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A paint crew gets a rush order to paint 80 houses in a new

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A paint crew gets a rush order to paint 80 houses in a new  [#permalink]

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New post 21 Aug 2006, 00:38
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A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week?

(A) 0.8(80 – y)
(B) 0.8 + 0.0025y
(C) 80/y – 1.25
(D) 80/1.25y
(E) 80 – 0.25y
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Re: A paint crew gets a rush order to paint 80 houses in a new  [#permalink]

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New post 19 May 2016, 10:50
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Original Time taken to finish the job with normal rate= 80/x hrs

time taken to paint y houses with rate of x houses/week= y/x
Time taken to paint remaining (80-y) houses with rate of 1.25 houses/week = 80-y/1.25x

Total new time = y/x + 80-y/1.25x

Fraction = New time/original time
y/x + 80-y/1.25x/80/x= .25y +80/80 = 0.8 + 0.0025y

B is the answer :)
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New post 21 Aug 2006, 01:19
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Time for painting houses at mixed rate= (t1)
80-y/1.25x + y/x = 320-4y/5x + y/x or 320-4y/5x + 5y/5x

Simplified = 320 + y/5x

Time for painting houses at constant rate = 80/x (t2)

t1/t2 = 320+y/5x * x/80 ----> 32+y/400

or 0.8 + 0.0025y
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New post 21 Aug 2006, 05:09
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gmatornot wrote:
Rush Paint Job

A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week?

(A) 0.8(80 – y)
(B) 0.8 + 0.0025y
(C) 80/y – 1.25
(D) 80/1.25y
(E) 80 – 0.25y


original time = 80/x weeks
new time = y/x + (80-y)/1.25x
ratio=y/x + (80-y)/1.25x : 80/x = 1.25y+80 - y:100 = 0.25y+80:100
ratio = 0.8 0.0025y
answer = B
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New post 21 Aug 2006, 07:49
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(B) 0.8 + 0.0025y

Pretty straightforward question...

@ x houses/wk , 80 houses
time = 80/x weeks ... (1)

@ x houses/wk, y houses
time = y/x weeks ... (2)

@ 1.25x houses/wk, (80-y) houses,
time = (80-y)/1.25x weeks...(3)

Adding (2) & (3):
time = (y+320)/5x .... (4)

Ratio of (4)/(1)

= ((y+320)/5x) x (x/80)
= 0.08 + 0.0025y

Answer B.
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New post 21 Aug 2006, 09:54
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B.

BY TAKING ARBITARARY VALUE FOR X AND Y,

LETS SAY Y = 20,
X = 4,

and then plug in number of x and y, upi get 0.85, which is the same as in B.
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New post 22 Aug 2006, 21:33
Answer is B.. Great job everyone !
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Re: A paint crew gets a rush order to paint 80 houses in a new  [#permalink]

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New post 20 May 2013, 08:55
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I am working on a similar problem with a few differences. Can anyone help?

"A painting crew painted 80 houses. They painted the first y houses at a rate of x houses per week. Then more painters arrived and everyone worked together to paint the remaining houses at a rate of 1.25x houses per week. How many weeks did it take to paint all 80 houses in terms of x and y?

The answer is (y+320)/5x

I decided to find the time it took by adding T1 and T2 (y/x and [80-y]\1.25x respectively) but I'm stuck past there. Does anyone know how I can solve this?
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Re: A paint crew gets a rush order to paint 80 houses in a new  [#permalink]

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New post 21 May 2013, 07:30
1
Quote:
I am working on a similar problem with a few differences. Can anyone help?

"A painting crew painted 80 houses. They painted the first y houses at a rate of x houses per week. Then more painters arrived and everyone worked together to paint the remaining houses at a rate of 1.25x houses per week. How many weeks did it take to paint all 80 houses in terms of x and y?

The answer is (y+320)/5x

I decided to find the time it took by adding T1 and T2 (y/x and [80-y]\1.25x respectively) but I'm stuck past there. Does anyone know how I can solve this?


To answer questions such as this one, I do a chart that helps me organizing information

R * T = W
Few Painters x T(1) = y
--------------------------------------
All painters 1.25x T(2) = (80 - y)/1.25x

What is the question? The question ask you to find how many weeks did it take to paint all 80 houses in terms of x and y? So in other terms what is T(1)+T(2)?

Simply rearrange the term T(1) and T(2)
T(1) + T (2) = y/x + (80 - y)/1.25x
T(1) + T (2) = (1.25y + 80 - y)/1.25x ==> I put everything under the same denominator
T(1) + T (2) = (0.25y + 80)/1.25x
T(1) + T (2) = (0.25y*4 +80*4)/1.25x*4 ==> Multiply by 4/4
T(1) + T (2) = (y+320)/5x

Here the "hard" part is to rearrange the term so that it matches the correct answer.

Hope it helps,
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Re: A paint crew gets a rush order to paint 80 houses in a new  [#permalink]

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New post 15 Aug 2014, 07:13
This may be a case of bad numberpicking but it worked for me.

80 houses. Let's say that y=40 houses are painted at the rate x=10 houses per week <=> 1/10 week per house. 40*1/10 = 4 houses per week will be painted at this rate.
80-y = 80-40 = 40 houses are to be painted at the faster rate. X*1,25=12,5 houses per week <=> 1/12,5 weeks per house * 40 houses = 40/12,5 = 80/25 = 320/100 = 3,2 weeks.
Which means finishing all houses at normal rate => 2*4 = 8 weeks. Faster rate = 4+3,2 = 7,2 weeks.
7,2/8 = 9/10 = 0,9. Insert y=40 in equations and it is clear that only (B) gives us 0,9.

0,8 + 0,0025*40 = 0,8 + 0,1 = 0,9.
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A paint crew gets a rush order to paint 80 houses in a new  [#permalink]

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New post 19 May 2016, 10:10
WholeLottaLove wrote:
I am working on a similar problem with a few differences. Can anyone help?

"A painting crew painted 80 houses. They painted the first y houses at a rate of x houses per week. Then more painters arrived and everyone worked together to paint the remaining houses at a rate of 1.25x houses per week. How many weeks did it take to paint all 80 houses in terms of x and y?

The answer is (y+320)/5x

I decided to find the time it took by adding T1 and T2 (y/x and [80-y]\1.25x respectively) but I'm stuck past there. Does anyone know how I can solve this?


After doing calculations and looking at my answer and the available answer choices, I came to the following conclusion.
It took me around 10 minutes to solve the problem. Well, math is not my background but no excuses, since GMAT doesn't tests your math skills. I wonder what will I do in exam. I can't waste 10 minutes for a question and I have roughly 1 month for the exam. Anyway,
the answer is not \(\frac{(y+ 320)}{5x}\) rather \(\frac{(y + 320)}{5x}\) /\(\frac{80}{x}\)

Now, tricky part is here.

\(\frac{(y + 320)}{5x}\) \(*\) \(\frac{x}{80}\)

= \(\frac{y + 320}{400}\)

=\(\frac{y}{400}\) + \(\frac{320}{400}\)

=\(\frac{1}{400}\) \(*\) \(y + 0.8\)
\(\frac{1}{400}\)= 0.0025

therefore,

ANS. is B

Bunuel Is there a simple way to solve?
chetan2u Is there a simple way to solve?
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Re: A paint crew gets a rush order to paint 80 houses in a new  [#permalink]

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New post 19 Mar 2017, 14:25
Hello everyone, thank you for the asnwers. However, I do not understand the last step of the equation. Why do we divide new time on the original time? I did not really get the question tbh...
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Re: A paint crew gets a rush order to paint 80 houses in a new  [#permalink]

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New post 15 May 2018, 10:39
MBAlad wrote:
Time for painting houses at mixed rate= (t1)
80-y/1.25x + y/x = 320-4y/5x + y/x or 320-4y/5x + 5y/5x

Simplified = 320 + y/5x

Time for painting houses at constant rate = 80/x (t2)

t1/t2 = 320+y/5x * x/80 ----> 32+y/400

or 0.8 + 0.0025y



Hello pushpitkc

could you write step by step how we from this 80-y/1.25x + y/x we get this 320 + y/5x is this combined rate ?

and how we get this t1/t2 = 320+y/5x * x/80 ----> 32+y/400 :? is this combined time ?

thank you :-)
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A paint crew gets a rush order to paint 80 houses in a new  [#permalink]

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New post 15 May 2018, 11:34
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dave13 wrote:
MBAlad wrote:
Time for painting houses at mixed rate= (t1)
80-y/1.25x + y/x = 320-4y/5x + y/x or 320-4y/5x + 5y/5x

Simplified = 320 + y/5x

Time for painting houses at constant rate = 80/x (t2)

t1/t2 = 320+y/5x * x/80 ----> 32+y/400

or 0.8 + 0.0025y



Hello pushpitkc

could you write step by step how we from this 80-y/1.25x + y/x we get this 320 + y/5x is this combined rate ?

and how we get this t1/t2 = 320+y/5x * x/80 ----> 32+y/400 :? is this combined time ?

thank you :-)


Hey dave13

\(\frac{80-y}{1.25x} + \frac{y}{x} = \frac{4(80-y) + 5y}{5x}\) (Here, the LCM of 1.25x and x is 5x)

This can be further simplified as \(\frac{320 - 4y + 5y}{5x} = \frac{320 + y}{5x}\)

If the painters had painted the houses at a constant rate, time taken would be 80/x

So, the ratio of the time taken is \(\frac{\frac{320+y}{5x}}{\frac{80}{x}}\) , which when simplified becomes \(\frac{320+y}{5x}* \frac{x}{80} = \frac{320 + y}{400}\)

Hope this helps you!
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New post 24 Jul 2018, 03:38
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Not fair, not fair!

How did we get from here:

(y/x)+(80-y)/1.25x : 80/x
.
.
.
.
.
.
To here: 0.8 + 0.0025y

I'll really appreciate a break down of the solving please, thanks!
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A paint crew gets a rush order to paint 80 houses in a new &nbs [#permalink] 24 Jul 2018, 03:38
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