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14 May 2018, 22:40
I think this is a highquality question and I agree with explanation.



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01 Jun 2018, 00:01
I think this is a highquality 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: M2711
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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, ...
BunuelHow do we get that x is odd? Please help in understanding
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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, ...
BunuelHow 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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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 10:30amFrom 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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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 viseversa). 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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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 viseversa). 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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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 viseversa). 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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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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Re: M2711
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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: M2711
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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 viseversa). 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 BunuelFrom 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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