Problem Solving

Randy can finish a job in a hrs and Alvin can finish the same in b hrs.If they both work together which of the following represents the portion of the job that Randy will not have to complete?
ab/(a+b)
(a-b)/(a+b)
b/(b-a)
a/(a+b)
b/(a+b)

abhirup1711 wrote:Randy can finish a job in a hrs and Alvin can finish the same in b hrs.If they both work together which of the following represents the portion of the job that Randy will not have to complete?
ab/(a+b)
(a-b)/(a+b)
b/(b-a)
a/(a+b)
b/(a+b)

Time and rate are RECIPROCALS.
Since the time ratio for Randy and Alvin = a:b, the rate ratio for Randy and Alvin = b:a.
Implication:
For every b units produced by Randy, a units are produced by Alvin.
Thus, of every a+b units, the fraction produced by Alvin = a/(a+b).

The correct answer is D.

Alternate approach:

Let the job = 10 units.

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

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

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

A quick scan of the answers reveals that only D works:
a/(a+b) = 5/(5+2) = 5/7.

The correct answer is D.

Hi Mitch,
I am not sure, where I am going wrong.

My approach
----------
Randy can do 1/a job in 1hr.
Alvin can do 1/b job in 1 hr,
If both work together, they can do (1/a + 1/b) job in 1 hr.

Portion of job Randy will not have to do = portion of Job Alvin does = (1/b) / (1/a + 1/b) = a/(a+b)

So answer should be C.

Thanks
-sks74