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Lindsay can paint 1/x of a certain room in 20 minutes. What

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by swerve » Sat Sep 29, 2018 9:27 am

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Answers

A

B

C

D

E

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Difficulty—

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 respectives rates?

A. 1/3x
B. 3x/(x - 3)
C. (x - 3) / 3x
D. x / (x - 3)
E. (x - 3) / x

The OA is C.

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lindsay

by GMATGuruNY » Sat Sep 29, 2018 3:19 pm
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 respectives rates?

A. 1/3x
B. 3x/(x - 3)
C. (x - 3) / 3x
D. x / (x - 3)
E. (x - 3) / x
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.
Last edited by GMATGuruNY on Sun Sep 30, 2018 9:48 am, edited 1 time in total.
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by deloitte247 » Sun Sep 30, 2018 9:36 am
$$In\ 20\ \min utes\ lindsay\ can\ paint\ \frac{1}{x}$$
$$In\ 60\ \min utes\ \left(1hr\right)lindsay\ can\ paint\ =\frac{\left(\frac{1}{x}\cdot\ 60\right)}{20}$$
$$=\frac{\left(\frac{60}{x}\right)}{20}$$
$$=\frac{60}{x}\cdot\frac{1}{20\ }=\frac{3}{x}in\ 60\ \min utes\ \left(1hour\right)$$
$$If\ lindsay\ can\ paint\ \frac{3}{x}part\ of\ the\ room\ in\ 1\ hour,\ let\ the\ remaining\ part\ of\ the\ room\ be\ =y.$$
Bearing in mind that Joseph paints the remaining pat of the room .
Therefore, Joseph can paint y in 60 minutes,
$$\frac{3}{x\ }+y=x\ \left(total\ room\right)$$
$$y\ =\ x-\frac{3}{x}$$
$$If\ Joseph\ work\ for\ 60\ \min utes\ =\ x-\frac{\left(3\right)}{x},\ then,$$
$$20\ \min utes\ =\ \frac{\left(\left(\frac{x}{1}-\frac{3}{x}\right)\cdot\frac{20}{1}\right)}{60}$$
$$=\ \frac{\left(\frac{x}{1}-\frac{3}{x}\right)}{3}\ =\ \left(\frac{x}{1}-\frac{3}{x}\right)\cdot\frac{1}{3}$$
$$\frac{\left(x-3\right)}{3x}$$

Option C is CORRECT.
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by Brent@GMATPrepNow » Sun Sep 30, 2018 10:33 am
swerve 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 respectives rates?

A. 1/3x
B. 3x/(x - 3)
C. (x - 3) / 3x
D. x / (x - 3)
E. (x - 3) / x

The OA is C.

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Given: Lindsay can paint 1/x of a certain room in 20 minutes
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]
= (x-3)/3x

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Sat Oct 13, 2018 5:34 pm
swerve 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 respectives rates?

A. 1/3x
B. 3x/(x - 3)
C. (x - 3) / 3x
D. x / (x - 3)
E. (x - 3) / x

We are given that Lindsay can paint 1/x of a room in 20 minutes; thus, she can paint 3/x of a room in 60 minutes (or in 1 hour). Thus, her hourly rate is 3/x room/hr. We are also given that when she works with Joseph, they can paint the entire room in 1 hour. If we let total work = 1 and j = the number of hours it takes Joseph to paint the room by himself, then Joseph's rate = 1/j room/hr. We can create the following equation and isolate j:

work of Lindsay + work of Joseph = 1

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

3/x + 1/j = 1

Multiplying the entire equation by xj, we obtain:

3j + x = xj

x = xj - 3j

x = j(x - 3)

x/(x - 3) = j

Since j = x/(x - 3) and 1/j = Joseph's rate, then Joseph's rate, in terms of x, is (x - 3)/x.

Since 20 minutes = 1/3 of an hour, and since work = rate x time, Joseph can complete:

[(x - 3)/x](1/3) = (x - 3)/(3x) of the job in 20 minutes.

Alternate Solution:

Since Lindsay and Joseph, working together, can paint the entire room in 1 hour, then in 20 minutes, they can paint 1/3 of the room. If we let r be the fraction of the room that Joseph can paint in 20 minutes, then it must be true that:

1/x + r = 1/3

r = 1/3 - 1/x

r = x/(3x) - 3/(3x)

r = (x - 3)/(3x)

Answer: C

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