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Two painters, Ray and Taylor, are painting a fence. Ray paints at a un

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Two painters, Ray and Taylor, are painting a fence. Ray paints at a un  [#permalink]

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New post 16 Oct 2016, 21:59
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Question Stats:

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Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long?

a) 320
b) 400
c) 450
d) 500
e) 580

Please assist with above problem.
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Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un  [#permalink]

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New post 16 Oct 2016, 22:06
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Ray can paint 40 feet every 160 minutes..
Then Ray can paint 40/160 feets per minute.
Taylor can paint 50 feet every 125 minutes..
Then Taylor can paint 50/125 fret per minutes
Then in one minute both together can paint (40/160)+(50/125) feets.. nothing but 13/20 feets per minute..
Time taken for 260 feet 260/(13/20) or (260*20)/13 = 400 minutes
Answer B

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Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un  [#permalink]

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New post 16 Oct 2016, 22:12
alanforde800Maximus wrote:
Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long?

a) 320
b) 400
c) 450
d) 500
e) 580

Please assist with above problem.


Efficiency of Ray = \(\frac{40}{160}\) => \(\frac{1}{4}\)
Efficiency of Taylor = \(\frac{50}{125}\) => \(\frac{2}{5}\)

Combined efficiency is \(\frac{1}{4}\) + \(\frac{2}{5}\) = \(\frac{13}{20}\)

Time required to paint a 260 meters wall is \(260*\frac{20}{13}\) = \(400\) minutes.

Hence answer will be (B) 400
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Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un  [#permalink]

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New post 16 Oct 2016, 22:31
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RAY : 40 feet ->160 minutes | TAYLOR : 50 feet -> 125 minutes
Thus, 1 feet -> 4 minutes | 1 feet -> 2.5 minutes
Thus Total One day work, 1/4+1/2.5 = 65/100
Thus for Total Work(260feet) : 100/65 * 260 = 400 minutes

Answer is B
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Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un  [#permalink]

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New post 17 Oct 2016, 08:57
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alanforde800Maximus wrote:
Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long?

a) 320
b) 400
c) 450
d) 500
e) 580

Please assist with above problem.


Let's determine how much each person can paint in ONE MINUTE

Ray paints at a uniform rate of 40 feet every 160 minutes
We have (40 feet)/(160 minutes)
Divide top and bottom by 160 to get: (0.25 feet)/(1 minute)
So, Ray paints 0.25 feet every ONE MINUTE

Taylor paints at a uniform rate of 50 feet every 125 minutes
We have (50 feet)/(125 minutes)
Divide top and bottom by 125 to get: (0.4 feet)/(1 minute)
So, Taylor paints 0.4 feet every ONE MINUTE

So, working TOGETHER, the can paint 0.25 + 0.4 feet every ONE MINUTE
In other words, they can paint 0.65 feet every ONE MINUTE

How many minutes will it take for them to paint a fence that is 260 feet long?
Time = output/rate
= 260/0.65
= 400 minutes
Answer:

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Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un  [#permalink]

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New post 05 Dec 2017, 18:39
alanforde800Maximus wrote:
Two painters, Ray and Taylor, are painting a fence. Ray paints at a uniform rate of 40 feet every 160 minutes, and Taylor paints at a uniform rate of 50 feet every 125 minutes. If the two painters paint simultaneously, how many minutes will it take for them to paint a fence that is 260 feet long?

a) 320
b) 400
c) 450
d) 500
e) 580


We are given that Ray paints at a uniform rate of 40 feet every 160 minutes. Thus, the rate of Ray is 40/160= 1/4 ft/min.

We are also given that Taylor paints at a uniform rate of 50 feet every 125 minutes. Thus, the rate of Taylor is 50/125= 2/5 ft/min.

We need to determine the time it will take to paint a fence, that is 260 feet long, when Ray and Taylor work simultaneously.

To determine the time to paint a 260-foot-long fence, we can use the combined work formula:

Work done by Ray + Work done by Taylor = 260 feet (the total work completed)

Because Ray and Taylor are working simultaneously, we can let the time they both work together be t minutes. We now can express the individual work done by Ray and Taylor. We must remember that work = rate x time.

Work done by Ray = (1/4)t

Work done by Taylor = (2/5)t

(1/4)t + (2/5)t = 260

We can multiply the entire equation by 20 to cancel out the fraction and we have:

5t + 8t = 5,200

13t = 5,200

t = 400 minutes

Answer: B
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Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un  [#permalink]

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New post 24 Jan 2018, 20:54
Hi All,

The 'key' to most 'rate' questions is to find a simple unit measure that's comparable for all the entities involved. In this prompt, we're dealing with 'total feet painted per volume of minutes.' Since the question asks for the number of MINUTES it takes to paint a fence, we should organize our information in terms of 'feet painted per minute.'

Based on the information in the prompt...
Ray paints 40 feet every 160 minutes = 1/4 foot per minute
Taylor paints 50 feet every 125 minutes = 2/5 foot per minute

In TOTAL, they paint 1/4 + 2/5 = 5/20 + 8/20 = 13/20 feet per minute

While that measurement might seem 'weird', we can convert it...

(13/20 feet per minute)(20 minutes) = 13 feet every 20 minutes

The prompt asks for the time it would take to paint a 260 foot long fence....

260/13 = (20 units of 13 feet) x 20 minutes = 400 minutes

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Re: Two painters, Ray and Taylor, are painting a fence. Ray paints at a un   [#permalink] 24 Jan 2018, 20:54
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