The age of a father is twice that of his elder son. Ten years hence, the age of the father will be three times that of his younger son. If the difference between the ages of the two sons is $15\text{years}$, the current age of the father is:

Let the present age of the Father be $F$, the Elder son be $E$, and the Younger son be $Y$.
From the given conditions, we can form the following linear equations:

1. The age of the father is twice that of the elder son:

$$F=2E⟹E=\frac{F}{2}\text{— (i)}$$

2. Ten years hence, the father's age will be three times that of the younger son:

$$F+10=3(Y+10)$$

$$F+10=3Y+30$$

$$3Y=F-20⟹Y=\frac{F-20}{3}\text{— (ii)}$$

3. The difference between the ages of the two sons is $15\text{years}$:

$$E-Y=15\text{— (iii)}$$

Substitute the expressions for $E$ and $Y$ from equations (i) and (ii) into equation (iii):

$$\frac{F}{2}-\frac{F-20}{3}=15$$

Multiply the entire equation by the least common multiple ($\text{LCM}$) of $2$ and $3$, which is $6$, to clear the fractions:

$$6\times \left(\frac{F}{2}\right)-6\times \left(\frac{F-20}{3}\right)=15\times 6$$

$$3F-2(F-20)=90$$

$$3F-2F+40=90$$

$$F+40=90$$

$$F=90-40=50\text{years}$$

Therefore, the present age of the father is $50\text{years}$.