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If 17 labourers can dig a ditch 20 m long in 18 days, working 8 hours

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If 17 labourers can dig a ditch 20 m long in 18 days, working 8 hours  [#permalink]

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New post 15 Feb 2019, 08:47
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Question Stats:

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If 17 laborers can dig a ditch 20 m long in 18 days, working 8 hours a day, how many minimum numbers of more laborers should be engaged to dig a similar ditch 39 m long in a maximum of 6 days, each laborer working 9 hours per day.
1. 89
2. 84
3. 78
4. 72
5. 68
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If 17 labourers can dig a ditch 20 m long in 18 days, working 8 hours  [#permalink]

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New post Updated on: 16 Feb 2019, 19:56
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\(\frac{Man*Days*Hours}{Work-done}\)

\(\frac{17*18*8}{20} = \frac{(x+17)*9*6}{39}\)

x+17 = \(\frac{17*18*8*39}{(20*9*6)}\) = \(\frac{17*8*13}{20}\) = 26*17/5 = 88.4 = 89

Min numbers of labors required are x+17 = 89, x = 72

D is the answer.
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Originally posted by Chethan92 on 15 Feb 2019, 09:49.
Last edited by Chethan92 on 16 Feb 2019, 19:56, edited 1 time in total.
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Re: If 17 labourers can dig a ditch 20 m long in 18 days, working 8 hours  [#permalink]

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New post 16 Feb 2019, 17:48
Afc0892 wrote:
\(\frac{Man*Days*Hours}{Work-done}\)

\(\frac{17*18*8}{20} = \frac{(x+2)*9*6}{39}\)

x+17 = \(\frac{17*18*8*39}{(20*9*6)}\) = \(\frac{17*8*13}{20}\) = 26*17/5 = 88.4 = 89

Min numbers of labors required are x+17 = 89, x = 72

D is the answer.


Hi

Could you please explain the x+2? I'm a little confused on where that came from.

Thank you,
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Re: If 17 labourers can dig a ditch 20 m long in 18 days, working 8 hours  [#permalink]

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New post 16 Feb 2019, 19:19
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awjpca wrote:
Afc0892 wrote:
\(\frac{Man*Days*Hours}{Work-done}\)

\(\frac{17*18*8}{20} = \frac{(x+2)*9*6}{39}\)

x+17 = \(\frac{17*18*8*39}{(20*9*6)}\) = \(\frac{17*8*13}{20}\) = 26*17/5 = 88.4 = 89

Min numbers of labors required are x+17 = 89, x = 72

D is the answer.


Hi

Could you please explain the x+2? I'm a little confused on where that came from.

Thank you,


Hey! Sorry, it should have been x+17. Typo error

Posted from my mobile device
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Re: If 17 labourers can dig a ditch 20 m long in 18 days, working 8 hours  [#permalink]

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New post 17 Feb 2019, 01:05
cfc198 wrote:
If 17 laborers can dig a ditch 20 m long in 18 days, working 8 hours a day, how many minimum numbers of more laborers should be engaged to dig a similar ditch 39 m long in a maximum of 6 days, each laborer working 9 hours per day.
1. 89
2. 84
3. 78
4. 72
5. 68


-------Labourers-------Length-----------Days---------Hours

Earlier------17------------20----------------18-------------8

Now---------x-------------39----------------6---------------9

x = \(17 *\frac{39}{20} * \frac{18}{6} * \frac{8}{9}\)

(In this method We copy down 17 as it is because the unknown is the new number of labourers

In the Length column earlier the length was less and now the length is more.

So, the number of labourers will be more. So we do \(\frac{More length}{Less length}= \frac{39}{20}\)for this column

and so on for the Days column and the Hours column)

x = 88.4

Since x is the number of men, minimum value of x = 89

Additional number of men required = 89-17 = 72

Choice D
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Re: If 17 labourers can dig a ditch 20 m long in 18 days, working 8 hours  [#permalink]

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New post 18 Feb 2019, 18:05
cfc198 wrote:
If 17 laborers can dig a ditch 20 m long in 18 days, working 8 hours a day, how many minimum numbers of more laborers should be engaged to dig a similar ditch 39 m long in a maximum of 6 days, each laborer working 9 hours per day.
1. 89
2. 84
3. 78
4. 72
5. 68


The rate of 17 laborers is 20/(18 x 8) = 5/36 meter per hour. We can create a proportion where n is the number of people needed for the 39 m long ditch:

17/(5/36) = n/[39/(6 x 9)]

(36 x 17)/5 = 54n/39

Dividing both sides by 18, we have:

(2 x 17)/5 = 3n/39

34/5 = n/13

5n = 442

n = 88.4

Since we need to guarantee that it will take no more than 6 days to finish the job, we round 88.4 up to 89; thus, we need a total of 89 laborers, which means we need to hire 89 - 17 = 72 more laborers.

Answer: D or 4
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Re: If 17 labourers can dig a ditch 20 m long in 18 days, working 8 hours   [#permalink] 18 Feb 2019, 18:05
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