Number Series Questions and answers For SBI Clerk Preliminary exam 2021

NUMBER AND MISSING SERIES QUESTIONS WITH THE LATEST PATTERN-

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There will be five questions in your Banking exam, and from it, you will be found from this number series topic. out of five questions, you will easily solve two questions in the preliminary exam, but you will need to practice a lot for the rest of the three questions. So we have decided to start a practice session for you.   

Q1.     0.5          1             2           4          8          ?

a. 16 

b.15

c.14

d.None

e.15.50

Sol-  1= 0.5 * 2

         2= 1*2 

         4= 2 *2

         8= 4 *2

         ?= 8 * 2= 16 ,option(a) is correct. Logic= next term = 2 times of previous term.

 Note- * means multiplication    

Q2. 32    144   504    1260    1890    ? 

a.     a.   945

b.    b. 2145

c.    c.   2560

d.    d. 815

e.   e.  915

Sol-To deal with such type of questions just think about how second the term we can bring from the first term

So 32 ×? = 144

 Or, ? = 144/32= 9/2 now we got the first number which is 9/2 so we will think around it.

So option(a) is correct

 Q3.   90        55        75          142.50         ?            862.50   

  a.285

  b.325

  c.470

  d.855

  e.270  

Sol-  Option(b) is correct.

Q4.    5    12     39      160       ?        4836      

a.850

b.750

c.800

d.805

e. 820  

Sol-  option(d) is correct.

Q5.   1      ?      7       31       151      911  

a. 4

b.5

c.2

d.3.5

e.3

Sol- Whenever in question initial numbers are missing then just focus on the last few terms. Try to find out logic based on series.

31 = 7 × 4 +3 

151= 31 ×5 -4

911= 151×6 +5

similarly,

7=?×3-2   or, 9= 3?  or, ? = 3 

so, option (e) is correct.

Q6. ?    16    64   8    128   4    

a.32

b.4

c.8

d.16

e.64

Sol- Once again the initial first number is missing so, not to worry just focus on the last few terms. 

 64 = 16 × 4   ; 8 = 64 ÷ 8   ;  128 = 8 × 16 ;  4 = 128 ÷ 32  so pattern is multiplication and division alternatively so , 32 ÷ 2 = 16 

option(a) is correct.

Q7.  2     30     130    350     ?      1342    

a.712

b.720

c.742

d.738

e.750

Sol- When you go through the common difference method, it will take more time, and not sure that whether it will be solved or not. 

So, just obey my suggestion always remember when you see any number close to the cube of a natural number i.e., close means either greater or smaller by out of following  1,2,3,4 and 5 then try to make a pattern. 

2 = (1)³ + 1 

30 = (3)³ +3 

130 =(5)³ + 5 

Now you got the pattern next number will be 7 and his next will be  9.

? = (9)³ +9  = 738. 

Q8. 821    784      743      ?      653     600    

 a.700

 b. 699

c. 721

d. 671

e.702

Sol- In a first look whenever you get numbers are increasing or decreasing by single or double-digit then go with the common difference method, it will solve your given series. The common difference means next term - previous term. 

Q9. 2      3    27       74     ?    237   

a.148

b.144

c.134

d.178

e.179

Sol-  Series based on double common difference just go ahead to solve.

Q10.  182  186    190    198   222   ?    

a. 308

b. 318

c.328

d.316

e.None

Sol-  Correct option is (b).

Q11. ?   3450    1728    580    150    36  

a. None

b.3454

c.3548

d.3448

e. 3624

Sol- Again in this question the very first number is missing, which makes it very difficult but you need to focus on the last few terms for getting clue/logic present in this series. One thing, you can notice here numbers are getting decreasing so fast, it is reflecting series is based on division so try to frame a logic based on division and some other operations.

3450 ÷ 2 +3 = 1728

1728 ÷ 3 +4 = 580

580 ÷ 4 +5 = 150

150 ÷5 +6 = 36

Similarly the first term should be   ?/1  +2  = 3450

                                                       Or, ?/1 = 3450-2= 3448

                                                        Or, ?= 1×3448= 3448

Option(d) is correct.

Q12.    1210     1222     1242   1269    1302     ? 

a. 1424

b.1360

c.1348

d.1340

e.None

     

Sol- Here we can see numbers are increasing with a margin of two digits number hence we should focus the logic based on common difference or maybe double the common difference to get the logic.

First c.d =  12            20             27              33  

second c.d=       8                 7              6               Here should be 5  so the next number will be = 33 

+5=38 which will be present in series of the first c.d.

so the missing number ? = 1302 + 38 = 1340                        

Q13.  102      105       114       141       ?        465 

a.222

b.218

c.336

d.none

e.228

Sol-   Series are based on common differences, we can get this logic by observing numbers are increasing not suddenly.

First c.d= 3       9        27         ?       ?  

in the first c.d, we can see the next numbers are getting after multiplying the previous number by 3.

 so, the next two numbers will be 81  and  243   

so, in the original series, the missing number will be = 141+81 =    222 

option(a) is correct. 

Q14.   16    ?     8    12      24    60  

a.8

b.12

c.16

d.10

e.4

Sol -   Initial terms are missing just focus on the last few terms 

12=8 × 1.5

24= 12 ×2

60=24 ×2.5

So, we see ,we are multiplying 1.5,2 and 2.5 i.e., just increases by 0.5 so ,

?  = 16 ×0.5 = 8 

Q15. 201   199   192   175   143   ?  

a. 89

b.91

c.93

d.103

e.None  

Sol-  Based on common difference(c.d). Here numbers are decreasing. 

First c.d =  2         7        17         32     ? 

second c.d=     5         10        15         20        ? (in first c.d) = 32 + 20 = 52  

so , next no. in original series will be = 143- 52 = 91

hence option(b) is correct.     

Q16.   ?          8          14        34       76         148    

a. None

b.12

c. 4

d.16

e.8 

Sol- Concentrate on the last few terms, here numbers are increasing in a mostly two-digit number so go with the common difference method.

 First c.d = ?      6            20              42            72      

second c.d =  ?          14           22            30   

from the second common difference we can see numbers are increasing by 8  so first number in the second common difference will be = 14 -8 = 6  

so, First common difference =  0       6        20       42      72    

so, ?(missing number in original series) = 8-0 = 8