Suppose the larger pump alone can empty the tank in L hours, and the smaller pump can finish the job in S hours, then each hour the large pump empties 1/L portion of the tank, and the small pump empties 1/S per hour
Working together for three hours, they empty the whole tank, which is 100% of it, so 3/L+3/S=100%=1
Larger pump can empty the tank in 4 hours less than the smaller one, so L=S-4
replace L: 3/(S-4)+3/S=1
Make the denominator the same to solve for:
3S/[S(S-4)] +3(S-4)/[S(S-4)]=1
(3S+3S-12)/[S(S-4)]=1
(3S+3S-12)=[S(S-4)]
S^2-10s+12=0
use the quadratic formula to solve for S
S is about 8.6
The answer is not whole hour.
Dear Amjelugo, we can subtract 4 from 50 to get $46. 4/6 hr= $8, 11/6 hr= $22, and 5/6 hr=$10. 10+22+8=$40 46-40=$6 $6=3/6 hr. Enrique walked Mrs. Camacho's dog and pick up her groceries for 3 hr.
Answer:
A
Step-by-step explanation:
B IS WRONG
C IS WRONG
