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by GmatKiss » Fri Aug 12, 2011 1:46 am
A master builder is building a new house. He gets 3 apprentices who EACH work 2/3 as fast as he does. If all 4 work on it together, they should finish it in what fraction of the time that it would have taken the master builder working alone?

a)4/7
b)1/3
c)2/3
d)3/4
e)4/3

[spoiler]OA: B[/spoiler]
Last edited by GmatKiss on Fri Aug 12, 2011 5:50 am, edited 1 time in total.
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by bubbliiiiiiii » Fri Aug 12, 2011 2:04 am
let the rate of work of master be x.

rate of work of each apprentice is 2x/3.

rate of work of 3 apprentices is 2x/3 * 3= 2x.

Combined rate of all four is 3x.

Let the piece of work be n.

then,

if master at rate x can do the work in n units then the team at rate 3x can do in n/3.

Thus, B

Please let me know if the approach is correct or not.
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by pemdas » Fri Aug 12, 2011 2:22 am
i also think it's b

the rate of three workers is 2x/3 and the rate of builder is x
(2x/3)*3 + x = work, given One work, work=1 ==> (2x/3)*3+x=1, x=1/3
the speed of builder is 1/3 work per time unit and the combined speed of three workers is (1/3)*(2/3)*3 or 2/3 work per time unit. Builder and three workers have combined speed of 1 work per time unit.

So if all 4 work they do one work in 1 time unit. If builder does one work alone, he spends 3 time units. The relationship is 1/3.
GmatKiss wrote:A master builder is building a new house. He gets 3 apprentices who EACH work 2/3 as fast as he does. If all 4 work on it together, they should finish it in what fraction of the time that it would have taken the master builder working alone?

a)4/7
b)1/3
c)2/3
d)3/4
e)4/3
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by GMATGuruNY » Fri Aug 12, 2011 3:38 am
GmatKiss wrote:A master builder is building a new house. He gets 3 apprentices who EACH work 2/3 as fast as he does. If all 4 work on it together, they should finish it in what fraction of the time that it would have taken the master builder working alone?

a)4/7
b)1/3
c)2/3
d)3/4
e)4/3
Let work = 18 units.
Let rate for master builder = 3 units per hour.
Then rate for each apprentice = (2/3)*3 = 2 units per hour.
Combined rate for master builder and 3 apprentices = 3 + (3*2) = 9 units per hour.
Time for all workers = w/r = 18/9 = 2 hours.
Time for master builder alone = w/r = 18/3 = 6 hours.
Time for all workers/Time for master builder alone = 2/6 = 1/3.

The correct answer is B.
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by bubbliiiiiiii » Fri Aug 12, 2011 4:20 am
Thanks mitch for acknowledging the answer and providing us with a plug-in approach.

A question here:

Why/How can we reach the answer by using the work done per day concept to reach the answer. I tried to do that and got 11/3 as my answer.
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by GMATGuruNY » Fri Aug 12, 2011 4:33 am
bubbliiiiiiii wrote:Thanks mitch for acknowledging the answer and providing us with a plug-in approach.

A question here:

Why/How can we reach the answer by using the work done per day concept to reach the answer. I tried to do that and got 11/3 as my answer.
Let x = master builder rate.
Time for master builder = 1/x.

Rate for each apprentice = (2/3)x.
Rate for 3 apprentices = 3*(2/3)x = 2x.
Combined rate for master builder and 3 apprentices = x + 2x = 3x.
Time for all workers = 1/(3x).

Time for all workers/Time for master builder alone = [1/(3x)] / (1/x) = 1/(3x) * x = 1/3.
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by bubbliiiiiiii » Fri Aug 12, 2011 4:52 am
Thanks again for response. :)

I was approaching it this way.

Let x = master builder rate.
Time for master builder = 1/x.

Rate for each apprentice = (2/3)x.
Time for each apprentice = 1/[(2/3)x] = 3/(2x)

Time for all workers = 1/x+3*[3/(2x)] {Time for master = 3 * Time for each apprentice}

1/x + 9/2x = 11/(2x).

What is wrong with bold faced statements. When do we use that approach?
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by GMATGuruNY » Fri Aug 12, 2011 6:31 am
bubbliiiiiiii wrote: Time for each apprentice = 1/[(2/3)x] = 3/(2x)

Time for all workers = 1/x+3*[3/(2x)] {Time for master = 3 * Time for each apprentice}
Times can't be added in this manner. When elements work together, more work is produced in less time, so the amount of time needed is reduced.

If it takes John 2 days to produce 4 widgets and Mike 3 days to produce 4 widgets, it will NOT take John and Mike 2+3 = 5 days to produce 4 widgets: since John and Mike will be working together, the time will be reduced considerably.
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