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 i**n **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