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
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
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