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# M11 #19

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Senior Manager
Joined: 16 Jul 2009
Posts: 262
Followers: 3

Kudos [?]: 93 [1] , given: 3

M11 #19 [#permalink]  19 May 2010, 07:41
1
KUDOS
Working at their normal 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?

25
28
29
30
32

It follows from the stem that in 3 days one digger can dig 5 meters of trench in hard soil. Therefore, his digging speed on hard soil is $$\frac{5}{3}$$ meters per day. On soft soil his speed would be $$\frac{5}{3}*\frac{6}{5}= \frac{6}{3}$$ meters per day or 6 meters in 3 days. If one digger can dig 6 meters of a trench in 3 days then we need $$\frac{180}{3}$$ = 30 diggers to dig the entire 180-meter long trench in 3 days.

Here how do we arrive at the fraction $$\frac{6}{5}$$?
Manager
Joined: 08 May 2010
Posts: 143
Followers: 0

Kudos [?]: 56 [1] , given: 39

Re: M11 #19 [#permalink]  19 May 2010, 07:49
1
KUDOS
I would do the problem in a different way. The 20 diggers dig 100 meters in 3 days in hard soil. They then dig 20% faster in soft soil. So this should be 120 meters in 3 days. Now I set up a proportion 120meters for 20 diggers compared with 180 meters for how many workers? The days and soil are now the same. Cross multiplying I get 20(180) = 120x ==> x= 20(180)/120, x= 30.

I don't know whether this is correct but it seems to make sense.
Thanks
Skip
Senior Manager
Joined: 16 Jul 2009
Posts: 262
Followers: 3

Kudos [?]: 93 [0], given: 3

Re: M11 #19 [#permalink]  19 May 2010, 22:51
Amazing one Skip! You've indeed skipped.. all the steps given in the solution and made it a under 1-minute solution..Kudos for you..
CIO
Joined: 02 Oct 2007
Posts: 1217
Followers: 92

Kudos [?]: 754 [0], given: 334

Re: M11 #19 [#permalink]  19 May 2010, 23:05
$$\frac{6}{5}$$ is 1.2, which comes from "20% faster in soft soil". $$\frac{5}{3}*\frac{6}{5}$$ is the same as $$\frac{5}{3}*1.2$$
abhi758 wrote:
Working at their normal 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?

25
28
29
30
32

It follows from the stem that in 3 days one digger can dig 5 meters of trench in hard soil. Therefore, his digging speed on hard soil is $$\frac{5}{3}$$ meters per day. On soft soil his speed would be $$\frac{5}{3}*\frac{6}{5}= \frac{6}{3}$$ meters per day or 6 meters in 3 days. If one digger can dig 6 meters of a trench in 3 days then we need $$\frac{180}{3}$$ = 30 diggers to dig the entire 180-meter long trench in 3 days.

Here how do we arrive at the fraction $$\frac{6}{5}$$?

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Math Expert
Joined: 02 Sep 2009
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Kudos [?]: 42286 [0], given: 6012

Re: M11 #19 [#permalink]  03 May 2012, 21:09
Expert's post
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 180/6=30 diggers are required.

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Re: M11 #19   [#permalink] 03 May 2012, 21:09
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# M11 #19

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