## Reasoning problems with their Solutions

#### Problem # 1

Starting from 5 and moving clockwise  we have these numbers 5, 10,?,50,122.
so 5 can be written  as 2^2+1 starting from 1st prime  number i. e.  Square of 1st  prime number  and plus 1

10 can be written  as 3^2+1 , Square  of 2nd prime number  and plus 1.

50 can be written as 7^2 +1,Square  of  4th prime number  and plus 1.

Similarly 122 can be written as 11^2 +1, Square  of  5th prime number  and plus 1.

Hence ? can be replaced as 5^2 +1, Square  of  3rd prime number ( 3 )  and plus 1.
So
Therefore  Required  number  is 26.

#### Problem # 2

In this figure largest numbers are appearing in last row. so we should search the relation column wise .
In 1st column we have to search the relationship  between  9 and 6 to give 45. so in 2nd column we have to find relationship between 12 and 7 to give 95. And same logic we shall apply in 3rd columns.

### 1st Column

In 1st columns if we add and subtract both numbers  then multiply  it then we shall have 45 as follows
we have two numbers in 1st column  9 and 6 .
(9+6)×(9-6) = (15 ) × ( 3 ) = 45

### 2nd Column

And again in 2nd  columns if we add and subtract both numbers   of 2nd column and then multiply  these with each others  then we shall have 45 like this
we have two numbers in 1st column  9 and 6 .
(12+7 ) × ( 12 - 7) = (19  ) × ( 5 ) = 95

### 3rd Column

Same logic can be applied for 3rd column ,we add and subtract both numbers   of 3rd column and then multiply  these with each others
( 8 + 3 ) × ( 8 -3 ) = (11 × 5 ) = 55
Hence 55 shall replace  "?" in the given figure

#### Problem # 3

Since largest numbers  in all the three rows lie in 1st column of the given figure.
It means we have to search a relation between  20 and 7 to give 28 in 1st row. and similar relation must be between 35 and 12 to give 84 .

### 1st Row

So in 1st row if we multiply both numbers with each others and divide the result  with 5 we shall get 28 like this 20 × 7 =140
Now divide 140 with 5 such that 140/5 =  28

### 2nd Row

So in 2nd row if we have to multiply both numbers with each others and divide the result  with 5 we shall get 28 like this 35 × 12 = 420 .

### 3rd Row

Now divide 420 with 5 such that 420/5 =  84
Same logic we have to apply in 3rd row to get answer 45 . So If we  multiply  9 with 45 we shall get 275 then we have to divide 405 with 5 to give  , It will replace ? question mark as follows
9 × 25 = 225 / 5 = 45
So  45 shall be the right number to replace question  mark .

#### Problem # 4

Since largest numbers appears on 3rd columns therefore the solution must be row wise, if we treat a and b as 1st and 2nd numbers then 3rd number which is our desired numbers must be equal to (a-1)×(b) i . e. product/ multiplication  of (a-1)  and b. Hence 3rd element of 1st row must be
(8-1)×(3) = 7×3 = 21
3rd element of 2nd row must be
(6-1)×(5) = 5×5 = 25
So 3td element of 3rd row must be
(12-1)×(2) = 11×2= 22
So 22 will replace "?" Question  mark.

#### Problem # 5

In 1st column 5 × ( 6 +7 ) = 5 × 13 = 65
In 2nd column 4 × ( 3 +2 ) = 4 × 5 = 20
So same formula will be used in 3rd column
In 3rd  column.  9 × ( ? +4 ) = 45
( ? +4 ) = 45/9 =5
? = 5 - 4  = 1
So required answer is   " 1 ".

#### Problem # 6

Multiplying 1st three elements of all the columns to get 4th elements
1 × 8 × 9 = 72
3 × 6 × 5 = 90
2 × 7 × ? = 56
This implies ? = 4
So ? will be replaced by 4

#### Problem # 7

In 1st column square of three numbers i.e.
( 1 + 4 + 2 )^2  = 7^2  = 49
In 2nd column square of three  numbers i.e.   ( 4 + 2 +2 ) ^2 =  ( 8)^2 = 64
In 3rd column the square of  ( ? + 5 + 3 ) ^2  must be 169
this implies  ( ? + 8 )^2 = 169
( ? + 8 ) = 13
? = 5

#### Problem # 8

If we calculate the sum of 1st column ,2nd column , 2nd row and 3rd row ,then it is found 25 in all the cases .so total of all rows and all the columns must be 25 . It can be seen that if we put "11"  in place of question mark then total of all the rows and columns is  25

#### Problem # 9

Since biggest numbers are in the fourth columns of every row. So if we mutiply 1st three numbers and then add 4th number to it ,we shall have 5th number in every row .
( 4 × 3 × 2 )  +  8 = 24 + 8 = 32
( 5 × 3 × 1 ) + 9 = 15 + 9 = 24
( 7 × 3 × 3 ) + 7 = 63 + 7 = 70
Similarly when we multiply first three numbers and then adding force number to it in the last row we shall have fourth number in the last row like this
( 2 × 9 × 4 ) + 12 = 72 + 12 = 84
So 84 will replace question mark " ?"

#### Problem # 10

1st Row
multiply 3 and 2 then add one less than the 2nd number to it
( 3 × 2 ) + ( 2 - 1 ) = 6 + 1 = 1
2nd Row
multiply 5 and 4 then add one less than the 2nd number to it
( 5 × 4  ) + ( 4 - 1 ) = 20 + 3 = 23
3rd Row
multiply 7 and 6 then add one less than the 2nd number to it
( 7 × 6 ) + ( 6 - 1 ) = 42 + 5 = 47
4th Row
( 9 × 8 ) + ( 8 - 1) = 72 + 7 = 79
5th Row

( 10 × 9 ) + ( 9 - 1 ) = 90 + 8 = 98
So 98 will replace. " ? "

## FINAL WORDS

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## How to find proportion of Four numbers

If A : B = 2 : 3, B : C = 4 : 5 and      C : D = 6 : 7,    what is A  : D  ?

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## What is next terms in the series ?

Let us solve next number in sequence puzzle
15,29,56,108,208,????
Here 1st term is 15, To find out 2nd term if we multiply the 1st term With 2 and subtract 1 from it, i.e (15 *2) -1  = 29
To find 2nd term we have to multiply the previous term with 2 and subtract 2 (double the previous time) from it
I.e 29*2 - 2 = 56
To find 3rd term we have to multiply the previous term with 4   and subtract 2  (double the previous time) from it
I.e 56*2 - 4 = 108
To find 4th term we have to multiply the previous term with 8 (double the previous time) and subtract 2 from it
I.e 108*2 - 8 = 208
Similarly 5th term can be calculated as
multiply the previous term with 2 and subtract  subtract 2 (double the previous time) from it
I.e 208*2 - 16 = 400
Next term 400*2 - 32 = 768

Next term  768*2 - 64 = 1472
Next term  1472*2 - 128 = 2816
Or we can write it in another form as follows
15+(15–1) = 29
29+(29–2) = 56
56+(56–4) = 108
108+(108–8) = 208
208+(208–16) = 400
400+(400–32) = 768

768+(768–64= 1472

1472+(1472-128) =2816

And similarly this trend follows

## What’s the next number in the sequence 0, 2, 6, 12?

We can get the answer in two  ways

### 1st way

1 * (1–1) = 0 ( 1st term)

2 * (2–1) = 2 (2nd term)

3 * (3–1) = 6 (3rd term)

4 * (4–1) = 12 (4th term)

5 * (5–1) = 20 (5th term)

Next term must be  20
Similarly
6 * (6–1) = 30  (6th term)

7 * (7–1) = 42  (7th term)

### 2nd Way

1^2 –1 = 0

2^2 – 2 = 2

3^2 – 3 = 6

4^2 – 4 = 12

5^2 – 5 = 20
Next term must be  20
Similarly

6^2 – 6 = 30

7^2 – 7 = 42

## Can you solve this  Series: 30, 31, 28, 33, 26, 35, … What number should come next?

We can split this sequence into two series by picking alternate numbers from the given sequence.
30, 31, 28, 33, 26, 35, …
1st  one is 30,  28, 26, … and 2nd one is  31, 33,  35, …
Now after careful analysing the 1st sequence ,it next terms should be  with decrease of  2 ,we get other terms as follows
30,  28, 26, 24,22,20 …
And after careful analysing the 2nd  sequence ,it next terms should be  with increase of  2 ,we get other terms as follows
31, 33,  35, ,37,39,41…

26-2 = 24.
So the text terms should be 30, 31, 28, 33, 26, 35, 24,37,22

24 is the next number in the required sequence

____________________________________

## FINAL WORDS

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## Appeal

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## HOW TO ADD 1 TO 1000 NUMBERS VERY EASILY

If you want to add 1 to 1000 then follow these two steps

2 Multiply last number and it's next numbers then divide by 2 . And thats it  .

अगर आप 1 से लेकर किसी भी अंतिम संख्या को जोड़ना चाहते हो तो सबसे पहले अंतिम संख्या में एक जोड़ दें फिर अंतिम संख्या और उससे आगे बाली संख्या को गुणा करके 2 से भाग दे दें ,आप इस तरह से कितनी भी शंख्ययाओं का जोड़ एक दम से कर सकते हो

### उदाहरण

If we have to add 1st 50 numbers then  Multiply 50 with 51 and divide the product with 2 and it is the sum of 1st 50 natural numbers

अगर हम 1 से 50 को जोड़ना चाहें तो 50 और 51 गुना करके 2 से भाग देंगे
1 +2+3+4+5+6+7+……+50 =(50×51)÷2 = 25×51=1275
अगर हम 1 से 100 को जोड़ना चाहें तो 100 और 101 गुना करके 2 से भाग देंगे
1+2+3+4+5+…….+100=( 100×101) ÷2 =50×101=5050
अगर हम 1 से 1000 को जोड़ना चाहें तो 1000 और 1001 गुना करके 2 से भाग देंगे
1+2+3+4+5+…….+1000 =(1000×1001)÷2 = 500×1001= 500500
अगर हम 1 से 2000 को जोड़ना चाहें तो 2000 और 2001 गुना करके 2 से भाग देंगे
1+2+3+4+5+…….+2000=( 2000×2001)÷2 =1000×2001 = 2001000 Share: