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work n Time

by sixpointer » Sat Oct 23, 2010 1:01 pm
17. A man starts a piece of work . Starting from the second day
onwards everyday new man joins .With every new man joining the
work that each man can do per day doubles .The work is completed in 5
days .On which day would they have completed the work ,if the work
that each of them could do per day had remained constant.?
1.10th 2.12th 3. 15th 4. 18th 5.16th
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by Stuart@KaplanGMAT » Sat Oct 23, 2010 8:28 pm
sixpointer wrote:17. A man starts a piece of work . Starting from the second day
onwards everyday new man joins .With every new man joining the
work that each man can do per day doubles .The work is completed in 5
days .On which day would they have completed the work ,if the work
that each of them could do per day had remained constant.?
1.10th 2.12th 3. 15th 4. 18th 5.16th
Let's say that on day 1, the man does x work.

On day 2, there are 2 men, each of whom do 2x work. So, 4x work gets done.

On day 3, there are 3 men, each of whom do 4x work. So, 12x work gets done.

On day 4, there are 4 men, each of whom do 8x work. So, 32x work gets done.

Finally, on day 5, there are 5 men, each of whom do 16x work. So, 80x work gets done.

Total: x + 4x + 12x + 32x + 80x = 129x work.

If the work remains constant, then the progression is:

x + 2x + 3x + 4x + 5x + .... + nx, on which n is the last day. We need that series to sum to 129.

Since we have a series of consecutive numbers, we could use the formula:

sum = # of terms * average of terms

but that's going to be some ugly math:

sum = n * (first term + last term)/2

129 = n * (1 + n)

129 = n + n^2

0 = n^2 + n - 129... some ugly quadratic!

So, rather than actually do all that ugly math, it's much quicker to backsolve at this point:

Starting with B or D, I'd choose D since it seems like we need a big number.

If n=18, then our sum is:

sum = 18 * (1+18)/2

sum = 18 * 9.5

sum = way too much!

Try B:

if n = 12, then we have:

sum = 12 * (1 + 12)/2

sum = 12 * 6.5

sum = way too small!

Accordingly, the answer has to be (C) 15.
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by goyalsau » Sat Oct 23, 2010 8:51 pm
Dear Stuart

you did all the right thing but didn't notice that 15 will be = 120
and we need 129 and there is one option as 16.

So its 16 not 15

option E
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by Stuart@KaplanGMAT » Sat Oct 23, 2010 9:26 pm
goyalsau wrote:Dear Stuart

you did all the right thing but didn't notice that 15 will be = 120
and we need 129 and there is one option as 16.

So its 16 not 15

option E
Ahh, my mistake was that I didn't look at choice E at all - on the GMAT, the choices are almost always in ascending or descending order, so I assumed (bad me!) that E was greater than D.

If I had noticed that E was lower than D, that would have affected the way that I backsolved the problem (I would have rearranged the answers so that I only had to check 2 of them).
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by sixpointer » Sat Oct 23, 2010 10:58 pm
Stuart Kovinsky wrote:
sixpointer wrote:17. A man starts a piece of work . Starting from the second day
onwards everyday new man joins .With every new man joining the
work that each man can do per day doubles .The work is completed in 5
days .On which day would they have completed the work ,if the work
that each of them could do per day had remained constant.?
1.10th 2.12th 3. 15th 4. 18th 5.16th
Let's say that on day 1, the man does x work.

On day 2, there are 2 men, each of whom do 2x work. So, 4x work gets done.

On day 3, there are 3 men, each of whom do 4x work. So, 12x work gets done.

On day 4, there are 4 men, each of whom do 8x work. So, 32x work gets done.

Finally, on day 5, there are 5 men, each of whom do 16x work. So, 80x work gets done.

Total: x + 4x + 12x + 32x + 80x = 129x work.

If the work remains constant, then the progression is:

x + 2x + 3x + 4x + 5x + .... + nx, on which n is the last day. We need that series to sum to 129.

Since we have a series of consecutive numbers, we could use the formula:

sum = # of terms * average of terms

but that's going to be some ugly math:

sum = n * (first term + last term)/2

129 = n * (1 + n)

129 = n + n^2

0 = n^2 + n - 129... some ugly quadratic!

So, rather than actually do all that ugly math, it's much quicker to backsolve at this point:

Starting with B or D, I'd choose D since it seems like we need a big number.

If n=18, then our sum is:

sum = 18 * (1+18)/2

sum = 18 * 9.5

sum = way too much!

Try B:

if n = 12, then we have:

sum = 12 * (1 + 12)/2

sum = 12 * 6.5

sum = way too small!

Accordingly, the answer has to be (C) 15.

With every new man joining the work each Man can do per day double
Let each man can do one unit work

1st day ->1 unit

2nd day->2+1

3rd day ->4+2+1

4th day ->8+4+2+1

5th day ->16+8+4+2+1

Where Iam doing mistake ??:(

Please explain.

OA is 5th (16th)
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by pesfunk » Sun Oct 31, 2010 12:04 am
I think it should be:

1st day ->1 unit

2nd day->2+2

3rd day ->4+4+4

4th day ->8+8+8+8

5th day ->16+16+16+16+16

No of units completed = 1 + 4 + 12 + 32 + 80 = 129

No of days to complete the job in 2nd condition:
In 15 days no of units completed will be 15*16/2 = 120
In 16 days no of units completed will be 17*16/2 = 136

=> Work will be completed on 16th day
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