Lindsay can paint 1/x of a certain room in 20 minutes. What fraction of the same room can Joseph paint in 20 minutes if the two of them can paint the room in an hour, working together at their respective rates?
A. 1/3x
B. 3x/(x - 3)
C. (x - 3) / 3x
D. x / (x - 3)
E. (x - 3) / x
Help, anybody?
Work Rate Problem - 5
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Given: Lindsay can paint 1/x of a certain room in 20 minutesLindsay can paint 1/x of a certain room in 20 minutes. What fraction of the same room can Joseph paint in 20 minutes if the two of them can paint the room in an hour, working together at their respective rates?
A. 1/3x
B. 3x/(x - 3)
C. (x - 3) / 3x
D. x / (x - 3)
E. (x - 3) / x
So, in 1 HOUR, Lindsay can paint 3/x of the room
Given: Lindsay and Joseph can paint the room in 1 HOUR.
During that one hour, Lindsay can paint 3/x of the room.
So, during that 1 HOUR, Joseph must paint the rest (whatever Lindsay did not paint)
So, during the 1 HOUR, the fraction of the room that Joseph paints = 1 - 3/x
= x/x - 3/x
= (x-3)/x
So, (x-3)/x = the fraction of the room that Joseph paints in one hour.
Since 20 minutes = 1/3 of an hour, Joseph can paint (1/3)[(x-3)/x] of the room in 20 minutes.
(1/3)[(x-3)/3x] = [spoiler](x-3)/3x = C[/spoiler]
Cheers,
Brent
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Or pick a number. Say x = 6. If Lindsay can paint 1/6 of a room in 20 minutes, then she can paint the full room in 120 minutes, or two hours. So her rate is 1 room/ 2 hours, or 1/2. If together they can paint the room in an hour, then the combined rate is 1 room/1 hour, or 1. We know that rates are additive in work problems. So if Lindsay's rate is 1/2, Joseph's rate is J, and their combined rate is 1, we know that 1/2 + J = 1. And J = 1/2.saadishah wrote:Lindsay can paint 1/x of a certain room in 20 minutes. What fraction of the same room can Joseph paint in 20 minutes if the two of them can paint the room in an hour, working together at their respective rates?
A. 1/3x
B. 3x/(x - 3)
C. (x - 3) / 3x
D. x / (x - 3)
E. (x - 3) / x
Help, anybody?
If Joseph's rate is 1/2, and he works for 20 minutes (or 1/3 of an hour) then his total work is (1/2)* (1/3) = 1/6. (Or we can simply see that Joseph's rate is the same as Lindsay's, and we already knew she could do 1/6 of a room in 20 minutes.)
Now just plug in x = 6 to the answer choices and see what gives us 1/6. Only C will work.
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Let the room = 20 units.
Let x = 4.
Since Lindsay paints 1/x of the room in 20 minutes -- the equivalent of 1/3 hour -- Lindsay's rate = (1/4) * 20 = 5 units per 1/3 hour = 15 units per hour.
Since Lindsay and Joseph paint the entire room in 1 hour, their combined rate = 20 units per hour.
Thus, Joseph's rate = (combined rate) - (Lindsay's rate) = 20-15 = 5 units per hour.
In 20 minutes -- the equivalent of 1/3 hour -- the amount of work produced by Joseph = r*t = 5 * (1/3) = 5/3 units.
Thus, the fraction painted by Joseph = (5/3)/20 = 1/12. This is the target.
Now plug x=4 into the answers to see which yields the target value of 1/12.
Only C works:
(x-3)/3x = (4-3)/(3*4) = 1/12.
The correct answer is C.
Let x = 4.
Since Lindsay paints 1/x of the room in 20 minutes -- the equivalent of 1/3 hour -- Lindsay's rate = (1/4) * 20 = 5 units per 1/3 hour = 15 units per hour.
Since Lindsay and Joseph paint the entire room in 1 hour, their combined rate = 20 units per hour.
Thus, Joseph's rate = (combined rate) - (Lindsay's rate) = 20-15 = 5 units per hour.
In 20 minutes -- the equivalent of 1/3 hour -- the amount of work produced by Joseph = r*t = 5 * (1/3) = 5/3 units.
Thus, the fraction painted by Joseph = (5/3)/20 = 1/12. This is the target.
Now plug x=4 into the answers to see which yields the target value of 1/12.
Only C works:
(x-3)/3x = (4-3)/(3*4) = 1/12.
The correct answer is C.
Last edited by GMATGuruNY on Sun Sep 30, 2018 10:09 am, edited 1 time in total.
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Help is on the way! You're asking about a lot of rate problems, so why not take a look at some excellently worked examples?saadishah wrote:Help, anybody?