Problem Statement: Two water taps can fill a tank in 8 hours. The tap with a larger diameter takes 10 hours less than the tap with a smaller diameter to fill the tank separately. We need to find the time it takes for each tap to separately fill the tank.

Let's assume:
Let the smaller tap take x hours to fill the tank separately.
Therefore, the larger tap will take (x - 10) hours to fill the tank separately.

Understanding the Problem:
The rate at which the taps fill the tank is inversely proportional to the time taken. In other words, if a tap takes less time to fill the tank, it has a higher filling rate.

Using the Concept of Work:
We can use the concept of work to solve this problem. The work done by a tap is given by the product of its filling rate and the time taken. Since the work done by both taps together is to fill the tank, we can write the equation as follows:

1/x + 1/(x - 10) = 1/8

Simplifying the Equation:
To simplify the equation, we can take the least common multiple (LCM) of the denominators, which is 8(x)(x - 10). Multiplying all terms by this LCM, we get:

8(x - 10) + 8x = (x)(x - 10)

Simplifying further:

8x - 80 + 8x = x^2 - 10x

16x - 80 = x^2 - 10x

Rearranging the equation:

x^2 - 26x + 80 = 0

Solving the Quadratic Equation:
To solve the quadratic equation, we can factorize it or use the quadratic formula. Factoring the equation, we get:

(x - 20)(x - 4) = 0

This gives us two possible solutions: x = 20 and x = 4. However, x cannot be equal to 4 because the larger tap would take negative time (4 - 10 = -6), which is not possible. Therefore, x = 20 is the only valid solution.

Final Answer:
Hence, the smaller tap takes 20 hours to fill the tank separately, while the larger tap takes (20 - 10) = 10 hours to fill the tank separately.