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M27-11

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Intern
Joined: 30 May 2017
Posts: 9

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14 May 2018, 22:40
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 29 Apr 2018
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01 Jun 2018, 00:01
I think this is a high-quality question and I agree with explanation. Please paraphrase the questions, started working simultaneously and independently seems like they started working at the same time but independently on separate cars.
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Status: EAT SLEEP GMAT REPEAT!
Joined: 28 Sep 2016
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01 Jun 2018, 09:28
Quote:

Bonnie and Clyde, when working together, complete the painting of the car in $$\frac{xy}{x+y}$$ hours (sum of the rates equal to the combined rate or reciprocal of total time: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{T}$$. From that we can get that $$T=\frac{xy}{x+y}$$). Now, if $$x=y$$, then the total time would be: $$\frac{x^2}{2x}=\frac{x}{2}$$, since $$x$$ is odd, then this time would be $$\frac{odd}{2}$$: 0.5 hours, 1.5 hours, 2.5 hours, ...

Bunuel

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02 Jun 2018, 03:51
Quote:

Bonnie and Clyde, when working together, complete the painting of the car in $$\frac{xy}{x+y}$$ hours (sum of the rates equal to the combined rate or reciprocal of total time: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{T}$$. From that we can get that $$T=\frac{xy}{x+y}$$). Now, if $$x=y$$, then the total time would be: $$\frac{x^2}{2x}=\frac{x}{2}$$, since $$x$$ is odd, then this time would be $$\frac{odd}{2}$$: 0.5 hours, 1.5 hours, 2.5 hours, ...

Bunuel

Bonnie can paint a stolen car in $$x$$ hours, and Clyde can paint the same car in $$y$$ hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both $$x$$ and $$y$$ are odd integers, is $$x=y$$?
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Location: Finland
Concentration: Accounting, Entrepreneurship
WE: Account Management (Entertainment and Sports)

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25 Sep 2018, 00:46
I approached statement 1 the same way as arhumsid did before:

"S1: $$x^2+y^2 \lt 12$$
Since we already know that x and y are Odd integers, there's not much left to check for in the above condition and since adding two squares will very soon pass such a small number as 12, we can do the manual work here-
x=1; y=1 => Yes (both equal and sum of squares less than 12)
x=1; y=3 => No (both unequal and still the sum of squares less than 12)"

=>statement 1 is insufficient alone to answer

Statement 2: Bonnie and Clyde complete the painting of the car at 10:30am

From the given details we know that Bonnie and Clyde worked for 45minutes to finish painting one car. We need to prove whether x=y.

If x=y=1 then Bonnie and Clyde working together would finish painting a car in 1/2hr (30min < 45min)
If x=y=3 then Bonnie and Clyde working together with the same 3hrs/car rate would finish painting the car in 1.5hr (90min > 45 min)

We took the smallest pozitive odd integers to make x and y hr/car rate equal and we saw that with 1hr/car=x=y Bonnie and Clyde would finish painting the car less than 45 min and with a 3hr/car rate they would finish painting car more than 45 min.

If the precondition wouldn't hold on odd integer condition, x=y=45min*2=1.5hr. Because of the given precondition, x and y cannot be equal.

=> Therefore, Bonnie and Clyde must work with different hr/car rate => statement 2 is sufficient alone to answer
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Joined: 24 Apr 2016
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16 Oct 2018, 19:48
Bunuel wrote:
Official Solution:

Bonnie and Clyde, when working together, complete the painting of the car in $$\frac{xy}{x+y}$$ hours (sum of the rates equal to the combined rate or reciprocal of total time: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{T}$$. From that we can get that $$T=\frac{xy}{x+y}$$). Now, if $$x=y$$, then the total time would be: $$\frac{x^2}{2x}=\frac{x}{2}$$, since $$x$$ is odd, then this time would be $$\frac{odd}{2}$$: 0.5 hours, 1.5 hours, 2.5 hours, ...

(1) $$x^2+y^2 \lt 12$$. It's possible that $$x$$ and $$y$$ are odd and equal each other if $$x=y=1$$, but it's also possible that $$x=1$$ and $$y=3$$ (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am. They complete the job in $$\frac{3}{4}$$ of an hour (45 minutes), since it's not $$\frac{odd}{2}$$ then $$x$$ and $$y$$ are not equal. Sufficient.

Why does it matter whether they complete the job in $$\frac{odd}{2}$$ in order for $$x=y$$?
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Joined: 02 Sep 2009
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16 Oct 2018, 20:54
dracobook wrote:
Bunuel wrote:
Official Solution:

Bonnie and Clyde, when working together, complete the painting of the car in $$\frac{xy}{x+y}$$ hours (sum of the rates equal to the combined rate or reciprocal of total time: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{T}$$. From that we can get that $$T=\frac{xy}{x+y}$$). Now,if $$x=y$$, then the total time would be: $$\frac{x^2}{2x}=\frac{x}{2}$$, since $$x$$ is odd, then this time would be $$\frac{odd}{2}$$: 0.5 hours, 1.5 hours, 2.5 hours, ...

(1) $$x^2+y^2 \lt 12$$. It's possible that $$x$$ and $$y$$ are odd and equal each other if $$x=y=1$$, but it's also possible that $$x=1$$ and $$y=3$$ (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am. They complete the job in $$\frac{3}{4}$$ of an hour (45 minutes), since it's not $$\frac{odd}{2}$$ then $$x$$ and $$y$$ are not equal. Sufficient.

Why does it matter whether they complete the job in $$\frac{odd}{2}$$ in order for $$x=y$$?

This is explained in highlighted part above.
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Re: M27-11 &nbs [#permalink] 16 Oct 2018, 20:54

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