Question 232146
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Let *[tex \Large x] represent the number of hours it would take A to fill the jar by itself.  Since A pumps twice as fast as B, it would take B two times longer to fill the jar, so *[tex \Large 2x] represents the number of hours it would take B to fill the jar by itself.


Since *[tex \Large x] is the amount of time for A to fill the jar, *[tex \Large \frac{1}{x}] is the amount of the jar that A can fill in 1 hour.  Likewise, B can fill *[tex \Large \frac{1}{2x}] in one hour.  And together they can fill:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{2x}]


of the jar in 1 hour.


Add 'em up:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2\ +\ 1}{2x}\ =\ \frac{3}{2x}]


And we know  it takes 5 hours to fill the jar working together, so we can conclude that together they can fill *[tex \Large \frac{1}{5}] of the jar in 1 hour, therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{2x}\ =\ \frac{1}{5}]


Solve for *[tex \Large x] to get the time for A to fill the jar by itself (Hint cross-multiply). B is twice that value.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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