Anil and Beena are friends, and the difference between their ages is $5$ years. Anil’s father Dinesh is three times as old as Anil, and Beena is twice as old as her sister Charu. The ages of Dinesh and Charu differ by $45$ years. If Beena is older than Anil, then find the sum of the present ages of both Dinesh and Charu.

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**Solution:**

1. Let Anil's age be $A$, Beena's age be $B$, Dinesh's age be $D$, and Charu's age be $C$.

2. Given that Beena is older than Anil by $5$ years:
   $$B = A + 5$$

3. Dinesh is three times as old as Anil:
   $$D = 3A$$

4. Beena is twice as old as Charu:
   $$B = 2C \implies C = \frac{B}{2} = \frac{A + 5}{2}$$

5. The difference between Dinesh's and Charu's ages is $45$ years:
   $$D - C = 45$$

6. Substitute $D$ and $C$ from above:
   $$3A - \frac{A + 5}{2} = 45$$

7. Multiply both sides by $2$ to clear the denominator:
   $$6A - (A + 5) = 90$$

8. Simplify:
   $$6A - A - 5 = 90 \implies 5A - 5 = 90$$

9. Add $5$ to both sides:
   $$5A = 95$$

10. Divide both sides by $5$:
    $$A = 19$$

11. Calculate Dinesh's age:
    $$D = 3 	imes 19 = 57$$

12. Calculate Charu's age:
    $$C = \frac{19 + 5}{2} = \frac{24}{2} = 12$$

13. Find the sum of Dinesh's and Charu's ages:
    $$D + C = 57 + 12 = 69$$

**Answer:** $69 	ext{ years}$

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