Number Series – Logical Reasoning questions

This post is all about number series reasoning questions. All you have to do is identify the pattern and do the required calculation to find the missing number in the series. All the questions here are fully solved so that you can grasp the concept in depth and then can solve the questions without any external help.

This type of questions are part of logical reasoning syllabus which is asked in competition exams like GMAT, CAT, CMAT, NMAT, SNAP, SSC-CGL, SSC-CHSL, RRB, IBPS, NDA, AFCAT etc

Number Series Questions

(Q1)  Find the missing number in the series
4117, 5138, 6159, 7170, ?

         (a) 7138       (b) 7659

         (c) 8191      (d) 8179

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 +21 in all consecutive elements

Now we select such an option such that when subtract to 7170 it gives +21.

Here we find 8191 is only option which gives our required series.

(Q2)  Find the missing number in the series
4, 8, 12, 24, 36, ?  

        (a) 72           (b) 48

        (c) 60            (d) 144

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 in first two differences we get +4 and next two difference we get +12 which is a multiple of 4

That is 4×3=12

Then 12×3=36 will be our next difference

Now we select such an option such that when subtract to 36 it gives +36.

Here we find 72 is only option which gives our required series.

(Q3)  Find the missing number in the series
2, 7, 27, 107, ?

      (a) 327         (b) 427

      (c) 227         (d) 127

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.

But Here we don’t get any particular series so leave this difference method

Now if we multiply 4 and subtract 1 in first element we get second elemenet

And if 7 is multiplied  by 4 subtracted by 1 we get our third element

Now we select such an option such that it gives 107×4-1 =427         

Here we find 427 is our answer

Q4)  Find the number pattern
242, 393, 4164, ?

      (a) 5525       (b) 5255

      (c) 5235       (d) 5325

Solution-Here, there are only four elements of series

So we are not going to take difference of such elements

Here we see that for our first element middle digit is multiplication of first and last digit

And we follow the same pattern for our second and third element

So here we find our last element by selecting option as 5225

Q5)  1, 2, 4, 3, 9, 4, 16, 5, ?, ?

          (a) 6, 22     (b) 21, 9

          (c) 25, 6      (d) 30, 8

Solution –Here we see that there is a square series between pairs

That is third element is square of second element And fifth element is square of fourth element and so on

Here we get a different type of series in which first and last element are not taken in calculation

So as pattern our last element should be square of 5

So 25, 6 are our answer

Q.6)  255, 366, 479, 684, ?

 (a) 891         (b) 125

 (c) 216          (d) 343

Solution-here we see that in each element our first and last digit forms squares of number 5,6,7,8

So we selecting such an option that our last element should have first and last digit to form 9² that is 81

So here 891 is our answer

Q7.)  2, 15, 4, 47, 7, 118, 11, ?, ?

 (a) 260, 15   (b) 252, 16

(c) 250, 17   (d) 254, 16

Solution- here by taking alternate difference of elements we get a series of addition of prime number 2,3,4,5

And our second element is square of third element with minus 1 and here we are also getting a new series in similar way

So our second last element should be 16²-4=252

So here 252, 16 are our answers.

Q.8)   8, 29, 113, 449, ?

(a) 673          (b) 984

 (c) 1484       (d) 1793

Solution- here we see that the difference between two elements is very large

So it is a multiplicative series

When our each element is multiplied by 4 and subtracted by 3 we get our next element

And we get our last element as 179

Q. 9)  6341, 5432, ____, 3614

 (a) 4253       (b) 4614

 (c) 4532       (d) 4523

Solution-here we get a special type of series in which all four digits of consecutive element are in alternate decreasing and increasing order to each other

If subtract 1 from all even digits we get our next element digits

And if we add 1 in all odd digits we get our next element digits

And in next step we do just opposite operation to even and odd digits

Here we get a -1 and+1 series of alternate digits

So our next element of series should be

5-1=4

4+1=5

3-1=2

2+1=3

We get as 4523

Q.10)  69, 72, 78, 87, ?, 114

(a) 111          (b) 99

 (c) 93                        (d) 96

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 the difference is simply a table of 3

So our next difference should be +12 and +15

We selecting such an option whose difference with 87 should be +12

And difference with 114 should be +15

 So here our answer should be 99