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Doug can paint a room in 5 hours. Dave can paint the same room in 7 ho

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Doug can paint a room in 5 hours. Dave can paint the same room in 7 ho  [#permalink]

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New post 27 Mar 2019, 00:20
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Question Stats:

57% (02:06) correct 43% (02:10) wrong based on 53 sessions

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Doug can paint a room in 5 hours. Dave can paint the same room in 7 hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let t be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by t?


(A) \((\frac{1}{5} + \frac{1}{7})(t + 1) = 1\)

(B) \((\frac{1}{5} + \frac{1}{7})t + 1 = 1\)

(C) \((\frac{1}{5} + \frac{1}{7})t = 1\)

(D) \((\frac{1}{5} + \frac{1}{7})(t - 1) = 1\)

(E) \((5 + 7)t = 1\)

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Doug can paint a room in 5 hours. Dave can paint the same room in 7 ho  [#permalink]

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New post Updated on: 27 Mar 2019, 01:45
Bunuel wrote:
Doug can paint a room in 5 hours. Dave can paint the same room in 7 hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let t be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by t?


(A) \((\frac{1}{5} + \frac{1}{7})(t + 1) = 1\)

(B) \((\frac{1}{5} + \frac{1}{7})t + 1 = 1\)

(C) \((\frac{1}{5} + \frac{1}{7})t = 1\)

(D) \((\frac{1}{5} + \frac{1}{7})(t - 1) = 1\)

(E) \((5 + 7)t = 1\)


rate of D = 1/5 and D= 1/7
together rate ; 1/5+1/7
time = t-1 ; as it includes lunch break
work = 1
so 1= (1/5+1/7)* (t-1)
\((\frac{1}{5} + \frac{1}{7})(t - 1) = 1\)
IMO D
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Originally posted by Archit3110 on 27 Mar 2019, 01:27.
Last edited by Archit3110 on 27 Mar 2019, 01:45, edited 1 time in total.
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Re: Doug can paint a room in 5 hours. Dave can paint the same room in 7 ho  [#permalink]

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New post 27 Mar 2019, 01:42
Rate of doing work of dave=1/7
Rate of doing work of Doug=1/5
Total time taken including lunch to complete work =t
Total time taken excluding lunch to complete work=t-1
Therefore,1/7+1/5=1/(t-1)
(1/7+1/5)(t-1)=1

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Re: Doug can paint a room in 5 hours. Dave can paint the same room in 7 ho  [#permalink]

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New post 05 Apr 2019, 07:29
Hello,

Could someone explain why the one-hour lunch break is considered in the original ((1/5)+(1/7)) combined rate calculation? I do not understand how D is correct since that would mean we need to remove the 1 hour from their combined rate (given) which does not include the one hour in the first place.


Any insight would be great! :)

KHow

Bunuel wrote:
Doug can paint a room in 5 hours. Dave can paint the same room in 7 hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let t be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by t?


(A) \((\frac{1}{5} + \frac{1}{7})(t + 1) = 1\)

(B) \((\frac{1}{5} + \frac{1}{7})t + 1 = 1\)

(C) \((\frac{1}{5} + \frac{1}{7})t = 1\)

(D) \((\frac{1}{5} + \frac{1}{7})(t - 1) = 1\)

(E) \((5 + 7)t = 1\)
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Doug can paint a room in 5 hours. Dave can paint the same room in 7 ho  [#permalink]

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New post 09 Apr 2019, 02:01
[quote="KHow"]Hello,

Could someone explain why the one-hour lunch break is considered in the original ((1/5)+(1/7)) combined rate calculation? I do not understand how D is correct since that would mean we need to remove the 1 hour from their combined rate (given) which does not include the one hour in the first place.


Any insight would be great! :)

KHow


read the highlighted part
Doug can paint a room in 5 hours. Dave can paint the same room in 7 hours.
Doug and Dave paint the room together and

take a one-hour break for lunch. ----------------- here they take one hour lunch break

Let t be the total time, in hours, required for them to complete the job working together, including lunch. --------- here total time include lunch also


Which of the following equations is satisfied by t?
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Doug can paint a room in 5 hours. Dave can paint the same room in 7 ho   [#permalink] 09 Apr 2019, 02:01
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