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Kyra and Sage, working together can paint a room in 5 hours. If Melora

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Kyra and Sage, working together can paint a room in 5 hours. If Melora  [#permalink]

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New post 20 Mar 2018, 01:05
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Re: Kyra and Sage, working together can paint a room in 5 hours. If Melora  [#permalink]

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New post 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



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Re: Kyra and Sage, working together can paint a room in 5 hours. If Melora  [#permalink]

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New post 20 Mar 2018, 01:45
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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...

\(1/5 + 1/x = 1/4\)
\(1/x = 1/4 - 1/5\)
\(1/x = (5-4)/20\)
\(x = 20\)
Hence E.

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Re: Kyra and Sage, working together can paint a room in 5 hours. If Melora  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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*:
\(\frac{1}{M}=(\frac{1}{4}-\frac{1}{5}) = (\frac{(5-4)}{20})=\frac{1}{20}\)

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:\(\frac{1}{A}-\frac{1}{B} =\frac{(B-A)}{AB}\)
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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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 = 20

The answer is E.
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Re: Kyra and Sage, working together can paint a room in 5 hours. If Melora  [#permalink]

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New post 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   [#permalink] 05 Apr 2018, 11:05
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