The average of present age of A, B and C is 29 years. If age of A is two years less than that of B and age of B is eight years less than C, then find the present age of A.

The correct option is 25 years.

Let us solve this problem step-by-step by setting up algebraic equations based on the information provided in the question.

Step 1: Express the sum of their present ages.
We are given that the average present age of A, B, and C is 29 years.
The average of three numbers is calculated as the sum of the numbers divided by 3.
Therefore,

$\frac{\text{Age of A}+\text{Age of B}+\text{Age of C}}{3}=29$

Multiplying both sides by 3 gives the total sum of their ages:

$\text{Age of A}+\text{Age of B}+\text{Age of C}=29\times 3=87$

Step 2: Express all ages in terms of A's age.
Let the present age of A be $A$ years.
1. We are given that the age of A is 2 years less than that of B:
$A=B-2\Rightarrow B=A+2$
2. We are also given that the age of B is 8 years less than C:
$B=C-8\Rightarrow C=B+8$
Substituting $B=A+2$ into the equation for C:
$C=(A+2)+8=A+10$

Step 3: Substitute these expressions into the sum equation and solve for A.

$A+B+C=87$

$A+(A+2)+(A+10)=87$

$3A+12=87$

Subtract 12 from both sides:

$3A=87-12$

$3A=75$

Divide by 3:

$A=\frac{75}{3}=25$

Hence, the present age of A is 25 years.