if a boy is 4 years less than twice his sisters age and the sum of the age is 11 how old is he
go through each of the possibilities of getting to 11. 1+10 2+9 3+8 4+7 and so on. think of what you are looking for, twice the girls age will be four more than the boys age.
or, using algebra
b+s=11
b=2s-4
2s-4+s=11
3s=15
s=5
so, b=6
check:
6=2*5-4
To solve this problem, let's assign variables to the ages. Let's say the boy's age is represented by 'B', and his sister's age is represented by 'S'.
According to the problem, we know that the boy is 4 years less than twice his sister's age:
B = 2S - 4
We are also given that the sum of their ages is 11:
B + S = 11
Now we can solve this system of equations to find the values of B and S.
Substitute the value of B from the first equation into the second equation:
(2S - 4) + S = 11
Combine like terms:
3S - 4 = 11
Add 4 to both sides of the equation:
3S = 15
Divide both sides of the equation by 3:
S = 5
Now that we know S = 5, we can substitute this value back into the second equation to find B:
B + 5 = 11
Subtract 5 from both sides of the equation:
B = 6
Therefore, the boy is 6 years old.