Brothers and Sisters

If each boy has as many brother as sisters, it implies, there are x girls, and x+1 boys (counting the one boy whose perspective is shown here). Now if, every girl has twice as many brother as sisters, it gives the equation: 2(x-1) = x+1

Solving for x, we get x=3. Thus, Girls = x = 3 Boys = x+1 = 4

Total = 3+4 = 7

no. of girls =x

no. of boys=y

from the first problem statement-

y-1=x

from the second-

2(x-1)=y

solving the linear equation in two variables-

x=3

y=4

therefore total no. of children =3+4=7

First we have to consider that when any boy or girl compares his/her brothers with his/her sisters , he excludes himself /herself (during that comparison) from the group that he/she belongs to , be it brothers (boys) group or sisters (girls) group .

Then, Let the boys group be $y$ and the girls group be $x$ .

And when any boy compares his brothers (other boys) with his sisters (all girls) , he finds that the ratio is : $y - 1 : x \\ \ \ 1 \ \ : 1$ This implies that : $y - 1 = x \ \ \ \ \ \ \ \ \ \ \ , \ \ \ \ \ \ \ y = x + 1 \ \ \ \ \ \ \Rightarrow (1)$ And when any girl compares her brothers (all boys) with her sisters (other girls) , she finds that the ratio is : $y : x - 1 \\ 2 : \ \ 1 \ \$ This implies that : $y = 2 (x - 1) = 2x - 2$ Substituting in $(1)$ : $2x - 2 = x + 1 \ \ \ \ , \ \ \ \ 2x - x = 2 + 1 \ \ \ , \ \ \ \ x = \boxed{3}$

Substituting in $(1)$ again : $y = x + 1 \ \ \ , \ \ y = (3) + 1 = \boxed{4}$

Therefore, $y + x = 4 + 3 = \boxed{7}$ Which is the sum of the number of brothers (boys) and sisters (girls) .

It says 7 children in the beginning guys...

b is the number of boys and g is the number of girls

Each boy has b - 1 brothers and g sisters Each girl has b bothers and g - 1 sisters

Therefore b -1 = g and g - 1 = b/2

=> b - 2 = b/2

=> b = 4

=> 4 - 1 = g = 3

b + g = 4 + 3 = 7

Result: 7

I thought alex has 4 bro and sis. so belly has 8.removing belly it is 7.

Let S be the total number of sisters and B be the total number of brothers

S=B-1

B=2(S-1)

S=B-1 --> B=S+1

S+1=2(S-1)

S+1=2S-2

1+2=2S-S

3=S

B=S+1

B=3+1=4 S=3; B=4 ---> S+B=7

1.Each boy has as many sisters as brother. 2.Each girl has twice as many brothers as sisters. Let total number of boys=x total number of girls=y then x-1=y. according to first statement. x=2(y-1) according to second statement. On solving both the equations. x=4 , y=3 x+y = 7

the sum of total is in first row..... it's so simpal pure logic.......

M = male, F = female

M - 1 = F and M = 2(F - 1) = 2F - 2

Therefore 2F - 2 - 1 = F so F = 3 and M = 4 and M + F = 7

From the perspective of a boy excluding himself, there are as many girls as there are remaining boys, or g=b-1. Likewise, from the perspective of a girl excluding herself, there are twice as many boys as there are remaining girls, or b=2(g-1). Doing the math:

b=2g-2

b=g+1 (First equation)

g+1=2g-2

g+3=2g

g=3

3+1=b=4

b+g=4+3=7

I took a slightly different logic It might be useful in competitive exams. Consider the no of girls as x thus according to first condition there are x+1boys. Totally2x+1 children. So the answer must be odd which brings down to option 5 or 7. But 5 cant be possible so the answer is 7