Chapter 11 QuaDraTiC appliCaTionS 405

Worker Quantity Rate Time

Pipe I (^115)
t− 3
t −^53
Pipe II 1 1
t
t
Together 1 1
2
2
The equation to solve is^111
tt 53 2
.
– 
+  =  The LCD is 2 tt() .– ^53
111
2
2 5
3
1 2 5
3
(^12)
(^53)
(^53)
tt
tt
t
tt
t
−
+=
 −



⋅
−
+−



⋅=ttt
tt tt
tt
 −



⋅
+−




=−




+−
5
3
1
2
225
3
5
3
22110
3
5
3
4 10
3
5
3
34 10
3
3 5
3
2
2
2
=−
−=−
 −



=−

tt
ttt
tt t

−=−
=−+
=− −
12 10 35
03 17 10
0532
2
2
ttt
tt
()tt()
t
t
−=

50
5
320
32
2
3
t
t
t
−=

=
(t =^23 cannot be a solution because t −^53 would be negative.)
Pipe II can fill the reservoir in 5 hours and Pipe I can fill it in
5 5
3
5
1
3
3
5
3
15
3
5
3
10
3
–  =  ⋅ –  =  –  =  =  313 hours or 3 hours 20 minutes.