one pipe can fill a pool 1.5 times faster than a second pipe . when both pipes are opened , they fill the pool in four hours . how long would it take to fill the pool if only the slower pipe is used ?
"say the rate of the slower pipe is r pool / hour , then the rate of the faster pipe would be 1.5 r = 3 r / 2 . since when both pipes are opened , they fill the pool in four hours , then their combined rate is 1 / 4 pool / hour . thus we have that r + 3 r / 2 = 1 / 4 - - > r = 1 / 10 pool / hour - - > time is reciprocal of rate thus it ' s 10 / 1 = 10 hours . answer : e ."
divide(inverse(divide(inverse(add(const_3, const_2)), add(const_1, 1.5))), 1.5)