one pipe can fill a pool 1.25 times faster than a second pipe . when both pipes are opened , they fill the pool in five hours . how long would it take to fill the pool if only the faster pipe is used ?
say the rate of the slower pipe is r pool / hour , then the rate of the faster pipe would be 1.25 r = 5 r / 4 . since when both pipes are opened , they fill the pool in five hours , then their combined rate is 1 / 5 pool / hour . thus we have that r + 5 r / 4 = 1 / 5 - - > r = 4 / 45 pool / hour , faster pipe fills at 1.25 r which is 5 / 4 * 4 / 45 = 1 / 9 - - > time is reciprocal of rate thus it ' s 9 / 1 = 9 hours . answer : e .
divide(inverse(divide(inverse(add(const_3, const_2)), add(const_1, 1.25))), 1.25)