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2 men and 3 boys can do a piece of work in 10 days while 3 men and 2

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2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 16 Apr 2017, 06:20
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2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days . In how many days can 2 man and 1 boy do the work?

A. 15
B. 18
C. 12.5
D. 10
E. 16
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Re: 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 16 Apr 2017, 21:49
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Chemerical71 wrote:
2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days . In how many days can 2 man and 1 boy do the work?

A. 15
B. 18
C. 12.5
D. 10
E. 16


    • 2 men and 3 boys can do a piece of work in 10 days.
      o Thus, 20 men and 30 boys can do a piece of work in 1 day......(i)

    • 3 men and 2 boys can do the same work in 8 days.
      o Thus, 24 men and 16 boys can do the same work in 1 day....(ii)

    • Equating (i) and (ii) we get -
      o 20 men + 30 boys = 24 men + 16 boys
      o 4 men = 14 boys
      o 2 men = 7 boys

    • Substituting this in equation (i) we get

      o 10 boys can do a piece of work in 10 days.
      o But we need to find out in how many days 2 men and 1 boy can do the work, which is equivalent to 8 boys.

         8 boys can do the same work in (10*10/8) = 12.5 days.

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Re: 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 16 Apr 2017, 09:53
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1
Chemerical71 wrote:
2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days . In how many days can 2 man and 1 boy do the work?

A. 15
B. 18
C. 12.5
D. 10
E. 16



Two points before coming to the solution..
1) under no circumstances, it is sub-600 level.
2) the choices are always in ascending order, so may not be a very authentic source although the Q is ok.

Let m men can complete the job on their own and B boys can complete the work.
So \(\frac{2}{m}+\frac{3}{b}=\frac{1}{10}\)..
\(\frac{2*3}{m}+\frac{3*3}{b}=\frac{3}{10}\)..(i)

\(\frac{3}{m}+\frac{2}{b}=\frac{1}{8}\)...
\(\frac{2*3}{m}+\frac{2*2}{b}=\frac{2}{8}\)...(ii)

Subtract i from ii..
\(\frac{9}{b}-\frac{4}{b}=\frac{3}{10}-\frac{2}{8}\)..
\(\frac{5}{b}=\frac{1}{20}\)...
Or b=20*5=100..
Substitute in i to get value of m..
\(\frac{2}{m}+\frac{3}{100}=\frac{1}{10}\)..
\(\frac{2}{m}=\frac{1}{10}-\frac{3}{100}\)..
\(\frac{2}{m}=\frac{7}{100}\)..

We are looking for \(\frac{2}{m}+\frac{1}{b}=7/100+1/100=8/100\)..
So time taken=100/8=12.5..
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Re: 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 16 Apr 2017, 12:45
Let x be men and y boys
(2y+3x)/xy=1/10
(3y+2x)/xy=1/8
Through elimination you will get
X=2/7y
Therefore 10/y=1/10
Y rate=1/100
X rate=3.5/100
So rate of 2x+y=8/100 or 1/12.5 per day. Therefore it will be completed in 12.5 day

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2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 16 Apr 2017, 14:10
1
2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days . In how many days can 2 man and 1 boy do the work?

A. 15
B. 18
C. 12.5
D. 10
E. 16

let m and b=rates for 1 man and 1 boy respectively
multiplying,
6m+9b=3/10 and
6m+4b=1/4
subtracting,
b=1/100
m=3.5/100
let d=days
d(7/100+1/100)=1
d=12.5 days
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Re: 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 16 Apr 2017, 23:58
chetan2u wrote:
Chemerical71 wrote:
2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days . In how many days can 2 man and 1 boy do the work?

A. 15
B. 18
C. 12.5
D. 10
E. 16



Two points before coming to the solution..
1) under no circumstances, it is sub-600 level.
2) the choices are always in ascending order, so may not be a very authentic source although the Q is ok.

Let m men can complete the job on their own and B boys can complete the work.
So \(\frac{2}{m}+\frac{3}{b}=\frac{1}{10}\)..
\(\frac{2*3}{m}+\frac{3*3}{b}=\frac{3}{10}\)..(i)

\(\frac{3}{m}+\frac{2}{b}=\frac{1}{8}\)...
\(\frac{2*3}{m}+\frac{2*2}{b}=\frac{2}{8}\)...(ii)

Subtract i from ii..
\(\frac{9}{b}-\frac{4}{b}=\frac{3}{10}-\frac{2}{8}\)..
\(\frac{5}{b}=\frac{1}{20}\)...
Or b=20*5=100..
Substitute in i to get value of m..
\(\frac{2}{m}+\frac{3}{100}=\frac{1}{10}\)..
\(\frac{2}{m}=\frac{1}{10}-\frac{3}{100}\)..
\(\frac{2}{m}=\frac{7}{100}\)..

We are looking for \(\frac{2}{m}+\frac{1}{b}=7/100+1/100=8/100\)..
So time taken=100/8=12.5..
thank you for reply..I have been preparing for gmat since few months.For quant, i am doing manhattan, veritas and some gmat club solution mainly you , Bunnel after completing official sources.. I think this problem is consistent with gmat quant that's why i have posted. :-D
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Re: 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 11 Sep 2018, 07:52
10*(2M+3B)=8*(2M+16B)

we get 2M=7B.

We have to find 2M+1B= 7B+1B

putting this in the first equation we get-- replacing 2M by 7B
10B takes 10 days

Therefore 1B takes 100 days .

hence 8B takes 100/8 = 12.5 days
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Re: 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 14 Sep 2018, 17:47
Chemerical71 wrote:
2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days . In how many days can 2 man and 1 boy do the work?

A. 15
B. 18
C. 12.5
D. 10
E. 16


Let m = the rate of 1 man and b = the rate of 1 boy. We can create the equations:

2m + 3b = 1/10

and

3m + 2b = 1/8

Multiplying the first equation by -2 and the second by 3, we have:

-4m - 6b = -2/10

and

9m + 6b = 3/8

Adding the equations together, we have:

5m = 3/8 - 2/10

5m = 3/8 - 1/5

5m = 15/40 - 8/40

5m = 7/40

m = 7/200

Substitute m = 7/200 into 2m + 3b = 1/10, we have

2(7/200) + 3b = 1/10

7/100 + 3b = 1/10

3b = 10/100 - 7/100

3b = 3/100

b = 1/100

Letting x = the number of days needed to complete the work by 2 men and 1 boy, we have:

x(2(7/200) + 1/100) = 1

x(8/100) = 1

x = 100/8 = 12.5

Answer: C
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2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 24 Sep 2018, 04:05
Chemerical71 wrote:
2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days . In how many days can 2 man and 1 boy do the work?

A. 15
B. 18
C. 12.5
D. 10
E. 16


Dear EMPOWERgmatRichC

I tried to use same concept that you used in the following but ended up choosing wrong answer:
https://gmatclub.com/forum/if-12-men-an ... l#p2137361


3 men and 2 boys would take a total of 8 days to complete a task.

If we HALVE the number of workers again, we would again DOUBLE the amount of time needed to complete the task:

1.5 men and 1 boys would take a total of 16 days to complete a task.

The question ask for 2 men and 1 boy. With using extra 0.5 man , it should take little less than 16, which is 15 in answer choices.
It is the close number to 16. However, the answer is (12.5)

Can you help please?

Thanks
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Re: 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 24 Sep 2018, 11:25
1
Hi Mo2men,

You CAN use that same logic here, but you have to be a bit more focused on the 'impact' that each man has on the overall calculation/rate. The prompt gives us two pieces of data to work with:

2 men and 3 boys can do a piece of work in 10 days
3 men and 2 boys can do the same work in 8 days

Consider the second piece of information relative to the first piece. We remove 1 boy from the job, so we LOSE that boy's work output. Adding 1 extra man 'makes up' for that loss AND then cuts the total down from 10 days to 8 days. Thus, that 1 man clearly has a big impact on the total time; by himself, he clearly represents MORE than a 2 day decrease in time needed to complete the job (again, he's also making up for the lost productivity from losing that 1 boy). Thus, 1/2 of a man would account for MORE than a 1 day decrease in time needed to complete the job.

You are absolutely correct that it would take 1.5 men and 1 boy a total of 16 days to complete the task. Adding that extra 1/2 of a man would decrease that total by MORE than 1 day though, so the correct answer CANNOT be 15 days (it has to be something less than that). Logically, 12.5 days makes far more sense.

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Re: 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2  [#permalink]

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New post 24 Sep 2018, 12:33
Chemerical71 wrote:
2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days . In how many days can 2 man and 1 boy do the work?

A. 15
B. 18
C. 12.5
D. 10
E. 16

Let´s repeat EXACTLY the same approach we used here: https://gmatclub.com/forum/if-12-men-an ... 11-20.html

\(\begin{array}{*{20}{c}}
{\left( {\text{I}} \right)} \\
{\left( {{\text{II}}} \right)}
\end{array}\,\,\begin{array}{*{20}{c}}
{\left[ {2\,\,{\text{men}}\,\,\, \cup \,\,\,{\text{3}}\,\,{\text{boys}}} \right]\,\,\, - \,\,{\text{1}}\,\,{\text{work}}\,\,\, - \,\,\,10\,\,\,{\text{days}}} \\
{\left[ {3\,\,{\text{men}}\,\,\, \cup \,\,\,{\text{2}}\,\,{\text{boys}}} \right]\,\,\, - \,\,{\text{1}}\,\,{\text{work}}\,\,\, - \,\,\,8\,\,\,{\text{days}}}
\end{array}\)

\({\text{?}}\,\,\,{\text{:}}\,\,\,\left[ {2\,\,{\text{men}}\,\,\, \cup \,\,\,1\,\,{\text{boy}}} \right]\,\,\, - \,\,{\text{1}}\,\,{\text{work}}\,\,\, - \,\,\,?\,\,{\text{days}}\)


Let "task" be the fraction of this work that one man can do in 1 day, hence:

\(1\,{\text{man}}\,\,\, - \,\,\,1\,\,{\text{day}}\,\,\, - \,\,\,1\,\,\,{\text{task}}\)

Let k (k>0) be the fraction of the "task" defined above that one boy can do in 1 day (where k may be between 0 and 1, or equal to 1, or greater), hence:

\(1\,{\text{boy}}\,\,\, - \,\,\,1\,\,{\text{day}}\,\,\, - \,\,\,k\,\,\,{\text{task}}\)


Now the long-lasting benefit of this "structure": everything else becomes easy and "automatic":

\(\left( {\text{I}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left[ {2\,\,{\text{men}}\,\,\, \cup \,\,\,{\text{3}}\,\,{\text{boys}}} \right]\,\, - \,\,{\text{10}}\,\,{\text{days}}\,\,\, - \,\,\,2 \cdot 10 \cdot 1 + 3 \cdot 10 \cdot k\,\,\,{\text{tasks}}\,\, = \,\,\,1\,\,{\text{work}}\,\)

\(\left( {{\text{II}}} \right)\,\,\, \Rightarrow \,\,\,\,\left[ {3\,\,{\text{men}}\,\,\, \cup \,\,\,{\text{2}}\,\,{\text{boys}}} \right]\,\, - \,\,{\text{8}}\,\,{\text{days}}\,\,\, - \,\,3 \cdot 8 \cdot 1 + 2 \cdot 8 \cdot k\,\,\,{\text{tasks}}\,\, = \,\,\,1\,\,{\text{work}}\)

Therefore: \(20 + 30 \cdot k = 24 + 16 \cdot k\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,k = \frac{2}{7}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\,{\text{work}}\,\,{\text{ = }}\,\,\,{\text{28}}\frac{4}{7}\,\,\,{\text{tasks}}\)


\(?\,\,\,\,:\,\,\,\,\left[ {2\,\,{\text{men}}\,\,\, \cup \,\,\,1\,\,{\text{boy}}} \right]\,\,\,\, - \,\,\,{\text{1}}\,\,{\text{day}}\,\,\, - \,\,\,2 \cdot 1 + 1 \cdot \frac{2}{7} = 2\frac{2}{7}\,\,{\text{tasks}}\)

And we finish in "high style", using UNITS CONTROL, one of the most powerful tools of our method:

\(?\,\,\, = \,\,\,28\frac{4}{7}\,\,\,{\text{tasks}}\,\,\,\left( {\frac{{1\,\,{\text{day}}}}{{2\frac{2}{7}\,\,{\text{tasks}}}}\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\,\, = \,\,\,\,\frac{{7 \cdot 28 + 4}}{{2 \cdot 7 + 2}} = \frac{{200}}{{16}} = \frac{{80 + 16 + 4}}{8} = 12\frac{1}{2}\,\,\,\,\,\left[ {{\text{days}}} \right]\)
Obs.: arrows indicate licit converter.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 &nbs [#permalink] 24 Sep 2018, 12:33
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