The present age of a woman is 30years older than her daughter. 15years ago she was twice as old as her daughter. How old is her daughter at present. How old would the woman be 12years from now.
W-30=D
w-15=2(D-15)
subtract the equations:
-30+15=D-2D+30
D=30+15=45
W=75
since w=75 then then the daughter will be: D=75-30
=45
and the women will be 75-12=63
To solve this problem, we can use algebraic equations to represent the given information and find the solution.
Let's first define variables:
- Let "W" represent the present age of the woman.
- Let "D" represent the present age of her daughter.
According to the problem, the present age of the woman is 30 years older than her daughter. This can be represented as:
W = D + 30
The problem also states that 15 years ago, the woman was twice as old as her daughter. We can express this information as:
(W - 15) = 2(D - 15)
Now, we have two equations:
1) W = D + 30
2) (W - 15) = 2(D - 15)
We can now solve these equations simultaneously to find the values of W and D.
Let's substitute the value of W from equation 1 into equation 2:
((D + 30) - 15) = 2(D - 15)
Simplifying, we get:
D + 15 = 2D - 30
Moving all terms to one side, we have:
2D - D = 15 + 30
D = 45
Therefore, the present age of the daughter is 45 years.
To find the present age of the woman, we can substitute the value of D into any of the two equations we formed earlier. Let's use equation 1:
W = D + 30
W = 45 + 30
W = 75
Therefore, the present age of the woman is 75 years.
Now, let's calculate how old the woman would be 12 years from now:
Present age of the woman + 12
75 + 12
= 87
So, the woman would be 87 years old 12 years from now.