Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo : Problem Solving (PS)

ezhilkumarank

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

08 Aug 2010, 21:14

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Answer is C -- \((x-1/3x)\)

Given details:

Rate at which L work --1/x
Rate at which L&J work -- x

Rate at which J alone works -- \((x-1/x)\)

Rate at which J alone works in 20 min. -- \((x-1/x) * (20/60) => (x-1/x)*(1/3)\)

On a side note, maybe this post needs to be moved to the Problem solving section.

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

09 Aug 2010, 09:54

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bunuel u always hit the nail to the point.......very easy explanation....thanks

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

28 Oct 2011, 16:31

Just wondering,

if the job took say, 2.5 hours to complete, how would the answer change?

I understand your working though Bunuel, it's very clear.

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shadowgun

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

02 May 2013, 12:00

Mindreko wrote:

Just wondering,

if the job took say, 2.5 hours to complete, how would the answer change?

I understand your working though Bunuel, it's very clear.

If the job takes Joseph and Lindsay 2.5 hours together to complete.

and Lindsay take as regular 1/x hours to complete.

Then the rate at which Jospeh works would be = 1/2.5 - 1/x
= (x - 2.5) / (2.5x)

Hence, in 20 minutes

2.5 hours = 150 mins, so in 20 mins he would complete : 2 / 15 * (x - 2.5 ) / (2.5 x )

Richard0715

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

08 May 2013, 17:17

if you use a smart number 2 for x, then A would be correct. I got this problem wrong two times so far in my studies and make the same mistake. Can someone explain why they would not think to use the number 2 as a smart number going into this problem?

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

08 May 2013, 21:51

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Richard0715 wrote:

if you use a smart number 2 for x, then A would be correct. I got this problem wrong two times so far in my studies and make the same mistake. Can someone explain why they would not think to use the number 2 as a smart number going into this problem?

Have you checked Option C using x=2 ?

\(\frac{(x-1)}{3x}\) ----------> \(\frac{(2-1)}{3X2}\) ----------> \(\frac{1}{6}\) ------> Same as Option A

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

05 Aug 2013, 01:50

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pjagadish27 wrote:

Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joseph, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?

A)1/3x

B)x/(x-3)

C)(x – 1)/3x

D)x/(x-1)

E)(x – 1)/x

I used substitution of values for the variable x, when x=2 i get an answer and when x=3 i get another answer! Please highlight my mistake. :cry:

LINDSAY ==> 1/X of room in 1 hr
LINDSAY + JOSEPH ==> FULL ROOM i.e 1 ROOM IN 1 HR
JOSEPH 1 HR WORK + LINDSAY 1 HR WORK = FULL ROOM PAINTING = 1
JOSEPH 1 HR WORK + 1/X = 1
JOSEPH 1 HR WORK = 1- 1/X = (X-1)X
THEREFORE JOSEPH WORK IN 20 MINUTES(1/3 OF HOUR) =\((1/3)*((X-1)/X)\) = \((X-1)/3X\)

Hope this helps

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

05 Aug 2013, 18:12

Let Joseph rate of work in 1 hr be y
Given : Lindsay rate of work in 1 hr as 1/x

Together,

1/x + y = 1

So Joseph rate of work would be

y = 1 - 1/x ==> x-1/x

In 20 min(1/3 of an hour), it would be 1/3(x-1/x) ==> x-1/3x
Option C

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

30 Aug 2013, 16:58

1/x th of work-----1 hr
1 full work----------? hr --------x hrs by L to complete full work

Now xy/x+y=1
we have to find y i.e., hrs taken by J to complete full work

divide numerator and denominator of L.H.S with y
we get x divided by (x+y)*1/y=1
----x=(x+y)*1/y
----x=x/y+1
----x-1=x/y
----x-1/x=1/y
reciprocal both sides
----x/x-1=y

Now it takes x/x-1 hrs to complete 1 full work by J

then in 1/3 hrs i.e., 20 min----- ? work J completes

=1/3*1 whole divided by (x/x-1)
=1/3*(x-1/x)
=x-1/3x

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

Updated on: 10 Dec 2014, 02:48

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Rate of Lindsay \(= \frac{1}{60x}\)

Rate of Joseph \(= \frac{1}{60} - \frac{1}{60x} = \frac{x-1}{60x}\)

Work done by Joseph in 20 Minutes \(= \frac{x-1}{60x} * 20 = \frac{x-1}{3x}\)

Originally posted by PareshGmat on 10 Dec 2014, 02:37.
Last edited by PareshGmat on 10 Dec 2014, 02:48, edited 1 time in total.

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

10 Dec 2014, 04:06

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Expert Reply

Similar question to practice: https://gmatclub.com/forum/lindsay-can-p ... 56880.html

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

10 Jan 2015, 18:17

Best approach for me is to plug in #s.

Say x=2, so Lindsay paints 1/2 of the room in one hour. Use D=RT for each Lindsay, Joseph and for both. (D:Work, R:Rate, T:Time) Say D=6

Consider Lindsay only: Lindsay paints 1/2 of D=6, meaning 3. Her rate is then R(Lindsay)=3 if T=1.

Consider Lindsay and Joseph: D=RT, D=6 and T=1, R becomes 6. R here is combined i.e., R= R(Lindsay)+R(Joseph)
R(Joseph) = 3 since R(Lindsay)=3.

Consider Joseph only: D=RT, R=3 and T= 20 min or 1/3 hr
D becomes 1. In other words, Joseph has painted 1/6 of the room in 20 min or 1/3 hr. 1/6 is target value. Plugin x=2 in Answer choices to match the target, C works.

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

30 Jan 2017, 05:00

1) First, let's find Lindsay's rate: \(r*1=\frac{1}{x}, r=\frac{1}{x}\)
2) Now let's find Lindsay and Joseph's rate when working together: R*1=1; R=1
3) Joseph's rate is equal to Lindsay and Joseph's rate together - Lindsay's rate: \(R-r=1-\frac{1}{x}\)
4) Joseph's work in 20 minutes is equal to his rate (per hour) divided by 3. \((1-\frac{1}{x})/3=\frac{x-1}{x}*\frac{1}{3}=\frac{x-1}{3x}\)

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

30 Jan 2017, 13:34

Jinglander wrote:

Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joseph, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?

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

let 1/j=J's rate
1/j=1-(1/x)
1/j*1/3=1/3-1/3x=(x-1)/3x
C

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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

01 Feb 2017, 09:16

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Jinglander wrote:

Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joseph, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?

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

We are given that Lindsay's rate to paint a room is 1/x. We are also given that when she works with Joseph, they can paint the 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. We can create the following equation and isolate j.

work of Lindsay + work of Joseph = 1

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

1/x + 1/j = 1

Multiplying the entire equation by xj, we obtain:

j + x = xj

x = xj - j

x = j(x - 1)

x/(x-1) = j

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

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

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

Answer: C

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Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

12 Apr 2018, 16:57

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Jinglander wrote:

Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Joseph, working together at their respective rates, can paint the room in one hour, what fraction of the room can Joseph paint in 20 minutes?

A. \(\frac{1}{3x}\)

B. \(\frac{x}{(x-3)}\)

C. \(\frac{(x-1)}{3x}\)

D. \(\frac{x}{(x-1)}\)

E. \(\frac{(x-1)}{x}\)

If the algebraic equation eludes you, pick an unusual number for \(x\). Rates are in \(\frac{rooms}{hr}\)

Let \(x = 6\)
L's rate = \(\frac{1}{x}=\frac{1}{6}\)

L and J's combined rate =
\((\frac{1}{6}+\frac{1}{J})=\frac{1}{1}\)

J's rate: \(\frac{1}{J}=(\frac{1}{1} - \frac{1}{6}) = \frac{5}{6}\)

Work competed in 20 minutes = \(\frac{1}{3}\) hour?
\(RT= W\)
In 20 minutes, J completes
\((\frac{5}{6}*\frac{1}{3}) =\frac{5}{18}\) of a room

Using x = 5, find the answer* that yields \(\frac{5}{18}\)

A. \(\frac{1}{3x}=\frac{1}{(3*6)}=\frac{1}{18}\) - NO

B. \(\frac{x}{(x-3)}=\frac{6}{(6-3}=\frac{6}{3}=2\) - NO

C. \(\frac{(x-1)}{3x}=\frac{(6-1)}{18}=\frac{5}{18}\) - MATCH

D. \(\frac{x}{(x-1)}=\frac{6}{(6-1)}=\frac{6}{5}\)- NO

E. \(\frac{(x-1)}{x}=\frac{(6-1)}{6}=\frac{5}{6}\) - NO

Answer C

*1) if your answer is a fraction, do not reduce it (your answer is based on an assigned value for x - don't depart from that value); and 2) with this method you have to check all the answer choices

B

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Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

26 Jul 2018, 11:30

I don't think this is a good question since you can plug the number 2 in and you would get A or C as the answer.

Can someone comment their opinion on the question quality?

gmatclubot

Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]

26 Jul 2018, 11:30

GMAT Problem Solving (PS) Questions

Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo