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IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Bank Exams MCQ


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30 Questions MCQ Test Mock Tests for Banking Exams 2024 - IBPS RRB PO (Scale 1) Mains Mock Test - 4

IBPS RRB PO (Scale 1) Mains Mock Test - 4 for Bank Exams 2024 is part of Mock Tests for Banking Exams 2024 preparation. The IBPS RRB PO (Scale 1) Mains Mock Test - 4 questions and answers have been prepared according to the Bank Exams exam syllabus.The IBPS RRB PO (Scale 1) Mains Mock Test - 4 MCQs are made for Bank Exams 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for IBPS RRB PO (Scale 1) Mains Mock Test - 4 below.
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IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 1

Read the statement given below and the questions that follow. Considering all the details to be true, mark the suitable option as your answer

Statement: The students these days do not have the habit of reading and that is the reason their knowledge in many areas is not as profound as it used to be earlier as they depend on others to tell them the important information or events.

Course of Action:

I. Schools should promote reading as a habit in students right from their early classes

II. Parents must help the children come closer to reading by making their personal library at home

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 1
As per the condition, only I is the correct answer as it will help in addressing the problem that has been given. II can be eliminated as it is an extreme solution. Personal library for everyone is a bit unreasonable, so this can be ruled out.

Hence, option (b) is correct.

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 2

Read the statement given below and the questions that follow. Considering all the details to be true, mark the suitable option as your answer

Statements:

A. People rush to cancel weddings, demand refunds; wedding industry takes another hit

B. With yellow alert being sounded in Delhi amid rising COVID-19 cases, government allows only 20 guests at marriage events

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 2
Due to rise in Covid cases, the people rush to cancel their weddings and this is best explained in the order that goes like due to B→A is happening.

So, the best option is (c).

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IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 3

Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and Give an answer.

(A) If the data in statement I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question.

(B) If the data in statement II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question.

(C) If the data either in statement I alone or in statement II alone is sufficient to answer the question.

(D) If the data given in both statements I and II together are not sufficient to answer the question.

(E) If the data in both statements I and II together are necessary to answer the question.

Eight persons from P to W are sitting in a linear row facing north. Who sits second to the left of T?

Statement I: R sits fourth to the left of S, who is not sitting at any of the extreme ends. Two persons sit between R and V. As many persons sit to the left of S as to the right of U.

Statement II: T sits third to the right of P. Three persons sit between T and W, who sits at one of the extreme ends. W is sitting adjacent to Q.

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 3
If the data in both statements I and II together are necessary to answer the question.

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 4

Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and Give an answer.

(A) If the data in statement I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question.

(B) If the data in statement II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question.

(C) If the data either in statement I alone or in statement II alone is sufficient to answer the question.

(D) If the data given in both statements I and II together are not sufficient to answer the question.

(E) If the data in both statements I and II together are necessary to answer the question.

Eight persons from I to P are sitting around a square table in such a way that four persons sit at the corners and the remaining sit in the middle of the sides of the table, all are facing the centre. Who sits immediate right of P?

Statement I: L sits third to the right of K and sits on the corners. One person sits between L and J. P sits second to the right of J. I sits immediate right of N. M is not sitting adjacent to L.

Statement II: N and L face each other. Two persons sit between J and N. M sits third to the right of O, who sits adjacent to L. P sits immediate left of K.

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 4
If the data either in statement I alone or in statement II alone is sufficient to answer the question.

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 5

Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and Give an answer.

(A) If the data in statement I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question.

(B) If the data in statement II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question.

(C) If the data either in statement I alone or in statement II alone is sufficient to answer the question.

(D) If the data given in both statements I and II together are not sufficient to answer the question.

(E) If the data in both statements I and II together are necessary to answer the question.

Seven persons from A to G are living in a 7-storey building numbered 1 to 7 from bottom to top respectively. Who lives immediately above A?

Statement I: E lives three floors below A and immediately above G. The number of persons living below G is one less than the number of persons living above F. As many persons live between C and B as between B and D.

Statement II: Two floors are between D and C, who is living immediately below F. One person lives between F and G, who lives on an even numbered floor. As many persons live above E as below B, who lives above A.

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 5
If the data given in both statements I and II together are not sufficient to answer the question.

From the statements given, we can form two possibilities in each statement.

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 6

Study the following information carefully and answer the questions given below

If,

A%B (24m) means point B is 26m north of point A

A@B (38m) means point B is 35m south of point A

A#B (4m) means point B is 7m east of point A

A$B (25m) means point B is 23m west of point A

AB?C means C is the mid-point of A and B

P$M (30m); KR?C; T@U (15m); R%K (16m); C$F (12m); MP?T; U#R (14m)

If J@T (5m), then what is the shortest distance and direction of J with respect to M?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 6
Following diagram shows the final arrangements of the points given.

If T is 2m south of J then the shortest distance between J and M is =√142+√42=10√2.
IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 7

Study the following information carefully and answer the questions given below

If,

A%B (24m) means point B is 26m north of point A

A@B (38m) means point B is 35m south of point A

A#B (4m) means point B is 7m east of point A

A$B (25m) means point B is 23m west of point A

AB?C means C is the mid-point of A and B

P$M (30m); KR?C; T@U (15m); R%K (16m); C$F (12m); MP?T; U#R (14m)

Q. What is the distance between M and Q, if P#Q (5m)?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 7
Following diagram shows the final arrangements of the points given.

If Q is 8m east of P then the distance between M and Q is 36m.

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 8

Study the following information carefully and answer the questions given below

If,

A%B (24m) means point B is 26m north of point A

A@B (38m) means point B is 35m south of point A

A#B (4m) means point B is 7m east of point A

A$B (25m) means point B is 23m west of point A

AB?C means C is the mid-point of A and B

P$M (30m); KR?C; T@U (15m); R%K (16m); C$F (12m); MP?T; U#R (14m)

If T@D (6m), then the distance between U and D is the same as between __ and __?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 8
Following diagram shows the final arrangements of the points given.

If D is 3m south of T, then the distance between U and D is 9m which is same as between K and C.

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 9

If “1” is added to the odd positioned digits(from left) and “2” is subtracted from the even positioned digits(from left) of each number and all the digits are added within each number. Then which of the following number yield a resultant as a prime number?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 9
a) 36728415 ->44809223 -> 32

b) 27456982 -> 35537790 -> 39

c) 13628753 -> 21709561 -> 31

d) 74328519 -> 82409327 -> 35

e) 52837614 -> 60918422 -> 32

Option c forms a prime number after performing the operations.

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 10

If the vowels of the following words are changed to the second preceding letters and consonants are changed with its complementary pair as per English alphabets, and then all the letters in the words are arranged in alphabetical order, then which of the following will be the fifth letters from the right end of the newly formed words?

I) WORKING

II) FRIENDSHIP

III) ABSOLUTE

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 10
I) WORKING -> DMIPGMT -> DGIMMPT

II) FRIENDSHIP -> UIGCMWHSGK -> CGGHIKMSUW

III) ABSOLUTE -> YYHMOSGC ->CGHMOSYY

It is clear that I, M and K are the fifth letters from the right end of the newly formed words.

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 11

In the given questions, the relationship between different elements is shown in the statements followed by some conclusions. Find the conclusion which is logically follows.

Statements:

S > R ≥ G < H = E ≤ O; K > D ≤ G > Q > N = P

Conclusions:

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 11
a) K > D ≤ G < H; ->from the statements it is clear that H < K -> is False

b) S > R ≥ G > Q; ->from the statements it is clear that S < Q -> is False

c) N < Q < G ≤ R < S; ->from the statements it is clear that N < S -> is True

d) P = N < Q < G ≤ R; ->from the statements it is clear that P > R -> is False

e) D ≤ G < H = E ≤ O; ->from the statements it is clear that D ≤ O -> is False

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 12

In the given questions, the relationship between different elements is shown in the statements followed by some conclusions. Find the conclusion which is logically follows.

Statements:

D > W ≤ G = R < I ≤ T; C = N < J < R ≥ P < K

Conclusions:

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 12
a) K > P ≤ R < I ->from the statements it is clear that I > K -> is False

b) C = N < J < R = G ≥ W ->from the statements it is clear that W < C -> is False

c) W ≤ G = R < I ≤ T -> from the statements it is clear that T < W -> is False

d) G = R ≥ P ->from the statements it is clear that G ≥ P -> is True

e) W ≤ G = R > J ->from the statements it is clear that J > W -> is False

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 13

Study the following information carefully and answer the below questions

In a family of three-generation has eight members and two married couples. E is the only daughter of L. V is the only son of O. S has three kids. R has only one brother. L is the father-in-law of K’s father. V has only one sibling. H is the brother-in-law of O and vice versa. V is the grandchildren of S. No single parent has a child.

If R is married to A, then how A is related to H?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 13

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 14

Study the following information carefully and answer the below questions

In a family of three-generation has eight members and two married couples. E is the only daughter of L. V is the only son of O. S has three kids. R has only one brother. L is the father-in-law of K’s father. V has only one sibling. H is the brother-in-law of O and vice versa. V is the grandchildren of S. No single parent has a child.

How many female members are in the family?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 14

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 15

Study the following information carefully and answer the below questions

In a family of three-generation has eight members and two married couples. E is the only daughter of L. V is the only son of O. S has three kids. R has only one brother. L is the father-in-law of K’s father. V has only one sibling. H is the brother-in-law of O and vice versa. V is the grandchildren of S. No single parent has a child.

How R is related to K?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 15

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 16

Direction: Two statements are followed by two Conclusions I and II. You have to consider the statements to be true, even if they seem to be at variance from commonly known facts. You are to decide which of the given conclusions can definitely be drawn from the given statements and indicate your answer accordingly.

Statements:

1. The demand for clothes from non-Indian brands is increasing more rapidly than indigenous clothing brands.

2. The above mentioned non-Indian brands are looking for a way to reduce the cost of their products.

Conclusions:

I. The non-Indian clothing brands should manufacture their products in India.

II. A new product from indigenous clothing brands would reduce the current high demand for non-Indian clothing products.

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 16
The statements clearly state that non-Indian brands are trying to find a way to meet the demand of the market while keeping the cost low. Therefore, installing a manufacturing plant in India would increase the supply as well as reduce the retail price of the clothing products. The second conclusion does not completely comply with the above statements as the demand for products from indigenous brands is not as high as that of non-Indian brands. It may be an assumption but cannot be qualified as a conclusion of the above statements. So, only conclusion I follows.

Hence, the correct option is (A).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 17

Direction: The question below consists of a question and three statements numbered I, II and III given below it. You have to decide whether the data provided in the statements are sufficient to answer the question.

Q. What is the code for ‘right’ in the coded language?

I. ‘op pq te’ means ‘she is right’ and ‘pq qr sm’ means ‘he is sincere’.

II. ‘nm pu sf’ means ‘they are dancing’ and ‘pt sn pq’ means ‘sky is clear’.

III. ‘op fr pq’ means ‘she is dancing’ and ‘te nm ln’ means ‘they answered right’.

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 17
As per the data in the question,

I. ‘op pq te’ means ‘she is right’ and ‘pq qr sm’ means ‘he is sincere’.

pq = is,

Thus, Statement I alone is not sufficient to answer the question.

II. ‘nm pu sf’ means ‘they are dancing’ and ‘pt sn pq’ means ‘sky is clear’.

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

III. ‘op fr pq’ means ‘she is dancing’ and ‘te nm ln’ means ‘they answered right’.

Thus, Statement III alone is not sufficient to answer the question.

Now, let us check with Statement I and II,

I. ‘op pq te’ means ‘she is right’ and ‘pq qr sm’ means ‘he is sincere’.

II. ‘nm pu sf’ means ‘they are dancing’ and ‘pt sn pq’ means ‘sky is clear’.

pq = is

Thus, Statement I and II are not sufficient to answer the question.

Now, let us check with Statement II and III,

II. ‘nm pu sf’ means ‘they are dancing’ and ‘pt sn pq’ means ‘sky is clear’.

III. ‘op fr pq’ means ‘she is dancing’ and ‘te nm ln’ means ‘they answered right’.

nm = they, pq = is

Thus, Statement II and III are not sufficient to answer the question.

Now, let us check with Statement I and III,

I. ‘op pq te’ means ‘she is right’ and ‘pq qr sm’ means ‘he is sincere’.

III. ‘op fr pq’ means ‘she is dancing’ and ‘te nm ln’ means ‘they answered right’.

pq = is , op = she, te = right

Thus, Statement I and III are sufficient to answer the question.

So, data of the Statement I and III are sufficient to answer the question.

Hence, the correct option is (B).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 18

How many such pairs of letters are there in the word “AUTONOMOUS” each of which has as many letters between them in the word (in both forward and backward directions), as they have between them in the English alphabetical series?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 18
The given word

AUTONOMOUS

The given word can be represented as follows,

So, we can see that 7 such pairs are there.

Forward: NO, NS, OS

Backward: NO, OT, OU, TU

Hence, the correct option is (D).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 19

Direction: Study the following information carefully and answer the Question based on it:

Input: 962437 794528 465783 213945 753169 236185

Step 1: 734269 825497 647538 129354 961357 321658

Step 2: 162063 243524 21854 22720 54335 6640

Step 5 is the last step of rearrangement. As per the rules followed in the above steps, find out in each of the following questions the appropriate steps for the given input.

Input for the given Question,

Input: 576218 415893 642739 717340 931621 842163.

What is the Addition of 3rd element from left end in step 2 and 3rd element from right end in step 1?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 19
The Digits are arranged in such a way that,

In step 1 = if digit starts with odd number then it will be reverse and if it starts with even number then it will be reverse in pair of two.

Suppose in the given input it is 962437 starts with odd number i.e., 9. So in step 1 it will be reversed and we get 734269. In case of 465783 it will be 647538.

In step 2 = At first, the even digit numbers are arranged in descending order then the odd digit numbers are arranged in their respective order. After that multiplication of two-two digits of those are taken.

Ex – In step 1, the even digits numbers are = 825497 and 647538 now arranged them in descending order we get 825497 and 647538 so the multiplication form of 825497 is 8 × 2 = 16 5 × 4 = 20 and 9 × 7 = 63 i.e., 162063.

(This is the 1st number of step 2).

We get 162063 and 243524 from step 1(825497 647538 respectively).

Then the multiplications of odd numbers are arranged according their positional value.

21854 comes form 734269, 22720 from 129354, 54335 from 54335 and 6640 from 6640.

In step 3 = Difference of Addition of odd and even digits within the number.

Ex – 162063, 6 + 6 + 2 = 14, 3 + 1 = 4, 14 – 4 = 10.

In step 4 = Square of 1, 2, 3….so on and then add that Square in the numbers in Ascending order means Square of 1 is added in 1 and Square of 2 added in 4 and so on.

In step 5 = Addition of two numbers, than again add the digits of the number obtained.

Ex – 2 + 8 (1st and 2nd number of step 4) = 10 then 1 + 0 = 1 (1st number of step 5).

And 8 + 17 (2nd and 3rd number of step 4) = 25 = 2 + 5 = 7 (2nd number of step 5).

So, the solution of given input will be like this-

Input: 576218 415893 642739 717340 931621 842163

Step 1: 812675 148539 467293 43717 126139 481236

Step 2: 81235 32218 241427 1277 44027 2627

Step 5 is the last step of rearrangement.

3rd element from left end in step 2 is 241427 and 3rd element from right end in Step 1 is 43717.

241427 + 43717 = 285144.

Hence, the correct option is (B).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 20

Direction: Study the following information carefully and answer the Question based on it:

Input: 962437 794528 465783 213945 753169 236185

Step 1: 734269 825497 647538 129354 961357 321658

Step 2: 162063 243524 21854 22720 54335 6640

Step 5 is the last step of rearrangement. As per the rules followed in the above steps, find out in each of the following questions the appropriate steps for the given input.

Input for the given Question,

Input: 576218 415893 642739 717340 931621 842163.

What is the product of 4th letter from right end and 4th letter from left end in Step 4?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 20
The Digits are arranged in such a way that,

In step 1 = if digit starts with odd number then it will be reverse and if it starts with even number then it will be reverse in pair of two.

Suppose in the given input it is 962437 starts with odd number i.e., 9. So in step 1 it will be reversed and we get 734269. In case of 465783 it will be 647538.

In step 2 = At first, the even digit numbers are arranged in descending order then the odd digit numbers are arranged in their respective order. After that multiplication of two-two digits of those are taken.

Ex – In step 1, the even digits numbers are = 825497 and 647538 now arranged them in descending order we get 825497 and 647538 so the multiplication form of 825497 is 8 × 2 = 16 5 × 4 = 20 and 9 × 7 = 63 i.e., 162063.

(This is the 1st number of step 2).

We get 162063 and 243524 from step 1(825497 647538 respectively).

Then the multiplications of odd numbers are arranged according their positional value.

21854 comes form 734269, 22720 from 129354, 54335 from 54335 and 6640 from 6640.

In step 3 = Difference of Addition of odd and even digits within the number.

Ex – 162063, 6 + 6 + 2 = 14, 3 + 1 = 4, 14 – 4 = 10.

In step 4 = Square of 1, 2, 3….so on and then add that Square in the numbers in Ascending order means Square of 1 is added in 1 and Square of 2 added in 4 and so on.

In step 5 = Addition of two numbers, than again add the digits of the number obtained.

Ex – 2 + 8 (1st and 2nd number of step 4) = 10 then 1 + 0 = 1 (1st number of step 5).

And 8 + 17 (2nd and 3rd number of step 4) = 25 = 2 + 5 = 7 (2nd number of step 5).

So, the solution of given input will be like this-

Input: 576218 415893 642739 717340 931621 842163

Step 1: 812675 148539 467293 43717 126139 481236

Step 2: 81235 32218 241427 1277 44027 2627

Step 5 is the last step of rearrangement.

4th letter from left end is 20 and 4th letter from right end is 12 in Step 4.

12 × 20 = 240

Hence, the correct option is (A).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 21

Direction: Study the given information carefully and answer the following questions below.

The data given are in some coded format, throughout and proceed further

P#Q means P is to North of Q.

P¥Q means P is to South of Q.

P$Q means P is to West of Q.

P&Q means P is to East of Q.

P #@ Q means P is 8m North of Q.

P&µ Q means P is 4m East of Q.

Point K is to 3m South of point C.

P$µK; Q#µK; R$µQ; S¥@R; T&@S; U¥µT

If Z¥&S and Z is 4m away from the midpoint of T and U, then which of the following is possibly correct?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 21

P #& Z → P is North East to Z → False

S ¥ Z → S is South to Z → False

K #& Z → K is North East to Z → False

Z ¥ C → Z is South to C. → Possibly True

So, Z ¥ C is possibly true.

Hence, the correct option is (D).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 22

Direction: Study the given information carefully and answer the following questions below.

The data given are in some coded format, throughout and proceed further

P#Q means P is to North of Q.

P¥Q means P is to South of Q.

P$Q means P is to West of Q.

P&Q means P is to East of Q.

P #@ Q means P is 8m North of Q.

P&µ Q means P is 4m East of Q.

Point K is to 3m South of point C.

P$µK; Q#µK; R$µQ; S¥@R; T&@S; U¥µT

If J#&K and point J lies on the line TU and RQ, then which of the following will be true?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 22

I. C$@J → C is West to J with 8m → False

II. J#µP → J is North of P with 4m distance. → False

III. J & R → J is East of R → True

So, J & R will be True.

Hence, the correct option is (C).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 23

Direction: In a certain number system there are only two notations to represent numbers: # and %.

0 is represented by % and 1 by #. The subsequent numbers are represented in the following manner:

2 is represented as #%,

3 is represented as ##,

4 is represented as #%%,

5 is represented as #%# and so on.

Based on this coded language, answer the following questions.

What is the resultant of the following expression?

###%# ÷ #% × ##%% + ##% × #%#

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 23
According to the given information,

So, this entire code language is based on only 2 symbols % and #, i.e. 0 and 1.

Logic:

On observation, we notice that starting from the rightmost digit, each digit is multiplied with corresponding power of 2, starting with 20, and then all these terms are added to obtain the number in the decimal system.

[The rightmost term is multiplied with 20, the second term from right is multiplied with 21, the third term from right is multiplied with 22 and so on..]

2 → 10 ⇒ (0 × 20) + (1 × 21) = (0 × 1) + (1 × 2) = 0 + 2 = 2

3 → 11 ⇒ (1 × 20) + (1 × 21) = (1 × 1) + (1 × 2) = 1 + 2 = 3

4 → 100 ⇒ (0 × 20) + (0 × 21) + (1 × 22) = (0 × 1) + (0 × 2) + (1 × 4) = 0 + 0 + 4 = 4

5 → 101 ⇒ (1 × 20) + (0 × 21) + (1 × 22) = (1 × 1) + (0 × 2) + (1 × 4) = 1 + 0 + 4 = 5 and so on.

Based on this logic,

‘###%#’ ↔ 11101

11101 ⇒ (1 × 20) + (0 × 21) + (1 × 22) + (1 × 23) + (1 × 24)

= (1 × 1) + (0 × 2) + (1 × 4) + (1 × 8) + (1 + 16)

= 1 + 0 + 4 + 8 + 16 = 29

‘#%’ ↔ 10

10⇒ (0 × 20) + (1 × 21)

= (0 × 1) + (1 × 2)

= 0 + 2 = 2

'##%%' ↔ 1100

1100 ⇒ (0 × 20) + (0 × 21) + (1 × 22) + (1 × 23)

= 0 + 0 + 4 + 8 = 12

'##%' ↔ 110

110 ⇒ (0 × 20) + (1 × 21) + (1 × 22)

= 0 + 2 + 4 = 6

'#%#' ↔ 101

101 ⇒ (1 × 20) + (0 × 21) + (1 × 22)

= 1 + 0 + 2 = 5

The resultant of the expression is: 29 ÷ 2 × 12 + 6 × 5 = 204

Now, we need to convert 204 into the given code language.

204 = 128 + 64 + 8 + 4

∴ 204 = (0 × 20) + (0 × 21) + (1 × 22) + (1 × 23) + (0 × 24) + (0 × 25) + (1 × 26) + (1 × 27)

∴ 204 ⇒ 11001100

Converting 0s and 1s to %s and #s respectively, we get

204 = ##%%##%%

So, the resultant of the expression "###%# ÷ #% × ##%% + ##% × #%#" is ##%%##%%.

Hence, the correct option is (C).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 24

Direction: In a certain number system there are only two notations to represent numbers: # and %.

0 is represented by % and 1 by #. The subsequent numbers are represented in the following manner:

2 is represented as #%,

3 is represented as ##,

4 is represented as #%%,

5 is represented as #%# and so on.

Based on this coded language, answer the following questions.

What will be the square of "##%##"?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 24

According to the given information,

So, this entire code language is based on only 2 symbols % and #, i.e. 0 and 1.

Logic:

On observation, we notice that starting from the rightmost digit, each digit is multiplied with corresponding power of 2, starting with 20, and then all these terms are added to obtain the number in the decimal system.

[The rightmost term is multiplied with 20, the second term from right is multiplied with 21, the third term from right is multiplied with 22 and so on..]

2 → 10 ⇒ (0 × 20) + (1 × 21) = (0 × 1) + (1 × 2) = 0 + 2 = 2

3 → 11 ⇒ (1 × 20) + (1 × 21) = (1 × 1) + (1 × 2) = 1 + 2 = 3

4 → 100 ⇒ (0 × 20) + (0 × 21) + (1 × 22) = (0 × 1) + (0 × 2) + (1 × 4) = 0 + 0 + 4 = 4

5 → 101 ⇒ (1 × 20) + (0 × 21) + (1 × 22) = (1 × 1) + (0 × 2) + (1 × 4) = 1 + 0 + 4 = 5 and so on.

Based on this logic,

‘###%#’ ↔ 11101

11101 ⇒ (1 × 20) + (0 × 21) + (1 × 22) + (1 × 23) + (1 × 24)

= (1 × 1) + (0 × 2) + (1 × 4) + (1 × 8) + (1 + 16)

= 1 + 0 + 4 + 8 + 16 = 29

‘#%’ ↔ 10

10⇒ (0 × 20) + (1 × 21)

= (0 × 1) + (1 × 2)

= 0 + 2 = 2

'##%%' ↔ 1100

1100 ⇒ (0 × 20) + (0 × 21) + (1 × 22) + (1 × 23)

= 0 + 0 + 4 + 8 = 12

'##%' ↔ 110

110 ⇒ (0 × 20) + (1 × 21) + (1 × 22)

= 0 + 2 + 4 = 6

'#%#' ↔ 101

101 ⇒ (1 × 20) + (0 × 21) + (1 × 22)

= 1 + 0 + 2 = 5

The resultant of the expression is: 29 ÷ 2 × 12 + 6 × 5 = 204

Now, we need to convert 204 into the given code language.

204 = 128 + 64 + 8 + 4

∴ 204 = (0 × 20) + (0 × 21) + (1 × 22) + (1 × 23) + (0 × 24) + (0 × 25) + (1 × 26) + (1 × 27)

∴ 204 ⇒ 11001100

Converting 0s and 1s to %s and #s respectively, we get

729 = #%##%##%%#

So, the square of "##%##" is "#%##%##%%#".

Hence, the correct option is (B).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 25

Direction: In a certain number system there are only two notations to represent numbers: # and %.

0 is represented by % and 1 by #. The subsequent numbers are represented in the following manner:

2 is represented as #%,

3 is represented as ##,

4 is represented as #%%,

5 is represented as #%# and so on.

Based on this coded language, answer the following questions.

How the resultant for the given expression will be coded?

39 × 5 + 114 × 2 − 125

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 25

According to the given information,

So, this entire code language is based on only 2 symbols % and #, i.e. 0 and 1.

Logic:

On observation, we notice that starting from the rightmost digit, each digit is multiplied with corresponding power of 2, starting with 20, and then all these terms are added to obtain the number in the decimal system.

[The rightmost term is multiplied with 20, the second term from right is multiplied with 21, the third term from right is multiplied with 22 and so on..]

2 → 10 ⇒ (0 × 20) + (1 × 21) = (0 × 1) + (1 × 2) = 0 + 2 = 2

3 → 11 ⇒ (1 × 20) + (1 × 21) = (1 × 1) + (1 × 2) = 1 + 2 = 3

4 → 100 ⇒ (0 × 20) + (0 × 21) + (1 × 22) = (0 × 1) + (0 × 2) + (1 × 4) = 0 + 0 + 4 = 4

5 → 101 ⇒ (1 × 20) + (0 × 21) + (1 × 22) = (1 × 1) + (0 × 2) + (1 × 4) = 1 + 0 + 4 = 5 and so on.

Based on this logic,

‘###%#’ ↔ 11101

11101 ⇒ (1 × 20) + (0 × 21) + (1 × 22) + (1 × 23) + (1 × 24)

= (1 × 1) + (0 × 2) + (1 × 4) + (1 × 8) + (1 + 16)

= 1 + 0 + 4 + 8 + 16 = 29

‘#%’ ↔ 10

10⇒ (0 × 20) + (1 × 21)

= (0 × 1) + (1 × 2)

= 0 + 2 = 2

'##%%' ↔ 1100

1100 ⇒ (0 × 20) + (0 × 21) + (1 × 22) + (1 × 23)

= 0 + 0 + 4 + 8 = 12

'##%' ↔ 110

110 ⇒ (0 × 20) + (1 × 21) + (1 × 22)

= 0 + 2 + 4 = 6

'#%#' ↔ 101

101 ⇒ (1 × 20) + (0 × 21) + (1 × 22)

= 1 + 0 + 2 = 5

The resultant of the expression is: 29 ÷ 2 × 12 + 6 × 5 = 204

Now, we need to convert 204 into the given code language.

204 = 128 + 64 + 8 + 4

∴ 204 = (0 × 20) + (0 × 21) + (1 × 22) + (1 × 23) + (0 × 24) + (0 × 25) + (1 × 26) + (1 × 27)

∴ 204 ⇒ 11001100

Converting 0s and 1s to %s and #s respectively, we get

298 = #%%#%#%#%

So, the resultant of the expression "39 × 5 + 114 × 2 − 125" will be coded as ‘#%%#%#%#%'.

Hence, the correct option is (A).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 26

Direction: Study the following information carefully and answer the Question based on it:

Input: 962437 794528 465783 213945 753169 236185

Step 1: 734269 825497 647538 129354 961357 321658

Step 2: 162063 243524 21854 22720 54335 6640

Step 5 is the last step of rearrangement. As per the rules followed in the above steps, find out in each of the following questions the appropriate steps for the given input.

Input for the given Question,

Input: 576218 415893 642739 717340 931621 842163.

What is the difference between 2nd letter from right end in Step 5 and 2nd letter from left end in step 3?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 26

The Digits are arranged in such a way that,

In step 1 = if digit starts with odd number then it will be reverse and if it starts with even number then it will be reverse in pair of two.

Suppose in the given input it is 962437 starts with odd number i.e., 9. So in step 1 it will be reversed and we get 734269. In case of 465783 it will be 647538.

In step 2 = At first, the even digit numbers are arranged in descending order then the odd digit numbers are arranged in their respective order. After that multiplication of two-two digits of those are taken.

Ex – In step 1, the even digits numbers are = 825497 and 647538 now arranged them in descending order we get 825497 and 647538 so the multiplication form of 825497 is 8 × 2 = 16 5 × 4 = 20 and 9 × 7 = 63 i.e., 162063.

(This is the 1st number of step 2).

We get 162063 and 243524 from step 1(825497 647538 respectively).

Then the multiplications of odd numbers are arranged according their positional value.

21854 comes form 734269, 22720 from 129354, 54335 from 54335 and 6640 from 6640.

In step 3 = Difference of Addition of odd and even digits within the number.

Ex – 162063, 6 + 6 + 2 = 14, 3 + 1 = 4, 14 – 4 = 10.

In step 4 = Square of 1, 2, 3….so on and then add that Square in the numbers in Ascending order means Square of 1 is added in 1 and Square of 2 added in 4 and so on.

In step 5 = Addition of two numbers, than again add the digits of the number obtained.

Ex – 2 + 8 (1st and 2nd number of step 4) = 10 then 1 + 0 = 1 (1st number of step 5).

And 8 + 17 (2nd and 3rd number of step 4) = 25 = 2 + 5 = 7 (2nd number of step 5).

So, the solution of given input will be like this-

Input: 576218 415893 642739 717340 931621 842163

Step 1: 812675 148539 467293 43717 126139 481236

Step 2: 81235 32218 241427 1277 44027 2627

Step 5 is the last step of rearrangement.

2nd letter from right end in Step 5 is 8 and 2nd letter from left end in Step 3 is 8

8 – 8 = 0

Hence, the correct option is (E).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 27

Direction: Study the following information carefully and answer the Question based on it:

Input: 962437 794528 465783 213945 753169 236185

Step 1: 734269 825497 647538 129354 961357 321658

Step 2: 162063 243524 21854 22720 54335 6640

Step 5 is the last step of rearrangement. As per the rules followed in the above steps, find out in each of the following questions the appropriate steps for the given input.

Input for the given Question,

Input: 576218 415893 642739 717340 931621 842163.

What is the resultant if 3rd letter from right end in Step 4 is divided by 3rd letter from left end in Step 3?

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 27
The Digits are arranged in such a way that,

In step 1 = if digit starts with odd number then it will be reverse and if it starts with even number then it will be reverse in pair of two.

Suppose in the given input it is 962437 starts with odd number i.e., 9. So in step 1 it will be reversed and we get 734269. In case of 465783 it will be 647538.

In step 2 = At first, the even digit numbers are arranged in descending order then the odd digit numbers are arranged in their respective order. After that multiplication of two-two digits of those are taken.

Ex – In step 1, the even digits numbers are = 825497 and 647538 now arranged them in descending order we get 825497 and 647538 so the multiplication form of 825497 is 8 × 2 = 16 5 × 4 = 20 and 9 × 7 = 63 i.e., 162063.

(This is the 1st number of step 2).

We get 162063 and 243524 from step 1(825497 647538 respectively).

Then the multiplications of odd numbers are arranged according their positional value.

21854 comes form 734269, 22720 from 129354, 54335 from 54335 and 6640 from 6640.

In step 3 = Difference of Addition of odd and even digits within the number.

Ex – 162063, 6 + 6 + 2 = 14, 3 + 1 = 4, 14 – 4 = 10.

In step 4 = Square of 1, 2, 3….so on and then add that Square in the numbers in Ascending order means Square of 1 is added in 1 and Square of 2 added in 4 and so on.

In step 5 = Addition of two numbers, than again add the digits of the number obtained.

Ex – 2 + 8 (1st and 2nd number of step 4) = 10 then 1 + 0 = 1 (1st number of step 5).

And 8 + 17 (2nd and 3rd number of step 4) = 25 = 2 + 5 = 7 (2nd number of step 5).

So, the solution of given input will be like this-

Input: 576218 415893 642739 717340 931621 842163

Step 1: 812675 148539 467293 43717 126139 481236

Step 2: 81235 32218 241427 1277 44027 2627

Step 5 is the last step of rearrangement.

3rd letter from right Step 4 is 20 and 3rd letter from left end in Step 3 is 4.

20 ÷ 4 = 5

Hence, the correct option is (D).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 28

Direction: In the following question, three statements are given followed by four conclusions I, II, III and IV. You have to consider the given statements to be true even, if they seem to be at variance with commonly known facts. Read all the conclusions and decide which of the following logically follows from the given statements disregarding the commonly known facts.

Statement:

All king are smart.

No smart is bold.

Some bold are stones.

Conclusions:

I. No king are bold.

II. Some stones are not smart.

III. Some stones are bold.

IV. All bold being king is a possibility.

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 28
The least possible Venn diagram is given below

1) No king are bold → True.

2) Some stones are not smart → True.

3) Some stones are bold → True.

4) All bold being king is a possibility → False

So, the correct answer correct answer is Only I, II and III follows.

Hence, the correct option is (A).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 29

Direction: In the following question, three statements are given followed by four conclusions I, II, III and IV. You have to consider the given statements to be true even, if they seem to be at variance with commonly known facts. Read all the conclusions and decide which of the following logically follows from the given statements disregarding the commonly known facts.

Statement

Only a few Table are empty.

Only a few empty are pocket.

All pocket are imports.

Conclusion

I. All Table being empty is a possibility

II. No imports is a table

III. All Empty cannot be Pocket

IV. No Pocket is Table

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 29
The least Possible Venn diagram is shown below

Conclusions:

I. All table being empty is a possibility → False. (Only few table is empty is given)

II. No imports is a table → False(No definite relation is given)

III. All Empty cannot be Pocket → True(Only few empty is pocket given)

IV. No Pocket is Table → False (No definite relation is given)

So, the correct answer is only III follows.

Hence, the correct option is (A).

IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 30

Direction: In the following question, three statements are given followed by four conclusions I, II, III and IV. You have to consider the given statements to be true even, if they seem to be at variance with commonly known facts. Read all the conclusions and decide which of the following logically follows from the given statements disregarding the commonly known facts.

Statements:

Only a few Tiger are Cat.

No Cat is Mouse.

All Mouse are Parrot.

Conclusions:

I. All Tiger are Cat is a possibility.

II. Some Parrot can never be Cat.

III. Some Tiger are not Cat.

IV. All Parrot can be Tiger.

Detailed Solution for IBPS RRB PO (Scale 1) Mains Mock Test - 4 - Question 30
The least Possible Venn diagram is shown below

Conclusions:

I. All Tiger are Cat is a possibility → False (as only few tiger is cats given)

II. Some Parrot can never be Cat → True (Parrot which are mouse can never be cat so true)

III. Some Tiger are not Cat → True. (As only few tiger is cat given)

IV. All Parrot can be Tiger → True.

So, the correct answer is II, III and IV follows.

Hence, the correct option is (B).

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