Respuesta :
[tex]\frac{1}{2}[/tex] of the water takes [tex]\frac{1}{2}[/tex] of the fuel and empty tank takes [tex]\frac{1}{6}[/tex] of the fuel, the
fraction that would will be spent if there was no leak is [tex]\underline{\dfrac{11}{18}}[/tex] of the fuel
How can the fraction of the fuel spent be found?
The given parameters are;
The volume of oil consumed = [tex]\mathbf{\dfrac{1}{2}}[/tex] of the fuel tank
Volume consumed if the water tank was empty = [tex]\dfrac{1}{6}[/tex] of the fuel tank
Let w represent the weight of the water that fills the tank.
Therefore;
[tex]The \ average \ weight \ carried \ by \ oil = \dfrac{w + \dfrac{w}{2} }{2} = \mathbf{\dfrac{3 \cdot w}{4}}[/tex]
The amount of fuel consumed by the water that leaked = [tex]\dfrac{1}{2} - \dfrac{1}{6} = \dfrac{1}{3}[/tex]
Let x represent the amount of oil used to carry only the water if there
was no leak, therefore;
[tex]\mathbf{\dfrac{\frac{3 \cdot w}{4} }{w} }= \dfrac{\frac{1}{3} }{x}[/tex]
Which gives;
[tex]\mathbf{\dfrac{3 \cdot w}{4} \times x}} = \dfrac{1}{3} \times w }[/tex]
[tex]x = \dfrac{4}{3 \times 3} = \dfrac{4}{9}[/tex]
The total amount of fuel that will be consumed without leakage is
therefore;
[tex]Total \ amount \ of \ fuel = \dfrac{4}{9} + \dfrac{1}{6} = \mathbf{ \dfrac{11}{18}}[/tex]
The amount of oil consumed if there was no leakage is therefore;
- [tex]\underline{\dfrac{11}{18} \ of \ the \ fuel \ tank}[/tex]
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