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Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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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. \(\frac{x}{(x3)}\) C. \(\frac{(x1)}{3x}\) D. \(\frac{x}{(x1)}\) E. \(\frac{(x1)}{x}\)
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Originally posted by Jinglander on 08 Aug 2010, 13:46.
Last edited by Bunuel on 18 Dec 2017, 23:39, edited 4 times in total.
Added the OA.



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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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08 Aug 2010, 21:14
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Answer is C  \((x1/3x)\) Given details: Rate at which L work 1/x Rate at which L&J work  x Rate at which J alone works  \((x1/x)\) Rate at which J alone works in 20 min.  \((x1/x) * (20/60) => (x1/x)*(1/3)\) On a side note, maybe this post needs to be moved to the Problem solving section.
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Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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09 Aug 2010, 03:12
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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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09 Aug 2010, 09:54
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]
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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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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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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 )



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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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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]
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08 May 2013, 21:51
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 ? \(\frac{(x1)}{3x}\) > \(\frac{(21)}{3X2}\) > \(\frac{1}{6}\) > 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]
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05 Aug 2013, 01:29
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Well, I am not a fond of trying values or of backsolving methods. So I will present different method to solve this one  Percent Method. First some theory.When we say a person can finish the given task in X hours, we can also say that he can finish \(\frac{100}{X}%\) task in one hour (Whole task always equals to 100%) When we say another person can finish the same task in Y hours, we can also say that he can finish \(\frac{100}{Y}%\) task in one hour. Finally we can say that they can finish \((\frac{100}{X} + \frac{100}{Y})%\) task in one hour. We will try this method in a simple question Q : A can finish certain work in 10 days. B can finish the same work in 20 days. In how many days can they finish the work working together? A can finish certain work in 10 days > He can complete 10% of the work in a day B can finish the same work in 20 days. > He can complete 5% of the work in a day Working together they can complete (10+5)% work in a day. Now that we know Total work always equals 100% and that they are finishing 15% work in a day working together, So we can say that they can complete the total work in \(\frac{100}{15}\) (i.e. 6.66) days. Back to your question..........Lindsay can paint \(\frac{1}{X}\) of a certain room in one hour. > This simply states that Lindsay can paint the room in X hours > Lindsay can paint \((\frac{100}{X})%\) of the room in one hour Lindsay and Joseph, working together at their respective rates, can paint the room in one hour > Working together they can paint the 100% of the room in one hour Equation is ( Rate of Lindsay of one hour + Joseph Rate of of one hour) = Rate of Lindsay+Joseph of one hour \(\frac{100}{X}\) + Joseph Rate of of one hour = 100 > Joseph's Rate of one hour = \(100  \frac{100}{X}\) > Joseph's Rate of one hour = \(\frac{100(X  1)}{X}\) > We can rephrase this as Joseph is completing \(\frac{(100(X1))}{X}%\) of 100% room in one hour > In Fraction He is completing \(\frac{(100(X1))}{100X}\) in one hour > \(\frac{(X1)}{X}\) what fraction of the room can Joseph paint in 20 minutes? > what fraction of the room can Joseph paint in \(\frac{1}{3}\) hour? > \(\frac{1}{3} \frac{(x1)}{x}\) > \(\frac{(x1)}{3x}\) Option CHope that Helps.
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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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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/(x3) C)(x – 1)/3x D)x/(x1) 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. 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 = (X1)X THEREFORE JOSEPH WORK IN 20 MINUTES(1/3 OF HOUR) =\((1/3)*((X1)/X)\) = \((X1)/3X\) 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]
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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 ==> x1/x
In 20 min(1/3 of an hour), it would be 1/3(x1/x) ==> x1/3x Option C



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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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30 Aug 2013, 16:58
1/x th of work1 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 x1=x/y x1/x=1/y reciprocal both sides x/x1=y
Now it takes x/x1 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/x1) =1/3*(x1/x) =x1/3x



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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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Rate of Lindsay \(= \frac{1}{60x}\) Rate of Joseph \(= \frac{1}{60}  \frac{1}{60x} = \frac{x1}{60x}\) Work done by Joseph in 20 Minutes \(= \frac{x1}{60x} * 20 = \frac{x1}{3x}\)
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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]
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10 Dec 2014, 04:06



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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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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]
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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: \(Rr=1\frac{1}{x}\) 4) Joseph's work in 20 minutes is equal to his rate (per hour) divided by 3. \((1\frac{1}{x})/3=\frac{x1}{x}*\frac{1}{3}=\frac{x1}{3x}\)



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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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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/(x3) C. (x1)/3x D. x/(x1) E. (x1)/x let 1/j=J's rate 1/j=1(1/x) 1/j*1/3=1/31/3x=(x1)/3x C



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Re: Lindsay can paint 1/x of a certain room in one hour. If Lindsay and Jo [#permalink]
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01 Feb 2017, 09:16
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/(x3) C. (x1)/3x D. x/(x1) E. (x1)/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/(x1) = j Since j = x/(x1) 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]
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12 Apr 2018, 16:57
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. \(\frac{x}{(x3)}\)
C. \(\frac{(x1)}{3x}\)
D. \(\frac{x}{(x1)}\)
E. \(\frac{(x1)}{x}\) 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. \(\frac{x}{(x3)}=\frac{6}{(63}=\frac{6}{3}=2\)  NO C. \(\frac{(x1)}{3x}=\frac{(61)}{18}=\frac{5}{18}\)  MATCH D. \(\frac{x}{(x1)}=\frac{6}{(61)}=\frac{6}{5}\) NO E. \(\frac{(x1)}{x}=\frac{(61)}{6}=\frac{5}{6}\)  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
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