+28 A father and his son can dig a well if the father works 6 hours and his son works 12 hours, or they can do it if the father works 9 hours, and the son works 8 hours. How long will it take for the son to dig the well alone?
the answer is aparently 20, but i still dont know how to come up with it. pls help
Let's say that the father can dig a well in f hours, and the son can dig a well in s hours. We know that the father and son can dig a well if the father works 6 hours and the son works 12 hours, or they can do it if the father works 9 hours and the son works 8 hours. This means that:
61+121=f1+s1
and
91+81=f1+s1
We can solve this system of equations to find that f=18 and s=36. Therefore, it will take the son 36 hours to dig the well alone.
F=Father, S=Son
F/ 6 + S / 12 =F/9 + S/8, solve for S in terms of F
S = (4F) / 3
Since his father worked a total of: 6 + 9 = 15 hours, the son could do the same job in:
S =( 4*15) / 3 = 20 hours.
It is also obvious that when the father works: 6+9=15 hours, the son works: 12 + 8=20 hours.
+396 Think in terms of shovels.
If the father can dig, on average, f shovels per hour and it takes W shovels to dig the well, it would take the father, working alone, W/f hours to dig the well.
Similarly if the son can dig, on average, s shovels per hour, it would take him, working alone W/s hours to dig the well, (and notice in passing that this is the number to be calculated).
Now, the well can be dug if the father works for 6hrs and the son for 12 hrs, so
6f + 12s = W
and similarly
9f + 8s = W.
Multiply the first equation by 3 and the second by 2,
18f + 36s = 3W
18f + 16s + 2W,
and subtract,
20s = W,
so W/s = 20.
