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)
Here 227 is our answer
Q2. Find the missing number in the number series
100, 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
Here 7 is our answer
(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
Here 192 is our answer
(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
Here 20 is our answer
(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.
Here 10 is our answer
(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