The sum of two times the present age of A and three times the present age of B is 106 years. Four times the present age of B exceeds three times the present age of A by 11 years. What will be the sum of the ages (in years) of A and B, 4 years from now?

Correct option is D

Given:

Sum of two times the present age of A and three times the present age of B is 106 years.

Four times the present age of B exceeds three times the present age of A by 11 years:

Solution:

Let the present age of A be $x$ years and present age of B be $y$ years.

From 1st condition

​$$2x + 3y = 106 \quad \text{(Equation 1)}$$ 

From 2nd condition  

​$$4y = 3x + 11$$ 

​$$y = \frac{3x + 11}{4} \quad \text{(Equation 2)}$$ 

Substitute $$y =\frac{3x + 11}{4}$$​ into Equation 1: 

​$$2x + 3 \left( \frac{3x + 11}{4} \right) = 106$$ 

$$2x + \frac{9x + 33}{4} = 106$$ 

$$\frac{8x+9x + 33}{4}= 106$$ 

$$8x + 9x + 33 = 106×4=424$$ 

$$17x + 33 = 424$$ 

$$17x= 391$$ 

​$$x = \frac{391}{17} = 23$$ 

​Now, substitute $x=23$ into Equation $$y =\frac{3x + 11}{4}$$ 

​$$y = \frac{3(23) + 11}{4}$$ 

$$y = \frac{69 + 11}{4} = \frac{80}{4} = 20$$ 

​So, the present age of A is $23$ years and  the present age of B is $20$ years

4 years from now , the age of A will be 23 + 4 = 27 years and the age of B will be 20 + 4 = 24 years 

The sum of their ages 4 years from now is: 27 + 24 = 51 years

Thus, the sum of the ages of A and B, 4 years from now, is 51 years