New Pattern Reasoning Questions for IBPS Clerk 2018
Reasoning Mixed Questions Based on IBPS PO Prelims. Reasoning Mixed Questions for IBPS Clerk Prelims 2018. Welcome to the www.letsstudytogether.co online Reasoning section. If you are preparing for IBPS, RBI and other competitive exams, you will come across a section on the Reasoning Ability. Here we are providing you a quiz on “Reasoning Mixed Questions Based on IBPS PO Prelims 2018” based on the latest pattern for your daily practice.
The pattern of the Reasoning section was changed nowadays, questions asked on new pattern. So we are providing you “New Pattern Reasoning Questions for IBPS Clerk 2018” for banking exams such as IBPS PO, IBPS Clerk, SBI Clerk, IBPS RRB Officer, IBPS RRB Office Assistant, IBPS SO, SBI SO and other competitive exams.
Reasoning Mixed Questions Based on IBPS PO Prelims  Set – 4
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Question 1 of 10
1. Question
In the question below, relationship between some elements is shown in the statements (s). These statements are followed by two conclusions. Read the statements and give answer.
Statements: L > M ≥ N = P; A < M = B; P < Q ≤ R > S
Conclusions:
I. B ≥ R
II. M > Q
Correct
D. Neither conclusion I nor II is true
Given statement: L > M ≥ N = P; A < M = B; P < Q ≤ R > S
On combining statements, we get,
B = M ≥ N = P < Q ≤ R and M ≥ N = P < Q
Conclusions:
I. B ≥ R: False (As, B = M ≥ N = P < Q ≤ R, we cannot establish any relationship between B and R)
II. M > Q: False (As, M ≥ N = P < Q, we cannot establish any relationship between M and Q.)
Incorrect
D. Neither conclusion I nor II is true
Given statement: L > M ≥ N = P; A < M = B; P < Q ≤ R > S
On combining statements, we get,
B = M ≥ N = P < Q ≤ R and M ≥ N = P < Q
Conclusions:
I. B ≥ R: False (As, B = M ≥ N = P < Q ≤ R, we cannot establish any relationship between B and R)
II. M > Q: False (As, M ≥ N = P < Q, we cannot establish any relationship between M and Q.)

Question 2 of 10
2. Question
In each of these questions, relationship between some elements is shown in the statements (s). These statements are followed by two conclusions. Read the statements and give answer.
Statements: G = H < I ≥ J; I < F < E; F > L ≥ M = O
Conclusions:
I. E > G
II. O > J
Correct
A. Only conclusion I is true
Given statement: G = H < I ≥ J; I < F < E; F < L ≤ M = O
On combining statements, we get,
E > F > I > H = G and J ≤ I < F < L ≤ N = O
Conclusions:
I. E > G: True (As, E > F > I > H = G)
II. O > J: False (As, J ≤ I < F > L ≤ N = O, so, no relation between O and J can be established)
Incorrect
A. Only conclusion I is true
Given statement: G = H < I ≥ J; I < F < E; F < L ≤ M = O
On combining statements, we get,
E > F > I > H = G and J ≤ I < F < L ≤ N = O
Conclusions:
I. E > G: True (As, E > F > I > H = G)
II. O > J: False (As, J ≤ I < F > L ≤ N = O, so, no relation between O and J can be established)

Question 3 of 10
3. Question
Directions (35) : Answer the questions based on the information given below.
45369 52817 14658 29564 38149
If in each of the given numbers, the highest digit of the number is added to the second lowest digit of the number, which number will yield the lowest value?
Correct
B. 52817
45369 52817 14658 29564 38149
After adding the highest digit with the second lowest digit we get,
13 10 12 13 12
The lowest number is 10, which is represented by 52817.
Incorrect
B. 52817
45369 52817 14658 29564 38149
After adding the highest digit with the second lowest digit we get,
13 10 12 13 12
The lowest number is 10, which is represented by 52817.

Question 4 of 10
4. Question
Directions (35) : Answer the questions based on the information given below.
45369 52817 14658 29564 38149
After arranging all the digits in the ascending order within the number, if the digits which are at the same place in the numbers after the arrangement is added, which number will yield the highest value?
Correct
A. 45369
45369 52817 14658 29564 38149
After arrangement
34569 12578 14568 24569 13489
After adding the digits which are at the same place after the arrangement
15 2 13 13 9
The highest value is 15 which is represented by 45369.
Incorrect
A. 45369
45369 52817 14658 29564 38149
After arrangement
34569 12578 14568 24569 13489
After adding the digits which are at the same place after the arrangement
15 2 13 13 9
The highest value is 15 which is represented by 45369.

Question 5 of 10
5. Question
Directions (35) : Answer the questions based on the information given below.
45369 52817 14658 29564 38149
If sum of all the digits at the odd positions are subtracted from the sum of all the digits at the even positions in the numbers, which number will yield second highest value? (Example: in 45369, 4 is at odd position and 5 is at even position)
Correct
E. 38149
45369 52817 14658 29564 38149
After subtracting the sum of all digits in the odd position from the sum of all the digits in the even position, we get:
5 17 6 4 1
So, the second highest digit is 1 which is represented by 38149.
Incorrect
E. 38149
45369 52817 14658 29564 38149
After subtracting the sum of all digits in the odd position from the sum of all the digits in the even position, we get:
5 17 6 4 1
So, the second highest digit is 1 which is represented by 38149.

Question 6 of 10
6. Question
When the vowels of the word ULTRAMICROSCOPIC are replaced by their second successive letters in the alphabetical series and the consonants are replaced by their immediate preceding letters in the alphabetical series, then which letter is the second to the right of the fourth letter from the left side after rearrangement?
Correct
B. L
ULTRAMICROSCOPIC
After replacement, we get W K S Q C L K B Q Q R B Q O K B
The letter which is second to the right of the fourth from the left side after rearrangement is L.
Incorrect
B. L
ULTRAMICROSCOPIC
After replacement, we get W K S Q C L K B Q Q R B Q O K B
The letter which is second to the right of the fourth from the left side after rearrangement is L.

Question 7 of 10
7. Question
Directions (78) : Answer the questions based on the information given below.
A number arrangement machine when given an input line of numbers rearranges them following a particular rule in each step. The following is an illustration of input and rearrangement
Input: 48 33 57 62 25 69 84 77
Step I: 21 29 52 53 65 66 73 88
Step II:6 22 14 16 22 24 20 32
Step III: 1 2 4 1 2 4 0 2
And step III is the last step of the above input. As per the rules followed in the above steps, answer the questions.
Input: 32 23 49 64 55 76 90 82.
How many odd numbers are there in the step III?
Correct
D.6
In Step I, 4 is added to all even numbers and 4 is subtracted from all odd numbers and then arranged in an ascending order.
In step II, twice of the sum of the digits of all numbers are obtained.
In step III, every number is divided by 5 and remainder is written.
Given input: 32 23 49 64 55 76 90 82
Step I: 19 36 45 51 68 80 86 94
Step II: 20 18 18 12 28 16 28 26
Step III: 0 3 3 2 3 1 3 1
Thus, Step III is the output.
So, there are six odd numbers in the step III.
Incorrect
D.6
In Step I, 4 is added to all even numbers and 4 is subtracted from all odd numbers and then arranged in an ascending order.
In step II, twice of the sum of the digits of all numbers are obtained.
In step III, every number is divided by 5 and remainder is written.
Given input: 32 23 49 64 55 76 90 82
Step I: 19 36 45 51 68 80 86 94
Step II: 20 18 18 12 28 16 28 26
Step III: 0 3 3 2 3 1 3 1
Thus, Step III is the output.
So, there are six odd numbers in the step III.

Question 8 of 10
8. Question
Directions (78) : Answer the questions based on the information given below.
A number arrangement machine when given an input line of numbers rearranges them following a particular rule in each step. The following is an illustration of input and rearrangement
Input: 48 33 57 62 25 69 84 77
Step I: 21 29 52 53 65 66 73 88
Step II:6 22 14 16 22 24 20 32
Step III: 1 2 4 1 2 4 0 2
And step III is the last step of the above input. As per the rules followed in the above steps, answer the questions.
Input: 32 23 49 64 55 76 90 82.
Which element is the third to the left of sixth element from the left end in the last but one step?
Correct
C. 18
In Step I, 4 is added to all even numbers and 4 is subtracted from all odd numbers and then arranged in an ascending order.
In step II, twice of the sum of the digits of all numbers are obtained.
In step III, every number is divided by 5 and remainder is written.
Given input:32 23 49 64 55 76 90 82
Step I: 19 36 45 51 68 80 86 94
Step II: 20 18 18 12 28 16 28 26
Step III: 0 3 3 2 3 1 3 1
Thus, Step III is the output.
So, 18 is third to the left of sixth element from the left end in the step II.
Incorrect
C. 18
In Step I, 4 is added to all even numbers and 4 is subtracted from all odd numbers and then arranged in an ascending order.
In step II, twice of the sum of the digits of all numbers are obtained.
In step III, every number is divided by 5 and remainder is written.
Given input:32 23 49 64 55 76 90 82
Step I: 19 36 45 51 68 80 86 94
Step II: 20 18 18 12 28 16 28 26
Step III: 0 3 3 2 3 1 3 1
Thus, Step III is the output.
So, 18 is third to the left of sixth element from the left end in the step II.

Question 9 of 10
9. Question
In the question below, relationship between some elements is shown in the statements (s). These statements are followed by two conclusions. Read the statements and give answer.
Statements: A < C ≤ D; M > N ≥ O = P; C = S < T ≤ P.
Conclusions:
I. M > A
II. N ≥ S
Correct
A. Only conclusion I is true
Given statement: A < C ≤ D; M > N ≥ O = P; C = S < T ≤ P.
On combining statements, we get,
M > N ≥ O = P > S = C > A and N ≥ O = P ≥ T > S
Conclusions:
I. M > A: True (As, M > N ≥ O = P > S = C > A)
II. N ≥ S: False (As, N ≥ O = P ≥ T > S, so N > S)
Incorrect
A. Only conclusion I is true
Given statement: A < C ≤ D; M > N ≥ O = P; C = S < T ≤ P.
On combining statements, we get,
M > N ≥ O = P > S = C > A and N ≥ O = P ≥ T > S
Conclusions:
I. M > A: True (As, M > N ≥ O = P > S = C > A)
II. N ≥ S: False (As, N ≥ O = P ≥ T > S, so N > S)

Question 10 of 10
10. Question
When the digits divisible by 2 of the number 9467391275754 are replaced by their immediate succeeding digit and the remaining digits are left unchanged, then what is the highest number formed by the, 3^{rd} 4^{th}, 5^{th}, 6^{th} and 7^{th} digits of the rearranged number?
Correct
C. 97731
9467391275754
After replacement, we get 9 5 7 7 3 9 1 3 7 5 7 5 5
From the 3^{rd}, 4^{th}, 5^{th}, 6^{th} and 7^{th} digits of the rearranged digits i.e. 7, 7, 3, 9 and 1, the maximum number formed is
97731.
Incorrect
C. 97731
9467391275754
After replacement, we get 9 5 7 7 3 9 1 3 7 5 7 5 5
From the 3^{rd}, 4^{th}, 5^{th}, 6^{th} and 7^{th} digits of the rearranged digits i.e. 7, 7, 3, 9 and 1, the maximum number formed is
97731.
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