Juan has twice as many brothers as sisters. His sister Maria has 5 times as many brothers than sisters. How many children are in the family?

To solve this problem, we can use algebra. Let's assume that Juan has x sisters. According to the given information, Juan has twice as many brothers as sisters. So, Juan has 2x brothers.

Now, let's consider Maria. The problem states that Maria has 5 times as many brothers as sisters. Since Maria is a sister, we have to subtract Juan's sisters from Maria's total siblings. So, Maria has 5(x - 1) brothers. Here we subtract 1 because Maria is not counted as her own sibling.

To find the total number of children in the family, we need to count both Juan's and Maria's siblings. Therefore, the total number of children is x + 2x + (x - 1) + 5(x - 1).

Simplifying the expression, we get:
x + 2x + (x - 1) + 5(x - 1) = 9x - 5

Since we don't have any other information about the number of children in the family, we cannot solve for the specific value of x. However, we can conclude that the family has 9x - 5 children in total.

if there are b boys and g girls,

b-1 = 2g
5(g-1) = b

5g-5 = 2g+1
3g = 6
g = 2
so, b=5

There are b+g=7 children