Let Pool A and Pool B= \(a\) when full.
Let Pipe X have a rateof \(2x\) , and therefore Pipe Y has a rateof \(x\) .
This means that it'll take Pipe X, alone,\(\frac{a}{2x}\) to empty thepool.
After \(t\) hours, Pipe X willhave \(a-2xt\) to pump out and Pipe Y willhave \( a -xt\) left to pump out.
As we are told," the remaining volume of water in Pool A is two-thirds that of PoolB":
\( a-2xt =\frac{2}{3} (a -xt)\)
\( 3a - 6xt = 2a -2xt\)
\( a =4xt\)
Plugging this backinto \(\frac{a}{2x}\):
\(\frac{4xt}{2x}\)
\(2t\)
ANSWERD
...
Let Pipe X have a rateof \(2x\) , and therefore Pipe Y has a rateof \(x\) .
This means that it'll take Pipe X, alone,\(\frac{a}{2x}\) to empty thepool.
After \(t\) hours, Pipe X willhave \(a-2xt\) to pump out and Pipe Y willhave \( a -xt\) left to pump out.
As we are told," the remaining volume of water in Pool A is two-thirds that of PoolB":
\( a-2xt =\frac{2}{3} (a -xt)\)
\( 3a - 6xt = 2a -2xt\)
\( a =4xt\)
Plugging this backinto \(\frac{a}{2x}\):
\(\frac{4xt}{2x}\)
\(2t\)
ANSWERD
...
Statistics : Posted by Nidzo • on 04 Jul 2024, 01:30 • Replies 4 • Views 121
