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Kyra and Sage, working together can paint a room in 5 hours. If Melora
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20 Mar 2018, 01:05
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Kyra and Sage, working together can paint a room in 5 hours. If Melora helps Kyra and Sage paint the room, the three of them can paint the room in 4 hours. What amount of time (in hours) would it take Melora, working alone, to paint the room? (A) 6 (B) 8 (C) 10 (D) 15 (E) 20
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Re: Kyra and Sage, working together can paint a room in 5 hours. If Melora
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20 Mar 2018, 01:06
Bunuel wrote: Kyra and Sage, working together can paint a room in 5 hours. If Melora helps Kyra and Sage paint the room, the three of them can paint the room in 4 hours. What amount of time (in hours) would it take Melora, working alone, to paint the room?
(A) 6 (B) 8 (C) 10 (D) 15 (E) 20 17. Work/Rate Problems On other subjects: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative Megathread
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Re: Kyra and Sage, working together can paint a room in 5 hours. If Melora
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20 Mar 2018, 01:45
Bunuel wrote: Kyra and Sage, working together can paint a room in 5 hours. If Melora helps Kyra and Sage paint the room, the three of them can paint the room in 4 hours. What amount of time (in hours) would it take Melora, working alone, to paint the room?
(A) 6 (B) 8 (C) 10 (D) 15 (E) 20 Rate of painting room for K + S = 1/5 ( how much part of the room they can paint in one hour ) When M is working with K & S > New rate is 1/4 ( as they take 4 hrs to paint room , they paint 1/4th room in one hour) Let M alone finish paint job in x hours. So her rate is 1/x. So Rate of K&S + Rate of M = Rate of K&S&M together. We have... \(1/5 + 1/x = 1/4\) \(1/x = 1/4  1/5\) \(1/x = (54)/20\) \(x = 20\) Hence E. Please give Kudos if you liked the explanation...
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Re: Kyra and Sage, working together can paint a room in 5 hours. If Melora
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20 Mar 2018, 09:26
Bunuel wrote: Kyra and Sage, working together can paint a room in 5 hours. If Melora helps Kyra and Sage paint the room, the three of them can paint the room in 4 hours. What amount of time (in hours) would it take Melora, working alone, to paint the room?
(A) 6 (B) 8 (C) 10 (D) 15 (E) 20 Let the total work be 20 units So, Efficiency of Kyra and Sage is 4 units And, Efficiency of Melora , Kyra and Sage is 5 units So, We have ( k + s ) + m = 5 ; 4 + s = 5 So, Efficiency of m is 1 , Thus time required by Melora to paint the room is 20 Hours, Answer must be (E)
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Kyra and Sage, working together can paint a room in 5 hours. If Melora
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20 Mar 2018, 09:53
Bunuel wrote: Kyra and Sage, working together can paint a room in 5 hours. If Melora helps Kyra and Sage paint the room, the three of them can paint the room in 4 hours. What amount of time (in hours) would it take Melora, working alone, to paint the room?
(A) 6 (B) 8 (C) 10 (D) 15 (E) 20 In 1 hour, Kyra and Sage can paint 1/5 of the room In 1 hour, Kyra, Sage and Melora can paint 1/4 of the room Therefore, in 1 hour, Melora can paint 1/4  1/5 = 1/20 of the room Therefore Melora needs 20 hours to paint the room => Answer (E)



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Kyra and Sage, working together can paint a room in 5 hours. If Melora
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20 Mar 2018, 12:18
Bunuel wrote: Kyra and Sage, working together can paint a room in 5 hours. If Melora helps Kyra and Sage paint the room, the three of them can paint the room in 4 hours. What amount of time (in hours) would it take Melora, working alone, to paint the room?
(A) 6 (B) 8 (C) 10 (D) 15 (E) 20 The combined rate of two persons is given. When the third person, Melora, joins, the equation is simple. 1) Combined rate of Sage and Kyra = \(\frac{1}{5}\) Substitute that rate in the next step 2) Combined rate of Sage, Kyra, and Melora: Together all three can paint the room in 4 hours: \((\frac{1}{S}+\frac{1}{K})+\frac{1}{M}=\frac{1}{4}\), that is, \(\frac{1}{5}+\frac{1}{M}=\frac{1}{4}\)3) Melora's rate, working alone*: \(\frac{1}{M}=(\frac{1}{4}\frac{1}{5}) = (\frac{(54)}{20})=\frac{1}{20}\)4) M's time? Work is 1 (room). Rate is inversely proportional to time. Flip the rate fraction: \(\frac{1room}{20hrs} > \frac{20hrs}{1room}\)Melora takes 20 hours to paint the room alone. Answer E * Shortcuts Rates:\(\frac{1}{A}\frac{1}{B} =\frac{(BA)}{AB}\) Use LCM to see proof. For added rates, the shortcut is \(\frac{1}{A}+\frac{1}{B}=\frac{(A+B)}{AB}\)
Inverting (flipping) the rate fraction to get time \(W = 1\) room, \(r\) \(R * T = W\), so \(T = \frac{W}{R}\) \(T=\frac{1r}{\frac{1r}{20hrs}}= (1r * \frac{20hrs}{1r}) = 20 hrs\)
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Kyra and Sage, working together can paint a room in 5 hours. If Melora
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24 Mar 2018, 14:30
Solution Let the total amount of work done to paint the room is \(LCM (5,4) = 20\) units. • Kyra and Sage, working together can paint a room in \(5\) hours.
o Thus, Kyra and Sage complete \(4\) units of work in \(1\) hour. o \(K+S=4\)\((1)\)
• If Melora helps Kyra and Sage paint the room, the three of them can paint the room in \(4\) hours.
o Thus, Kyra, Sage, and Melora complete \(5\) units of work in \(1\) hour. o \(K+S+M=5\)\((2)\) By subtracting equation \((1)\) from equation \((2)\), we get: • \(M=1\) unit, OR, \(M\) completes \(1\) unit of work in \(1\) hour.
• Thus, to complete \(20\)units of work Melora will take \(\frac{20}{1}\)=\(20\) hours. Answer: E
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Kyra and Sage, working together can paint a room in 5 hours. If Melora
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24 Mar 2018, 18:54
Bunuel wrote: Kyra and Sage, working together can paint a room in 5 hours. If Melora helps Kyra and Sage paint the room, the three of them can paint the room in 4 hours. What amount of time (in hours) would it take Melora, working alone, to paint the room?
(A) 6 (B) 8 (C) 10 (D) 15 (E) 20 K and S paint 1/5 of the room in 1 hour M saves them 1 hour by painting 1/5 of the room in 4 hours 4/(1/5)=20 hours for M to paint room alone E



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Kyra and Sage, working together can paint a room in 5 hours. If Melora
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30 Mar 2018, 16:46
Bunuel wrote: Kyra and Sage, working together can paint a room in 5 hours. If Melora helps Kyra and Sage paint the room, the three of them can paint the room in 4 hours. What amount of time (in hours) would it take Melora, working alone, to paint the room?
(A) 6 (B) 8 (C) 10 (D) 15 (E) 20 Remember: Rate * Time = Work K = Kyra  S = Sage  M = Melora (K+S) * 5 = 1 > (K + S) working together at their own rate took 5 hours to paint 1 room. This means the rate at which (K + S) work is (1/5). (M+K+S) * 4 = 1 > (M + K + S) working together at their own rate took 4 hours to paint 1 room. This means the rate at which (M + K + S) work is (1/4). Now we need to figure out at what rate Melora (M) works at:(M + K + S) = (1/4) > This is the rate at which when (M + K + S) work together. (M) + (1/5) = (1/4) > We already know that (K + S) = 1/5 since this is the rate at which they work together. M = (1/4)  (1/5) M = (1/20) > This is the rate at which Melora (M) works alone. Now to solve for how long it will take Melora to paint 1 room:R * T = W > Our standard rate formula (1/20) * T = 1 > Plug in the rate for which Melora works (1/20) and solve for T (time) T = 20The answer is E.



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Re: Kyra and Sage, working together can paint a room in 5 hours. If Melora
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05 Apr 2018, 11:05
Bunuel wrote: Kyra and Sage, working together can paint a room in 5 hours. If Melora helps Kyra and Sage paint the room, the three of them can paint the room in 4 hours. What amount of time (in hours) would it take Melora, working alone, to paint the room?
(A) 6 (B) 8 (C) 10 (D) 15 (E) 20 We can let K, S, and M = the time it takes Kyra, Sage, and Melora to paint the room respectively and create the equations: 1/K + 1/S = 1/5 And 1/M + 1/K + 1/S = 1/4 Subtracting the first equation from the second we have: 1/M = 1/4  1/5 1/M = 5/20  4/20 = 1/20 So it would take Melora 20 hours to paint the room. Answer: E
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Re: Kyra and Sage, working together can paint a room in 5 hours. If Melora
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