remus weights less than only Kyle. Hence Kyle is the heaviest

followed by Remus at second place. Kyle > Remus > Sarah > (Alan, Barton)

3. Peter weights as much as Sarah but less than Dana. Hence Dana is heavier than everyone except Kyle and Remus. Kyle > Remus > Dana > Peter = Sarah > (Alan, Barton)

4. Alan does not weigh the minimum. Hence Alan is heavier than Barton. Kyle > Remus > Dana > Peter = Sarah > Alan > Barton

So the final arrangement is: Kyle > Remus > Dana > Peter = Sarah > Alan > Barton Hence we can conclude that Dana is the 3rd heaviest individual.

43. According to the question, we have 7 individuals – Alan, Barton, Dana,

Kyle, Peter, Remus and Sarah.

Considering the information provided in the question: 1. Sarah weights more than Barton and Alan. Hence Sarah > (Alan, Barton).

2. Remus weights less than only Kyle. Hence Kyle is the heaviest followed by Remus at second place. Kyle > Remus > Sarah > (Alan, Barton)

3. Peter weights as much as Sarah but less than Dana. Hence Dana is heavier than everyone except Kyle and Remus. Kyle > Remus > Dana > Peter = Sarah > (Alan, Barton)

4. Alan does not weigh the minimum. Hence Alan is heavier than Barton. Kyle > Remus > Dana > Peter = Sarah > Alan > Barton

So the final arrangement is: Kyle > Remus > Dana > Peter = Sarah > Alan > Barton Hence we can conclude that: Only one person is heavier than Remus Peter weighs more than Barton No one weighs less than Barton No one is heavier than Kyle Hence the answer is ‘No one weighs less than Barton’.

44. Given statements: U ≤ V; W < V; W > X; X = Y; Y ≥ Z; S ≥ V; A < U; W

≥ B

On combining: A < U ≤ V > W > X = Y ≥ Z; S ≥ V; W ≥ B

Conclusions:

I. A ≤ S → False (as A < U ≤ V and S ≥ V → A < U ≤ V ≤ S → A < S)

II. Y > B → False (as W > X = Y and W ≥ B → B ≤ W > X = Y → thus clear relation between Y and B cannot be determined)

III. Z < V→ True (as V > W > X = Y ≥ Z → V > Z)

Therefore, only conclusion III follows.

45. Given statements: A = O; A > X; A < R; T > R; X ≥ Y; O ≥ P

On combining: P ≤ O = A < R < T; A > X ≥ Y

Conclusions:

I. R > X → True (as A < R and A > X → X < A < R → X < R)

II. P ≤ T → False (as P ≤ O = A < R < T → P < T)

III. Y < O → True (as O = A and A > X ≥ Y → O > X ≥ Y → O > Y)

Therefore, only conclusion I and III follows.

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

Conclusions:

I. Some baskets are mugs → It’s possible but not definite, hence false.

II. Some buckets are not bowls → Clearly true as buckets which are mugs cannot be bowls.

III. No baskets are cups → It’s possible but not definite, hence false.

IV. Some baskets are bowls → It’s possible but not definite, hence false.

Hence only conclusion II follows.


Conclusions:

I. No paper is a magazine → It’s possible but not definite, hence false.

II. Some books are not magazines → Clearly true as books which are brown cannot be magazines.

III. Some boxes are books → It’s possible but not definite, hence false.

IV. All magazines being boxes is a possibility → Possibility is true.

Hence only conclusion II and IV follow.


Conclusions:

I. Some woods are glass → Clearly true.

II. Some statues are glass → It’s possible but not definite, hence false.

III. Some rocks are not figures→ It’s possible but not definite, hence false.

Hence only conclusion I follows.


Conclusions:

I. Some machines are smart → Clearly true.

II. Some machines are not smart→ It’s possible but not definite, hence false.

III. Some programs are sentient→ It’s possible but not definite, hence false.

IV. No machine is a program → It’s possible but not definite, hence false.

Hence only conclusion I follows.


Conclusions:

I. Some Diesel are programs → It’s possible but not definite, hence false.

II. All diesels are oils→ It’s possible but not definite, hence false.

III. Some diesels are oils→ It’s possible but not definite, hence false.

IV. Some fluids are program→ It’s possible but not definite, hence false.

Hence none of the given options follows.

51. Seven people: D, E, F, G, H, J and K

Anniversary in months: February, March, April, June, September, October and November.

Colour: Blue, Red, Green, Yellow, White, Orange and Purple.

1) J has an anniversary in the month which has more than 30 days.

Thus J may have anniversary either in March or in October.

2) Only one person has an anniversary between J and the one who likes Blue.

Case 1

Month People Colour

February

March J

April

June Blue

September

October

November

Case 2

Month People Colour

February

March

April

June Blue

September

October J

November

3) Both K and F have an anniversary in one of the months after the one who likes Blue.

4) K has an anniversary immediately before F.

Thus case 2 gets eliminated and case 1 gets 2 sub cases.

Case 1a

Month People Colour

February

March J

April

June Blue

September K

October F

November

Case 1b

Month People Colour

February

March J

April

June Blue

September

October K

November F

5) The one who likes Green has an anniversary in the month which has less than 30 days.

The only possible month is February.

6) Only three people have an anniversary between the one who likes Green and the one who likes Purple.

7) Only two people have an anniversary between K and the one who likes Yellow.

Case 1a

Month People Colour

February Green

March J Yellow

April

June Blue

September K Purple

October F

November

Case 1b

Month People Colour

February Green

March J

April Yellow

June Blue

September Purple

October K

November F

8) G has an anniversary immediately after the one who likes Yellow.

9) Only two people have an anniversary between G and H.

Thus case 1a gets eliminated.

Case 1b

Month People Colour

February H Green

March J

April Yellow

June G Blue

September Purple

October K

November F

10) D has an anniversary immediately before the one who likes Red.

The only month possible for D is September.

11) F does not like Orange.

Thus J likes Orange and F likes White.

Case 1b

Month People Colour

February H Green

March J Orange

April E Yellow

June G Blue

September D Purple

October K Red

November F White

Thus K has an anniversary in month October.







Case 1

Month People Colour

February

March J

April

June Blue

September

October

November

Case 2

Month People Colour

February

March

April

June Blue

September

October J

November




Case 1a

Month People Colour

February

March J

April

June Blue

September K

October F

November

Case 1b

Month People Colour

February

March J

April

June Blue

September

October K

November F





Case 1a

Month People Colour

February Green

March J Yellow

April

June Blue

September K Purple

October F

November

Case 1b

Month People Colour

February Green

March J

April Yellow

June Blue

September Purple

October K

November F




Case 1b

Month People Colour

February H Green

March J

April Yellow

June G Blue

September Purple

October K

November F





Case 1b

Month People Colour

February H Green

March J Orange

April E Yellow

June G Blue

September D Purple

October K Red

November F White

Thus F likes white color.







Case 1

Month People Colour

February

March J

April

June Blue

September

October

November

Case 2

Month People Colour

February

March

April

June Blue

September

October J

November




Case 1a

Month People Colour

February

March J

April

June Blue

September K

October F

November

Case 1b

Month People Colour

February

March J

April

June Blue

September

October K

November F





Case 1a

Month People Colour

February Green

March J Yellow

April

June Blue

September K Purple

October F

November

Case 1b

Month People Colour

February Green

March J

April Yellow

June Blue

September Purple

October K

November F




Case 1b

Month People Colour

February H Green

March J

April Yellow

June G Blue

September Purple

October K

November F





Case 1b

Month People Colour

February H Green

March J Orange

April E Yellow

June G Blue

September D Purple

October K Red

November F White

Here the pattern is

Green is related to April (month of marriage of second person to marry after one who likes green);

Yellow is related to September (month of marriage of second person to marry after one who likes yellow)

Thus Purple must be related to November.







Case 1

Month People Colour

February

March J

April

June Blue

September

October

November

Case 2

Month People Colour

February

March

April

June Blue

September

October J

November




Case 1a

Month People Colour

February

March J

April

June Blue

September K

October F

November

Case 1b

Month People Colour

February

March J

April

June Blue

September

October K

November F





Case 1a

Month People Colour

February Green

March J Yellow

April

June Blue

September K Purple

October F

November

Case 1b

Month People Colour

February Green

March J

April Yellow

June Blue

September Purple

October K

November F




Case 1b

Month People Colour

February H Green

March J

April Yellow

June G Blue

September Purple

October K

November F





Case 1b

Month People Colour

February H Green

March J Orange

April E Yellow

June G Blue

September D Purple

October K Red

November F White

Thus E and F have their anniversary in months April and November

respectively.







Case 1

Month People Colour

February

March J

April

June Blue

September

October

November

Case 2

Month People Colour

February

March

April

June Blue

September

October J

November




Case 1a

Month People Colour

February

March J

April

June Blue

September K

October F

November

Case 1b

Month People Colour

February

March J

April

June Blue

September

October K

November F





Case 1a

Month People Colour

February Green

March J Yellow

April

June Blue

September K Purple

October F

November

Case 1b

Month People Colour

February Green

March J

April Yellow

June Blue

September Purple

October K

November F




Case 1b

Month People Colour

February H Green

March J

April Yellow

June G Blue

September Purple

October K

November F





Case 1b

Month People Colour

February H Green

March J Orange

April E Yellow

June G Blue

September D Purple

October K Red

November F White

Thus three people have an anniversary between the months in which H and D

have an anniversary.

56. Eight family members: L, M, N, O, P, Q, R and S.

1) P sits second to the right of the M.

2) M faces the center.

3) Only two people sit between P and N.

4) N is the daughter of L.

5) No female is an immediate neighbour of N.

6) S is not an immediate neighbour of M.

7) S is the wife of Q.

8) Q sits third to the right of S.

9) Both the immediate neighbours of N face opposite directions.

10) N’s brother sits to her immediate right.

11) Neither M nor Q is the brother of N.


12) L’s wife sits to the immediate right of P.

13) L sits second to the left of his wife.

14) O is neither an immediate neighbour of S nor P.

Thus there is no place left for O in case 1, hence it gets eliminated.

Thus R is L’s wife and O is N’s brother.

15) Only three people sit between L and his brother.

Thus Q is L’s brother.

16) N’s husband sits to the immediate right of R.

Thus P is N’s husband.

17) P and Q face a direction opposite to that of L.

Thus we get the final arrangement.

Thus only one person i.e. N sits between M and R’s son i.e. O when counted

from the left of M.







If M is the father of R, then L is M’s son in law and the position of M with

respect to L is third to the left.







Thus the only correct statement regarding R is ‘R sits second to the right of Q’.







Thus O sits exactly between N and L when counted from the left of L.







Thus P’s father-in-law i.e. L sits third to the right of P.

61. In Row 1: P, Q, R and S (facing south)

In Row 2: A, B, C, and D (facing north)

Places: Haryana, Bengaluru, Kolkata, Delhi, Jaipur, Mumbai, Pune and Patna

1) The person from Haryana is second to the right of the person who faces C.

2) The immediate neighbour of person from Haryana faces the person from

Kolkata.

So we get the following 2 cases.

3) Only one person sits between person from Kolkata and B.

4) One who faces B is second to the left of P.

5) One who is immediate neighbour of B faces person from Bengaluru.

6) A person facing the person from Bengaluru is second to the right of the

person from Delhi.

7) There is only one person between the person from Bengaluru and S.

8) R faces the person from Patna.

9) R is neither from Bengaluru nor Jaipur.

Thus P is from Jaipur.

10) The person who is immediate neighbour of D is from Pune.

11) D is not an immediate neighbour of B.

Thus case 1 gets eliminated.

Thus pair RS is seated at the extreme ends in one of the two rows.






Kolkata.






person from Delhi.








The only incorrect statement is ‘R is at the extreme right end of the row’ as R is

at the extreme left end of the row.






Kolkata.






person from Delhi.


Here C is the odd one as except C all are seating at extreme ends.






Kolkata.






person from Delhi.








Thus the only incorrect pair is R – Haryana.






Kolkata.






person from Delhi.


Thus A belongs to Delhi.

Quantitative Aptitude

66. Laws of Indices:-

1-: am × an = a{m+n}

2-: am ÷ an = a{m-n}

3-: [(am )n] = amn

4-: (a)(1/m) = m√a

5-: (a)(-m) =1/am

6-: (a)(m/n) = n√am

7-: (a)0 = 1

Now, the given expression:

(4 × 4)3 ÷ (4)5 × (2 × 8)2 = (4)?

⇒ (22 × 22)3 ÷ (22)5 × (2 × 24)2 = (16)?

⇒ (2)12 ÷ (2)10 × (2)8 = (2)? × 4

⇒(2)?= 210 ÷ 4

⇒ ? = 2.5

67. 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,

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,

√529 ÷ 46 × 6.4 + (8)3 – 252 = ?

⇒ 23 ÷ 46 × 6.4 + (8)3 – 252 = ?

⇒ ? = 3.2 + 512 – 252

⇒ ? = 263.2

68. 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,

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,

14% of 250+ ?% of 300 = 125

⇒35 + ?% of 300 = 125

⇒ 3× ? = 90

⇒ ? = 30

69. 150.08% of 1168.03 ÷ 7.96 + 18.956 = 29.231 ×?

Here, 150.08 ≈ 150; 1168.03 ≈ 1168; 7.96 ≈ 8; 18.956 ≈ 19; 29.231 ≈ 29

Taking the approximate values we have,

150% of 1168 ÷ 8 + 19 = 29 × ?

⇒ 1168 × 1.5 ÷ 8 + 19 = 29 × ?

⇒ 219 + 19 = 29 × ?

⇒ 29 × ? = 238

⇒? = 238/29 = 8.20 ≈ 8

Hence, the value is 8.

70. 179.86 × 18.01 - ?2 = 2984

Here, 179.86 ≈ 180 and 18.01 ≈ 18

Taking the approximate values, we have

⇒ 180 × 18 - ?2 = 2984

⇒ 3240 - ?2 = 2984

⇒ ?2 = 3240 – 2984

⇒ ?2 = 256

⇒ ? = 16

71. From the table,

The total number of employees in Scale IV = 45 + 125 + 155 + 65 + 35 + 55 = 480

The number of employees are deployed in ‘operations’ department in Scale IV = 125

The required percentage =125

480× 100 = 26.04

∴ 26.04% of Scale IV officers are deployed in ‘operations’ department.

72. From the table,

Total number of employees in ‘Personnel’ department = 220 + 125 + 85 + 45 + 30 + 8 = 513

In Personnel department, the number of employees in Scale II = 125


513× 100 = 24.36

∴ Out of the total number of employees in ‘Personnel’ department, approximately

24.36% employees are in Scale II.

73. From the table,

In public Relations department, the number of employees in Scale II = 155

In public Relations department, the number of employees in Scale I = 180

So, in public Relations department, the number of employees in Scale II is (180 – 155) = 25 less than that in scale I.


180× 100 = 13.89

∴ In public Relations department, the number of employees in Scale II is less than

that in Scale I by 13.89%.

74. From the table,

Total number of employees in Scale VI = 8 + 15 + 30 + 20 + 10 + 8 = 91

Total number of employees in Scale I = 220 + 625 + 650 + 350 + 320 + 180 = 2345


2345× 100 = 3.88

∴ Total number of employees in Scale VI is 3.88% of the total number of employees

in Scale I.

75. From the table,

The total number of employees in Scale III = 85 + 280 + 225 + 150 + 100 + 120 = 960

The total number of employees in Scale IV = 45 + 125 + 155 + 65 + 35 + 55 = 480

∴ The ratio between the total number of employees in Scale III and Scale IV

respectively = 960:480 = 2:1

76. Allan invested three fourth of the money that Matt invested and for twice the time as Matt.

Suppose Matt invested Rs. T for N years. So, Allan invested Rs. 3T/4 for 2N years.

⇒ Ratio of amounts invested multiplied by time for Matt and Allan = (TN) : (6TN/4) = 2 : 3

Let profit earned by Matt be Rs. P and that by Allan be Rs. (P + 300).

We know, ratio of amounts invested multiplied by time is same as ratio of profits.

⇒ P/(P + 300) = 2/3

⇒ 3P = 2P + 600

⇒ P = 600

∴ Total profit = P + P + 300 = Rs. 1500

77. Let the age of Nancy be N years.

⇒ Age of Nancy’s brother = (N + 5) years

The sum of their ages is three fourth of their father’s age.

⇒ Father’s age is 4/3 of sum of ages of Nancy and Nancy’s brother.

⇒ Father’s age = (4/3)(N + N + 5) years = (4/3)(2N + 5) years

After 4 years, father’s age will be twice the age of Nancy’s brother.

⇒ [(4/3)(2N + 5) + 4] = 2(N + 5 + 4)

⇒ 8N + 20 + 12 = 6N + 54

⇒ N = 22/2 = 11

∴ Age of Nancy 7 years from now = (11 + 7) years = 18 years

78. Let the speed of the 210 m long train = x km/h

As the trains are travelling in opposite directions,

Relative speed of the first train w.r.t the second train = Speed of train 1 + Speed of train 2

⇒ Relative speed = (70 + x)km/h = (70 + x) × 5/18 m/s ---(1)

Thus,

Distance covered by train 1 = Length of Train 1 + Length of train 2

⇒ Distance covered by train 1 = 140 + 210 = 350 m

Time taken = 7 seconds

∴ Relative speed × (Time taken) = Total distance covered

⇒ (70 + x) × (5/18) × 7 = 350

⇒ (70 + x) × (5/18) = 50

⇒ (70 + x)/18 = 10

⇒ x = 110 km/h

Thus,

The speed of the second train is 110 km/h

79. The pattern of given series is:

→ 24,

→ 50 = (24 × 2) + 2,

→ 98 = (50 × 2) – 2,

→ 198 = (98 × 2) + 2,

→ 394 = (198 × 2) – 2,

→ ? = (394 × 2) + 2,

→ ? = 790


→ 100,

→ 49 = (100 × 0.5) – 1,

→ 47 = (49 × 1) – 2,

→ 67.5 = (47 × 1.5) – 3,

→ 131= (67.5 × 2) – 4,

→ ? = (131 × 2.5) – 5,

→ ? = 322.5


→ 33,

→ 43 = 33 + 10,

→ 65 = 43 + (10 + 12),

→ 99 = 65 + (10 + 24),

→ 145 = 99 + (10 + 36),

→ ? = 145 + (10 + 48),

→ ? = 203


→ 68 – 19 = 49 (7 x 7)

→ 102 – 68 = 34 (*36 = 6 x 6 )

→ 129 – 102 = 27 (*25 = 5 x 5)

→ 145 - 129 = 16 (4 x 4)

→ 154 – 145 = 9 (3 x 3)

If we replace 102 by 104, then 104 – 68 = 36 & 129 – 104 = 25.

So, the wrong number is 102.


→ 6 = 1 × 5 + 1

→ 26 = 6 × 4 + 2

→ 81 = 26 × 3 + 3

→ 166 = 81 × 2 + 4

→ 171 = 166 × 1 + 5

Hence, 172 is wrong in the series.

84. Let’s assume that x litres of water is mixed with y litres of soda.

Cost price of soda = Rs. 13/Litre

Cost of y litre soda = Rs. 13y

Total quantity of the mixture = (x + y) litre

The cost price of the mixture = Rs. 13y [∵ Water is assumed available free of cost]

The selling price of (x + y) litres of mixture = Rs. 14.40 × (x + y)

The cost price of the mixture =SellingPrice×100

100+Profit%= 13𝑦

⇒(14.40) × (x + y) × 100

100 + 20= 13y

⇒ 1440x + 1440y = 13y × 120

⇒ 1440x = 1560y – 1440y

⇒ 1440x = 120y

⇒ x/y = 1/12

∴ Water and soda should be mixed in 1 : 12 ratio.

85. Let’s assume the cost price of first box to be Rs. x

And, the cost price of second box to be Rs. y

According to the given information,

⇒ x + y = 1035 -------Equation (1)

∵ first box is sold at 25% profit,

Selling price of first box = x + (25% of x) = 1.25x

∵ second box is sold at 18% loss,

Selling price of second box = y – (18% of y) = 0.82y

As per the given information, selling prices of both the boxes are same.

⇒ 1.25x = 0.82y

⇒ x = 0.82y/1.25

Substituting the value of x in Equation (1), we get,

⇒82𝑦

125+ 𝑦 = 1035

⇒207𝑦

125= 1035

⇒ 𝑦 =1035 × 125

207= 625

Substituting in Equation (1), we get,

x = 1035 – 625 = 410

∴ the selling prices of two boxes are Rs. 625 and Rs. 410 respectively.

86. We know that, average = Sum of all quantities/Number of quantities

Average age of family = Sum of ages of all family members/Number of members

∴ 16 = Sum of ages of 4 family members five years ago/4

⇒ Sum of ages of 4 family members 5 years ago = 4 × 16 = 64

In five years, each of these 4 members’ ages have increased by 5.

∴ Sum of present ages of 4 family members = Sum of ages of 4 family members 5 years ago + (5 × 4)

∴ Sum of present ages of 4 family members = 64 + 20 = 84

Let the age of the baby be x years.

∴ Sum of present ages of all family members including baby = 84 + x

According to the information given in the question, average present age of family members including baby, is 1 more than what it was 5 years ago excluding baby, i.e. 16 + 1 = 17

∴ Average = 17

⇒ 17 = (84 + x)/5

⇒ 84 + x = 17 × 5 = 85

⇒ x = 1

∴ the age of the baby is 1 year.

87. I. 16x2 + 20x + 6 = 0

⇒ 8x2 + 10x + 3 = 0 [Dividing both sides by 2 ]

⇒ 8x2 + 6x + 4x + 3 = 0

⇒ 2x(4x + 3) + 1(4x + 3) = 0

⇒ (4x + 3)(2x + 1) = 0

Then, x = - ¾ or x = - ½

II. 10y2 + 38y + 24 = 0

⇒ 5y2 + 19y + 12 = 0 [Dividing both sides by 2 ]

⇒ 5y2 + 15y + 4y + 12 = 0

⇒ 5y(y + 3) + 4(y + 3) = 0

⇒ (y + 3)(5y + 4) = 0

Then, y = - 3 or y = - 4/5

So, when x = - ¾ , x > y for y = - 3 and x > y for y = - 4/5

And when x = - ½ , x > y for y = - 3 and x > y for y = - 4/5

∴ We can clearly observe that x > y.

88. I. x2 + 11x + 30 = 0

⇒ x2 + 5x + 6x + 30 = 0

⇒ x(x + 5) + 6(x + 5) = 0

⇒ (x + 5)(x + 6) = 0

Then, x = - 5 or x = - 6

II. y2 + 7y + 12 = 0

⇒ y2 + 4y + 3y + 12 = 0

⇒ y(y + 4) + 3(y + 12) = 0

⇒ (y + 4)(y + 3) = 0

Then, y = - 4 or y = - 3

So, when x = - 5, x < y for y = - 4 and x < y for y = - 3

And when x = - 6, x < y for y = - 4 and x < y for y = - 3

∴ We can clearly see that x < y.

89. I. x2–16 = 0

⇒ x2 = 16

⇒ x = ± 4

Then, x = + 4 or x = - 4

II. y2- 9y + 20 = 0

⇒ y2 – 5y – 4y + 20 = 0

⇒ y(y – 5) – 4(y – 5) = 0

⇒ (y – 5)(y – 4) = 0

Then, y = + 5 or y = + 4

So, when x = + 4, x < y for y = + 5 and x = y for y = + 4

And when x = - 4, x < y for y = + 5 and x < y for y = + 4

∴ We can clearly observe that x ≤ y.

90. I. x - √121= 0

⇒ x = √121 = 11

Then, x = + 11

II. y2 – 121 = 0

⇒ y2 = 121

⇒ y = ± 11

Then, y = + 11 or y = - 11

So, when x = + 11, x = y for y = + 11 and x > y for y = - 11

∴ So, we can observe that x ≥ y.

91. I. x2 – x – 12 = 0

⇒ x2 – 4x + 3x – 12 = 0

⇒ x(x – 4) + 3(x – 4) = 0

⇒ (x – 4)(x + 3) = 0

Then, x = + 4 or x = - 3

II. y2 + 5y + 6 = 0

⇒ y2 + 3y + 2y + 6 = 0

⇒ y(y + 3) + 2(y + 3) = 0

⇒ (y + 3)(y + 2) = 0

Then, y = - 3 or y = - 2

So, when x = + 4, x > y for y = - 3 and x > y for y = - 2

And when x = - 3, x = y for y = - 3 and x < y for y = - 2

∴ So, we can observe that no clear relationship cannot be determined between x and

y.

92. From the given data,

College A B C D E F Total

Number of students taking Commerce (in thousand)

40 25 17.5 35 37.5 30 185

Number of students taking Commerce from all six colleges (in thousands) = 185

Total number of colleges = 6

Average number of students taking Commerce (in thousands) = 185/6 = 30.83


Number of students taken Commerce in college E (in thousands) = 37.5

Number of students taken Science in college E (in thousands) = 27.5

Required ratio = 37.5 : 27.5 = 15 : 11


College A B C D E F Total

No. of students taking Arts (in thousands)

22.5 50 40 35 50 40 237.5


Stream Number of student in college B (in

thousands)

Arts 50

Science 45

Commerce 25

Total 120

Number of students taking Science in college B (in thousands) = 45

Total Number of students in college B (in thousands) = 120

Required % =45

120× 100 = 37.5%


Number of students taking Commerce in college B (in thousand) = 25

Number of students taking Commerce in college A (in thousand) = 40

Number of students taking Commerce in college C (in thousand) = 17.5

Total number of students taking Commerce in college A and C together (in thousand) = 17.5 + 40 = 57.5

Required number of students (in thousand) = 57.5 – 25 = 32.5

97. Aditya can do 50% more work than Radhika in the same time.

⇒ Aditya’s efficiency is 1.5 times of Radhika’s efficiency

Radhika alone do a piece of work in 30 hrs.

∴ Part of work done by Radhika in one hour = 1/30

∴ Part of work done by Aditya in one hour = (2/3) × (1/30) = 1/20

Working together, part of work finished by both in one hour =1

30+

1

20=

2+3

60=

1

12

∴ they’ll take 12 hours to finish the entire work.

⇒ In order to finish twice the original work, they’ll require 12 × 2 = 24 hours.

98. If rectangular sheet is folded along its length to from a right circular cylinder, the height of cylinder will be same as breadth of rectangle, and circumference of base will be same as length of rectangle.

If height of cylinder is H cm, and radius is R cm, then

H = breadth = 10 cm

And, 2πR = length = 44 cm

⇒ 2 × (22/7) × R = 44

⇒ R = 7 cm

∴ Volume of cylinder = πR2H = (22/7) × 7 × 7 × 10 = 1540 cubic cm

99. There are 3 vowels in word TESTBOOK, E, O and O. If there are vowels at first and last position, this is possible in 3 ways.

E at first and O at last, O at first and E at last, and O at both first and last.

In each of the three cases, the remaining six letters that have to come in between will be different except that T will be occur twice.

⇒ Remaining 6 letters can be arranged in 6!/2! Ways, i.e., 360 ways.

∴ Total number of ways in which this can be done = 3 × 360 = 1080

100. Simple Interest = (P × R × T)/100 Where, P = Principal, R = % rate of interest, T = time period in year

As per given information, Simple interest = 60% of principal

∴ 0.6P = (P × R × 4)/100

⇒ R = 15

Now let's find out the compound interest of Rs. 48,000 after 2 years at 10% P = Rs. 48,000 T = 2 years

R = 15%

Amount after 2 years = 𝑃(1 +𝑅

100)𝑇 = 48000(1 +

15

100)2 = 48000(

115

100)2 = 48000(

23

20)2

= 63480

Compound Interest = Rs.63480 - Rs.48000 = Rs.15480

Get the Question Paper to this Live Leak here:

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Detailed Syllabus - IBPS PO 2017 Prelims

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