20 minutes is 1/3 of an hour. In an hour, Laura would paint
3 * 1/x = 3/x;
Joseph, then, would have painted the rest:
1 - 3/x, or (x-3)/x.
So in 20 minutes Joe would do
(x-3)/3x.

Given
rLaura = 1/x room in 20 minute

rL = 1/(x/3) room/hr

Let rJ = 1/j be Joseph's rate

1/(x/3) + 1/j = 1

or 3/x + 1/j = 1
or 3j+x = xj

or x/(x-3) = j

Therfore, rJ = 1/(x-3)/x room/hr

60 minutes : (x-3)/x room
20 : (x-3)/3x

Answer : C : (x-3)/3x

Good one haas and brain... i was stuck on how to start this one out but you guys are right on the nose..


C.

braindancer's solution is much more elegant and intuitive... like that!!

Thanks brain/.

3(1/x) +3(1/y) = 1

(1/y) = 1/3 - 1/x

1/y = (x-3)/3x

3-C

I did it in a simpler way

Laura's 1 hour work = 3/x
Joseph's one hour work = y

Laura' sand Joseph's 1 hour work = 1 ( as given they completed full work in one hour).

So 3/x + y = 1
i.e y = (x-3)/x

So Joseph's 20 minutes work = (x-3)/3x
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Laura can paint 1/x of room in 20 mins
Hence in 1 min she can paint 1/20x of room

Let Joseph paints 1/J of the room in 1 min

Hence
1/20x + 1/J = 1/60
1/J = x-3/60x
Hence J= 60x/x-3

Therefore in 60x/x-3 mins joseph paints 1 room
in 20 mins he can paint = 20 * (x-3) / 60x = (x-3)/3x of the room

Rate of Laura \(= \frac{1}{20x}\) (Per Min)

Consider rate of Joseph \(= \frac{1}{j}\) (Per Min)

Both combined perform the task in 60 minutes

\(\frac{1}{20x} + \frac{1}{j} = \frac{1}{60}\)

\(\frac{1}{j} = \frac{1}{60} - \frac{1}{20x}\)

\(\frac{1}{j} = \frac{x-3}{60x}\) (Per Min)

So in 20 minutes, work done by Joseph

\(= \frac{x-3}{60x} * 20\)

\(= \frac{x-3}{3x}\)

Answer = C

The Question is a good candidate for plugin...Here is how I would approach

Let us assume value of x for laura to be 12 so we have that laura does 1/12th of room in 20 minutes or to paint one complete laura takes 240 minutes

Let Joseph take y minutes to paint 1 room so we have
Also given that
1/240 +1/y = 1/60 ----> Solve for y and we get y =240/3 or 80 minutes.
(Here we should choose the value of x in such a way that 1/y = \((1/60- 1/x)\geq{0}\) so that y is not negative.


So Joseph paints 1 room in 80 minutes so in 20 minutes, he will paint 1/4th of the room.

Plug in value of x in the above options and we get answer C.
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This is a very easy question...

it is given in the question that L completes the (1/x)th of a work in 20 minutes. So she completes 3/xth of a work in 60 minutes.

the job is finished by both of them in 1 hour.. so in 60 minutes the other fraction of work i.e (1- (3/x)) is done by Joseph.

So Joseph does (x-3)/x of a work in 60 minutes. So in 20 minutes he would have done only (1/3)rd of it = (1/3)((x-3)/x)

which is (x-3/3x)--> Choice C

Options with variables are often done by plugging in numbers.
Both working together can paint the room in 1 hr so if their individual rates were equal, each would take 2 hours alone.
2 hours is 120 mins so in 20 mins each would complete 120/20 = 1/6th of the room alone.
So if x = 6 (Laura completes 1/6th of the room in 20 mins), the correct option will give 1/6. (Joseph will also paint 1/6th of the room if their rates are same)
If you put x = 6 in the options, only option (C) will give 1/6

Answer (C)
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J's rate=1-3/x
J's time=1/3 hour
(1-3/x)(1/3)=x-3/3x

Together they paint 1 room in 60 min. So in 20 min they paint (1/3) room. Out of this Laura painted (1/x) room, so Joseph must have painted (1/3) - (1/x) room => (x – 3)/(3x) room

Convert everything in terms of 20 min.

Let the part of the room painted by Joseph in 20 min be 1/y

In 20 min, we have part of work done as 1/x +1/y=20/60=1/3

1/y=1/3 - 1/x = (x-3)/(3x)
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