Two water taps together can fill a tank in 1 7/8 hours. The tap with a longer diameter takes 2 hours less than the tap with a smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Let's assume that the tap with the smaller diameter takes x hours to fill the tank separately. Therefore, the tap with the larger diameter will take x - 2 hours to fill the tank separately.

We are given that when both taps are open together, they can fill the tank in 1⅞ hours, which can be expressed as 15/8 hours.

To solve this problem, we can set up the following equation based on the rates at which the taps fill the tank:

1 / x + 1 / (x - 2) = 8 / 15

To simplify this equation, we can multiply both sides by the least common multiple (LCM) of the denominators, which is 15(x)(x - 2):

15(x - 2) + 15x = 8(x)(x - 2)

15x - 30 + 15x = 8x^2 - 16x

30x - 30 = 8x^2 - 16x

Rearranging the equation:

8x^2 - 46x + 30 = 0

Dividing the entire equation by 2 to simplify it:

4x^2 - 23x + 15 = 0

Now, we can solve this quadratic equation for x using factoring or the quadratic formula.

Factoring:
4x^2 - 23x + 15 = (4x - 3)(x - 5) = 0

Setting each factor to zero:
4x - 3 = 0 --> x = 3/4
x - 5 = 0 --> x = 5

Since x represents the time taken by the smaller tap, it cannot be negative. Therefore, x = 5 is the valid solution.

Hence, the tap with the smaller diameter takes 5 hours to fill the tank separately, and the tap with the larger diameter takes 5 - 2 = 3 hours to fill the tank separately.