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

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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?

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

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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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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?

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.

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

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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?

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

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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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29 Sep 2014, 02:25

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

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19 Jul 2017, 16:16

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

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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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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?

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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