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M27-11

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Re M27-11  [#permalink]

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New post 14 May 2018, 22:40
I think this is a high-quality question and I agree with explanation.
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New post 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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Re: M27-11  [#permalink]

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New post 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

How do we get that x is odd? Please help in understanding
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New post 02 Jun 2018, 03:51
Adi93 wrote:
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

How do we get that x is odd? Please help in understanding


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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M27-11  [#permalink]

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New post 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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New post 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.

Answer: B


Why does it matter whether they complete the job in \(\frac{odd}{2}\) in order for \(x=y\)?
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New post 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.

Answer: B


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  [#permalink]

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New post 17 Oct 2018, 16:52
Thank you!

Bunuel wrote:
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.

Answer: B


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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New post 19 Jan 2019, 03:58
@Bunuel:- is this reasoning correct- "they started at the same time and ended at the same time. Therefore, we can certainly infer that the time they took was same- x=y"?
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New post 20 Jan 2019, 02:48
Manat wrote:
@Bunuel:- is this reasoning correct- "they started at the same time and ended at the same time. Therefore, we can certainly infer that the time they took was same- x=y"?


No... x and y are the individual times Bonnie and Clyde need to paint the car. The question asks whether x = y...
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Re: M27-11  [#permalink]

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New post 25 Feb 2019, 07:28
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.

Answer: B



Bunuel

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, ...

could you please explain me your assumption (I highlighted it in red). I cannot reconstruct your process of thought here, as the questions never says that the x = y.
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Re: M27-11   [#permalink] 25 Feb 2019, 07:28

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