## Ten most important questions of Reasoning Analogy in missing numbers in various figures

Ten most Important questions of Reasoning Analogy of missing numbers have been discussed with their Solutions in an easy manners. These questions are very very important for competitive exams like SSC CGL , SSC CHSL And RRB NTPC Etc Because such types of problems are frequently asked in these exams.

** Problem #1**

Moving clock wise from top right number , Double the top right number to get bottom right number and add 2 to get bottom left number, now again double this number to get top left number in all the three figures . ##
Hence in option (4) 24 is right answer

Hence in option (4) 24 is right answer

##
** ****Problem #2**

Multiply all the numbers which are on the corners of the given triangles to get the number which is on the centre of particular triangle.

In 1st triangle multiply all the numbers on the corner of triangle

1 × 2 × 3 = 6

Again In 2nd triangle multiply all the numbers on the corner of triangle

2 × 5 × 3 = 30

Similarly In 3rd triangle multiply all the numbers on the corner of triangle

4 × 8 × 5 = 160

##
Hence in option (3) 160 is right answer

Hence in option (3) 160 is right answer

**Problem #3**

##
If we add the top and bottom numbers and the subtract from it the sum of left and right terms we shall get the middle number in all the figures.

in 1st figure
(3+7) - (2+5) = 10 - 7 = 3

In 2nd figure
(9+9) - (7+8) =18 -15 = 3

In 3rd figure
(8+8) - (6+6) = 16 -12 = 4

Hence in option (2) 4 is right answer

If we add the top and bottom numbers and the subtract from it the sum of left and right terms we shall get the middle number in all the figures.

in 1st figure

in 1st figure

(3+7) - (2+5) = 10 - 7 = 3

In 2nd figure

In 2nd figure

(9+9) - (7+8) =18 -15 = 3

In 3rd figure

In 3rd figure

(8+8) - (6+6) = 16 -12 = 4

Hence in option (2) 4 is right answer

**Problem # 4**

Multiply the top numbers and bottom numbers row wise the take the difference of it .

In 1st figure

( 5 × 6 ) - ( 3 × 8) = 30 - 24 = 6

2nd figure

( 10 × 4 ) - ( 2 × 7 ) = 40 - 14 =26

3rd figure

( 9 × 7 ) - ( 6 × 8 ) = 63 - 48 = 15

##
Hence in option (2) 15 is right answer

Hence in option (2) 15 is right answer

**Problem #5**

##

Cross multiply the numbers to get middle number in all the figures .

## In 1st figure

## 12 × 5 = 15 × 4 = 60

## In 2nd figure

3 × 14 = 6 × 7 = 42

In 3rd figure

26 × 3 = 13 × 6 = 78

Hence in option (3) 78 is right answer

Hence in option (3) 78 is right answer

**Problem #6**

##
Take cube roots of all the numbers which are on outer side

all the ellipse ( given figure ) and then add these numbers to get middle number in the figure .

In 1st figure cube roots of 1, 64, 27 and 8 are 1 , 4 , 3 and 2 respectively.

Now add these numbers. i.e 1 + 4 + 3 +2 = 10
In 2nd figure cube roots of 8, 125, 64 and 27 are 2 , 5 , 4 and 3 respectively.

Now add these numbers. i.e 2 +5 + 4 +3 = 14
Similarly In 3rd figure cube roots of 27 , 216 , 125 and 64 are 3 , 6 , 5 and 4 respectively.
Now add these numbers. i.e 3 + 6 + 5 + 4 = 18

Hence in option (4) 18 is right answer

Take cube roots of all the numbers which are on outer side

all the ellipse ( given figure ) and then add these numbers to get middle number in the figure .

all the ellipse ( given figure ) and then add these numbers to get middle number in the figure .

In 1st figure cube roots of 1, 64, 27 and 8 are 1 , 4 , 3 and 2 respectively.

Now add these numbers. i.e 1 + 4 + 3 +2 = 10

In 2nd figure cube roots of 8, 125, 64 and 27 are 2 , 5 , 4 and 3 respectively.

Now add these numbers. i.e 2 +5 + 4 +3 = 14

Similarly In 3rd figure cube roots of 27 , 216 , 125 and 64 are 3 , 6 , 5 and 4 respectively.

Now add these numbers. i.e 3 + 6 + 5 + 4 = 18

Hence in option (4) 18 is right answer

**Problem #7**

In all the figure add both numbers which are 2nd and 3rd rows then multiply this result with the number which is on the right side of given figure in the 1st row to get the number on the left side of 1st row of given figure.

In 1st figure 1 + 2 =》3 × 9 = 27

In 2nd figure 2 + 3 =》 5 × 7 = 35

In 3rd figure x + 4 =》 (x+4) × 4 = 36

4x + 16 = 36

4x = 36 - 16 = 20

x = 5

Hence in option (3) 5 is right answer

**Problem #8**

####
__1st Figure__

__1st Figure__

From 1st figure we can see that lower number is thrice (3 times) of sum ( addition ) of both numbers which are on upper portion of figure i. e . 18 + 9 = 27 .

Now multiply it with 3 , we get 27 ×3 =81 , which is the lower number in 1st figure.

####
__3rd Figure__

__3rd Figure__

And for third figure add both the numbers 24 and 7 and then multiply the sum with 3 like this 24 + 7 = 31 and 31 × 3 = 93

####
__2nd Figure__

Similarly ? in 2nd figure can be found by adding 23 and 8 and then Multiply it with 3 as follows __2nd Figure__

**23 + 8 = 31 × 3 = 93 ,**Hence 93 is the required number.

Hence in option (3) 93 is right answer

**Problem #9**

From 1st two figures if we multiply top and bottom terms and then subtract the result of product of left and right terms from it we shall get the number in the middle of the circles in both the given figures.

( 6 × 7 ) - ( 4 × 5 ) = 20 ( Middle Number in 1st Figure )

( 9 × 7 ) - ( 6 × 3 ) = 45 ( Middle Number in 2nd Figure )

Similarly middle number of 3rd figure can be calculated as follows

( 8 × 8 ) - ( 4 × 6 ) = 40 ( Middle Number in 3rd Figure )

Hence in option (3) 40 is right answer

**Problem #10**

If we add the left and right numbers and the subtract from it the sum of top and bottom terms we shall get the middle number in all the figures

( 5 + 6 ) - ( 4 + 7) = 0

( 7 + 6 ) - ( 8 + 4 ) = 1

similarly 3rd term can be calculated as follows

( 11 + 2 ) - ( 0 + 2 ) = 11

Hence in option (3) 11 is right answer

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