# Number Series Reasoning Questions

This is the third post about the number series reasoning question for students aiming for the competitive exams like GMAT, CAT, NMAT, SNAP, SSC, SSC-CGL, SSC-CHSL, IBPS, SBI PO, SBI clerk, NDA, NABARD exams.

All the questions are fully solved so that the readers understand the concept fully and can start the practice on their own.

## Number Series Reasoning Solved Questions

### (Q1) Find the missing number 3, 11, 38, 102, ?, 443

(a)227                      (b)237

(c)247                      (d)217

Solution-  here we see that there is no such a big difference in consecutive elements so it is not a multiplication or division series

So lets we take a difference of consecutive elements

By simplifying the difference in cubes we get a required series as shown

By selecting such an option which gives our fifth element as difference of +(5³)

### Q2. Find the missing number in the number series100, 52, 28, 16, 10 ,?

(a)5                            (b)7

(c)8                            (d)9

Solution-Here we see that the element are in decreasing order with large difference between them

So it is division series of 2 or multiplication series of ½

So here we multiply with  ½  with addition of  2  we get a required series

By selecting such an option which gives our sixth element as 10× +2

### (Q3) 5760, 960, ?, 48, 16 , 8

(a)240                (b)192

(c)160                (d)120

Solution-  Here we see that our first element is very large number and find that it is decreasing order series with big differences

So in order to avoid difficulty in solving here we starts from last element of series and we find that it is multiplicative series with consecutive integers with (2,3,4,5,6)

By selecting such an option which gives our third element as 192×5

### (Q4) 2807, 1400, 697, 346, 171, 84, 41, ?

(a)22                         (b)19

(c)20                         (d)21

Solution- Here we see that second element is almost half of first element, and third element is half of second and so on.

If we subtract last digit of element from itself and divided by 2 we get our next element of series.

And by this method we get our required series

By selecting such an option which gives our last element as (41–1)× 1/2

### (Q5) 15, 30, ?, 40, 8, 48

(a)10                           (b)20

(c)30                           (d)15

Solution-Here we see that there is increment and decrement in the series

So here second element is obtained by multiplying 2 in first element

And fifth element is obtained just dividing fourth element by 5

Then sixth element is obtained by multiplying 6 in fifth element

so here we find that our third element is obtained by dividing 30 by 3
we get an 10 and after multiplying it by 4 we get 40 which is our next element of series.

### (Q6)   1438, 1429, 1417, 1402, ?

(a) 1378          (b) 1384

(c) 1387          (d) 1392

Solution-Here we simply solve by taking difference of first element with second element then difference of second element with third element and so on.

Here we see that there is a difference of 3 in consecutive elements And  we get a decreasing order of -3 series

Now we select such an option such that when subtract to 1402 it gives -18. Here we find 1384 is only option which gives our required series

### (Q7)    2460, 3570, 4680, ?

(a) 8640        (b) 5670

(c) 5970        (d) 5790

Solution–Here we simply solve by taking difference of first element with second element then difference of second element with third element and so on

Here we see that there is a difference of +1110  in all consecutive elements

Now we select such an option such that when 4680 subtract from it gives +1110.

Here we find 5790 is only option which gives our required series.

### (Q8)   7714, 7916, 8109, ?

(a) 8311        (b) 8312

(c) 8509        (d) 8515

Solution-Here we see that there is only 4 elements in series

So first we take a difference between consecutive elements

Here we don’t find any particular series so we are taking difference by selecting options
So first we select 8311
If we subtracting first and second element and third and last fourth element we get +202 in both cases

Here we find 8311 is only option which gives our required series

### (Q 9)   2, 5, 9, 19, 37, ?

(a) 73          (b) 75

(c) 76           (d) 78

Solution–Here we simply solve by taking difference of first element with second element then difference of second element with third element and so on.

Here we don’t get any particular series so leave this difference method
Now if we multiply 2 and add 1 in first element we get second element
And if 5 is multiply by 2 subtracted by 1 we get our third element

Now we select such an option such that it gives 37×2+1 =75          .
Here we find 75 is our answer.

### (Q10)   110, 132, 156, ?, 210

(a) 162                    (b) 172

(c) 182          (d) 192

Solution-Here we simply solve by taking difference of first element with second element then difference of second element with third element and so on.

Here we see that there is a difference of 2 in consecutive elements
And  we get a increasing order of +2 series .

For the above series we add 156 + 26 ==> 182, which is the required number