Alphanumeric Series Questions for SBI PO 2020 PDF

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Alphanumeric Series Questions for SBI PO 2020 PDF

Download SBI PO Alphanumeric Series Questions & Answers PDF for SBI PO Prelims and Mains exam. Top-15 Very Important SBI PO Questions with solutions for Banking Exams.

Question 1: Which of the following will come next in the following series ?
9 1 9 8 2 9 8 7 3 9 8 7 6 4 9 8 7 6 5 5 9 8 7 6 5 4 6 9 8 7 6 5 4

a) 7

b) 8

c) 3

d) 2

e) 4

Instructions

Study the following arrangement carefully and answer the questions given below :
1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

Question 2: If all the even digits are deleted from the above arrangement which of the following will be tenth from the right end of the arrangement ?

a) 5

b) 3

c) 1

d) 7

e) 9

Question 3: How many such 4s are there in the above arrangement each of which is immediately preceded by a digit which has a numerical value more than four ?

a) None

b) One

c) Two

d) Three

e) More than three

Question 4: Which of the following will come in the place of the question mark (?) in the following series based on the English alphabetical order ?
ZA BY XC DW ?

a) VF

b) EU

c) UE

d) EV

e) VE

Question 5: What should come next in the following letter series based on English alphabet ?
CEA IKG OQM ?

a) STW

b) WUS

c) SWU

d) UWS

e) None of these

Instructions

Study the following arrangement carefully and answer the questions given below :
1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

Question 6: Which of the following is ninth to the left of twenty first from the left end of the above arrangement ?

a) 7

b) 5

c) 6

d) 8

e) None of these

Question 7: How many such 5s are there in the above arrangement each of which is immediately preceded by an odd digit and immediately followed by an even digit ?

a) None

b) One

c) Two

d) Three

e) More than three

Question 8: What should come next in the following letter series ?
PQRSTABCDEPQRSABCDEPQRSABCDPQ

a) R

b) T

c) A

d) B

e) None of these

Question 9: If in the English alphabet, all letters at odd numbered positions are written in serial order from left to right followed by the letters at even numbered positions written in reverse order, which letter will be sixth to the left of seventeenth letter from left ?

a) D

b) B

c) V

d) U

e) None of these

Instructions

Study the following arrangement carefully and answer the questions given below :
1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

Question 10: How many such 1s are there in the above arrangement each of which is immediately followed by a perfect square ?

a) None

b) One

c) Two

d) Three

e) More than three

Question 11: What should come next in the following letter series ?
ACEGIKMOBDFHJLNACEGIKMBDFHJLA

a) B

b) C

c) F

d) D

e) None of these

Question 12: The positions of how many alphabets will remain the same if each of the alphabets in the word DETRIMENT is rearranged in the alphabetical order from left to right?

a) None

b) One

c) TWO

d) Three

e) More than three

Question 13: The sum of the squares of two odd numbers is 11570. The square of the smaller number is 5329. What is the other number?

a) 73

b) 75

c) 78

d) 79

e) 80

Question 14: If the digits in the number 79246358 are arranged in descending order from left to right, what will be the difference between the digits which are third from the right and second from the left in the new arrangement?

a) 1

b) 2

c) 3

d) 4

e) 5

Question 15: Which of the following will come next in the following series ?
0 9 0 1 9 0 1 2 9 0 1 2 3 9 0 1 2 3 4 9 0 1 2 3 4 5 9 0 1 2 3 4 5

a) 0

b) 6

c) 9

d) 7

e) 4

Arrangement : 9 1 9 8 2 9 8 7 3 9 8 7 6 4 9 8 7 6 5 5 9 8 7 6 5 4 6 9 8 7 6 5 4

The pattern followed is that in alternate groups of numbers, one digit starting from ‘1’ in increasing order are written and in the other group digits in decreasing order are written with one digit appended after each new group.

9 , 1 , 9 8 , 2 , 9 8 7 , 3 , 9 8 7 6 , 4 , 9 8 7 6 5 , 5 , 9 8 7 6 5 4 , 6 , 9 8 7 6 5 4 3

Thus, next number = 3

=> Ans – (C)

Arrangement : 1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

If all the even digits are deleted, then

= 1 5 1 5 3 5 7 9 5 1 1 5 7

10th from right end = 5

=> Ans – (A)

Arrangement : 1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

4’s which are immediately preceded by a digit which has a numerical value more than four = (>4) (4)

= 1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

Thus, there are two such 4’s.

=> Ans – (C)

Series : ZA BY XC DW ?

The pattern followed is that alphabets starting from end are written in alternate order to those of starting from top

= (Z)A , B(Y) , (X)C , D(W)

Thus, next term = VE

=> Ans – (E)

Based on the English alphabet, the above arrangement follows the following pattern :

The first alphabet in each word is

C –(+6)–> I –(+6)–> O –(+6)–> U

The second alphabet is

C –(+2)–> E

$\therefore$ U –(+2) –> W

And the third alphabet is

C –(-2)–> A

$\therefore$ U –(-2) –> S

Thus, UWS wil be the next letter in the series.

Arrangement : 1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

21st from left end = 6

9th to the left of 6 = 6

=> Ans – (C)

Arrangement : 1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

5’s which are immediately preceded by an odd digit and immediately followed by an even digit

= (odd) (5) (even)

1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

Thus, there are 3 such 5’s.

=> Ans – (D)

The above series is the combination of two series i.e. :

PQRST & ABCDE

After that, last alphabet of the first series is removed => PQRS & ABCDE

Then ,the last alphabet of the second series is removed => PQRS & ABCD

After that, S should be removed => PQR & ABCD

Clearly, the next letter is R.

If all letters at odd positions are written in serial order from left to right, then this series will be :

A C E G I K M O Q S U W Y

After this even positioned letters are written in reverse, the series becomes :

A C E G I K M O Q S U W Y , Z X V T R P N L J H F D B

Now, 17th letter from left end = T

6th letter to the left of T = U

Thus, Ans – (D)

Arrangement : 1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

1’s which are immediately followed by a perfect square

= (1) (perfect square)

= 1 5 8 4 2 1 5 2 3 4 5 6 7 8 9 5 1 4 1 5 6 8 7 4

Thus, there is only one such 1.

=> Ans – (B)

This series is made from a combination of two series from the first 15 alphabets of the English alphabet.

First series is : The letters at odd positions => A C E G I K M O

Second series is : The letters at even positions => B D F H J L N

Then, the last letter from each series is removed. Thus making it : A C E G I K M B D F H J L

Again, after removing the last letter(i.e. M & L) => A C E G I K B D F H J

Clearly, the next letter after A is ‘C’.

If all the alphabets of the word DETRIMENT are rearranged in alphabetical order, then

DETRIMENT $\Leftrightarrow$ DEEIMNRTT

Clearly, the position of only 3 alphabets have remain same, i.e. D, E(first), T(second).

let the two numbers be a and b , where b is the smaller number

$(a)^2$ + $(b)^2$ = 11570

$(a)^2$ + 5329 = 11570

$(a)^2$ = 6241 = 79 x 79

a = 79

Hence option D is correct

Number – ‘79246358’

If the digits are arranged in descending order from left to right, new number = 98765432

Difference between the digits which are third from the right and second from the left = $8-4=4$

=> Ans – (D)