Let Anil takes x number of days to complete the work

⇒ 1 day = 1/x of the work

Sunil takes 5 more days than Anil to complete the work

⇒ x + 5 days

⇒ 1 day = 1/(x + 5) of the work

Together:

1 day = 1/x + 1/(x + 5)

1 day = (x + 5 + x)/x(x+5)

1 day = (2x + 5) / (x² + 5x) of work done

Find 4 days of work done:

1 day = (2x + 5) / (x² + 5x)

4 days = 4(2x + 5) / (x² + 5x)

4 days = (8x + 20) / (x² + 5x) of work done

Find the amount of work left to be done:

Work left = 1 - (8x + 20) / (x² + 5x)

Work left = (x² + 5x - 8x - 20) / (x² + 5x)

Work left = (x² - 3x - 20) / (x² + 5x)

Solve x:

Give that Anil needs 5 days to complete the rest of the work

(x² - 3x - 20) / (x² + 5x) ÷1/(x + 5) = 5

(x² - 3x - 20) / (x² + 5x) × (x + 5) = 5

(x² - 3x - 20) /x = 5

x² - 3x - 20 = 5x

x² - 8x - 20 = 0

(x - 10) (x + 2) = 0

x = 10 or x = -2 (Rejected, number of days cannot be negative)

Find the number of days each take to complete the work

Anil = x = 10 days

Sunil = x + 5 = 10 + 5 = 15 days

Answer: Anil takes 10 days and Sunil takes 15 days to complete the work.