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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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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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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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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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Bunuel wrote: 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...Hi BunuelDoes " working simultaneously and independently" means working on the same job but at different parts of the job?? Also, per S2, we know both Bonnie and Clyde took 45 mins (3/4 of an hour) to complete the job. So, x = y Then, how come x is not equal to y?? Is it because clearly 3/4(0.75) is not an integer? And its stated in the Q stem that x and y are odd integers?? Can you please elaborate on this? Thanks



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22 Apr 2019, 01:14
JIAA wrote: Bunuel wrote: 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...Hi BunuelDoes " working simultaneously and independently" means working on the same job but at different parts of the job?? Also, per S2, we know both Bonnie and Clyde took 45 mins (3/4 of an hour) to complete the job. So, x = y Then, how come x is not equal to y?? Is it because clearly 3/4(0.75) is not an integer? And its stated in the Q stem that x and y are odd integers?? Can you please elaborate on this? Thanks 1. Working simultaneously and independently, means that they work on different parts of the job. For example, one is painting the front and another the back of the car. This ensure that their works does not overlap and they do the job in fastest way possible. 2. When analysing the stem we got that IF \(x=y\), then the total time would be: \(\frac{odd}{2}\): 0.5 hours, 1.5 hours, 2.5 hours, ... From (2) we got that they complete the job in \(\frac{3}{4}\) of an hour (45 minutes). Is \(\frac{3}{4}\) of an hour (45 minutes), \(\frac{odd}{2}\) hours: 0.5 hours, 1.5 hours, 2.5 hours, ... ? No, thus x does not equal y.
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