rate-work -any short method

Hi hey_thr67:

The key to doing this fast is realizing that as long as A and B works at their constant rates, the fraction of work done by A and B in any given time is the same. With that in mind, let’s do this one quickly.

Machine A can complete a certain job in x hours.
Hence, work done by A in one hour = 1/x.

Machine B can complete the same job in y hours.
Hence, work done by B in one hour = 1/y.

When they are working together work done by A and B in one hour = 1/x+1/y.

Now the work that B doesn’t do, is the work that is done by A. Hence the correct fraction of work done by A
= (work done by A in one hour) / (Work done by A and B together in 1 hour)
= (1/x) / (1/x+1/y) = xy/( x*(x+y)) = y/(x+y). Hence E

Let me know if this helps

hey_thr67 wrote: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?

A:
x - y
------
x + y

B:
x
-----
y – x

C:
x + y
-----
xy

D:

y
-----
x – y

E:
y
----
x + y

OA is E

Time and rate are RECIPROCALS.
Let x = 2 hours and y = 5 hours.
Since the ratio of A's time to B's time = 2:5, the ratio of A's rate to B's rate = 5:2.
The ratio of the rates implies the following:
Of every 7 units produced, 5 will be produced by A and 2 will be produced by B, implying that the fraction not produced by B = 5/7. This is our target.
Now we plug x=2 and y=5 into the answers to which yields our target of 5/7.
A quick scan of the answer choices reveals that only E works:
y/(x+y) = 5/(2+5) = 5/7.

The correct answer is E.

Note that the same reasoning can be applied directly to the variables themselves.
Since the ratio of A's time to B's time = x:y, the ratio of A's rate to B's rate = y : x.
The ratio of the rates implies the following:
Of every (x+y) units produced, y will be produced by A, implying that the fraction not produced by B = y/(x+y).