## Twelve Most Interesting figures problems of Reasoning analogy part 3

Discussed the most important  reasoning analogy problems and frequently asked picture problems in previous competitive  examinations along with their answers .

If square root of any number is added /subtracted infinite time ,then what will be the answers .

## Problem # 1

### (a) 5        (b) 25            (c) 3             (d) 49

Outermost numbers are the  product of Square roots of two  numbers attached to it.

Since 14 is attached with 4 and 49 , the square root of 4 and 49 are 2 and 7 respectively now multiplying 2 and 7 to get 14.

Since 12 is attached
with 36 and 4 , the square root of 36 and 4 are 6 and 2 respectively now multiplying 6 and  2  t get 12.

So 30 must be the product Square Root  of 36  and ? , so if we put 25 instead of Question Mark   ? .Then product of square root of 25 and the square root of 36 which will be  equal to 30.

Similarly product of square root of  49 and Square Root of ? must be equal to 12 . So we had already put 25 in place of ? to give product 35 .

Hence  (b) 25  is right option.

## Problem # 2

Starting from 3 and moving clockwise multiply 3 with 2 and subtracting 1 from it  to get 5 .
Multiplying 5 with 2 and subtracting 3 from result so obtained to get 8.
Multiplying 8 with 2 and subtracting 2 from it to get 13.
Now multiplying 13 with 2 and then subtracting 4 from  result so obtained to get 22.
At last multiplying 22 with 2 then subtracting 5 from the result so obtained to get 39 .
so (c) 39 is the right answer

## 1st Step

In  1st Row B ,D are F are written in a  gap of one letter .
In 3rd Row N ,P and R are written in a gap of one letter
In 2nd Row H and J are also written in a gap of one letter ,So J and ? must be written with a gap of one letter .This means after J the Lett with one gap will be L.

## 2nd Step

In 1st row  digits associated are 3 ,3 and 6 means last digit in the Row is sum of 1st two digits 3+3 = 6.
In 3rd row  digits associated are 7 ,9  and 16 means last digit in this Row is sum of 1st two digits 7+9 = 16.
In 2nd  row  digits associated are 5 ,6 and ? will be replaced by sum of other two digits 5+6 = 11 .

So (a) L11 is the right option

## Problem # 4

Here in this Figure  result will  be calculated column wise in any particular column. In first column the fourth element is calculated by dividing second element with third  element and then multiplying first and the results so obtained.

## Problem # 5

Add all the numbers in any particular figure and then subtract 2 from it to get the middle number .

In figure number 1 we have 0 + 6 + 4 + 2 = 12 minus 2 = 10 { middle number }
In figure second 6 + 2 + 10 + 8 = 26 - 2  = 24 .
Similarly In figure 3  we have 4 + 14 + 12 + 10 = 40 - 2 = 38 ( Question Mark )  which is are required number.

Therefore option ( C ) is  right  answer .

### Want to check more reasoning problems with their solutions ,click here,.

## Problem # 6

If we add all the numbers column wise then we get 99 in all the three columns so we shall  have 42 in place of question mark to  have total   99 in all the three in first column .
Because 40 + 24 + 35 = 99
In second column we have 30 + 35 + 34 = 99 . While in last column when we add ? + 30  +27 the sum will be 99 so desired number  will be 42.
The right option is ( a) 42

## Problem # 7

Find the sum of numbers which are in the Row ( horizontal ) then subtract this total from the sum which are in column ( vertical ) to find the number which is in the middle of the given figure .
In figure I  5 + 6 is equal to 11 and  4 + 7 which is also equal to 11 so difference of both the  sum is equal to zero which is the middle number

In figure II  7 + 6 is equal to 13 an  8+ 4 which is also equal to 12 so difference of both the  sum is equal to  13  - 12 1which is the middle number

In figure III  11 + 2 is equal to 13 and  0 + 2 which is also equal to  2 so difference of both the  sum is equal to  13 - 0 = 11 which will be the middle number.
Correct option is c(11)

## Problem # 8

Multiplying all the elements in 2nd row with 2 then add elements of 1st and 2nd row downward to get the elements in  3rd row
1st Column.    13  + ( 7×2 ) = 13 + 14 = 27
2nd Column 54 + (45 × 2)  = 54 + 90 = 144
3rd Column   ? + (2× 32) = 68
This implies ? = 68 - 64 = 4
So (4) is right option .

## Problem # 9

### (d) 49

To find the middle number in each figure add all the numbers except middle number then divide it with 4 to  get middle number .

In figure 1 add 24 + 32 + 40 + 36 = 132 which  when divided by 4 gives 33

In figure II add the numbers  27 +19 +  20 + 22 = 88 Which when divided by 4 gives 22

similarly in III  figure when we add all the numbers 6 + 16 + 14 + 1 2 = 48 which is when divided by 4  gives 12
,So option   ( b ) is right option

## Problem # 10

Sum of all the digits of 1st and 2nd rows are 17 .Similarly sum of all digits of 3rd row must be 4 +7 + ? = 17 .This implies  the value of  ? should be 6.
Hence the correct  option will be  ( a ) 6

## Problem # 11

Spilt this circle into two parts ,one above x axis and other below x axis , the sum of digits in both the semi circle must be equal to each  other .
This means ? + 4 + 5 + 6 = 11 + 9 + 3+7
? +15  = 30
? = 15
So option ( d ) 15 will be right option

## Problem # 12

This series consists of two series of alternative numbers ,first one consists of 2, 4, 8, 16,32 so on and second series consists of the number 6, 9, 13, 18, so on .
So our next number in the given series will be derived from the second series, .6, 9, 13, 18, so on .
Look at the difference between first two number 6 and 9 which is equal to 3 ,And  difference between second number (9) and third number (13 ) in 2nd series is 4 And  difference between third number ( 13 ) and fourth number ( 18 ) in the second series is 5 so difference between 18 and next number must be 6 it  means next number must be 24 because 24 - 18 = 6
Show the right option is option (a) answer

## FINAL WORDS

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#### 1 comment:

1. Very good technique of getting solution