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M11-19

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M11-19 [#permalink]

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New post 16 Sep 2014, 00:44
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Working at their normal constant rates, 20 diggers can dig a 100-meter long trench in hard soil in 3 days. A construction company ordered the diggers to dig a 180-meter long trench in soft soil. If diggers can work 20% faster in soft soil than in hard soil, how many diggers are required to complete this task in 3 days?

A. 25
B. 28
C. 29
D. 30
E. 32
[Reveal] Spoiler: OA

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Official Solution:

Working at their normal constant rates, 20 diggers can dig a 100-meter long trench in hard soil in 3 days. A construction company ordered the diggers to dig a 180-meter long trench in soft soil. If diggers can work 20% faster in soft soil than in hard soil, how many diggers are required to complete this task in 3 days?

A. 25
B. 28
C. 29
D. 30
E. 32


20 diggers can dig a 100-meter long trench in hard soil in 3 days;

1 digger can dig a 100/20=5-meter long trench in hard soil in 3 days;

1 digger can dig a 5*1.2=6-meter long trench in soft soil in 3 days;

Therefore, to dig 180-meter long trench in soft soil in 3 days \(\frac{180}{6}=30\) diggers are required.


Answer: D
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Collection of Questions:
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Re: M11-19 [#permalink]

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New post 28 May 2016, 06:36
My approach might be bit more cubersome than Bunuel's, but here it goes:

1) First of all: \(Rate = \frac{Amount}{Time}\). So, if 20 diggers use 3 days to complete a 100-meter trench in hard soil, their rate is \(\frac{1}{3}\). The rate for one digger is thus \(\frac{1}{3}\) * \(\frac{1}{20}\) = \(\frac{1}{60}\).

2) In soft soil, a digger can work 20% faster (\(\frac{6}{5}\)) than in hard soil, so the rate for 1 digger, digging a 100-meter trench in soft soil, would be \(\frac{1}{60}\) * \(\frac{6}{5}\) = \(\frac{1}{50}\). In other words, 1 digger would use 50 days to dig a 100-meter trench in soft soil.

3) However, we want to dig 180 meters in soft soil (not 100 meters), so the total number of days required for 1 digger to dig a 180-meter trench would be: \(50*1.8 = 90\) days. Now, we want to complete this 180-meter trench within 3 days, and thus require \(\frac{90}{3} = 30\) diggers. Answer choice D.

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Re: M11-19 [#permalink]

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New post 28 May 2016, 10:43
Bunuel wrote:
Working at their normal constant rates, 20 diggers can dig a 100-meter long trench in hard soil in 3 days. A construction company ordered the diggers to dig a 180-meter long trench in soft soil. If diggers can work 20% faster in soft soil than in hard soil, how many diggers are required to complete this task in 3 days?

A. 25
B. 28
C. 29
D. 30
E. 32



Since the days are same the ratio of work should be SAME as ratio of diggers...
in hard soild 100m means 100*1.2 = 120 m in soft soil..
so \(\frac{120}{180}= \frac{20}{w}........w = 20*\frac{180}{120} = 30\)
D
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M11-19 [#permalink]

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New post 28 May 2016, 11:09
Bunuel wrote:
Working at their normal constant rates, 20 diggers can dig a 100-meter long trench in hard soil in 3 days. A construction company ordered the diggers to dig a 180-meter long trench in soft soil. If diggers can work 20% faster in soft soil than in hard soil, how many diggers are required to complete this task in 3 days?

A. 25
B. 28
C. 29
D. 30
E. 32



20 persons -- dig 100 m trench in hard soil - in 3 days
1 person -- dig \(\frac{100}{20}\) trench in hard soil - in 3 days
1 person -- digs 5 trench in hard soil - in 3 days

1 person -- digs 5*1.2 trench in soft soil - in 3 days
1 person -- digs 6 m trench in soft soil - in 3 days

Required trench length : 180 m. 180 = 6 * 30.

So
30 persons -- dig 6*30 m trench in soft soil - in 3 days

Answer: 30 persons.

D is desired answer

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Re: M11-19 [#permalink]

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New post 04 Jul 2016, 21:09
This is a typical work rate problem.. 20% faster means Diggers can dig more.. 100m *1.2=120m in soft soil.

So 120 m -> 20 Digger -> 3 days
Ques. is 180m -> ? -> 3 days

Hence 120/20=180/w

W=30 . Hence D is the answer.

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Re: M11-19 [#permalink]

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New post 18 Sep 2016, 02:05
Could someone help me understand where am I going wrong with this approach:

Soft Soild vs Hard soil : 20% lesser effort
180m vs 100m
days remain the same

So, no. of people required will be :

\(20 *\frac{180}{100}*\frac{3}{3}*\frac{80}{100}\)

\(=28.8 = ~29 people\)

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Re: M11-19 [#permalink]

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New post 20 Feb 2017, 15:58
rajarams wrote:
Could someone help me understand where am I going wrong with this approach:

Soft Soild vs Hard soil : 20% lesser effort
180m vs 100m
days remain the same

So, no. of people required will be :

\(20 *\frac{180}{100}*\frac{3}{3}*\frac{80}{100}\)

\(=28.8 = ~29 people\)


I used the same approach but I think you mistake is the workers are now 20% faster i.e. their productivity is \frac{6}{5} the original, so I need \frac{5}{6} of the workers

Final solution \(20 *\frac{180}{100}*\frac{3}{3}*\frac{5}{6}\) =30

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Re: M11-19 [#permalink]

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New post 15 Oct 2017, 08:41
100 mts 3 days 20 diggers----hard soil
soft soil 20% faster,so now

120 mts 3 days 20 diggers
total land left to dig (180-120)mts ie 60 mts more

20 diggers dig 120 mts
1 digger covers 120/20=6mts

so 60 mts left can be covered by 60/6=10 so ten more diggers required

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Re: M11-19   [#permalink] 15 Oct 2017, 08:41
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