Question

X and Y work at constant rates. How many more hours does it take machine Y, alone, to fill an order of a certain size than machine X alone? I. X and Y together fill order in 2⁄3 the time that X alone does. II. Y alone does it in twice the time as X alone does.

Answer 1

Answer with explanation:

V=Volume of tank

It is given that, X and Y work at constant rates.

Let rate of doing work of X is x  hour.

And, Rate of Doing work of Y is y  hour.

Statement I----X and Y together fill order in 2⁄3 the time that X alone does.

              $\frac{V}{x+y}=\frac{2V}{3x}\\\\x+y=\frac{3x}{2}\\\\y=\frac{x}{2}$

                                         -------------------------------------------------(1)

Statement II----Y alone does it in twice the time as X alone does.

               $\frac{V}{x}=\frac{V}{\frac{y}{2}}\\\\y=\frac{2}{x}$

Substituting the value of y in 1

    $\rightarrow \frac{x}{2}=\frac{2}{x}\\\\\rightarrow x^2=4\\\\\rightarrow x=\pm 2\\\\\rightarrow x=2$

→x≠ -2, because Rate of doing work can't be negative.

Substituting the value of x in 1, gives

    $y=\frac{2}{2}\\\\y=1$

→Rate of doing work of X= 2 hour

→Rate of doing work of Y=1 hour