Respuesta :
container B needs to be filled faster in order for both containers to gain the same amount of water per minute , and by 1/6 water per minute .
Step-by-step explanation:
Here we have , two leaky containers are filling with water. water enters container a at a rate of 2/3 cup per 1/2 minute and leaks out at a rate of 1/4 cup per 3/4 minute. water enters container b at a rate of 3/4 cup per 1/2 minute and leaks out at a rate of 1/2 cup per 3/4 minute. We need to find which container needs to be filled faster in order for both containers to gain the same amount of water per minute . Let's find out:
Rate of Filling container A:
⇒ [tex]\frac{\frac{2}{3} }{\frac{1}{2} }[/tex]
⇒ [tex]\frac{2}{3} (2) = \frac{4}{3}[/tex]
Rate of leaking Container A:
⇒ [tex]\frac{\frac{1}{4} }{\frac{3}{4} }[/tex]
⇒ [tex]\frac{1}{4} (\frac{4}{3}) = \frac{1}{3}[/tex]
Amount of water in Container A in one minute :
⇒ [tex]\frac{4}{3} - \frac{1}{3} = 1[/tex]
No , following same for Container B
Rate of Filling container B:
⇒ [tex]\frac{\frac{3}{4} }{\frac{1}{2} }[/tex]
⇒ [tex]\frac{3}{4} (\frac{2}{1}) = \frac{3}{2}[/tex]
Rate of leaking Container B:
⇒ [tex]\frac{\frac{1}{2} }{\frac{3}{4} }[/tex]
⇒ [tex]\frac{1}{2} (\frac{4}{3}) = \frac{2}{3}[/tex]
Amount of water in Container B in one minute :
⇒ [tex]\frac{3}{2} - \frac{2}{3} = \frac{5}{6}[/tex]
Since , Rate of water per minute in container A is 1 and , Rate of water per minute in container B is 5/6 , container A fills faster by container B by
⇒ [tex]1-\frac{5}{6} = \frac{1}{6}[/tex]
Therefore , container B needs to be filled faster in order for both containers to gain the same amount of water per minute , and by 1/6 water per minute .
