The difference between the ages of two sisters is 10 years. If 15 years ago, the elder one was twice 10 years. If 15 years ago, the elder one was twice as old as the younger one, find their present ages.

Hint: In order to solve this type of problems related to age, we need to follow the following steps:-
a) Express what we don't know as a variable.
b) Create an equation based on the information provided.
c) Solve for the unknown variable.
d) Substitute our answer back into the equation to see if the left side of the equation equals the right side of the equation and hence, we will get our final result.

Complete step-by-step answer:
Let the present age of 1st sister be x;
Let the present age of 2nd sister be y;
So, difference between the ages of two sisters is 10 years;
⇒ (x – y) = 10 ……………..(i)
15 years ago;
Age of 1st sister = x – 15
Age of 2nd sister = y – 15
According to the question,
Before 15 years, elder was twice 10 years;
So,
 ⇒ x – 15 = 2 × 10
⇒ x – 15 = 20 …(ii)
⇒ x = 20 + 15 = 35
⇒ x = 35 Years
And, it's also given that: -
⇒ (x – 15) = 2(y – 15)
⇒ (35 – 15) = 2(y – 15)
⇒ 20 = 2y – 30
⇒ 20 + 30 = 2y
⇒ 2y = 50
∴ y = 25 years
∴Present age of 1st sister (elder one) =
x = 35 years.
    Present age of 2nd sister (younger one) =
Y = 25 years.



Note: In order to solve age related problems, we need no some of the important points which are as follows:-
a.) If the present age is y, then n times the present age = ny.
b.) If the present age is x, then age n years later/hence = x + n.
c.) If the present age is x, then age n years ago = x – n.
d.) The ages in a ratio a: b will be ax and bx.