10 years ago, a mother's age was 3×1/2 of her daughter's age and 10 years from now, the mother's age will be 2×1/4 times of her daughter. What will be the sum of the ages of mother and daughter at present?

Correct option is A

Step 1: Translate the Problem into Equations

Let the present ages of the mother and daughter be M and D years, respectively.
1. 10 years ago, the mother's age was $$\frac{7}{2}$$​ times the daughter's age:
M - 10 = $$\frac{7}{2}(D - 10)$$​
2. 10 years from now, the mother's age will be $$\frac{9}{4}$$​ times the daughter's age:
M + 10 = $$\frac{9}{4}(D + 10)$$​

Step 2: Simplify the Equations

From the first equation:
M - 10 = $$\frac{7}{2}(D - 10)$$​
Multiply through by 2:
2M - 20 = 7(D - 10)
Simplify:
2M - 20 = 7D - 70
2M - 7D = -50 (Equation 1)

From the second equation:
M + 10 = $$\frac{9}{4}(D + 10)$$​
Multiply through by 4:
4M + 40 = 9(D + 10)
Simplify:
4M + 40 = 9D + 90
4M - 9D = 50 (Equation 2)

Step 3: Solve the Equations

Using equations (1) and (2):
1. 2M - 7D = -50
2. 4M - 9D = 50

Multiply the first equation by 2:
4M - 14D = -100 (Equation 3)

Subtract Equation 3 from Equation 2:
(4M - 9D) - (4M - 14D) = 50 - (-100)
5D = 150
D = 30

Substitute D = 30 into Equation 1:
2M - 7(30) = -50
2M - 210 = -50
2M = 160
M = 80

Step 4: Calculate the Sum of Ages

The sum of the present ages of the mother and daughter is:
M + D = 80 + 30 = 110

Final Answer:

The sum of their ages is **110 years**.