Two water taps together can fill a tank in 75/8 hrs. The top of larger diameter takes 10 hr. less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Suppose that the faster pipe takes x hours to fill the tank

slower pipe will take x+10 hours to fill the tank

since faster tap (larger diameter tap) takes x hours to fill the tank,

portion of tank filled in  1 hour = 1/x

portion of tank filled in 75/8 hours = 75/8x --------------------(i)

similarly portion of tank filled by slower pipe in x+10 hours to fill the tank is 1/(x+10)

portion of tank filled in 75/8 hours = 75/[ 8* (x+10) ] -----------------------(ii)

it is given that the tank is filled in 75/8 hours

so

75/8x + 75/[ 8* (x+10) ] = 1

taking 75/8 common outside and moving it to the RHS we get

1/x + 1/(x+10) = 8/75

( x + 10 + x )/ x(x+10) = 8/75

75 * (2x + 10) = 8 * (x²+10x)

150x + 750 = 8x² + 80x

8x² - 70x - 750 = 0

dividing by 2

4x² - 35x - 375 = 0

now solve the equation and ignore the negative answer obtained !

you will get the time taken by the tap of larger diameter, add 10 to it and you will get the time taken by the smaller diametered tap

PLEASE APPROVE !