Actually, I think I just solved it:

After 2 hours of 1/8 (speed), the portion completed by Allen is 2/8 or 1/4, which leaves 3/4 unfinished. Two people's total speed is 1/8 + 1/16 which is 3/16. The remaining work, 3/4 divided by combined speed 3/16 equals four hours of work.

I still wonder if there's another faster approach and tips for overall strategy. My major problem with a lot of word problems such as this type of work problem is I lack a unified approach so I often find myself stuck on one problem for a long time. Thanks again!

Like this question looks different from the previous one, so I was stumped again:

If 4 pipes can fill a pool in 15 hours, how many more of these pipes are required to fill the same pool in only 10 hours?

It seems like you can't use the table (R X T=W) for all sorts of work problem, I even see some people use ratio to solve it, so I'm very confused. And I'm worried that if I see a different variety of work problem on the real test, I will blank out and don't know how to react, let alone attacking it. Figuring out what kind of strategy to deploy is critical because that's the part that often takes most of my time

Your approach to the first problem was exactly right. 3/4 / 3/16 = 4.

For the second:

You need to adjust the standard RT = W formula when there is more than one person/machine working.
The new formula is N*R*T=W where N is the number that are working.

So, 4*R*15 = W, so W = 60R
N*R*10 = W
N*R*10 = 60R
N*10 = 60
N = 6

Redhorsep wrote:Like this question looks different from the previous one, so I was stumped again:

If 4 pipes can fill a pool in 15 hours, how many more of these pipes are required to fill the same pool in only 10 hours?

It seems like you can't use the table (R X T=W) for all sorts of work problem, I even see some people use ratio to solve it, so I'm very confused. And I'm worried that if I see a different variety of work problem on the real test, I will blank out and don't know how to react, let alone attacking it. Figuring out what kind of strategy to deploy is critical because that's the part that often takes most of my time

Let rate for each pipe = 1 unit per hour.
Rate for 4 pipes = 4*1 = 4 units per hour.
In 15 hours, work produced = r*t = 4*15 = 60 units. This is the value of the pool.
To fill the pool in 10 hours, required rate = w/t = 60/10 = 6 units per hour.
Since each pipe produces 1 unit per hour, the number of pipes needed = 6.
Thus, given 4 pipes, we would need 2 more.

Redhorsep wrote:Hi, please help me with this work problem:

Working alone, Allen can build a fence in 8 hours, and Danny can build the same fence in 16 hours. If after working for alone for 2 hours, Allen is joined by Danny, how many more hours will it take the two of them to complete the work?

Can you also show your strategy in approaching this type of work problem? When one person/machine works for a certain amount of time, and then it's joined by another person/machine with different speed, or this person stopped and another machine/person takes over? Will drawing a table help?

Thanks!

Let fence = 16 units.
Rate for Allen = w/t = 16/8 = 2 units per hour.
Rate for Danny = w/t = 16/16 = 1 unit per hour.
Work produced by Allen in 2 hours = r*t = 2*2 = 4 units.
Work remaining = 16-4 = 12 units.
When elements work together, we add their rates.
Combined rate for Allen+David = 2+1 = 3 units per hour.
Time for Allen+David to produce remaining 12 units = w/r = 12/3 = 4 hours.

GMATGuruNY wrote:

Redhorsep wrote:Like this question looks different from the previous one, so I was stumped again:

If 4 pipes can fill a pool in 15 hours, how many more of these pipes are required to fill the same pool in only 10 hours?

It seems like you can't use the table (R X T=W) for all sorts of work problem, I even see some people use ratio to solve it, so I'm very confused. And I'm worried that if I see a different variety of work problem on the real test, I will blank out and don't know how to react, let alone attacking it. Figuring out what kind of strategy to deploy is critical because that's the part that often takes most of my time

Let rate for each pipe = 1 unit per hour.
Rate for 4 pipes = 4*1 = 4 units per hour.
In 15 hours, work produced = r*t = 4*15 = 60 units. This is the value of the pool.
To fill the pool in 10 hours, required rate = w/t = 60/10 = 6 units per hour.
Since each pipe produces 1 unit per hour, the number of pipes needed = 6.

If 4 pipes can fill a pool in 15 hours, how many more of these pipes are required to fill the same pool in only 10 hours?

Shouldn't the answer be 2?

robosc9 wrote:

GMATGuruNY wrote:

Redhorsep wrote:Like this question looks different from the previous one, so I was stumped again:

If 4 pipes can fill a pool in 15 hours, how many more of these pipes are required to fill the same pool in only 10 hours?

It seems like you can't use the table (R X T=W) for all sorts of work problem, I even see some people use ratio to solve it, so I'm very confused. And I'm worried that if I see a different variety of work problem on the real test, I will blank out and don't know how to react, let alone attacking it. Figuring out what kind of strategy to deploy is critical because that's the part that often takes most of my time

Let rate for each pipe = 1 unit per hour.
Rate for 4 pipes = 4*1 = 4 units per hour.
In 15 hours, work produced = r*t = 4*15 = 60 units. This is the value of the pool.
To fill the pool in 10 hours, required rate = w/t = 60/10 = 6 units per hour.
Since each pipe produces 1 unit per hour, the number of pipes needed = 6.

If 4 pipes can fill a pool in 15 hours, how many more of these pipes are required to fill the same pool in only 10 hours?

Shouldn't the answer be 2?

Yes, indeed. Given 4 pipes, we would need 2 more.