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# Perry, Maria, and Lorna are painting rooms in a college dormitory.

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Re: Perry, Maria, and Lorna are painting rooms in a college dormitory. [#permalink]
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The shortest time is possible when Lorna and Maria work together as their speed of doing work is more. Together in 1 hour, they do 90% of the work (50%+40%). 1/0.9 = 1 hour 7 minutes.

The longest time is possible when Lorna and Perry work together as their speed of doing work is less. Together in 1 hour, they do 73.33% of the work (40%+33.33%). 1/0.7333 = 1 hour 22 minutes.
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Re: Perry, Maria, and Lorna are painting rooms in a college dormitory. [#permalink]
The shortest time will be there when
both Lornaa and Maria work together . So in 1 hour, they complete 90% of the work,which is 1/0.9 = 1 hour 7 minutes.

The longest time will be there when both Lorna and Perry work together. So in 1 hour, they do 73.33% of the work,which is 1/0.7333 = 1 hour 22 minutes.

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Re: Perry, Maria, and Lorna are painting rooms in a college dormitory. [#permalink]
here I have a question how do you decide which one will take less time, calculating all will take time minustark

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Re: Perry, Maria, and Lorna are painting rooms in a college dormitory. [#permalink]
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Let's deal with the givens first.
Let's say there are 100 units of work.
Time * rate = work done
Let P, M, L represent the individual's rate.
They are all painting the same standard room (same amount of work)
3*P = 100
2*M = 100
2.5*L = 100
Now, you can find their rate.
P = 33.3
M = 50
L = 40
Now to find the shortest time, pick the 2 fastest people, obviously, those who paint the room in less time are the fastest. Those two people are M and L.
To find the time those two people will take to paint the same standard room we will do t*M + t*L = 100
Plug and chug, t*50 + t*40 = 100
t(50+40) = 100
90t = 100
t = 100/90 hours
10/9 hours is 1 hour and 7 mins

Now to find the longest time, pick the 2 slowest people (aka the people who took the longest amount of time). Those people are P and L.
To find the time those two people will take to paint the same standard room we will do t*P + t*L = 100
33.3t + 40t = 100
t(33.3 + 40) = 100
73.3t = 100
drop the .3, it adds complication and the times between answer choice are not very close.
t = 1.37 hours which is 1 hour and 22 minutes

Originally posted by hassan233 on 12 Jun 2021, 16:41.
Last edited by hassan233 on 05 Jul 2021, 11:35, edited 1 time in total.
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Re: Perry, Maria, and Lorna are painting rooms in a college dormitory. [#permalink]
Hi! Thinking of a quick way to solve this. Any thoughts on the below approach?
Shortest: 2 hours and 2.5 hours speeds: Maria's half room done in 1 hour, Lorma's half room done in 1 hour and 15 mins, but Maria will help Lorma once Maria is done, so that 15 mins timeframe is roughly split in half. Thus shortest is 1 hour + 15/2 = 1 hour and 7 mins (closest choice).

Longest: same approach: 3 hours and 2.5 hours = 1.5 hours and 1 hour 15 mins, but split remaining 15 mins in half which is roughly 7, and add to 1 hour 15 mins = 1h15m + 7 = 1h22min.

Does this approach make sense or am I setting myself up for failure? It seems intuitive to me and rather quick compared to summing up fractions.
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Re: Perry, Maria, and Lorna are painting rooms in a college dormitory. [#permalink]
chetan2u

Thank you so much for your reply! It was helpful. Maybe I'm confused or something...but why did you flip/invert the fraction of 11/15*60 --> 15/11*60? Don't we usually do that when we divide?

Same thing happened for 1/2.5 became 2/5 -- why? Isn't it suppose to be 5/2?
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Re: Perry, Maria, and Lorna are painting rooms in a college dormitory. [#permalink]
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sssaniaa wrote:
chetan2u

Thank you so much for your reply! It was helpful. Maybe I'm confused or something...but why did you flip/invert the fraction of 11/15*60 --> 15/11*60? Don't we usually do that when we divide?

Same thing happened for 1/2.5 became 2/5 -- why? Isn't it suppose to be 5/2?

Hi

11/15 is one hour work, that is 11/15th of the work is completed in one hour.
=> 11/15 in 1 hour
So 1 work in 1/(11/15) hour or 15/11 hour.

That is we flip one hour or one day work to get time taken to complete the entire work.
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Re: Perry, Maria, and Lorna are painting rooms in a college dormitory. [#permalink]
rabide wrote:
Perry, Maria, and Lorna are painting rooms in a college dormitory. Working alone, Perry can paint a standard room in 3 hours, Maria can paint a standard room in 2 hours, and Lorna can paint a standard room in 2 hours 30 minutes. Perry, Maria, and Lorna have decided that, to speed up the work, 2 of them will paint a standard room together.`

Select the value closest to the shortest time in which a 2-person team could paint a standard room, and select the value closest to the longest time in which a 2-person team could paint a standard room, with each person working at his or her respective rate. Make only two selections, one in each column.

I solved this in a slightly different way

First, there are 3 unique ways by which we can match 2 of them together such that they both paint together

1.MP (Maria and Perry)
2.ML (Maria and Loma)
3.PL (Perry and Loma)

We can then find the number of time in which each of these combination could take to finish painting

Speed= 1/Time

Combined speed of each of the above combination is

MP>>1/2+1/3=5/6
ML>>1/2+2/5=9/10
PL>>1/3+2/5=11/15

Combined time (Inverse of speed):

MP>> 1 min 12 secs
ML>> 1 min 7 secs (Shortest)
PL>> 1 min 22 secs (Longest)­
Re: Perry, Maria, and Lorna are painting rooms in a college dormitory. [#permalink]
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