Combined Work problems

1) Machines A and B always operate independently at their respective constant rates. When working alone, machine A can fill a production lot in 5 hours, and machine B can fill the same lot in x hours. When the two machines operate simultaneously to fill the production lot, it takes them 2 hours to complete the job. What is the value of x?

A step by step approach:

Machine A can fill a production lot in 5 hours
Machine B can fill a production lot in x hours

Machine A can fill 1/5 th of the same production lot in 1 hour
Machine B can fill 1/x th of the same production lot in 1 hour

Machine A+B can fill (1/5)+(1/x)th of the same production lot in 1 hour
Machine A+B can fill 2*((1/5)+(1/x))th of the same production lot in 2 hours

From the question, When the two machines operate simultaneously to fill the production lot, it takes 2 hours to complete the job. So the value of 2*((1/5)+(1/x)) = 1
2*((1/5)+(1/x)) = 1
((1/5)+(1/x)) = 1/2
1/x = (1/2)-(1/5)
1/x = 3/10
x = 10/3

2) Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?

A step by step approach:

Machine A can complete a certain production lot in x hours.
Machine B can complete 1/5th of the same production lot in y hours.

Machine A can complete a certain production lot in x hours.
Machine B can complete the same production lot in 5y hours.

Machine A can complete (1/x)th part of a certain production lot in 1 hour.
Machine B can complete (1/5y)th part of the same production lot in 1 hour.

Machine A and B, working together,can complete ((1/x)+(1/5y))th part of a certain production lot in 1 hour.
Machines A and B, working together, can complete 1/2 of the same production lot in z hours.

Machine A and B, working together,can complete ((1/x)+(1/5y))th part of a certain production lot in 1 hour.
Machines A and B, working together, can complete 1/2z of the same production lot in 1 hour.

So, ((1/x)+(1/5y))=(1/(2z))

I leave the third one for you to solve !!

Machines A and B always operate independently at their respective constant rates. When working alone, machine A can fill a production lot in 5 hours, and machine B can fill the same lot in x hours. When the two machines operate simultaneously to fill the production lot, it takes them 2 hours to complete the job. What is the value of x?

Let the job = 10 units.
Rate for A alone = w/t = 10/5 = 2 units per hour.
Rate for A+B together = w/t = 10/2 = 5 units per hour.
Thus, rate for B alone = 5-2 = 3 units per hour.
x = time for B alone = w/r = 10/3 hours.

Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?

Let the lot = 20 units.

Let x = 5 hours.
Rate for A alone = w/t = 20/5 = 4 units per hour.

Let y = 4 hours.
This is the time for B to produce 1/5 of the lot (4 units).
Thus, B's rate = w/t = 4/4 = 1 unit per hour.

When elements work together, add their rates.
z = the time for A+B to produce 1/2 of the lot = 10/(4+1) = 2 hours.

The question asks for the value of y (4 hours). This is our target.
Now we plug x=5 and z=2 into the answers to see which yields our target of 4.

Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours. If A and B work together at their respective rates to complete the job, which of the following represents the fraction of the job that B will not have to complete?

(x - y)/(x + y)

x/(y – x)

(x + y)/xy

y/(x – y)

y/(x + y)

Let the job = 10 units.

Let x = 5 hours.
Rate for A alone = w/t = 10/5 = 2 units per hour.

Let y = 2 hours.
Rate for B alone = w/t = 10/2 = 5 units per hour.

The combined rate for A+B = 2+5 = 7 units per hour.
Of the 7 units produced each hour, 2 will produced by A and 5 will be produced by B.
Thus, the fraction of the job produced by A = 2/7. This is our target.
Now we plug x=5 and y=2 into the answers to see which yields our target of 2/7.

A quick scan of the answers reveals that only E works:
y/(x+y) = 2/(5+2) = 2/7.

The correct answer is E.