kevincan
Three employees rake the beach each day at constant rates. Working together, employees A and B can rake the beach in 3 hours, whereas A and C can rake the beach in 2 1/2 hours . Working together, can A, B, and C rake the beach in less than 2 hours?
(1) B rakes faster than does A.
(2) Working alone, C can rake the beach in less than 5 hours.
A+ B = 3 hrs
A+C = (5/2) hrs.
A+B+C can rake is less than 2 hrs ?
Let’s assume the total work to be 30 units.
A+ B = (30/3) = 10 units
A+C = (30/(5/2)) = 12 units.
C -B = 2 units. Or C = B+2
The possible combinations of (A,B,C) are
(1,9,11)
(2,8,10)
(3,7,9)
(4,6,8)
(5,5,7)
(6,4,6)
(7,3,5)
(8,2,4)
(9,1,3)
Statement 1:
B rakes faster than does A.
As per the statement, B is more efficient than A.
So, initial four cases can be used.
(1,9,11)
(2,8,10)
(3,7,9)
(4,6,8)
If we add the efficiencies, the total sum lies between 21 and 18 inclusive.
For Combined hours equal to 2, we need the efficiency sum to be 15.
So,
sufficient. Statement 2:
Working alone, C can rake the beach in less than 5 hours.
If c can rake in less than 5 hrs, the efficiency = (30/5) = 6.
Efficiency should be greater than 6.
Apart from the four earlier cases of statement 1, we get another case (5,5,7)= sum of 17.
17 is greater than 15.
So, all work for less than 2 hrs.
Hence,
Sufficient Option D