Answer:
b) 9
Step-by-step explanation:
Given: Three machines operating independently, simultaneously, and at the same constant rate can fill a certain production order in [tex]36[/tex] hours.
To Find: If one additional machine were used under the same operating conditions, in how many fewer hours of simultaneous operation could the production order be filled.
Solution:
Let the time taken by one machine to fill a certain production alone be[tex]=\text{x}[/tex]
Time taken when three machine operate independently and simultaneously
[tex]\frac{1}{\text{x}}+\frac{1}{\text{x}}+\frac{1}{\text{x}}=\frac{1}{36}[/tex]
[tex]\frac{3}{\text{x}}=\frac{1}{36}[/tex]
[tex]\text{x}=108[/tex] [tex]\text{hours}[/tex]
Let time taken when one additional machine is used [tex]=\text{y}[/tex]
Time when when one additional machine is used
[tex]\frac{1}{108}+\frac{1}{108}+\frac{1}{108}+\frac{1}{108}=\frac{1}{\text{y}}[/tex]
[tex]\frac{4}{108}=\frac{1}{\text{y}}[/tex]
[tex]\text{y}=27[/tex] [tex]\text{hours}[/tex]
it takes [tex]27[/tex] [tex]\text{hours}[/tex] when one additional machine is used
Now,
fewer hours taken when one additional machine is used
[tex]=\text{number of hours taken when three machines are used}-[/tex][tex]\text{number of hours taken when one additional machine is used}[/tex]
[tex]36-27[/tex]
[tex]9[/tex] [tex]\text{hours}[/tex]
in [tex]9[/tex] fewer hours of simultaneous operation the production order can be filled if one additional machine is used
Hence option b) is correct.