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15 welders work at a constant rate they complete an order

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15 welders work at a constant rate they complete an order  [#permalink]

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New post 31 May 2013, 22:31
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15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A) 5
(B) 2
(C) 8
(D) 4
(E) 6
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 31 May 2013, 23:25
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gmatquant25 wrote:
15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A)5
(B)2
(C)8
(D)4
(E)6


\(15*r*3\) = 1 unit of work. Thus, rate of each worker =\(\frac{1}{45}\)
The amount of work done in 1 day with all 15 workers =\(\frac{1}{3}\). The amount of work left to be done by (15-9) workers = \(\frac{2}{3}\)

Let them take d days to complete this work-->\(6*r*d\) = 2/3--> \(6*\frac{1}{45}*d =\frac{2}{3}\)--> d= 5 days.
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 31 May 2013, 22:54
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gmatquant25 wrote:
15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A)5
(B)2
(C)8
(D)4
(E)6


LCM (15,3,9) = 45

Suppose Total work is 45 Units.

15 Welders = 45 Units / 3 days
15 Welders = 45/3 = 15 Units / day
1 Welder = 1 Unit/day

Day 1: everyone worked = 15 units, remaining 30 Units

After Day 1, I have 6 Welders (15 - 9), and I need to complete 30 Units.

1 Welder = 1 Unit/day
6 Welders = 6 Unit/day

for 30 Units, 30(total units)/ 6 (units per day) = 5 days
Answer A
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 31 May 2013, 23:29
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gmatquant25 wrote:
15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A)5
(B)2
(C)8
(D)4
(E)6


Assume the order to be welding 45 rods...

Each welder works at 1 rod per day

On first day 15 rods are completed.

So now 30 rods have to be completed by 6 workers...

So, the answer is 30/6 = 5 days
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 01 Jun 2013, 03:17
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gmatquant25 wrote:
15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A)5
(B)2
(C)8
(D)4
(E)6


Similar question to practice: 6-workers-should-finish-a-job-in-8-days-after-3-days-came-153377.html

Hope it helps.
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 01 Jun 2013, 21:37
gmatquant25 wrote:
15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A)5
(B)2
(C)8
(D)4
(E)6


1. We need to find out the time taken by 6 workers after day 1.
2. total no.of wokers * total time taken = time taken by 1 worker
3. Time taken by 1 worker = 15*3 = 45 days
4. But on day 1 fifteen workers had already worked finishing 1/3 of the job. So 6 workers have to finish only 2/3 of the job.
5. Total time taken by 6 workers can be got from formula used at (2). i.e., 6* total time taken = 45. Total time taken by 6 workers to finish the complete job is 45/ 6 = 7.5 days.
6. Time taken by 6 workers to finish 2/3 of the job is 2/3 * 7.5 = 5 days.

The answer is choice A
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 29 Jul 2013, 05:48
Hello Brunel is this right approach

Rate of 15 workers = 1/3 job / day
Rate of 1 worker = 1/45 job / day

as 15 workers work for single day and 6 workers work for more days (x)

So now equation becomes

1/3 + 6/45(x-3) = 1
Solving this we get x=8

So 6 workers need x-3more days ie = 5days
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 08 Sep 2013, 12:33
gmatquant25 wrote:
15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A) 5
(B) 2
(C) 8
(D) 4
(E) 6



sol:

15 welders rate of job = 1/3

W= 1 (1/3) = 1/3

Remaining Job = 1-(1/3) = 2/3

Job---days---number of welders

2/3------2----------15
2/3------X ---------15-9 = 6


X= 2* (15/6) =5 days
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 25 Sep 2013, 20:47
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Man Days = Work

15 people * 3 days = 45 units of work

15 people * 1 day = 15 units

Work left = 45 -15 = 30 units

People left to work = 15-9= 6

6 people * x days = 30

x=5
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 23 Jul 2017, 09:29
Let rate of each worker be \(\frac{1}{x}\)
For 15 workers the rate is \(\frac{15}{x}\)

\(\frac{15}{x}\) = \(\frac{1}{3}\)
Hence rate of each worker is \(\frac{1}{45}\)

After day 1, \(\frac{1}{3}\) of the work will be complete, hence \(\frac{2}{3}\) of the work remains
The number of workers has now dropped from 15 to 6 (Since 9 workers left)
Rate of 6 workers = \(\frac{6}{x}\) = 6 * \(\frac{1}{45}\) = \(\frac{2}{15}\)

W= \(\frac{2}{3}\) ; R = \(\frac{2}{15}\); T = \(\frac{W}{R}\)
T = 5

Answer is A
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 24 Jul 2017, 03:53
gmatquant25 wrote:
15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A) 5
(B) 2
(C) 8
(D) 4
(E) 6



15x1/1/3 = 6xt/2/3

solving this we get, t= 5

hope this helps ....
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 24 Jul 2017, 04:01
Total work = 15*3 = 45 man days
1 st day = 15*1 = 15 man days

Rem work = 45-15 = 30 man days

This can be completetd in 30/6 = 5 days
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 26 Jul 2017, 16:45
gmatquant25 wrote:
15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A) 5
(B) 2
(C) 8
(D) 4
(E) 6


We are given that the rate of 15 welders is 1/3. Thus, after 1 day, 2/3 of the job is left to complete for 6 welders. If we let x = the rate of 6 welders, we can create the proportion to determine their rate:

15/(1/3) = 6/x

15x = 2

x = 2/15

Thus, the 6 welders will complete the job in (2/3)/(2/15) = 30/6 = 5 days.

Answer: A
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Re: 15 welders work at a constant rate they complete an order  [#permalink]

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New post 21 Oct 2018, 08:36
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gmatquant25 wrote:
15 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order?

(A) 5
(B) 2
(C) 8
(D) 4
(E) 6


15 Welders were supposed to do the job in 3 days but 9 of them left after one day. If none let left then whey will work 2 more days.

So 15 welders will do job in 2 days (because they already worked one day)
6 welders will need (15 x 2)/6= 5 Ans a.
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