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Doctors Medicines MCQs Q&A

  1. 5xy/4(y -x)
  2. 5xy/(y - x)
  3. 4xy/5(y - x)
  4. xy/5(x - y)
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1 Answers

Option 3 : 4xy/5(y - x)

Let the volume of the tank is V.

pipe A can fill the tank in x hours.

⇒ Rate at which pipe A fill the tank = V/x

Pipe B can empty the tank in y hours.

⇒ Rate at which Pipe B empty the tank = -(V/y)

Rate at which both the pipes can fill the tank = V(1/x - 1/y)

⇒ In 1 hr Pipe A and B can fill V(1/x - 1/y) volume of tank.

⇒ To fill 4/5 V, time taken to fill the tank = 4/5 V × (1/V(1/x - 1/y)) = 4xy/5(y - x)

∴ Both the pipes can fill the tank in (4xy/5(y - x))
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Related Questions

Consider the following solution to the producer-consumer synchronization problem. The shared buffer size is . Three semaphores empty, full and mutex are defined with respective initial values of 0, and 1. Semaphore empty denotes the number of available slots in the buffer, for the consumer to read from. Semaphore denotes the number of available slots in the buffer, for the producer to write to. The placeholder variables, denoted by P, Q, R, and S, in the code below can be assigned either empty or full. The valid semaphore operations are: wait ( ) and signal ( ). Producer: Consumer: do {         wait (P) ;         wait (mutex) ;         //Add item to buffer         signal (mutex) ;         signal (Q) ; } while (1) ; do {          wait (R) ;          wait (mutex) ;          //Consume item from buffer          signal (mutex) ;          signal (S) ; } while (1) ;   Which one of the following assignments to P, Q, R and S will yield the correct solution?
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