Q:If 20 engineers and 20 workers can together construct a 20km road in 20 days , then 40 engineers and 40 workers together construct 40km road in how many days ?
Ans.20 days. The answer is correct. I have some confusion regarding the solution. The solution is given as 20 X 20 = 400 man days for 20km road.
i.e workers(20) X days(20) = 400 man-days , so for 40km road: 800 man-days are required. therefore 800/40 = 20 man-days.
My question is why are we multiplying only Workers(20) X Days(20) and not :
Engineers(20) X Workers(20) X Days (20) = 8000 man-days to determine the man days ? Why are we leaving out the engineers from the man-days calculation ?
Maybe I am missing some very simple concept!
Work Problem
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It was simpler to think this way
20 engineers and 20 workers can together construct a 20km road in 20 days
40 engineers and 40 workers together can construct 20km road in half of the days as manpower has been doubled which is inversely proportional to the amount of time taken = 10 days
Now 40 engineers and 40 workers together can construct 40km road in 20 Days [Work and Time are directly proportional.
Manpower x Time / Work = Constant
20 engineers and 20 workers can together construct a 20km road in 20 days
40 engineers and 40 workers together can construct 20km road in half of the days as manpower has been doubled which is inversely proportional to the amount of time taken = 10 days
Now 40 engineers and 40 workers together can construct 40km road in 20 Days [Work and Time are directly proportional.
Manpower x Time / Work = Constant
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Engineers and Worker may have different efficiency with which they are working therefore you can't take it the way you are taking(independently).My question is why are we multiplying only Workers(20) X Days(20) and not :
Engineers(20) X Workers(20) X Days (20) = 8000 man-days to determine the man days ? Why are we leaving out the engineers from the man-days calculation ?
Maybe I am missing some very simple concept!
You have to consider the smallest group of manpower as (1 Engineer and 1 worker) but you can't use their efficiency independently until the relation to convert the Engineer's work into Worker's work is mentioned in the question.
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Hi Gmatinsight,GMATinsight wrote:It was simpler to think this way
20 engineers and 20 workers can together construct a 20km road in 20 days
40 engineers and 40 workers together can construct 20km road in half of the days as manpower has been doubled which is inversely proportional to the amount of time taken = 10 days
Now 40 engineers and 40 workers together can construct 40km road in 20 Days [Work and Time are directly proportional.
Manpower x Time / Work = Constant
I do understand that manpower has been doubled . What I don't understand is that to find the manpower why aren't we multiplying engineer with workers i.e engineers X workers instead of just considering workers or engineers in isolation.i.e (20X20X20X2)/(40X40). Why are we excluding engineers from manpower calculation ?
Last edited by Haldiram Bhujiawala on Sat Jul 05, 2014 7:34 am, edited 3 times in total.
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The mistake is explained by this representation. (where you have used X sign it should be + sign as it's the total manpower)Engineers(20) X Workers(20) X Days (20) = 8000 man-days
(20 Engineers + 20 Workers) X Days (20)
= 20(1Engineer + 1Worker) X Days (20)
= 400 man-days
Required work = 40 KM = 800 Man Days
(40 Engineers + 40 Workers) X Days (x)
= 40(1Engineer + 1Worker) X Days (x)
= 800 man-days [Required work]
therefore, x = 800/40 = 20 Days
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I hope now you can see that where you are using muliplication sign between Engineers and Workers needs to be replaced by addition sign as
1 Engineer and 1 Worker working together means "1 Engineer+1 Engineer"
1 Engineer and 1 Worker working together means "1 Engineer+1 Engineer"
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Use the following equation:Haldiram Bhujiawala wrote:Q:If 20 engineers and 20 workers can together construct a 20km road in 20 days , then 40 engineers and 40 workers together construct 40km road in how many days ?
(workers)(time) / output = (workers)(time) / output
In the equation above:
Workers and time are INVERSELY PROPORTIONAL.
As the number of workers increases, the amount of time required to produce the same output decreases.
Workers and output are DIRECTLY PROPORTIONAL.
As the number of workers increases, the amount of output also increases.
Time and output are also DIRECTLY PROPORTIONAL.
As the amount of time increases, the amount of output also increases.
In the problem above:
(20+20 engineers and workers)(20 days) / (20km of road) = (40+40 engineers and workers)(x days) / (40km of road)
(40)(20) / 20 = (80)(x) / 40
40 = 20x
x = 20 days.
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The problem is flawed -- and not representative of a typical work problem on the GMAT.Haldiram Bhujiawala wrote: I do understand that manpower has been doubled . What I don't understand is that to find the manpower why aren't we multiplying engineer with workers i.e engineers X workers instead of just considering workers or engineers in isolation.i.e (20X20X20X2)/(40X40). Why are we excluding engineers from manpower calculation ?
It needs to make clear how the output of the engineers is related to the output of the workers.
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A MUCH easier way of thinking about this is that you have DOUBLED the work but also DOUBLED the rate, so your time will remain unchanged.
Algebraically, you could represent that this way:
W = R*T
2W = (2R)*T
Notice that T will be the same in the second equation as it was in the first, since you've multiplied both sides by 2 (and the question makes clear that the RATE is doubling on the right hand side, so the 2 and the R can be grouped together).
While the question is awkwardly phrased, it seems clear to me that it was designed to test your ability to notice this sort of elegant solution. (It reminds me of a speed round MATHCOUNTS question that you can answer in five seconds, without any equations, if you catch the trick.)
Algebraically, you could represent that this way:
W = R*T
2W = (2R)*T
Notice that T will be the same in the second equation as it was in the first, since you've multiplied both sides by 2 (and the question makes clear that the RATE is doubling on the right hand side, so the 2 and the R can be grouped together).
While the question is awkwardly phrased, it seems clear to me that it was designed to test your ability to notice this sort of elegant solution. (It reminds me of a speed round MATHCOUNTS question that you can answer in five seconds, without any equations, if you catch the trick.)